Resistors in parallel question

Discussion in 'General Electronics Chat' started by blmatrix13, Sep 16, 2012.

1. blmatrix13 Thread Starter New Member

Jul 25, 2011
2
0
Hi everyone,

I have a question about resistors in parallel...basically a little confused and would like some clarification.

I have a high powered LED I would like to power but have a very small footprint on the PCB as to where I can put the the passive components. I thought it would be easier to add extra resistors to beef up the current handling capabilities. To accomplish this I thought putting a bunch of smaller wattage resistors in parallel would work the best. However, I don't know that it would increase the current handling capacity.

Basically I need 33 ohms at 4 watts so I thought I would put in 4 (1) watt resistors in the values of (3) 130 ohm and (1) 150 ohm. Will this do what I am expecting it will do or does the current not increase with the additional resistors?

What I am asking is does (4) 1 watt resistors = 4 watts?

Jeff

2. JMac3108 Active Member

Aug 16, 2010
349
66
Jeff,

If all of your parallel resistors were the same value, then you could just add the wattages (4 x 1W resistors) and say that it was equal to a single 4W resistor. But yours are not all the same. The current spits through the resistors. Each of the 130 ohm resistors get an equal amount, but the 150 ohm resistor gets less because its larger. So you need to calculate the power dissipation in each resistor using either P = I^2R or P = V^2/R.

For example, if the voltage across your parallel resistors was 10V, then the power dissipated by each 130 ohm resistor would be P = 10V^2/130ohms = 0.77W. The power dissipated by the 150 ohm resistor would be P = 10V^2/150 = 0.67W.

3. crutschow Expert

Mar 14, 2008
13,475
3,361
You also need to derate the power of the resistors to allow for temperature above the design value of 25°C. I would not operate them above about 75% of their power rating for good reliability.

4. THE_RB AAC Fanatic!

Feb 11, 2008
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1,305
I'm worried about "4 watts" dissipated from the resistors on "very small footprint on the PCB" and I have a sneaking feeling you have that really hot "high powered LED" very close to that PCB too?

Let me guess, you will finally add; "and it's all in a tiny little space"??

5. Dodgydave AAC Fanatic!

Jun 22, 2012
5,141
767
Try four 120 ohm resistors at 1Watt each in parallel, that will give you

30 ohms at 4 Watts

What voltage supply are you using, as maybe a constant current device may be easier?

6. blmatrix13 Thread Starter New Member

Jul 25, 2011
2
0
Wow...great responses here...thanks everyone.

First off, I knew in my mind that the current was split across the resistors, I don't know what it didn't come to be when I was thinking about it...Thatis my fault.

RB...you are right in a sense. The LED is an RGB LED running at 350mA per color. It is mounted to a heatsink so I am not all that worried about heat, but it is definately on my mind. The LED is running at 12 volts which is why the resistors are so large.

Constant current is definately the idea I had originally but I have a piece of open source firmware for the ATMEGA88 that is driving 3 FETs that I would use. I am by no means a firmware guy so modifing it is not really something I am looking to do or know how to do. That is why I wanted to run the LED in this setup to make life easy.

There is no way to run the LED directly off the FET's...right? Without the limiting resistors?

Jeff

7. THE_RB AAC Fanatic!

Feb 11, 2008
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Yep, you can do the miracle of drawing 100mA from the 12v supply and giving 320mA to the LED.

It's a SMPS effect called a "buck" circuit, all it really needs is an inductor and a high speed schottky diode for each FET. You will get low heat and high energy efficiency.

So you already have 3 FETs to switch the LEDs with high speed PWM from your micro, so you just need to add 3 inductors and 3 diodes, and a minor change in the source code to limit the max PWM duty (which sets your max LED current).