Please Check My LED Power Math

Hi Guys, Please check my math:

A) I have 8 5mm 3v .02a LEDs.
B) I wired them in 2 parallel series. So LED bank has 4 bulbs in it. Each LED bank should require 12v and 80 milliamps
C) I have a 12v 1a wall wart to power everything
D) The 12v will supply will supply adequate power which won't drop from 12v since the 2 series of 4 LEDs are wired in parallel. The current of 1amp is split across both banks.

Resistor Math
1amp -> pass through 2 x 320 ohm resistor -> Remaining current .26a
.26amp -> pass through 200 ohm resistor -> .6amp to Bank 1 of 4 LEDs
.26amp -> pass through 200 ohm resistor -> .6amp to Bank 2 of 4 LEDs

Does my math sound right...would you guys choose a different resistor values or maybe use a potentiometer to control each set of LEDs?

I see no math to check.
What is 12-(3x4)? 0.

(12V - 3 x 3V)/.02 = 150 ohm, 3 LED string
(12V - 2 x 3V)/.02 = 300 ohm, 2 LED string

So, 4 strings of 2 = 80mA total.

Lots of reserve current for other parts of your project from the power supply.

Do you have 2 in series, or 4?

With 3V LEDs and a 12V supply, the 4th may not light if a transistor is being used to turn the strips off & on.
The transistor will have some voltage drop across it also.

aarg:
Do you have 2 in series, or 4?

he said parallel, not series.
each LED is 20mA, 4 would consume 80mA as he stated in his post.
he did not ask us to offer a better way, just to check his math.
he is confusing by saying he has two series of strings of parallel wired LED's
"4 LEDs are wired in parallel."

he has a second circuit, also with "4 LEDs are wired in parallel"

so he has a series of the same parallel wired devices.
if he wanted a suggestion, we would say to wire in series, not parallel.

but one thing we can do is REQUIRE a schematic. this is too confusing because of his use of words.

CrossRoads:
With 3V LEDs and a 12V supply, the 4th may not light if a transistor is being used to turn the strips off & on.
The transistor will have some voltage drop across it also.

2N7000 TO-92 FET has 9 ohms resistance. handles 200mW continuous. 500mW pulsed.
good for up to 60 vols.
the FET version of a PN2222

Thanks Dave! That’s exactly what I was looking for :slight_smile:

Hi,

You need to look at how the LEDs are arrenged and what current and voltage mean.

You are assuming that a 12V 1Amp supply will when connected to any load, output 12V AND 1A.

This is not the case, 12V 1Amp, refers to the power supply being able to provide a regulated 12Vdc from no load current up to a load current of 1Amp
The load will define how much current is required.

Using the diagram, I assume this is how you have your LEDs connected.
LED_P_S.jpg
First off, the current flowing through the four LEDs in series will be a total of 20mA as they share the same current flow, not 4 x 20mA = 80mA.
Looking at the leg A, the total voltage across it is 12V
You have 4 LED in series, so the sum of the drop across them at 20mA is 4 x 2 = 8V.
This means the voltage across the resistor VA at 20mA will be 12 - 8 = 4V.
The resisitor, RA, needs to drop 4V at 20mA, it value is given by Ohms Law.

VA = IA x RA

Rearrange it;

RA = VA / IA

Substituting values;
RA= 4/ 0.02 = 200R.

So RA and RB can be 200R, which will provide 4V drop at the 20mA current drawn by each of the LED legs.
The total current drawn from the 12V power supply will be the sum of LEG A and LEG B,20mA + 20mA = 40mA, this is much less than can be supplied by your supply so it will be okay to do so.

Hope this helps you… Tom… :slight_smile:
Sorry used 2V instead of 3V… Will not work with 12V suppy

This is what Crossroads was saying.

if you put 4 in SERIES, the voltage drop MAY be too great and you may not be able to get all the LED's to light fully due to the circuit with LED's resistor and transistor/FET.

is you change to two circuits with 3 and one circuit with 2, it would guarantee they work properly.

The attached is based on 1.2v drop and 20mA per LED. if your voltage drop is different, then thing would change.

looking back at your first post, you said the LED's have a 3 volt drop. my sketch shows 1.2volts to get that resistor.

I do not think you can put 4 LED's in series with a 12v supply and also use an FET as I show.

you can try, but also try to 3 in series with a properly sized resistor based on the voltage drop of your FET/transistor. and your LED's

Hi,

I wired them in 2 parallel series. So LED bank has 4 bulbs in it.

Is that;

  • 2 legs in parallel, each with 4 LEDs wired in series

OR

  • 2 legs in parallel, each with 4 LEDs and their series resistors in parallel.

OR

  • 2 legs of 4LEDs and their series resistors in parallel, the 2 legs then in series.

I know some forum members cringe when it post the following line.

Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png

jdlev can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png.

Thanks.. Tom... :slight_smile:
PS, OPPS I used 2V not 3V sorry. Will not work, with 12V supply.

Lesson here is that OP should have posted a schematic right at the start, to prevent any misunderstanding of the circuit.

It needn't be fancy: pen and paper, reasonable photo from mobile phone.