/diy/ - Do It Yourself

[Has the hobby of electronics d(...) No.534787]

So i need a 24v 220mA power supply for my led strip, and i dont want to go out today, so i decided to make one

I started with a 12 V 500mA transformer that i had lying around, made a rectifier bridge using 4 1N4001 i did place a 25V 1000uF Capacitor to filter and i finally got my rectified signal tested on the osciloscope

Now this is where im getting confused im placing my capacitors and diodes, and it doesnt matter how many i add up, the voltages are not getting multiplied im measuring 20.4v across all the capacitors, even when they are rated to 25v and yes i also meassured from the 1st red node to the last node (pic related)

any help? im quite confused about whats going on

cant voltage multipliers be used in DC?

[Has the hobby of electronics d(...) No.534788]

i already tried several type of capacitors, electrolitic and film ones, rated from 25v to 50v and from 500nF to 1000uF with no luck

[Anonymous No.534789]

Voltage multipliers rely on the negative voltage swing to charge up the capacitors.

I had a similar issue when I was trying to create a 100Vdc source. My recommendation, just use different transformer and use a 24v voltage rectifier.

I ended up using what's called a marx generator, essentially putting 5 caps in parallel with a 20vdc source and then I used switch to stack them in series to charge up 1 100v cap, and then switched back to the charging.

I haven't had the time available to complete this, but if you use a bunch of transistors and 2 555 timers(one in astable mode and one in monostable), you can get the switching down.

In my case the marx generator was the only route I could go and current wasn't a huge deal as I was dealing in the uA range. For you, again, I'd just pick up a new voltage regulator, keep it simple.

[Anonymous No.534791]

>So i need a 24v 220mA

>I started with a 12 V 500mA transformer that i had lying around, made a rectifier bridge using 4 1N4001 i did place a 25V 1000uF Capacitor to filter

You failed because you didn't have proper voltage being fed to the system. Trying to feed a 25V capacitor from 12v power source won't work.

You should have started with the 12V 500mA transformer, and made a 2:1 step-up transformer, and attached that on the output lines to get 2V.

[Anonymous No.534804]

You're feeding your voltage doubler with DC. Lose the diode bridge and the first capacitor, and try again.
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html

[Anonymous No.534807]

>>534791
>trying to feed a 25V capacitor from a 12V power source won't work
nigga wut

capacitors work at anything less than or equal to their rated voltage
>>534804
>>534789
this

[Anonymous No.534809]

>>534807

12V will only give 12V potential. there's no step-up in a transformer. Capacitors are essentially rapid-action batteries. That means you have to overcome nominal voltage with higher voltage.

I design LED power supplies. I design LED systems.

Caps should only be there as a decoupler from the rest of the electrical circuit, and nothing more, in such a simple design.

[Anonymous No.534815]

>>534809
I thought you were saying that you couldn't completely charge a 25v capacitor from a 12v supply. I understand that you'll only get 12v from any capacitor you charge to 12v. What OP is trying to build is a voltage doubler. So he wants to turn 12VAC into 24VDC, by using diodes and capacitors. But I don't think he understands a voltage doubler works, exactly, because he put it in series with a bridge rectifier.

[Anonymous No.534839]

>>534787
>can't into basic electronics
Capacitors *block* DC and *pass* AC!

[Anonymous No.534841]

>>534839
..so just wait until tomorrow and buy a 24V transformer!

[Anonymous No.534844]

>>534839

Not entirely true. They cannot pass constant DC. They can output rapid bursts of DC current, acting as PWM. Capacitors are used as the last step in many LED rectifying circuits as decoupling caps or bypass caps before power is sent to the LED in systems using PWM.

[Anonymous No.534930]

>>534789
here
>>534839
Within reason. Capacitors block low frequencies but will pass high frequencies. DC is 0 Hz, which is obviously the lowest frequency, but as soon as you remove the DC source the cap will discharge, you an still have caps block AC sources, which is the whole idea behind filter circuits.

OP, I was thinking further on it, you could get away with losing the bridge rectifier and instead using the voltage multiplier circuit straight from transformer. This will give you DC out, although again I would use either a voltage regulator or a zener diode to clamp the voltage straight at 24. Luckily, leds are current driven so they'll handle voltage fluctuations but I would at the very least use a cap to filter out some of the ripple voltage off the voltage multiplier.

[Anonymous No.534937]

>>534844
Sure, but: OP's schematic? 1000uF filter cap creates nice, clean, ripple-free D.C., just like the notes I put on his schematic. Any A.C. voltage on the other side of that first capacitor is going to be measured in millivolts. My analysis is supported by the fact that his circuit doesn't work.

OP needs to just go get a transformer with a 24V secondary and stop fooling around. Simplest solution == best solution!

[Anonymous No.534959]

>>534787
So you took AC, then you stepped it down to a voltage lower than what you wanted, so to get it higher you needed to use a voltage multiplier, but you rectified it when it needs AC. What were you thinking?

[Has the hobby of electronics d(...) No.534963]

>>534959
lol i was thinking about not going out today

Ok so heres a 2nd attemp, and i actually doubled the voltage with this one however i need to find a way to rectify an already half wave rectified signal

[Anonymous No.534979]

>>534963
Why do you have two diodes there? What's the point of the bridge rectifier? Why are you drawing wavy lines for DC signals?

[Has the hobby of electronics d(...) No.534980]

>>534979
the 1st 2 diodes are actually the voltage multiplier and still gives AC signla

[Anonymous No.534986]

>>534980
No, you have a half wave rectifier, with a second diode that does nothing. Then there are smoothing capacitors and another rectifier that should do nothing.

[Anonymous No.534992]

you do not need the rectifier
the multilayer voltage circuit will give you dc because of the diodes

just make this circuit

[Has the hobby of electronics d(...) No.535001]

>>534992
using what values for the caps?

[Anonymous No.535007]

>>535001
big ones

[Anonymous No.535031]

>>535001

more than 24 v

[Anonymous No.535130]

[Anonymous No.535140]

>>534963 this is NOT a voltage multiplier
>>534992 this IS

[Anonymous No.535142]

the bigger the caps, the more current the thing will be able to source. just a wild guess, but i'd say you need at least 1000uF to get a fairly flicker-free steady voltage at 200mA.

[Anonymous No.535176]

>>535142
Actually if you figure that the thing will be on a 60 Hz line, you get .016 seconds per cycle. Current is Joules per second, so .200A = (J/.016), solving for joules you have .0033 J. Energy in a capacitor is equal to .5Cv^2, So .0033 =.5C(24^2) Solve for capacitance.

You'd only need an 11uF capacitor, but I'd boost it a little to err on the safe side.

You sure as hell won't need a 1mF cap, that's a decent sized cap.

[Anonymous No.535182]

>>535176
Your calculation is totally off and the base reason is that current is not joules per second, but coulombs per second. 1000uF is a reasonable starting value.

[Anonymous No.535183]

>>535182
Ok, my bad, I was working from memory and I was off, even then, C=(q/v) so a .133 uF cap is sufficient

[Anonymous No.535202]

>>534963
what the fuck are you doing nigger
I don't think capacitors do what you think they do

[Anonymous No.535212]

>>535183
Now it is even more wrong.
Let's try this. A 1000uF capacitor discharges 16.7V during 16.7ms, if you're discharging it at 1A current. You know, i = C dv/dt. Thus, 200mA discharges it 3.3V in the same time.

[Anonymous No.535217]

>>535176
As you discharge the capacitor its voltage drops, and it will probably supply less current depending on whatever you're powering. With such a small capacitor you might as well not use one at all. The voltage will drop no matter what if you're supplying power with a capacitor, it's just a matter of having a large enough capacitance that it isn't significant.

[anus presley No.535223]

you can double the voltage using this circuit which is not a multiplier, but just two half-wave rectifiers. seems to me it should be more efficient than a multiplier.

[Anonymous No.535227]

>>535223
It is actually exactly the same circuit as the classic doubler, just drawn differently. I like that style, though.

[Anonymous No.535258]

....

Take a look at CW generator:

http://en.wikipedia.org/wiki/Cockcroft-Walton_generator

Also http://en.wikipedia.org/wiki/Voltage_doubler..

It is _easy_ to multiply or divide AC voltage with small losses.
It is _hard_ to multiply or divide _stabilized_ DC voltage with even moderate losses.


First get AC or pulsating DC voltage in the order of 24V, then try to make DC and stabilize it.

12V 500mA is enough to draw 24V 220mA if you multiply the voltage afterwards. But generally, you'd get less losses if done with 24V transformer

btw, 25V capacitors may be not enough.

[Anonymous No.536061]

compared to the others this is a shitton easier to understand
just sayin

[Anonymous No.536062]

>>536061
and that is not the image I posted