# Reverse engineering an LED pcb design

Discussion in 'General Electronics Chat' started by KatherineTurner, Dec 10, 2016.

1. ### KatherineTurner Thread Starter New Member

Nov 24, 2016
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Recently I got hold of an LED pcb that has two independent circuit and each consists of 6 high power 5730 LED beads (Vf=2.759V each) with two 151 SMD resistors (150 ohm) and a swtich. The whole thing is powered by 12V DC and draws 100mA when both circuits switch on and 50mA when only one of them is on.

With a hundred 5730 LED beads in hands, I'm trying to design a pcb with different layout but using the same circuit.
However, I still couldn't figure out how is it possible to power up 6 LEDs with only 12V in each circuit.

I've come up and simulated several combinations on connecting the LEDs with two resistors.

#1: This obviously wouldn't work, since 2.75883*6=16.55V > 12V

#2: This circuit successfully draws 50mA, but can only have 3 LEDs. I need 6.

#3: This works fine for 6 LEDs but now the circuit draws 100mA and I have two extra resistors.

None of them seems to meet the requirements of 6 LEDs, 2 resistors, 12A and 50mA.

Is there any other combination of circuit that would fit the criteria?

2. ### dannyf Well-Known Member

Sep 13, 2015
2,196
421
1 resistor + 3 led in a string, 2 strings in parallel.

3. ### BobTPH Senior Member

Jun 5, 2013
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341
In your third circuit, take out one of the two 150 Ohm resistors on each branch. That should approximately cut the current in half.

On the other hand, what are the LEDs rated for? If they can handle 50mA, maybe you only need half as many running at 50mA.

Bob

4. ### Dodgydave AAC Fanatic!

Jun 22, 2012
7,752
1,293
Make the resistors 150 ohms(omit R2, R4), that will give you 25mA per chain, total 50mA.

5. ### KatherineTurner Thread Starter New Member

Nov 24, 2016
13
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I have omitted one resistor from each branch.

But the simulation result showed different values.
Each branch draws 28.79mA. The circuit draws 57.58mA.

Actually, the LED can handle up to 150mA at 3.4V according to its package.

6. ### BobTPH Senior Member

Jun 5, 2013
1,462
341
If you must have exactly 25mA, then raise the value of the resistor.

If the LEDs are capable of 150mA, why do insist on running them at 25mA, when it will take 6 times as many of them to produce the same amount of light?

But

7. ### KatherineTurner Thread Starter New Member

Nov 24, 2016
13
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Simply because it will burn the resistors.

A small SMD resistors comes in the range of 1/20w ~ 1w.
With 150mA, the power at the resistors can easily exceed 1w.

I wish there is some kind of high current design without using switchmode converter.

8. ### djsfantasi AAC Fanatic!

Apr 11, 2010
4,344
1,601
Do you have a link to the LED datasheet?

9. ### KatherineTurner Thread Starter New Member

Nov 24, 2016
13
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Here are two datasheets I found on the internet.
More can be found by searching "5730 led".

10. ### djsfantasi AAC Fanatic!

Apr 11, 2010
4,344
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So you have LEDs that run at 150ma, but you don't want to because you want to use smaller resistors, is that correct? They have a Vfd of 3.0-3.4b, yet your schematic only accounts for 2.7B. Calculating the resistance or current using your schematic does not compute. You want a high current design (without a switchmode converter) but you don't want to use high current components.

My guess is that it can't be done.

11. ### KatherineTurner Thread Starter New Member

Nov 24, 2016
13
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Vf may seems like a constant, but it actually changes a little bit depending on the current applied.
You are right about the 3~3.4 Vf and that's when the current is around 150mA.
At small current, the Vf is approximatly 2.7V.

12. ### BobTPH Senior Member

Jun 5, 2013
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But what is the point of using smaller resistors at the expense of having 3 times as many resistors and LEDs?

At Vf of 3.4V we can calculate the resistor needed to get 150 mA:

R = (12 - 3 * 3.4) / 0.15 = 12 Ohms

And the power is I^2 * r = .15 ^2 * 12 = 0.27W

Bob

13. ### WBahn Moderator

Mar 31, 2012
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But your sim results assume certain behavior on the part of the components; behavior that may or may not be highly accurate. Your results are just a bit more than 10% off from the 50 mA you mentioned in the first post. Where did that 50 mA number come from? Actual measurement? Where did the 2.75883 V value for Vf come from? Is it really good six sig figs? What is the internal resistance of the voltage source that is actually being used, as opposed to the ideal 12 V source in the sim? What is the tolerance of the resistors?

Let's just use the Vf = 2.75883 V and see what a hand calculation would predict:

I = 2{[12 V - 3(2.75883 V)] / 150 Ω} = 49.6 mA

Remember that sim results are only as good as the models used. If you use simple models, then expect the sim results to just be in the ballpark (if that). In most such cases, anything better than 20% agreement is probably "in the ballpark".