/diy/ - Do It Yourself

>>1166187
Using anons current values here and assuming that OP's circuit is drawing 20mA...
Add a 1K pot in series with the 47 ohm resistor

math:
>What's the voltage across the resistor?
V = IR
Vr = 20mA * 47ohms
Vr = 0.94Volts

>What would the Resistor value need be to get 1mA?
R = V/I
R = 0.94v / 1mA
R = 940 ohms

This solution has better control.
>>1166141

It will draw .175A* 2/12 = .029A = 29mA

To get 12V from regular batteries you'd need 12V/1.5V = 8 batteries in series. Not small.
(most batteres like watch, AAA, D are 1.5Volts)

An option would be to use a couple batteries and a boost converter.

Another is to try a 9V battery, they'll probably light up, but only ~2/3 brightness

The problem is that the strips come with built in resistors that were calculated for 12V.
For only 2" (how many led's is that?), and size being important, you might be better off just assembling LED's yourself so you could select the resistors to work with a lower voltage. You'd need a soldering iron or some ingenuity.

What's there to modify?
>Output of the 555 can only source or sink a current of up to 200mA
It'll run on 5-12V

just add a series resistor to the LED

Here's the math for the resistor.
Hook every thing in series and double Vf since there's 2 diodes in series.

Use heat shrink tubing at the connections. If your paranoid, you could add a fuse.


Get specs for LED. You need If(forward current) and Vf (forward voltage)
Just making them up...
Vf = 2.8V
If = 25mA

Add a 20% safety factor for If
25mA * 0.8 = 20mA = 0.02 Amps

Vr = V - Vf
Vr = 5V - 2.8V
Vr = 2.2V = Voltage drop for resistor

R = Vr / If
R = 2.2V / 0.02A = 110 ohms

Since you prolly want it bright and it will be easy to replace, round down to 100 ohms

double check using 100 ohms
If = Vr / R
If = 2.2V / 100 ohms = .022A = 22mA = not the 20% safety you wanted but prolly fine.

>if the cellphone charger actually waste energy just for being pluged to the wall
yes

Why not just use a switch?

To power an led, you need a resistor to "drop" the extra voltage.
The Voltage it needs to drop is (5.5V - LEDforwardVoltage)
Use 80% or less of the LED maximum current.
R = V / I
The resister value = (5.5V - LEDforwardVoltage) / (LEDmaxCurrent * 0.8)
Round up to the nearest common value

>>402536

You should check out this thread OP
>>397136

You need a DC power supply preferably at least a few Volts higher than the forward voltage of the LED (Vf , usually 2 to 3 Volts) and a resistor for each LED. An old phone charger or power supply from something. It also needs to be a high enough mA rating (20mA per LED - can be less if the Voltage is 9V or higher). If you find one post it's specs and the LED specs and someone here will help you with the resister values and probably even draw you a sketch.

28ma
2.6V pass
not sure, but I'm gonna assume that means
28mA max current with a forward voltage drop of 2.6V

The first one won't light because 1.5V does not overcome the 2.6 Volt LED voltage "barrier"

Second:
3 Volts applied
2.6 Volts "dropped" by the LED
This leaves 0.4 Volts to be dropped by the resister. (.4+ 2.6 = 3)

The current through the resister can be found using the formula
I = V/R = 0.4V/200ohms = .002 = 2mA
The current through the LED will be the same.
I don't think this will be enough to see the light. maybe in a dark room.

The 3rd one should light but is not a good design because of no current limiting resister.
Excessive current could result. I've seen people say that it worked fine like that but I wouldn't do it.(at least for long)

The polarity of the LED has to be right. If you look inside the LED you'll see one big and one small piece. the big one goes toward the -negative side of the battery. Some also have a flat part at the biggest round part of the LED but I don't remember if it goes to + or -

Also: Those are light bulbs in your schematic, not LED's

SOL on the transformer

>http://www.amazon.com/Yesurprise-50pcs-10mm-White-Light/dp/B0092X7TF0/ref=sr_1_7?s=hi&ie=UTF8&qid=1358892716&sr=1-7&keywords=Super+bright+led+10mm
Forward Voltage (V):DC3.2~3.4
Forward Current (mA):20

Using computah 12V as supply voltage:

the diode drops 3.3v. put 3 in series and they drop 9.9 volts.
they need a series resistOr to drop the remaining (12 - 9.9) = 2.1V

20mA is a max value, use 80% of this or 16mA

R = V / I = 2.1V / 16mA = 2.1/.016 = 131 ohms.

up this to a standard value of 150 ohms.
check:
2.1V/150ohms = 14mA = acceptable

so, use a 150 ohm resistOr and 3 LED's in series hooked to 12V (yellow wire to drives)
Can any one paste my post from the other thread I just closed an hour ago?(DAMN) or a link to archive?
I might use it to make a LED infographic since this is a common question.

>>368506
>If I wire two 150ohm resistors in parallel, do they act as a 75ohm resistor
yes
3 in parallel is 50 ohms.ect...

>>368521
also: use 68 ohms, a standard value.

Here you go OP