/ohm/ Electronics General

[Anonymous No.915959]

who generally electronics here?

>I'm new to electronics, where do I get started?
There are several good books and YouTube channels that are commonly recommended for beginners and those wanting to learn more, many with advanced techniques. The best way to get involved in electronics is just to make stuff. Don't be afraid to get your hands dirty.

>Links to get started
http://pastebin.com/9UgLjyND

also, general electronics thread

[Anonymous No.916144]

>>915959

first for flyback transformers

[Anonymous No.916279]

>>915959
Is he trying to run continuity tests on smd components with a big ass probe? The thing is like twice the size of the pads.

[Anonymous No.916296]

Best oscilloscope for ~500€?

[Anonymous No.916298]

>>916279
He's doing what the photographer told him to do.

[Anonymous No.916406]

I recently bought one of pic related off of eBay.

How do you control the voltage on these?

If it helps the main chip doing the voltage increase is XL6009E1

[Anonymous No.916407]

>>916406

...turn the trim pot? It's the blue thing with the little screw on top.

If you mean how to control it EXTERNALLY, then remove the trim pot and add your control line to the pin it was connected to.

[Anonymous No.916408]

>>916407
Manually was what I was talking about.

Thanks.

[Anonymous No.916428]

What is the function of the resistor in a passive high pass filter?

It seems like it just opens an alternate path to ground, so attenuating the signal at the output (the lower the resistance, the more this happens).

[Anonymous No.916432]

>>916428
>What is the function of the resistor in a passive high pass filter?

It provides a current path for high-impedance inputs. Were you to leave it out and connect a capacitor directly to something (like an opamp) that has several MΩ of impedance on its inputs or greater, the input capacitance on that opamp would charge up long before the filter capacitor did. Adding a current path between the capacitor and input provides the necessary current path for the larger capacitor to charge and discharge as intended.

[Anonymous No.916439]

>>916432
What do you mean by the input capacitance? How is it related to input impedance?

[Anonymous No.916443]

>>916432
>>916439
Do you mean that, when charge is forced into one end of the capacitor, an equal amount is forced out of the other side - and that needs somewhere to go?

Either way, does the signal to be filtered have to be biased (so current only ever goes one way around the circuit).

[Anonymous No.916451]

>>916296

One from a university auction/dumpster for 20 quid or free

[Anonymous No.916454]

where is a good place to buy a set of starter components?

I've been working with electronics for some time professionally but I've been doing some mucking about at home with theory crafting and prototyping. Thing is at work we just have pretty much everything and what we don't have we are not allowed to purchase our own parts we have to submit a PO for so some bean counter can pay 5x fair market value and I have no idea where to source this stuff.

Surely there is something better than putting 500 kinds of caps resistors and transistors in digikey and buying 10 of each, like some kind of assortment kit. Anyone know a good source for cheap?

[Anonymous No.916456]

>>916454
I'd like to know as well. I bought a decent pack from Maplin for £20. It's decent enough, but hardly comprehensive. Lots of resistors but scarce on most other things.

[Anonymous No.916463]

>>916439

Electronic devices are either low or high impedance inputs, depending on their construction and the intent behind their design.

High impedance inputs, like most anything made with a CMOS process, would ideally have infinite impedance. Since this isn't an ideal world, this ends up being anywhere from hundreds of kΩ to the GΩ range, middling in, say, the tens to hundreds of MΩ. High enough to where they can be considered infinite without consequence for most typical use cases.

Ideally, these inputs would also have no capacitive properties, but, again, this is not an ideal world. They do have a very small amount of input capacitance (usually in the low pF range or lower, though this varies wildly). Like the input impedance, however, this can usually be ignored since it's so small.


Devices with these kinds of inputs are called voltage mode, voltage controlled, or something along those lines, as they respond to changes in voltage at their inputs, not changes in current. Because of their high input impedance, they cannot sink or source current, which is necessary for an RC filter to work. Thus, the resistor; it provides the current path to allow the filter to work, but generates a voltage across itself dependent on the current, which, in turn, allows the voltage-sensitive inputs of [whatever you're feeding the filtered signal into] to work.

So, yeah, pretty much:

>when charge is forced into one end of the capacitor, an equal amount is forced out of the other side - and that needs somewhere to go
>Either way, does the signal to be filtered have to be biased (so current only ever goes one way around the circuit).

No. In fact this wouldn't work if you tried. Capacitors build up charge in response to current flowing through them; if it only ever flowed one way, the capacitor would just charge to arbitrarily high voltage. Moreover, the entire point of a high-pass filter is to prevent current from going in one direction for too long.

[Anonymous No.916466]

>>916454
>Anyone know a good source for cheap?

eBay. Half of that site (probably more) is essentially a market front for China. There are plenty of electronics assortments all over the place. And you can find them dirt cheap, since that's where most components are produced in the first place.

Just poke around some seller's site and fill your cart up with whatever generic components you want. Compare prices, obviously, and my advice is to find one that doesn't offer free shipping, but does combined shipping. That should mean the price of shipping isn't amortized into the cost of individual listings and they should be cheaper for that reason.

However, while this is fine for generic parts (resistors, ceramic caps, switches, basic semiconductors like transistors and diodes, etc.), you might be a little wary of doing this for any proprietary stuff. Fake parts is a big thing in China.

[Anonymous No.916469]

>>916466

by fake do you mean counterfiet? or do you mean like straight up a piece of plastic with numbers and pins on it and no internal parts?

or both kek china

[Anonymous No.916477]

>>916463
>Moreover, the entire point of a high-pass filter is to prevent current from going in one direction for too long.
Ah yes, I was just looking at a graph of a square wave before and after filtering. I makes sense now. Each peak is like a momentary DC offset that the capacitor damps back to zero. A soon as the signal stops changing the offset correction kicks in.

But that was a cool post BTW, thanks.

[Anonymous No.916484]

>>916463
>>916477
But then how does the resistor have an effect on the cutoff?

[Anonymous No.916501]

>>916484

the resistor is something the capacitor can work against to provide attenuation. it's like having a voltage divider, where the reactive element (the cap in this case) acts like a frequency-dependent resistor.

[Anonymous No.916504]

>>916501
>it's like having a voltage divider
Wow, so it is. Didn't notice that. So the lower frequencies make the cap's resistance higher, as if it were the first resistor in a dividor.

[Anonymous No.916506]

>>916469

Mostly the former, but I suppose the second is a potential issue. Sellers will get shut down pretty fast for actual fake item claims, though, so I doubt you'd be too likely to end up with an empty package no matter what you buy.

[Anonymous No.916517]

>>916296
Hard to go wrong with the Rigol DS1054Z. Looks like the Tektronix 1052B is good too.

[Anonymous No.916536]

I want to build a timer circuit to turn a 12V pump on an off for an ebb and flow hydroponic system. I'm going to use a PIC16f628a, and I want a built in 12V PSU so that I don't need an external wall wart to power it.


What is the smallest/most efficient 127/220v in, 12v, 2a out switched mode psu schematic that can be easily built?

[Anonymous No.916600]

what are some good books for DIY radio and general wireless communication?

[Anonymous No.916636]

>>916536

the transformer in switching PSU is critical and often difficult to find. and why would anyone waster their time looking when you can get a ready made, UL listed, complete unit for $2 or less at the thrift store?

if your wall-wart bigotry is too severe, you can always take the circuitry out and use in your own case.

[Anonymous No.916646]

>>916636
>Switching power supply
>Water Pump

Dafuq?

[Anonymous No.916661]

>>916646
you know what a switching power supply is, right?

linear supply for 2A would need a huge transformer and be very inefficient compared to a smpsu.
since its for a pump at 2amps, any switching noise is utterly harmless.

>>916636
what he said.
laptops die all they time, but their powersupplies live longer.
go to a recyclingstation or something and grab a few.

if not, you could always grap a desktop psu and jumpstart it.
bam, 12v at more current than you'll ever need for almost nothing.

[Anonymous No.916666]

>>916661
Not him, but don't most water pumps run off of AC?

[Anonymous No.916670]

>>916279
The ground probe is laying around at the bottom if the picture and he's using tweezers in the other hand. Literally wtf.

[Anonymous No.916671]

>>916517
Also, free 100MHz hack

[Anonymous No.916696]

>>916666
Not 12V ones.

Electrically, a pump is just a motor; what the motor is driving doesn't matter. You can get AC motors and DC motors.

>>916536
Make sure that you have reasonable transient suppression. A switching PSU will be less tolerant than a linear PSU to an inductive load with flyback spikes.

But I strongly suggest you buy a PSU. If you're asking /ohm/ for advice, you probably shouldn't be dicking around with mains voltage.

Particularly given your application. One of the most common ways for weed growers to get busted is when their garden catches fire due to an electrical fault and the fire brigade reports it to the cops.

[Anonymous No.916697]

>>916428
> What is the function of the resistor in a passive high pass filter?
The cut-off frequency is determined by the product of R and C. If you just use the following stage as the resistance, then any variation in the resistance of the following stage will change the cut-off frequency.

Typically a passive filter would be used to drive something with high impedance (by using an active buffer stage if necessary).

High impedance inputs tend to mean impedance varies with frequency and temperature (you just try to make the minimum it as high as possible; you don't care if it's 100x that figure under some conditions).

It also means that the capacitance would need to be quite small, so parasitic capacitance could affect the frequency.

Basically, if you have a high, variable resistance, putting a lower fixed resistance in series with it results in the combined resistance being more stable.

[Anonymous No.916767]

>>916696
It's a 600ma pump.

I've driven a windshield pump off a 12v switched mode PSU, which was also powering an arduino. The psu didn't really care about the pump, since I was driving it with a tip122 with a flyback diode in parallel with the pump.

[Anonymous No.916794]

So I bought a soldering iron (max 60W, variable) to fix the fans on my 360. What are some good beginner projects that I could use it for?

[Anonymous No.916798]

>>916794

Check teh webs for circuits built around the 555 IC

[Anonymous No.916800]

>>916794
Also audio amp kits

[Anonymous No.916960]

>>916794
Make you own usb and audio cables

[Anonymous No.917057]

Do class D amplifiers always need a switching power supply? I want to buy one but it says "8-17 VDC supply 12v switching recommended". So now I'm slightly confused.

[Anonymous No.917059]

>>917057
Nope.

Some people don't like to use switching supplies for audio because of noise from the supply being coupled into the audio.

This is probably less of an issue for a class D amplifier given that the amplifier itself is based upon switching, so supply noise is probably less of an issue than the amplifier's own noise.

[Anonymous No.917060]

>>917059
Cool thank you

[Anonymous No.917075]

>>916794
Power supply kit - $20 on Amazon.

[Anonymous No.917088]

>>916536
Just do the timer circuit with a 555, dude.

[Anonymous No.917169]

>>916406
>>916407
it is analog feedback of the output line. The IC looks at the incoming voltage and trys to maintain it at a certain voltage. The pot simply makes it only get a relative sample and so scales the output acording to the sample.

I needed extra amperage from these, so i removed the pot from 1 and added a wire between two of them so one follow the other (master/slave)

[Anonymous No.917328]

>>916800
Any audio design with an lm380 should be ignored. It offers horrible audio quality

[Anonymous No.917363]

>>917328
>good beginner projects that I could use it for?

[Anonymous No.917418]

so I think I accidentally mcfucked my raspberry pi

By sheer coincidence,

if I were to say, hook a 12 volt fan to a 5 volt GPIO pin, would that have any potential of breaking something?

[Anonymous No.917428]

>>917363
None, it is meant for toy keyboards and the like not for audio amplifiers. I am all for learning things with basic components but teaching people things with crappy components while there are plenty of good alternatives around is just stupid. For example you can use a tda2003 (or tda2030 for more power) if you need a cheap amplifier. I even believe that those are cheaper.

Same goes for the lm741 in audio, almost any opamp that you'll find has better specs. I get that people are used to the lm741 and it has a history but we don't teach people to work with MS-DOS when we teach them how to use a computer either.

[Anonymous No.917430]

>>917418
Don't know about the protection of the gpio but I suppose that the fan draws wayyy too much current for a GPIO pin and might have blown up a few things.

[Anonymous No.917444]

>>915959
What equipment I should I get to start out? Any parts? A good beginer project?

[Anonymous No.917471]

Hey /ohm/ I've been following mjlorton's tutorials and trying to think of simple(ish) projects to do.

I have these bamboo desk plants I potted in voss bottles, my idea is to have them sit on a base with leds in them that shine up through the water but I want to make it so they're connected to a photoresistor that turns them on as the ambient light outside gets darker. Are there any similar projects I can look up to get an understanding. Probably powered by a wall wart.

[Anonymous No.917473]

>>917444
multimeter, soldering iron +solder +solder sponge +solder wic, breadboard, jumper wires

leds, resistors, transistors, capacitors, potentiometers, battery holders or a really cheap variable power supply ($20 or under)

Maybe an arduino if you want a basic intro to the programming side

[Anonymous No.917474]

>>917471

sounds like you want some solar patio lights. they're usually found for free in your neighbor's yard, or at the dollar store.

[Anonymous No.917475]

>>917474
Oh, didn't think of that, I'd rather DIY it for the learning experience though. I'll just look up DIY solar patio lights (but I guess it's so inexpensive that everyone just guts store bought ones). If not then I guess I'll just gut them myself. I noticed those lights are either on or off, what I'd like to do is have them fade in as it gets darker.

[Anonymous No.917479]

>>917475
>>917474
>>917471

So I've been looking into photoresistors and in my head I had it the wrong way around (light = lots of resistance, dark = not very much resistance).

Do they make photoresistors that work the other way around, looking into it I've only found people posting this circuit, I think I understand it but when the photoresistor is offering resistance wouldn't the current just flow to the ground and wouldn't that be a waste of electricity?

[Anonymous No.917484]

>>917479
Just make a voltage divider and hook up your load in parallel to the photoresistor.

[Anonymous No.917485]

>>917484
let me get this straight then

when they're in parallel the current will flow through the led when there's resistance in the photoresistor but it'll flow through the photoresistor when there isn't resistance in the photoresistor.

How does the voltage divider come into it?

So wouldn't the current still be flowing regardless of what works being done and isn't that a waste of electricity? Or am I just splitting hairs that don't need to be split?

[Anonymous No.917488]

>>916466
>However, while this is fine for generic parts (resistors, ceramic caps, switches, basic semiconductors like transistors and diodes, etc.), you might be a little wary of doing this for any proprietary stuff. Fake parts is a big thing in China.

To add to this, usually you can do your more typical shopping for modules (like LCD's, or temperature sensors) on websites like Sparkfun and Adafruit to get project ideas, then run to eBay and realize that Chinese bootleggers are selling those modules for like one tenth that price.

This is usually not an issue if you're just buying parts to prototype, or doing little home gadget projects, but if you actually want a design you can sell, you'll need to refine your supply.

Case in point, I have a genuine, Adafruit-bought Arduino for the explicit purpose of testing parts I'm buying from Chinese eBayers, as there will usually be some library and example code to use. I've yet to run into a problem with components, but I'm not running them under extreme conditions.

Also, this same principle for eBay applies to websites AliExpress and BangGood.

[Anonymous No.917490]

>>917485
Just thought of something, if power is flowing through something, say an led, will it only flow at the rate that the led consumes it or will it flow regardless (as if it were a short circuit) and the led just uses what it would use from that?

If I'm not making any sense, is it like a stream with a waterwheel catching the flow as it passes, or is it like a waterwheel in a pipe that stands in the way and only lets water pass as it turns?

I know these are stupid questions but I'm learning a lot from them.

[Anonymous No.917492]

Noob here. What pin is the ground pin on the actual board?

[Anonymous No.917494]

Hey ya'll, I've been curious about this lately, and was wondering if any of you guys have an answer.

Say I've got an amp that's stable down to 8 ohms, and distorts a 1v input at 8 on the volume knob. If I put a 4 ohm load on it, would it be safe to say you could turn it up to a 4 without any damage occurring to the amp? Half the volume at half the resistance should theoretically give the same amount of power going through the output transistors, correct?

If I explained that retarded, my apologies. Just something I've been wondering about lately.

[Anonymous No.917495]

>>917492
GND. Bottom left.

[Anonymous No.917504]

>>917490
The current through a diode (including an LED) is roughly exponential with voltage across it. In practical terms, that means that once you reach its forward voltage (typically 1.6-2V for a red LED), it behaves like a short circuit, so the current is determined by whatever resistance is in series with the LED.

[Anonymous No.917507]

>>917494
>Half the volume
Half the volume does not mean the same as half the power or half the output voltage. Volume setting "4" is much less than half of the power of volume "8". Audio pots are logarithmic, because your ears react to volume logarithmically.
>half the resistance should theoretically give the same amount of power going through the output transistors, correct?
Even if you replace "volume" by "power" or "output voltage", it isn't true. If it were half the output voltage / one fourth of the power, you'd have the same amount of current going through the output transistors, but since the output voltage would be lower, the voltage across the transistors would be higher, meaning higher power dissipation in them.

[Anonymous No.917510]

>>917507
I see. So is there any way to gauge what safe levels for half an amp's stable ohmage would be? Or is that a bunch of graduate level electrical theory?

[Anonymous No.917543]

>>917495
thank you!

[Anonymous No.917571]

>>917430
well fuck, I guess that's a result of my limited knowledge of electronics

I'd assumed it would just draw the highest amount the pin would supply and nothing more
I guess my home made case-fan-to-USB-laptop-fan is gonna fuck up my laptop eventually too. Shit.


if anyone else knows more about the pi, here's what happened.

It was running fine with the fan for a day before. Then, an unrelated bug from a piece of software forced me to hard reset the pi. When I went to plug it back in, I heard a pop and now the power and activity lights stay solid and nothing else happens. There's no visible damage, but obviously something is amiss.

[Anonymous No.917587]

how do you deal with varying supply voltage in general? when you have e.g. a 9V battery as supply, how do you assure you use it to the maximum?

for example, when you design your circuit for 9V it will obviously fail as soon as the battery reaches a certain voltage below 9V. now you could take a voltage regulator and design your circuit for lets say 5V, so you use the battery from 9V to 5V. would that be the right approach?

[Swifts !WiNDUPyWhk No.917605]

>>917587
I'd go with the 9V Battery with the circuit working around 5V.
You can also make a Battery monitor/Low voltage warning with an OP Amp that way, since you have the 5V to reference from.

[Anonymous No.917607]

>>917571
>I guess my home made case-fan-to-USB-laptop-fan is gonna fuck up my laptop eventually too. Shit.
USB is a consumer-facing technology and is protected against being shorted.

If you plug a load with too little resistance into it, it'll switch off, and Windows will lie to you and say "power surge on USB port", when it really means "voltage drop on USB port". Generally the entire hub shuts down, but modern laptops tend to have a hub for each port.

You can even dead-short USB without damaging it.

[Anonymous No.917614]

>>917587
Consider the expected minimum voltage and the expected maximum voltage, design your shit to work between those.
Batteries have datasheets like the other components and they specify the expected capacity vs. end of discharge voltage. For alkaline batteries it's typically 0.8V or 0.9V per cell, or 4.8V - 5.4V for 9V batteries. The extra capacity you get by going to 0.8V is usually pretty small, so if you need 5V, you might be willing to set the cut-off to 5.4V. Or you might be willing to sacrifice even more capacity, if 5.4V is a pain.

>>917510
There might be some simple rule of thumb (other than "don't use lower impedance speakers than the manufacturer says"), but I'm not aware of it.
That said, you need roughly 10 times more power to make shit sound twice as loud, so there's not that much point in trying to squeeze the every ounce of juice out of the amp. You could run it at, say, quarter power or so if you insist on using wrong speakers.

[Anonymous No.917619]

>>917605
This post definitely required a trip. Thank fuck you posted with one or we would have had no idea what you were talking about you faggot.

[Anonymous No.917620]

>>917605
>>917614
thanks

[Anonymous No.917632]

Can someone point me towards a decent guide for using the Hitachi 44780 display? All the ones I found just bandy about terms like CGRAM without explaining what it means, or just tell you how to wire it up then use someone else's code to operate it.

I just want to know how you tell it to display a character/string.

[Anonymous No.917731]

Hi!

If someone is interested, I just put a video of my DIY solder station for Hakko 907 irons online. Hope you like it.

https://youtu.be/qk1ttUM5XTQ

[Anonymous No.917754]

So I've been away from hobby electronics for quite a while. I've had some ideas and want to get back into it. I still have most of my tools, but the parts bins had dissipated over the years and a few moves. Does anyone out there do just generic hobbyist/experimenter parts pack? Something with just a good mix of miscellany. Resistors, caps, different families of transistors, 74xx, 40xx, 555/6, mixed op-amps, etc. Just something to cover the basics while setting up the shop.

[Anonymous No.917770]

>>917479

In that circuit R5 has a lot more resistance that R2, therefore the current will just flow thru R2 and the output voltage will always be

V1 * R2/(R1+R2)

it's not convenient to drive a led directly throw that circuit, since the current consumed by the LED will affect the output voltage

[Anonymous No.917771]

>>917770
Won't the current split itself proportionally to each resistor?

[Anonymous No.917773]

>>917731
Don't feel like watching but what are the yellow buttons for? Preset temps?

[Anonymous No.917778]

>>917771
yes
but the extra resistor is orders of magnitudes greater than the other two so its affect is minimal
do the calcs of the output voltage with and without r5 and see for yourself

if you want to use a photoresistor to drives something then you want the photoresistor as part of a voltage divider circuit, feed the output (middle of the voltage divider) into the base of an npn transistor, add your load for medium loads like leds or buzzer or something, add a relay with flywheel diode for different voltages like mains
choose your resistor value (and photoresistor) to minimise (wasted) current flow but with enough to turn on the transistor to let enough current flow

[Anonymous No.917795]

Beautiful circuit boards, /ohm/?

I love looking at old late 70's/early 80's boards, particularly memory sections, dem DIP packages all lined up neatly in rows oriented in the same direction.

[Anonymous No.917803]

>>917771
The current always takes the path with less resistance.

if you put a LED in parallel to R2 the current will flow directly through the LED putting its voltage to 0

if you put a resistor in series with the LED the output voltage of the "voltage divider" will undesirably vary

you have to somehow make the input to the LED have "infinite" resistance (Op amp, BJT, arduino,etc)

[Anonymous No.917805]

>>917803
>The current always takes the path with less resistance.
No it doesn't.
>what is the parallel resistance formula

[Anonymous No.917806]

>>916798
what's this do? Just scrolls through the leds?

[Anonymous No.917809]

>be building high altitude balloon
>first think of having an SD card to store data
>then fall back to RTC eeprom of 32K
>fall back arduino 1K memory

I love memory optimization :)

[Anonymous No.917811]

>>917805
Yes it does, I'm not saying that all the current goes that way, just that more current will go to the path with less recistance

The parallel resistance formula is just a way to find an equivalent circuit, it is not the real circuit

[Anonymous No.917812]

>>917811
>just that more current will go to the path with less recistance
Okay, it sounded like you were saying it was a complete "winner takes all" type scenario.

[Anonymous No.917836]

>>917571
USB is pretty foolproof. GPIO and the IO on, for example an arduino, is often not protected in any serious way and overloading them may result in damage. The surge of power needed to start the fan after the reboot may have killed the GPIO ( which I believe is coming straight from the CPU? )

Is anything abviously broken on your raspberry pi?

[Anonymous No.917872]

>>917836
no, it looks perfectly fine. nothing's blown, there's no scorch marks, there was no smoke, just a tiny pop.

[Anonymous No.917907]

>>917571
Should have used a powered switching transistor.

[Anonymous No.917927]

>>917811
A consequence of the parallel resistance formula* being about the reciprocal of a sum is that adding a resistor /always decreases/ the total resistance. Always. You could put a quadrillion gigaohms in parallel with your existing resistors, and the total resistance would still go down.

Adding a resistor adds a term, which must make the total larger. If the total must be larger then its reciprocal must be smaller.

Further, you're adding up reciprocals, so if the resistance you're adding is huge, its reciprocal must be tiny. If you've already added a small resistance, then adding a large resistance means you're adding a small number to a big number. The reciprocal of the existing small resistance is dominating the calculation.

This means you can intuit (and be correct in intuiting) that putting a massive resistance in parallel with a small resistance will change the total resistance to be smaller, but only by a very tiny amount.

* 1/Rt = 1/R1 + 1/R2 + ..... 1/Rn

[Anonymous No.917938]

>>917927
>putting a massive resistance in parallel with a small resistance will change the total resistance to be smaller, but only by a very tiny amount.
1 megohm // 0 ohm (aka dead short) = ~1 megohm?

I don't think so.

[Anonymous No.917952]

>>917938
How could you possibly read that from that?
Here's a worked example.

0 ohms can't exist in nature, and 1/0 can't exist in mathematics, so let's use 0.0000001 ohms instead.

Existing state (0.0000001 ohms)
1/Rt = 1/0.0000001
1/Rt = 10000000
Rt = 0.0000001

Now let's add a one-megaohm resistor:
1/Rt = 1/0.00000001 + (1/1000000)
1/Rt = 10000000 + (0.000001)
1/Rt = 10000000.000001
Rt = 0.0000001 (at least to seven significant figures)

The one-megaohm resistor added in parallel to the original resistance of 100 microohms has made so little difference that most calculators round it out and don't even bother to mention it.

[Anonymous No.917957]

>>917632
Check out Patrickhooddaniel on YouTube or his site newbiehack. All lcds are pretty much the same pins I think all you have to do is check the data sheet and match up the pins to the microcontroller. He has a code tutorial on YouTube thats really good and explains basic sending strings and individual characters. I have a newhaven display I got running with the tutorial.

[Anonymous No.917960]

>>917952
As far as any engineer cares, 0.0000001 is equal to 0.

[Anonymous No.917963]

>>917938
Fucking reading comprehension....
>Putting a massive resistance in parallel with a small resistance will change the total resistance to be smaller [than the small resistance already was], but only a very tiny amount
Got it?

Also I don't know why you two idiots are arguing, sounds to me like you're both saying pretty much the same thing in different ways.

[Anonymous No.917994]

>>917960
this

[Anonymous No.918031]

>>917957
I'll have a look. I spend last night putting bits of info together from various sites. I still have no idea what the deal is with the "duty line" is though. Apparently 0 means 1/8 or 1/11, which, again, means nothing to me and I can't find a simple explanation form.

[Anonymous No.918032]

>>917795
looks like a high end pcb, but damn that resistor bodge...

[Ken Rockwell No.918034]

Found this in a family members box 'o junk.

What is it, or what does it do.

[Anonymous No.918035]

>>918034
It's a clock, Mr. /p/

[Ken Rockwell No.918036]

>>918035
How is it a clock?

The big component is a magnetic switch thing, from what i remember seeing in some videos. How does that have anything to do with a clock?

[Anonymous No.918037]

>>918036
you count the clicks it makes. then its up to you to convert it to actual time.

[Anonymous No.918039]

>>918036
ignore the troll

can you post pics of the underside?

[Ken Rockwell No.918041]

>>918039
Thought something felt out of place.

Here is the underside.

[Anonymous No.918048]

>>918034
>>918041
few quick conclusions:

-the 2 wires and two resistors are not connected to your relay circuit
-you can figure out positive and negative rails through the LED (flat side on the bottom rim is negative)
-the relay powered something at those three cut off connections at the top closests to the 4 wires on the right

Don't have time to reverse engineer this right now but I suppose that the transistor will power the relay depending on the position of the potentiometer (and maybe some kind of sensor that was connected to the circuit. maybe a DIT thermostat or something

[Anonymous No.918049]

>>918034
Looks like a generic thermostat / twilight switch / humidity switch / whatever. The actual sensor determining the use seems to be missing.

[Anonymous No.918054]

>>917773
left: increase/decrease temp
right: 3 preset temperatures, correct.

[Anonymous No.918211]

This is an optocoupler - why the diode across the input? Protection against reverse bias? I didn't know LEDs needed that.

[Anonymous No.918224]

>>918211
maximum reverse voltage of an LED is in the range of 5 or so volts, so yes it is indeed protection for the LED in this case.

[Anonymous No.918226]

>>918211
probably a clamp diode
google it

[Anonymous No.918230]

>>918224
Would I need it, given that it's only going to be used with MIDI connections? So there's little chance of the LED being reverse biased.

[Anonymous No.918232]

>>918226
I found an example and it shows a capacitor in series too.

[Anonymous No.918236]

I want to make my coffee table essentially a large USB hub, and if I could, also power a few leds. Is my best bet to get a cheap USB hub and break out the ports to the edges of the table?

[Anonymous No.918238]

>>918230
It also provides some protection against ESD and other overvoltages.

[Anonymous No.918251]

>>918230
why not? a 1n4148 is under a cent per piece...

[Anonymous No.918252]

>>918236
Yep.

Be aware that if you take a hub out of its shielding, you make it a better RF antenna, which can lead to weird thingsl like devices intermittently disappearing, computers switching on every time the fridge thermostat switches, that sort of thing.

Just for charging stuff you'll be fine though.

If it was me making a USB-charging coffee table, I'd use an Orico multi-port charger instead of a USB hub.

If you're not just charging stuff, keeping the usb hub guts in a metal box grounded to the shell of the incoming USB B (and therefore the upstream host's case) will reduce RF interference. USB is designed so that the cables and the plugs form an uninterrupted RF-shield that completely encloses the power and the data wires.

[Anonymous No.918254]

>>918251
I've only got 2. I need to get some kind of selection box.

[Anonymous No.918276]

>>918230
if you are 100% sure the input signal wont damage the led then you can forget about it sure

>>918232
capacitor blocks dc, so if you want to remove dc from your signal then use the cap

[Anonymous No.918278]

>>918276
>capacitor blocks dc, so if you want to remove dc from your signal then use the cap

I mean the circuit for a clamp uses a cap, which is principal to its operation. The other diagram with the optocoupler has no cap so it can't be a clamp circuit.

[Anonymous No.918303]

>>918236
You could just make a few usb extension leads and run them to where you want.

[Anonymous No.918307]

When I wire up a yellow LED to a 4V supply, via a 330 resistor, I read about 5.9mA, when it should be about 12mA. When I take out the LED and just load the resistor, the current is more or less 12mA. What gives?

[Anonymous No.918311]

>>918254
http://www.ebay.com/itm/100Pcs-1N4148-IN4148-DO-35-switching-signal-Doide-/381374423772?hash=item58cbafe2dc:g:9G0AAOSw8cNUO3HE
here you go. Just a single buck for 100, free shipping.
Just buy some common components in bulk, such as 1n4148, 1n4007, bc547/2n3904, bc557/2n3906 etc

[Anonymous No.918313]

>>918307
According to Ohm's law, this circuit has a resistance of about 690. Is this right? I tried to measure the resistance of the whole circuit and then the LED itself. Though it lit up, the meter wouldn't give me a reading.

[Anonymous No.918314]

>>918307
An LED is not a linear component as the resistor is.
See the attached file for an example diode, over which a voltage is placed. As you can see, there is no current when there is no voltage, but when you start increasing the voltage, as soon as it reaches a certain threshold voltage, the current increases rapidly. This is the diode forward voltage.
In a regular LED with resistor circuit, you can assume that if you have for example 12mA flowing through the diode, the diode is dropping that forward voltage (for a yellow led ~2V, silicon diode ~0.7V).
Following kirchoffs voltage law, that means the rest of the voltage is dropped over the resistor (2V over the resistor, because 2V + 2V = 4V supply voltage). You can now apply ohms law (R=U/I) to find the appropriate resistor.

Also, if you apply I=U/R to a resistor of 330 ohms, you will find that the theoratical current is ~6mA, which agrees with what you found.

[Anonymous No.918316]

>>918313
see
>>918314
The LED is non linear and thus does not have a 'resistance'.

[Anonymous No.918317]

>>918316
>The LED is non linear and thus does not have a 'resistance'.
This is what I thought, which is what confused me. I guess I thought the LED would have no effect on the current, and it would be the same with or without the LED.

[Anonymous No.918318]

>>918317
have you read >>918314?
It does not have a 'direct' effect on the current, but i guess you could say that it influences the voltage distribution, and thus the voltage over the resistor, which does affect the current.

[Anonymous No.918319]

>>918314
>Also, if you apply I=U/R to a resistor of 330 ohms, you will find that the theoratical current is ~6mA, which agrees with what you found.
Do you mean voltage by U? If so, 4/330 = ~12mA.

[Anonymous No.918321]

>>918319
Oh you mean the forward voltage. Right.

[Anonymous No.918322]

>>918319
No, as discussed before, there is a 2V voltage drop over the diode, which leaves 2V voltage drop over the resistor to give the supply voltage of 4V. If we take U = 2V (voltage over resistor) and R=330ohms it gives indeed 6mA.

Here is a quick summation of ohms law:
U = voltage over the resistor
I = current through the resistor
R = the resistance
The keyword here is 'the resistor'

Ohms law only hold with the voltage over a resistor, you cant pick 4V because that is the voltage over the entire circuit, not just the resistor. As discussed, 2V drops over the LED which leaves 2V for the resistor.

[Anonymous No.918323]

>>918321
no no, not the forward voltage. In this case the supply voltage - forward voltage. What is left is the voltage over the resistor.

[Anonymous No.918329]

>>918323
Yeah that's what I meant sorry. I'm stoned and they were both 2V in the example. I knew what you meant though.
>>918322
Right, so because the resistor only has 2 across it (about half the supply), that means it only allows half the current it would have without the LED?

[Anonymous No.918330]

>>918329
>Right, so because the resistor only has 2 across it (about half the supply), that means it only allows half the current it would have without the LED?
It only allows as much current through it as ohms law dictates. In this case 2 volts, which conveniently is half the supply voltage of 4 volts thus in this case yes, half the current as without the led.

This would of course not be the case if the supply voltage was, for example, 12V.
That would mean 10V across the resistor etc etc

[Anonymous No.918331]

>>918330
>This would of course not be the case if the supply voltage was, for example, 12V.
>That would mean 10V across the resistor etc etc

Yeah.

[Anonymous No.918339]

Another thing - I measure 4v at the point where the PSU goes into the breadboard, but 3.89v at the components. Where did that go?

[Anonymous No.918347]

Say if I have 2x1.5=3v batteries powering an led, they'll power it for x amount of time.

But if I have 4x1.5v=6v batteries and halve the voltage with a resistor and power an led will it run for twice as long?

[Anonymous No.918350]

>>918339
Losses in the breadboard's copper traces.

Though that would also imply that you're pushing a crazy amount of current in order to have a voltage drop of .11v solely from copper losses.

>>918347
It all depends on how you arrange the batteries.

Use Ohm's law.

[Anonymous No.918352]

>>918350
Okay, so if I pair two batteries in parallel, do that twice so I have 3v with twice the capacity it'll last twice as long. But I don't quite understand what a resistor would do in what I first said if all the batteries were in series?

[Anonymous No.918353]

>>918352
The resistor just makes sure the current going through the led isn't so high it'll blow the whole shmack.

Lower operating current also means longer life, but also less brightness.

[Anonymous No.918356]

>>918339
Every wire is essentially a resistor. If a current runs through it, the voltage over that wire is given by again ohms law:
U = I R
>>918347
I assume you have them in series. Then no.
That is because power by your circuit is consumed by P = U*I.
Power in the battery is constant, so when you have double the batteries, you have double the power. However, you have twice the voltage, so plugging 2*U with the same current into the above equation also gives 2 times the power draw. That is because more power will be dissipated in the resistor and converted to heat.

[Anonymous No.918357]

>>918356
....Who uses U in place of V?

[Anonymous No.918358]

>>918356
>>918353

Ahhh that's what I was getting at, so the resistor consumes the voltage instead of holding it back and only letting it trickle through.

basically it's still flowing at 6v but the resistor eats some of it up before it reaches the led so it doesn't blow.

[Anonymous No.918359]

>>918352
>>918356
In the situation where instead of two batteries in series, you have two times two in parallel in series, then yes, you have twice the capacity btw.
>But I don't quite understand what a resistor would do in what I first said if all the batteries were in series?
If you have the same resistor, you again take the supply voltage, substract the forward voltage (2V) and calculate the current. You will find that with the same resistor and twice the voltage, there is about twice the current through it.

[Anonymous No.918361]

>>918358
Well done anon.

Have a cookie.

[Anonymous No.918362]

>>918357
Almost all schools, research institutions and universities in the EU.
Check the wiki page, it is right next to V as a symbol. That is because V is also often used for velocity, and thus U is used to remove any doubts about what it represents.
>>918358
You can see the resistance as literally a resistance. It restricts the amount of electrons (current) coming through.
With a water analogous situation:
A very large tank holding water is a voltage source (battery) and if you connect a small pipe to it, it of course restricts the flow of water a lot better than a larger pipe. That pipe is the resistance.
Also, if you increase the water level of the tank (i.e. increase the voltage), the pressure will rise, and there will be more flow for a given pipe (i.e. with a higher supply voltage more current for a given resistance)

[Anonymous No.918364]

>>918362
>EU
Ah. That explains it.

[Anonymous No.918366]

>>918364
Generally well accepted in most of the world FYI.
Although it doesnt matter at all. The equation stays the same.

[Ken Rockwell No.918368]

>>918048
Thanks for that

I appreciate it. The 4 resistors (2 + the wires) were put on by me when practising my soldering skills.

[Anonymous No.918369]

>>918362
>You can see the resistance as literally a resistance. It restricts the amount of electrons (current) coming through.

But the smaller pipe (resistor) doesn't consume electricity it holds it back and only lets it trickle through meaning it would take longer for all the water to flow through it, which contradicts
>>918358
>>918356
>>918353

[Anonymous No.918370]

>>918369
You'll find that there's a voltage drop across it.

Kerchov's voltage law must still hold true.

Or just build the damn thing and measure it.

[Anonymous No.918372]

>>918370
>>918369
That, and V=IR.

[Anonymous No.918374]

>>918369
No, the pressure on the other side of the smaller pipe is lower, corresponding with a lower voltage.

The more i read your comments the more i think you should not be doing this while high.

[Anonymous No.918376]

>>918374
>>918369
O yeah and the pipe does 'consume' electricity (water), namely by friction with the wall.

[Anonymous No.918377]

>>918376
And that friction would just be the resistor turning the current into heat?

[Anonymous No.918378]

>>918357
people in first year who are trying to show off
>mfw i mark people down for using u instead of v even though they are taught to.

[Anonymous No.918379]

>>918374
Not high, just stupid.

[Anonymous No.918380]

I hate it when people try to come up with dumb analogies for electrical principles.

[Anonymous No.918382]

Jesus what have I started?

[Anonymous No.918384]

>>918377
I think you almost got it.
Although stricly not the current turning into heat, because the current going into a resistor comes out the other end. Just as the flow with water. The same amount of water that goes in the pipe, comes out at the end right? It is just not the same speed (indicating energy has been lost). Energy is lost in the resistor and the pipe.

[Anonymous No.918385]

>>918278
>the circuit for a clamp uses a cap, which is principal to its operation.
no its not
not the kind of clamp i'm talking about.
i guess i mean a clipper?
but everyone i know calls that a clamp
i guess we are lazy or dumb or something

[Anonymous No.918386]

>>918378
>implying i am in first year
no, i am starting my master in February.
And you must be seriously autistic to do something like that.
>>918380
It is literally the same thing. This system equivelance is really handy, because if you are well thought in one domain, you can translate a problem from other domains to said domain and solve it. Also, you can convert them to a generalized problem which eases multi domain simulation. And It can even be extends to more complicated circuits such as transformers, capacitors and inductors. Lookup bondgraphs. Voltage, pressure and temperature for example are all efforts, and their respecitve flows are current, waterflow and temperatureflow.

[Anonymous No.918387]

>>918386
>And It can even be extends
derp, should be can even be extended.

[Anonymous No.918388]

>>918387
Has Anyone Really Been Far Even as Decided to Use Even Go Want to do Look More Like?

[Anonymous No.918392]

>>918388
Well the content is clear and i guess that is what matters.

[Anonymous No.918409]

>>918384
So electricity is like water but not "exactly" like water. I think I was just taking that analogy too literally.

[Anonymous No.918524]

Can someone please do a safety check for pic related please?

I'm going to be testing it out tomorrow morning.
The capacitor is 2700 uF and will be charged up to 40V. A 9.6V battery with a boost circuit will increase the voltage which will then connect to a capacitor.

The capacitor discharges into the solenoid coil to make it kick stronger.

The alligator clips shown and the toggle switch will have their bare wires electrical taped before the whole thing starts.

[Anonymous No.918525]

>>918524
It was made following this diagram.

[Anonymous No.918527]

>>918524

[Anonymous No.918530]

>>916279
not like he can even touch the the pads, since the ICs are in BGA packages

[Anonymous No.918543]

>>918385
*shrugs*

You said Google it, so I did.

[Anonymous No.918546]

>>918527
No flyback diode?

[Anonymous No.918610]

>>916279
Its a stock photo, those fuckers have no idea what to do.

[Anonymous No.918621]

>>918610
I know, right? Clearly, he should have used a socket wrench!

[Swifts !WiNDUPyWhk No.918622]

>>918621
>>918610
>running a CPU that hasn't been properly torqued to spec.

[Anonymous No.918630]

>>918622
>that hasn't been properly torqued to spec.
I don't own a torque wrench that measures that low and doubt many average Joe's do.

[Anonymous No.918641]

>>918622
> tripfags posting inane rubbish

Why am I not surprised.

[Anonymous No.918665]

>>918546
Where should I place it?

Sorry I'm new.

[Anonymous No.918669]

>>918665
I'd have thought across the solenoid, with the stripe on positive. Maybe solenoids don't need them like motors do, not sure.

[Anonymous No.918694]

I am driving some motors using a set of batteries. The whole system operates at 60V and 60 A peak. During braking the motors produce high transient voltages as their drivers are four quadrant ones (unfortunately for me). Any idea how to dump the excess power? I was thinking a TVS along with a high power resistor but as battery voltage drops the TVS won't work. Any ideas?

[Anonymous No.918697]

>>918694
No flyback diode?

[Anonymous No.918704]

>>918697
Won't work.

The motor driver is constructed in such a way that when braking it dumps the extra energy back to the supply. That is the supply goes from 60V to 70V and more.

[Anonymous No.918715]

Can somone pls help me?

I am trying to run a 1A phone USB charger off a lighting ring. I've made a spur from one of the LED GU10 lights (there are 4), but it only charges the phone when the lights are switched ON.

I'd like it to draw power all the time, regardless of whether the lights are on or not. What are my options?

[Anonymous No.918753]

I'm trying to work an HD44780 display with an Arduino. I'm sure I've wired it correctly but it shows just a top line (it's a 2x16) of black blocks.

I'm just trying to write an E to the display right now.

I'm trying to recreate the procedure from this tutorial here (that's where those hex values came from).

Is there anything wrong with the code?

[Anonymous No.918767]

>>918753
http://web.stanford.edu/class/ee281/handouts/lcd_tutorial.pdf

Should probably post the link.

[Anonymous No.918769]

>>918753
>>918767
Yes I know I probably should have used an enum, but I just wanna get something on the display before I refine it.

[Anonymous No.918776]

>>916296
Look up the Digilent Analog Discovery board.

[Anonymous No.918798]

>>918753
are the pins correct?
show hardware

[Anonymous No.918801]

>>918798
You want a schematic or a pic of the setup? Thing is, the connections to the Arduino don't seem to matter. I remove them all and just leave the backlight, contrast and power - and it's the same. So whether the code is wrong or right, it's not the problem right now.

[Anonymous No.918805]

I'm currently hunting down a 9v Relay Switch, since 12v won't cut it and 6v in my case won't work.... Has anyone got any specific suppliers that could help me out?

[Anonymous No.918807]

>>918798
>>918801
I just realised there's an example sketch, so I loaded it and it works. I'm glad the display isn't fucked - and the wiring was correct too.

Which just leaves the code.

Although there's this library, I'd still like to solve this issue.

[Anonymous No.918818]

>>918316

It has a dynamic resistance.

[Anonymous No.918822]

>>918807
>I'd still like to solve this issue.
For one, I want 8 bit mode - and I eventually want to use a shift register so I'd write my own class for that (as I did for an LED display), so figuring out what's wrong with my code is something I'd like.

I'll play around but if there are any suggestions in the meantime I'd be grateful.

[Anonymous No.918828]

>>918822
I got this from another forum:
>A 2-line moderate-width character LCD that is uninitialized shows black boxes on the top line, and not on the bottom.

Exactly what I was getting so I'm not initialising properly. I'm stumped for now and it's nearly my bedtime.

[Anonymous No.918833]

>>918828
uh.. did you wire in a pot for contrast adjustment?

[Anonymous No.918837]

>>918833
Yeah the wiring is fine. As I say, I run the Arduino example sketch and it works. Just now I commented out most of my code and replaced it with explicit pin writes (for the same values) and no progress.

Could there be something up with the instructions in >>918767

[Anonymous No.918839]

>>918837
>no progress.
Well, to be specific, when I upload my code it just "overlays" the "hello world" from the official Arduino code with a P and a corrupt character of random pixels.

[Anonymous No.918842]

>>918753
are you driving r/w line?
should be write, right?

[Anonymous No.918843]

>>918839
Now it's doing other things. Arbitrary things. So let's say "undefined" behaviour is what I'm getting. The last time I ran it, it somehow restored the previously corrupted display contents from when I uploaded the official (working) example code before

So I upload the official code, it works. Then I upload mine and it mildly corrupts the display (that still persists from the other sketch), then I upload my code again (just for shits) and restored the corruption it previously made. WTF?

[Anonymous No.918862]

>>918843
I had a look around the Arduino library code and found a few things. It uses write4bits () a lot behind the scenes. Each time calling pulseEnable ().

So it seems you can't just set the enable pin high, wait a bit, then set it low again. Why the fuck not?

I got a shift register to work on pretty much my first attempt, no hassle. The fuck is this bullshit about having to pulse the enable pin? Never read that on any of the online resources on the matter.

[Anonymous No.918867]

>>918862
So I put that in my own code and... it doesn't work. I dunno, maybe it's the bit masking code. Too tired to continue for now.

[Anonymous No.918868]

>>918842
It's grounded so it's not an issue. I think we can eliminate any wiring issues. It's my code.

[Anonymous No.918883]

>>918862
Why wouldn't you pulse the enable pin?

D'you think the bits will sink in better if you leave them to steep?

[Anonymous No.918886]

>>918753
Look at your for loop. Closely.

[Anonymous No.918898]

>>918805
>I'm currently hunting down a 9v Relay Switch, since 12v won't cut it and 6v in my case won't work.

you can just stick 4 diodes in series with a 6V relay to make it a 9V relay. or a small resistor, calculated in the usual way.

[Anonymous No.918925]

I need to knock up a quick light, there's a resistor before each led, could I add the sum of all the resistors into one larger resistor right after the battery?

[Anonymous No.918926]

>>918925
No, because then your LEDs are connected to each other by a wire, so all the current intended for all the LEDs will go through the one with the lowest forward voltage until its magic smoke comes out, at which point it all goes through the next-lowest one, and so-on until all your LEDs are dead.

[Anonymous No.918927]

>>918926
Is that because the resistors share the load at whatever ratio but the leds don't?

[Anonymous No.918929]

>>918927
Correct. LEDs are non-resistive loads, so you can't use resistor formulae when dealing with them.

[Anonymous No.918936]

>>915959

I have some questions about LEDs. I'm building a wicked vape and I'm gonna hot glue 50 of them to an Altoids case.

Each LED is going to be controled by a seperate Arduino. It's gonna hook up to my Iphone through the Ethernet shield and flash 20 different colors. Look for it on kickstarter

What do LEDs have to do with electricity? How do I stop myself from choking on them?

also, what is a resistor?

[Anonymous No.918937]

>>918936
>not using a PIC

[Anonymous No.918939]

>>917723
>reusing old/broken smartphones
anyone interested?

[Anonymous No.918941]

>>918936

[Anonymous No.918972]

[Anonymous No.918977]

Why did they change V to E but only in some places and only after E was well known, it only makes things more complicated for the sake of "correctness", and how does U come into it?

I just picked this up and they're already fucking with the terminology

[Anonymous No.918982]

> they're already fucking with the terminology

the eternal conflict between tradition and modernity. did you know that until around 60 years ago, they used mostly positive-ground circuits, where all the voltages were negative ones. how stupid was that!

[Anonymous No.918987]

>>918982
Isn't that the way if works on the atomic level? And I don't care as long as they pick one and stick with it.

[Anonymous No.918988]

>>918525
>>918527
>>918524
I tried this circuit today.

It does make the solenoid fire, but it doesn't look like the capacitor discharges all the way since leaving the switch on the discharge position causes the solenoid to remain in its maximum position. In fact, after discharging into the solenoid and disconnecting the power source my multimeter still shows a maximum charge on the capacitor. The capacitor voltage doesn't even decay while the multimeter is in place.

What is going on?

[Anonymous No.918989]

>>918977
> Why did they change V to E
What are you referring to exactly? I've never seen E used for Volts. I think you might be getting confused by a textbook that uses it for Electric field (volts per metre), Energy, or some other value.

> and how does U come into it
Are you sure it isn't a µ? That's the symbol for micro. 1µF = 1 micro Farad = 0.000001 Farads.

[Anonymous No.918991]

>>918989
Electromotive force

> and how does U come into it
Nevermind, the guy was just writing his "V" like "u"

[Anonymous No.918992]

>>918988
Also, the voltage is only present on the capacitor.

Measuring anywhere else on the circuit, even where the solenoid wire connects to the capacitor, results in 0V.

[Anonymous No.919007]

>>918988
>>918992
Are you sure the switch leads are what you think they are? Sometimes things are weird and the spdt common isn't in the center. Your symptoms would make sense if the switch wiring had the cap connected to the battery and solenoid at the same time when you set it to discharge.

[Anonymous No.919008]

>>919007
To be honest, I don't know about the switch pinouts. That looks like another thing I'll have to check.

[Anonymous No.919011]

>>919008
Never mind it looks like I had it right. Common was in the center.

[Anonymous No.919015]

>>919011
>>919007
Here's another weird thing: the capacitor discharges very slowly.

Like I said, it doesn't seem to lose voltage at all when I check the voltage with my multimeter. I took the capacitor off the breadboard and touched its leads with a very low value resistor (~200 ohms). I held the resistor there for around 20 seconds and it still did not entirely discharge. The voltage at that time was 10V even though it had been around 45 time constants which should have depleted the capacitor.

[Anonymous No.919016]

>>918982
Always fun when you find some really old textbook in the archives of your university library

[Anonymous No.919026]

>>919015
You measured the voltage after removing the resistor? The remaining voltage is due to dielectric absorption.
https://en.wikipedia.org/wiki/Dielectric_absorption

If you measured 10V after 20s with resistor connected, then you should check your resistor or scrape off oxidation from wires.

[Anonymous No.919030]

The board I designed has arrived.
But since work closes in about 90 minutes and the wrong multiplexer was ordered I can't test it.
Also the programming is not finished.
Also the input jack is wired wrong because I was given the wrong footprint model.
FUCK

[Anonymous No.919031]

>>919030
Bottom

[Anonymous No.919033]

>>919030
>wrong footprint model
Always. Check. Everything. Even in an $X billion company with a department devoted to library support we get burned by shit like this.

[Anonymous No.919041]

Wow, /ohm/ is moving pretty fast these days. It's been a week and it's almost time for a new thread.

Proud of you, /ohm/!

[Anonymous No.919050]

>>918883
I don't know. My experience with shift registers tells me that a simple in, out, put the kettle on, should do the trick.

It's all academic because pulsing didn't make a difference (to my code - for the LiquidCrystal.h file it's a different story it seems).

[Anonymous No.919052]

>>919050
What do you think the pulseenable function does?
It pulses the enable pin.
What is the difference between pulsing the pin and turning it on and off?
Pro tip: There is no difference.

[Anonymous No.919053]

>>918886
I assume you mean the second one. I can't see anything. Do you mean the declaration of the loop or the bit masking and stuff within it?

I hope you're not trolling me because I can't see it.

[Anonymous No.919054]

>>919041
Not only /ohm/ but a thread on /diy/, so proud

[Anonymous No.919056]

>>919052
>Pro tip: There is no difference.
As in there's no difference whether I pulse the pin or not, shit still ain't no work.

[Anonymous No.919059]

>>919056
If I had the arduino code I would look at it for you but I don't so I can't. Hard to say what's wrong with what you are doing

[Anonymous No.919060]

>>919059
Shit, I did some serial.prints to see if the loop executes. It doesn't. I replaced the B0 and B7 with 10 and 3 and it suddenly worked perfectly.

WTF? Why should that make a difference?

[Anonymous No.919066]

Why is 63/37 solder much more expensive than 60/40? The only difference is mixture ratio right?

[Anonymous No.919142]

>>919060
Is it because it thinks B0 means zero in binary? If so then wouldn't it flag an error when I tried to define B0 as 10?

[Anonymous No.919144]

>>919053
Do you want me to just give you the answer?

I'd rather you solved it yourself, because you won't learn as much from me just telling you.

[Anonymous No.919145]

>>919142
Just tested it out in the IDE. Seems like that's the case. Well, that's fucking irritating. Basically a bug in the software. Just for comparison I tried to #define 0xAA 55 and, rightly, it errored.

Seems like they forgot to disallow you from using binary literals as identifiers.

>>919144

Um yeah I've sorted it.

>>919060
>>919142

[Anonymous No.919150]

>>919144
And there's no need to condescend. I'm not some kid. Where applicable, I prefer to use hex so I'd essentially forgotten all about binary literals.

In any case, I'm not sure they're even standard notation for C/C++. I'm aware that some accept binary in the form of 0b... but, as I say, I never use binary notation.

[Anonymous No.919151]

Okay so I changed all those definitions to D0...D7 and it's back on track. All that hassle because I thought it would make more sense to use B for "bit" as opposed to D for "data".

So that's that, I guess. Thanks all for your input. I'm still pretty annoyed that the compiler let me declare a numeric identifier but, hey...

[Anonymous No.919157]

>>919150
Binary literals are new in C++14, but many have supported some syntax as an extension.

[Anonymous No.919159]

>>919157
Should add that C++14 binary literals use the '0b' or '0B' prefix, Arduino seems to use some custom horseshit.

[Anonymous No.919161]

>>919159
>Arduino seems to use some custom horseshit.
Quite. I messaged them to let them know. For all I know they might already know but I couldn't find it on their known issues list.

[Anonymous No.919183]

>>919026
Yes I measured after removing the resistor. It's a brand new capacitor.

[Anonymous No.919202]

>>918524
I think that orange wire is what is causing the problem since it looks like it allows current to go to the solenoid and the capacitor at the same time.

[Anonymous No.919204]

I got one of those Hantek USB oscilloscopes. I hooked up a probe to the built in square wave and turned the trimmer on the probe til the edges squared. Is the trimmer supposed to go around indefinitely, or have I broke it already? If it's only supposed to turn within a small range, then it offered virtually no resistance to being overturned.

[Anonymous No.919208]

>>919204
Does the trimmer affect the waveform? Once you've tuned it, you're not really going to touch the trimmer anymore.

[Anonymous No.919209]

>>919208
It did at first. Now it doesn't (but at least it's tuned correctly for now). So I've ruined it?

[Anonymous No.919210]

>>919204
>Is the trimmer supposed to go around indefinitely
Pretty much all of them do that.

[Anonymous No.919211]

>>919210
Fair enough. Turning it no longer has an effect on the sharpness of the edge though.

[Anonymous No.919214]

>>919211
Well, shit sucks, maybe.
I don't know about Hantek's probes, but at least the trimmers in my probes are so hard to reach that unless I'm really careful, the result is that the "trimmers" seem to have no effect. Maybe your probes have the same problem.

[Anonymous No.919219]

>>919214
It seems a little noisy too. Well it was less than 50 quid. It's a decent starter oscilloscope. At least now I can see what's going on in a circuit which is gonna be enormously helpful.

[Anonymous No.919228]

>>919204

the trimmers in scope probes are tiny variable capacitors, like 0-30pF. they usually dont have a stop, so they rotate 360-deg. they're also quite fragile, and you normally use a plastic or nylon screwdriver to adjust them. sounds like you might want to replace yours.

[Anonymous No.919229]

>>919228
I used a jeweller's screwdriver. Then I noticed the little tool in the bag, but that was metal tipped too. If they're meant to turn constantly then what could I have done to damage them?

[Anonymous No.919273]

>>919202
Yes that fixed it. It works as expected now and the capacitor even discharges fine.

[Anonymous No.919295]

If a transistor is essentially a variable resistor then why isn't its value expressed in terms of, say, ohms (between C and E) per mV at the base?

[Anonymous No.919296]

>>919295

wut

[Anonymous No.919297]

>>919296
It's a resistor of sorts, right? So why don't they say things like "this transistor increases by x ohms for every y milliamps"? Is it because it's not linear or something?

[Anonymous No.919298]

>>919297

It amplifies, it doesn't introduce resistance.

[Anonymous No.919299]

>>919295
Because it strictly isnt. The device is not even linear so even if it was the case, you could not express it as ohms per mV.
If you are interested in transistors your best bet is to either learn about them via good internet tutorials (i.e. university lectures and stuff) or in books, and not via arduino tutorials or something.
If you are into electronics, the book 'the art of electronics', the 3rd edition just recently caim out. It is well worth the time and money (it isnt that expensive for 1220 pages, only 70 euros where i live) to read. It will teach you a solid basis for electronics, and every aspect of it.
Because i reckon that you lack the foundation, or at least the proper resources to learn electronics tbqh.

[Anonymous No.919301]

>>919297
Ofcourse it is non-linear, otherwise it would not be able to amplify a signal.

[Anonymous No.919302]

>>919298
No, it does the opposite right? Its impedance goes down the higher the base current goes. It's a variable resistor with the "fader position" represented as current at the base.

[Anonymous No.919303]

>>919302

I don't see how you can refer to it as a variable resistor.

[Anonymous No.919304]

>>919302
see >>919299
also, the book is available from certain bodies of water where pirates that do not actually capture ships reside.
Or so i have heard.

[Anonymous No.919305]

>>919299
>Because i reckon that you lack the foundation, or at least the proper resources to learn electronics tbqh.
A bold statement. It was just something that occurred to me, seeing as it's often likened to a variable resistor.

>>919301
>Ofcourse it is non-linear, otherwise it would not be able to amplify a signal.

But it's amplifying by modulating a stronger current than rendered by the input (eg magnetic pulses on a tape). I don't see how non-linearity is essential for amplification, in the most general sense.

A relay "amplifies" a signal, in a sense. So would a linear pot connected to a high voltage device - a small movement of my hand causes a bigger change elsewhere.

[Anonymous No.919307]

>>919304
I was considering it but cheaped out and bought Electronics From The Ground Up. Don't like it. It gives virtually no explanation of how circuits actually work. The specifics and principles of how components interact to make things happen.

[Anonymous No.919310]

>>919305
>seeing as it's often likened to a variable resistor.
Which is wrong and stupid. Just go ahead and read the art of electronics, transistors are too complicated too explain in posts on 4chan, when the material is right there for you to read and learn.
>I don't see how non-linearity is essential for amplification, in the most general sense.
I dont think you know what a linear device means.
And a relay amplifies a signal in a sense, but it is also non linear.
And i dont really get what you are trying to say with the potentiometer.

[Anonymous No.919311]

>>919305
>A relay "amplifies" a signal, in a sense. So would a linear pot connected to a high voltage device - a small movement of my hand causes a bigger change elsewhere.
Or to put it another way, a lever "amplifies" and that's entirely linear.

[Anonymous No.919312]

>>919307
Exactly what you are missing in that book is the specialty of the art of electronics.
I found a magnet link for you, but shhhh dont tell the mods:
magnet:?xt=urn:btih:b156a04eb3a124e17972502564cc8d099112e9dc&dn=The+Art+of+Electronics+%283rd+Edition%29%2C+by+Horowitz+%26amp%3B+Hill&tr=udp%3A%2F%2Ftracker.openbittorrent.com%3A80&tr=udp%3A%2F%2Fopen.demonii.com%3A1337&tr=udp%3A%2F%2Ftracker.coppersurfer.tk%3A6969&tr=udp%3A%2F%2Fexodus.desync.com%3A6969

[Anonymous No.919314]

>>919311
A lever does not amplify, the energy put into the system is the same as on the other side out of the system. Distance traveled and force applied change, but they add up together to the same energy on both sides, thus it does not amplify energy and is not a strict amplifier.
Transistors do multiply energy (with the use of an external voltage source ofcourse) because in an extreme example: A mosfet has picoamps input current with a small voltage, it can amplify that into milliamps with volts swing on the output. Thus an increase in signal energy.

[Anonymous No.919317]

>>919310
>I dont think you know what a linear device means.
>And a relay amplifies a signal in a sense, but it is also non linear.
>And i dont really get what you are trying to say with the potentiometer.

Are you autistic or something? I'm saying that, as a transistor allows more or less current to flow through the collector-emitter, depending on the collector current, then it's comparable to a variable resistor, on some level.

Both control current flow according to some supplied parameter - in the case of the transistor it's the base current, in the case of the variable resistor it's the position of the fader.

Still with me?

My question was, then, given that you can express a variable resistor's ability to restrict current in terms of how its impedance is linked with the absolute physical position of the fader, why is the resistance of the transistor not expressed in terms of how the collector-emitter impedance changes with the base current?

I then went on to attempt to demonstrate how non-linearity is not a pre-requisite for amplification, in the most general sense of the word.

I might have chosen bad examples, but I honestly didn't expect to have to explain it in this detail to someone with rampant Asperger's.

I refer you to my lever example >>919311 - I was trying to make a general point that the act of amplification, per se, doesn't appear to me as requiring non-linearity.

If I'm using a miniscule amount of energy to control the current into a big motor or whatever, then that's amplification. Whether that energy is kinetic or electrical, I don't see how the act itself necessary requires non-linearity.

Of course, that's because I don't know much about transistors at this stage. Indeed my intention here was to discover to what extent the "electrically controlled pot" analogy held.

[Anonymous No.919318]

>>919317
>more or less current to flow through the collector-emitter, depending on the collector current
base, even.

[Anonymous No.919321]

>>919314
>A lever does not amplify
True, but think of it in terms of position. Like those old toys that let you scale up or down a drawing. That's the analogy I was getting at. It's scaling linearly. Like a linear pot does when you move the fader.

[Anonymous No.919324]

>>919317
>Are you autistic or something?
You seem to autistic. I am helping you, and trying to show you why it is wrong, yet you start insulting. Fine by me, this is my last reply learn that shit yourself like the rest of us you lazy fuck, you really know jack shit yet have the guts to insult someone trying to help.

It is not at all comparable to a transistor because: A - it is not linear B - Output current does not depend on voltage of the output terminals, which violates ohms law if it was a resistor.
According to you humans are plants on 'some level' because both have a cell membrane. Stop acting retarded just to pretend being right.

>My question was, then, given that you can express a variable resistor's ability to restrict current in terms of how its impedance is linked with the absolute physical position of the fader, why is the resistance of the transistor not expressed in terms of how the collector-emitter impedance changes with the base current?
And again, that is a retarded fucking question because they are not the same thing. Still with me? Probably not. You know jack shit about transistors and your question shows it. Even if you know the basics of transistors you would know that this is a completely irrelevant and maybe even a stupid question.
>I then went on to attempt to demonstrate how non-linearity is not a pre-requisite for amplification, in the most general sense of the word.
And i went to disprove it.
>I might have chosen bad examples, but I honestly didn't expect to have to explain it in this detail to someone with rampant Asperger's.
You seem to be the one that has problems with social interaction, insulting someone trying to help...
>I refer you to my lever example >>919311 - I was trying to make a general point that the act of amplification, per se, doesn't appear to me as requiring non-linearity.
See >>919314
you retard.

[Anonymous No.919325]

>>919317
>If I'm using a miniscule amount of energy to control the current into a big motor or whatever, then that's amplification. Whether that energy is kinetic or electrical, I don't see how the act itself necessary requires non-linearity.
You arent transforming energy in a lever. It does require linearity, next to electronics you should also study up physics.
>Of course, that's because I don't know much about transistors at this stage. Indeed my intention here was to discover to what extent the "electrically controlled pot" analogy held.
Then maybe you should stop acting like a dick and first read up yourself.
>>919321
It multiplies position, that does not mean it amplifies energy, so your analogy is stupid, irrelevant and a bad argument to begin with. It just proofs my point that you need non-linearity to have amplification. (And before you go down that road, amplification is always amplification of energy in physics, not other quantities)

[Anonymous No.919326]

>>919317
>why is the resistance of the transistor not expressed in terms of how the collector-emitter impedance changes with the base current?

because this line of thinking doesn't get you anywhere. all the design formulas involve currents and voltage gains, throwing ohms into the mix just adds confusion.

[Anonymous No.919327]

>>919310
>transistors are too complicated too explain in posts on 4chan
But yeah they are complicated. I keep thinking about it too much like a pot and run into trouble. It does seem a crazy just to scale up a signal. So because the beta value is fairly constant for a transistor, that means it will give the same CE current for a given BE current, even if the supply voltage doubled?

That, of course, is not variable resistor-like, for starters. It's harder to get a mental picture of what's going on.

[Anonymous No.919328]

>>919324
>And again, that is a retarded fucking question because they are not the same thing.
Right. That's that cleared up then. Basically all I wanted to know.

[Anonymous No.919329]

>>919327
>It does seem a crazy just to scale up a signal. So because the beta value is fairly constant for a transistor, that means it will give the same CE current for a given BE current, even if the supply voltage doubled?
Yes sortof. But please, learn the basics first, go to google, type in 'transistor amplifier tutorial' or 'transistor basics' or read the art of electronics or whatever, but you are missing the very basics needed for understanding. They give you plenty of mental pictures to work with as well.
>That, of course, is not variable resistor-like, for starters. It's harder to get a mental picture of what's going on.
As said before, it is NOT a variable resistor, get that line of thinking out of your head ASAP to prevent trouble later on.

[Anonymous No.919332]

>>919325
>It multiplies position, that does not mean it amplifies energy,
I know. You getting the wrong end of the stick here. With a lever you could technically scale up an A4 drawing into one that could be seen from space. It's a linear spatial "amplification".

Surely that's analogous to how a linear pot would operate on a voltage source?

I mean, okay, now it's been established that the variable resistor comparison is dumb, so that's that.

Okay here: imagine a device were to be invented that fit my description. An electrically controlled variable resistor - where there is a linear relationship between the voltage at its "gate" pin and the resistance between its two other pins.

One could use such a device to amplify a weak signal by feeding it to the "gate" pin and placing the device into a voltage divider, connected to a high voltage.

Even if this thing was completely linear, it would be amplifying.

Hence my general point about the pure concept of amplification not necessarily requiring a linear characteristic.

I accept that this is not the case for the way the transistor operates, but again I was essentially enquiring as to how similar or different the transistor is to this notion of simply modulating resistance with a signal.

It's not as simple as that - fine. It's something I wish to learn, yes.

[Anonymous No.919335]

>>919329
I've read about it. I've looked at the equations and whatnot but its still hard to gain an intuition for it. I understand capacitors, etc, but the transistor is weird.

Now I've got an oscilloscope I'm sure the added visual element will help.

[Anonymous No.919336]

>>919335
And, ironically, from what I've read, I think I grasp the dynamics of op-amps more than the transistor.

[Anonymous No.919339]

>>919033

Isn't that the point of prototype 1 or 3-off boards before committing to a final design?

[Anonymous No.919352]

I am considering making an air ion gun(s) to get plastic cling film to adhere to non-conductive workpieces temporarily. For this purpose I'm guessing I would need some device that can give off both positive and negative ions, so that I could charge the workpiece with one type and the cling film with the other (small fans would blow air through the electrodes). All of the ion generators I see only say "negative ion generator" tho, and so I don't know what they are doing.

These devices have a lower-voltage input pair of wires and a high-voltage output pair. Is the output pair ground & AC+, or is one wire a negative charge and the other a positive charge? I would want one to be +5kv and the other to be -5kv, for example....

I know of the ion air guns used to -remove- static charges, and so far none I've seen had any means of switching the polarity,,,, so I dunno what they are doing.

[Anonymous No.919353]

New thread, when?
I'm on my phone otherwise I'd start one.

[Anonymous No.919369]

>>919352
>I would want one to be +5kv and the other to be -5kv, for example....
Or I guess really: it would need three outputs, a ground, a positive high voltage and a negative high voltage....

[Anonymous No.919385]

>>919150
You started it, accusing me of trolling you when I told you exactly where to look.

[Anonymous No.919492]

>>919273
>>918527
>>918524
Now I'm looking into replacing the switch with 2 MOSFETs so I can control the circuit with an arduino.

Would this one be appropriate for the discharging side? Specifically, could it take the peak current?

>IRFZ44N
https://www.arrow.com/en/products/irfz44npbf/international-rectifier?gclid=CPCyx7mC9MkCFYdgfgodkT0MaQ&gclid=CPCyx7mC9MkCFYdgfgodkT0MaQ#page-1

If it helps, I measured the solenoid coil resistance as 8 ohms when the "ram" is not moving and around 6 or 7 ohms when the "ram" is moving. I'm not sure if that shows everything since the screen on the ohm meter refreshes slowly.

I can't find any data for the minimum voltage required on the gate pin. Will the 5V from an Arduino pin be good enough?

[Anonymous No.919508]

(Unrelated to the above discussion)

What is a MOSFET? How is it different from a regular transistor?

...And what does a transistor really do? I've heard it be defined as a pipe valve; the collector is inlet, base is the spigot/valve, and emitter is outlet. Is this accurate, or...?

[Anonymous No.919511]

Hey guys, I have a question, I have a arduino leonardo, and I was wondering if I use rfid module, and microsd card module can I use both of them on the leonardo, even though they use the same mosi/miso pins? If so, would I hook them up on a bread board where they go to the same pin or would I need to read the pin layout and find secondary mosi/miso pins and 3.3v pins?

[Anonymous No.919512]

>>919511
As long as each slave device has a separate chip select line. SPI devices disconnect from the data lines if their CS is not asserted.

[Anonymous No.919519]

>>919492
Another question related to the solenoid: I know the current drawn changes depending on the location of the metal ram.

When I measured the current traveling through the solenoid with a multimeter, it peaked at 0.45A then decayed to 0 like it should. This is the same value on the eBay listing. So does this mean that the peak current will be 0.45A regardless of the voltage?

I thought it would be much higher seeing how the capacitor is dumping everything at once.

>>919508
I'm not an expert so I can't give you a full answer, but I chose a MOSFET over a "normal" transistor to use as a switch since they can usually pass more current through than a normal transistor.

[Anonymous No.919528]

> So does this mean that the peak current will be 0.45A regardless of the voltage?

a solenoid is not exactly a resistor, but it's close enough that I=E/R still applies. if you double the voltage, you double the current.

[Anonymous No.919529]

> If it helps, I measured the solenoid coil resistance as 8 ohms when the "ram" is not moving and around 6 or 7 ohms when the "ram" is moving.

haha. reminds me of a guy on youtube that thinks he discovered 'negative resistance' coz when he spins a motor, the resistance value changes, and sometimes goes negative.

you're probably seeing the same effect. you're creating a generator, which is dumping current into your meter, which causes false readings.

[Anonymous No.919538]

>>919385
Alright, Sensitive Sebastian. I wasn't being serious then. I was actually glad to have it narrowed down for me, but as I'm sure you can see, I was looking at it for ages and there was still no apparent error.

And this is 4chan, after all.

[Anonymous No.919566]

What should I look into in order to program this? Can it be reflashed? I have an arduino, are there guides for any PAL programming with arduinos that are you recommend?

[Anonymous No.919570]

>>919566
Most EPROMmers should be able to program it, don't know offhand where you can find the development tools.

[Anonymous No.919607]

What are the chances of me McGrillin myself if I were to strip the wires off a 12 volt wall wart and plugging a 12 v case fan into it

[Anonymous No.919608]

>>919607
>strip the wires off a...
OF a
amazing how one letter changes the meaning of a sentence

[Anonymous No.919633]

>>919528
>>919529
I'm trying to determine the peak current which occurs when the solenoid starts moving.

Is it correct to use the resistance when the solenoid is at rest?

[Anonymous No.919640]

I want to drive a single LED light from a USB source, but I don't know how. It's rated for 20mA and has a voltage drop of 2.0V. Three options for a resistor, which I don't know which is the correct:
>100 Ohm, since 2V / .02A
>250 Ohm, since 5.0V (supplied) / .02A (rated)
>50 Ohm, since 5.0V (supplied) / .01A (supplied).

I'm thinking the second value (250) is the most correct, but I want to make sure before I burn something.

[Anonymous No.919669]

>>919640
100Ω is the correct answer.

[Anonymous No.919688]

>>919640
Those are all incorrect, unfortunately. If there's 5V across the supply and 2V across the LED, what voltage has to be across the resistor?

[Anonymous No.919699]

>>919669
I misread the question and thought the voltage drop across the resistor was supposed to be 2v.

>>919688
Trust this person.

[Anonymous No.919715]

I spent some time working on a Microchip PIC24fj192gb106 based board for an internship, and was keen to mess around with it more in my spare time, what is the best setup for doing so? Are there many development/prototype boards that allow for usage of most of the pins (need UART's and GPIO's)

[Anonymous No.919730]

>>919607
Under 30V is fine to play with, that's when you start feeling it at least.

[Anonymous No.919734]

>>919715
You don't need a dev board. Most PICs come in the DIP form and if they don't a breakout board is all you need. Connecting a PIC programmer is really simple. PICKIT2 clones are around $12. If the PIC has an internal oscillator you don't need any other hardware to make it run, and if it doesn't you just need a crystal and two capacitors.

[Anonymous No.919741]

don't mind me
just postan some monster caps from the 80s

[Anonymous No.919749]

>>919741
Are they wired in parallel?

[Anonymous No.919763]

>>919734
What about for 64 pin/square PIC's?

[Anonymous No.919765]

>>919749

>Error: Our system thinks your post is spam. Please reformat and try again.
the fuck

[Anonymous No.919814]

>>919688
3V, I assume, so 3 / 0.02 = 150 ohms. I forgot what "voltage drop meant", stupidly enough. English isn't my first language so I slip up sometimes.

[Anonymous No.919832]

I don't quite understand voltage dividers.

Why does the voltage drop at a ratio of the resistors, regardless of the size of the resistors?

[Anonymous No.919835]

Hey /ohm/.
Could you help me Identify this component?
I tried googling, but I wasn't realy able to find any info on it.

[Anonymous No.919841]

>>919832
Because total voltage is split among different "branches" of the circuit. Voltage into a node = voltage out of a node, and voltage must "add up" throughout the circuit. Therefore, if you have two resistors at say 1 and 10 ohm, or 10 and 100 ohm, the drop would be the same, based on the resistor ratio.

Someone back me up on this because I'm not 100% sure but fairly certain this is the right answer.

[Anonymous No.919842]

>>919832
What do you mean by size?

[Anonymous No.919843]

>>919841
You are thinking current into a node is equal to current out of a node.

[Anonymous No.919847]

>>919842
ohms

[Anonymous No.919849]

>>919841
But if you short a circuit all the voltage goes back to the source.

I guess the way it seems is that say you have 5v then it passes thorugh a resistor and drops to 4v then drops through anouther resisor to 3v then that just goes back to the batttery

[Anonymous No.919861]

>>919847
>>919832
Ah but the voltage drop is related to the size of the resistors (V=IR), just that if you have them setup as a voltage divider you can simplify the equation for the output voltage to be dependent on the resistor ratio.

Have a look at Kirchhoff's Current & Voltage Laws and try simplifying a voltage divider for yourself

[Anonymous No.919876]

>>919832
Because the total resistance has to drop the whole amount regardless of supply voltage, so how else should that be distributed over the individual resistors?

[Anonymous No.919960]

>>919843
Whoops, yeah.

>>919861
This, KVL / KCL is what you need

[Anonymous No.919982]

I bought some switches cause they seemed cheap
http://www.amazon.co.uk/gp/product/B00LO2TWM0?psc=1&redirect=true&ref_=oh_aui_detailpage_o05_s00

they arrived and have 3 metal bits, did I fuck up? only things I've made up till now is a desk fan and a flashlight

also is the art of electronics actually worth it for a newbie?

[Anonymous No.920021]

i posted here asking about software circuit sims (rather than online, so i can pirate them) and someone recommended multisim. i've tried that for a while but i'm not a huge fan, are there good alternatives?

>>919982
that's an SPDT switch, pic related should clarify. if you have a multimeter just measure resistance between the poles in each switch state to know which to connect.

[Anonymous No.920025]

>>920021
whoops i missed that it was an led switch, disregard my post

[Anonymous No.920113]

>>918936
This better a joke

[Anonymous No.920142]

>>920021
>multisim. i've tried that for a while but i'm not a huge fan, are there good alternatives?

nope, all the alternatives are the same or worse. it's all bullshit software designed like it's 1999.

[Anonymous No.920248]

>>920142
>so tonight we're gonna design software like it's 1999.

[Anonymous No.920368]

I was going through my basement and found this cool stuff. What should I do with it?

[Anonymous No.920372]

>>920368
never turn down free transformers

[Anonymous No.920374]

Got a few of these, also found a VFD
>>920368

[Anonymous No.920380]

What are these? http://www.ebay.com/itm/New-High-voltage-Boost-Step-up-Pulse-Power-Module-Generator-DC-3-6V-6V-to-400KV/141820466952?_trksid=p2047675.c100005.m1851&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D1%26asc%3D34875%26meid%3D7530adb10c7b4150a18494318e45d530%26pid%3D100005%26rk%3D3%26rkt%3D6%26sd%3D231791449804

I bought one from some greasy chinamen in the Shenzhen shithole and it kind of worked, but no were near 400kv. Can't find any information about them. I tried to disassemble mine with a dremel but I destroyed it.

[Anonymous No.920383]

>>920380
seems like it's for tasers, where precision doesn't matter as long as the voltage is high. boost converters store energy in an inductor, then switch off the source causing a voltage spike as the inductor discharges. by doing this repeatedly they get a series of high voltage spikes that can be smoothed out to a higher voltage than the input.

[Anonymous No.920399]

>>920383

when I shredded my old one it was completely potted with no ferrite cores

other sources have called them inverters

[Anonymous No.920403]

>>920399

Like this one http://www.ebay.com/itm/DC3v-6V-OUT-700-kv-700KV-high-voltage-generator-of-high-voltage-inverter-/172039149896?hash=item280e551948:g:R6AAAOSwYHxWHKSO

[Anonymous No.920599]

Anyone every build any robots? What did you make it do? Why?

[Anonymous No.920600]

>>919835
http://www.ti.com/product/TL081

did you really google it because this is the first thing that came up when i searched? it's an op amp.

[Anonymous No.920682]

>>919849
Still don't understand this, seems like they're saying when there's one resistor it drops by however much depending on the resistor (overall voltage), but when there's two resistors it drops completely at a ratio of the two?

[Anonymous No.920684]

>>920600
I switched up the numbers on multiple searches. I feel retarded now.

[Anonymous No.920697]

>>920682
Wait, nevermind, so the voltage drops completely even if there's one resistor.

[Anonymous No.920699]

>>920682
1. The current through both resistors is the same, therefore the ratio of the voltages across them is equal to the ratio of their resistances. V1/V2=R1/R2.

2. The total voltage drop across the two resistors is equal to the supply voltage. V1+V2=V.

Solving both gives you V1=V*R1/(R1+R2) and V2=V*R2/(R1+R2).

[Anonymous No.920761]

Hi /ohm/ I'm looking at getting into building my own fx units and don't know much about electronics I've bought a delay pedal kit and if possible I'd like to learn what the various components do in the kit to make the end product sound like it does.
The kit I've bought is a deep blue delay and the kit comes with a board layout (pic related) that I'd like to transfer over to a breadboard so I can try and understand what's going on.
Are there some basic electronic lessons you guys could recommend that are audio related to help me learn some basics rather than the typical non audio starter stuff like how to light a bulb etc. So far I've made a tone generator which works and all but I literally learn nothing from building it as it was just a matter of soldering the right parts to the printed board?

[Anonymous No.920972]

For a switching transistor, is it better to place it on the positive side or on the negative side?

[Anonymous No.921010]

>>920972
what do you mean with positive/negative side? side of what?

[Anonymous No.921139]

>>920761
Audio and non audio stuff are all a piece of the same pie so you'll learn audio by just reading the suggested books in the OP or checking out the channels in the OP.

If you want to jump right to it, start reading about active filters, AC, network analysis, and signal processing.

[Anonymous No.921216]

>>921010
of the component you're switching on or off.

[Anonymous No.921410]

>>921216
i'm still not sure what you are trying to say but i assume pic related. if you just want to turn an LED on and off for example, it shouldn't matter.