LM317 LED driver calculation verification

Thread Starter

Mega hurtz

Joined Dec 2, 2021
3
Hi fellas,

I'm designing a bespoke IR illuminator and was hoping someone might be able to look over and correct my calculations for me if they could please.

Using a commercially produced IR LED array (5x parallel Strings of 3x series LED's) I have obtained the following numbers downstream of the original control circuitry in order to guesstimate the power consumption of the LED's contained within.

Total voltage across series (3x LED per series) = 4.3v
Voltage across single LED = 1.43v
Makes sense so far as voltage is multiplied with series, but current remains the same.

Total current draw across parallel strings (5x parallel Strings) = 1.2amp
Current draw across single series string (same as single LED?) = 0.21amp
Makes sense so far as current is multiplied with parallels, but voltage remains the same.

Am I correct enough assuming the individual LED's are close enough to 1.43v and 210ma?

I will be wiring 6x of these LED's in series being powered by a LM317 and have come up with 8.58v @ 0.20amp

Using a LM317 driver circuit and 1.25 as the reference voltage I arrive at a resistor value of 6.25ohm (or the next highest available) to power these LED's

Is it really that simple or am I missing something?

I'm also trying to wrap my head around the power dissipation of the resistor and am unsure whether to use volts and resistance, or current and resistance to calculate this. They both produce extremely different answers (8.58v@6ohm = 12.3w vs 0.20a@6ohm = 0.24w).

Any help or pointers in the right direction would be much appreciated, thanks!

ericgibbs

Joined Jan 29, 2010
15,359
hi Mega,
Welcome to AAC.
Confirm,
So you have 5 parallel strings of 8.5V at 210mA.

What's the supply into the LM317 volts.?
E

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ericgibbs

Joined Jan 29, 2010
15,359
hi Mega,
This is what I see for your circuit, looks OK.
E

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BobTPH

Joined Jun 5, 2013
4,772
The calculation for the resistor us I^2 R, where I s the current through i.e. 1.2A. So I get:

1.2^2 x 6 = 9W

Edit: Oh, wait, I see what you are doing now, one LM317 per string at 0.2A.

If that is right, I get:

0.2^2 x 6 = 0.24W

Edit2: Your calculation of V x I is wrong because the voltage across the resistor is 1.2V and I again get 0.24W

Bob

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Thread Starter

Mega hurtz

Joined Dec 2, 2021
3
Hi guys, after lurking for a while I am glad to finally be a part of this forum. I have assembled a few known circuits from step by step instructions, but am still VERY new to designing my own.

hi Mega,
Welcome to AAC.
Confirm,
So you have 5 parallel strings of 8.5V at 210mA.

What's the supply into the LM317 volts.?
E
Hi Eric, I should have been a bit clearer in my initial post. To confirm.
I will be desoldering the individual LED's from the commercial illuminator (that came with its own power supply) in order to rearrange the LED's into several different housings, with each new housing containing a single string of 6x LED's in series, and each housing using a LM317 for power.

Before I hack apart the current array, I wanted to measure the operating parameters of the unknown LED's in their current configuration of 5x parallel strings, of 3x series LED's per string, in order to calculate what is required when used in their new configuration.
That's how I arrived at 1.43v and 0.21amp per LED as a reference point and a new single string of 6x LED's in series in their new configuration of 8.58v and 0.20amps.

My input voltage will be 13.5v+- (automotive application) and it looks from the schematic you have provided it should work OK, Thank you.

Edit2: Your calculation of V x I is wrong because the voltage across the resistor is 1.2V and I again get 0.24W

Bob
Ahh, now I see! The voltage used for the calculation should not be LED circuit voltage, but the voltage between output/adjust, which is the 1.2v reference voltage from the adjust pin? Thank you Bob.

It's amazing how simple it is!

Audioguru again

Joined Oct 21, 2019
4,536
210mA is a very high current for an LED unless it is in a power package and you mount it on a heatsink.
What is the LED part number? (maximum allowed continuous current of one of your LEDs without a heatsink at 32 degrees C ambient temperature).
Is the heatsink for the LM317 suitable for its heating?

Thread Starter

Mega hurtz

Joined Dec 2, 2021
3
210mA is a very high current for an LED unless it is in a power package and you mount it on a heatsink.
What is the LED part number? (maximum allowed continuous current of one of your LEDs without a heatsink at 32 degrees C ambient temperature).
Is the heatsink for the LM317 suitable for its heating?
Yes, it does seem high. The current draw was confirmed (I think) by using the supplied power supply to power them. 1.2amps current was measured across the complete string of 5x LED's in parallel and approx 0.21amps was measured across a single LED string so that's what I'm working with. Each string of parallel LED's contains 3x LED's in series so voltage across the array was 4.3v, but as I understand it, the extra LED's in the series increase voltage, not amperage correct?

The LED's themselves have no markings on them and they are already part of a commercially produced array so I am trying to back calculate their values. I can accidentally smoke a few of them if needs be.

They are also Infrared LED's if that makes any difference to current draw?

Edit: Using what I have attempted to figure out so far, each LED is approx 286mw of power. They are surface mount and look a little like this, except of course for being IR.

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Audioguru again

Joined Oct 21, 2019
4,536
I do not know the size of the LED package. Its leads or its bottom carries its heat away.