# Is power rating for 100 mw 1W too much?

#### clangray

Joined Nov 4, 2018
261
I recently came across a finding I don't understand: 100mw power dissipation for a (4) LED circuit. I don't have a problem with the power value itself (100mw) but the recommendation to go with 1 Watt power instead of 1/2w or even 1/8w. Should the rating be changed from 1mw resistor to something else? My understand is that the maximum power per resistor should not exceed Power x 2, for example .04 x 2 = .08( < .25) is covered by resistor with .25power and up.

 Number of LEDs 4 Supply voltage, Vs 5V LED forward voltage, Vf 1.25V LED forward current, If 20mA Individual Diode Power Dissipation 25mW Total Diode Power Dissipation 100mW Recommended Resistor Wattage 1 Watt Resistors

#### dl324

Joined Mar 30, 2015
16,162
I recently came across a finding I don't understand:
It doesn't make much sense to me either.

It would be helpful if you drew a schematic showing how the LEDs are wired.

A Vf of 1.25V@20mA isn't credible.

Power dissipation in the diodes has nothing to do with power dissipation in the current limiting resistors.

Post where you got this information so others can be wary of information from that source.

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#### atferrari

Joined Jan 6, 2004
4,730
Should the rating be changed from 1mw resistor to something else?
The above does not make sense either. Where that 1mw comes from?

#### Papabravo

Joined Feb 24, 2006
20,609
If the diodes are wired in parallel to the +5V supply, then you need 1 resistor per diode. the value of the resistor, should be calculated as follows:

$$R\;=\;(5\text{V}\;-\;1.25\text{V})/20 \text{ ma}\;=\;187.5\Omega$$

The nearest 1% value is 187Ω. The power it will dissipate is:

$$P_{d}\;=\;(20\text{ ma})^2\cdot 187\Omega\;\approx\;75\text{ mw}$$

On that basis, I would say that that a quarter watt resistor would do the job.

If the diodes are in series then you might be able to get away with no resistor at all, but that could be risky.

#### WBahn

Joined Mar 31, 2012
29,519
What is it you are trying to find the power for? We need a schematic of the circuit to provide useful feedback.

If the LEDs are in parallel (not a good idea, but it's done more than it should be), then the total current draw is 80 mA. The total voltage drop across the resistor would be 3.75 V. This would give a power dissipation in that single resistor of 300 mW. A common rule of thumb is to double that, which would take you to 600 mW. The smallest power rating that is at least that would normally be a 1 W resistor. While a 500 mW resistor wouldn't meet the 2x rule of thumb, it would probably be good enough.

But this is assuming that all four LEDs are in parallel and placed in series with a single current limiting resistor.

Also, what LEDs are you using that only have a 1.25 V forward voltage? I'm guessing that must be IR LEDs?

If each LED has it's own current limiting resistor (which is the proper way to do it), then each one would only dissipate 75 mW. Doubling that would be 150 mW. It would probably be okay to use 1/8 W resistors, but 1/4 W resistors would be a better choice.

What is the duty cycle of these LEDs? If they are only on for short periods of time (say, a few seconds followed by at least that much time being off), the going with the smaller power rating is pretty safe. But if they are on for long durations (say ten seconds or more), then go with the higher rating.

#### clangray

Joined Nov 4, 2018
261

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#### clangray

Joined Nov 4, 2018
261
Per request I've added a screenshot above showing the recommendation of 1W resistors.

#### WBahn

Joined Mar 31, 2012
29,519
That appears to just be some website that someone put together. Who knows what they are basing their numbers on.

#### Ian0

Joined Aug 7, 2020
8,947
Probably they have shares in a company that makes 1W resistors.

(1.25V would be appropriate for infra-red LEDs, expecially those in opto-isolators)

#### Alec_t

Joined Sep 17, 2013
14,009
What colour LEDs are you using?

#### clangray

Joined Nov 4, 2018
261
What colour LEDs are you using?
Am using IR Leds
I = 16.8 ma

ok so what is power rating each resistor and all 4?
P = I^2 * R
= (.0168A)^2 * 150 ohms(above schematic)
= .0423w per 150ohm/resistor
total dissipation = .0423w * 4 = 1.69w

Is the power rating the highest wattage of the individual resistors or the total circuit? The 100ma in schematic above is total I think.

#### dl324

Joined Mar 30, 2015
16,162
Am using IR Leds
That would have been helpful to mention in your first post. Most people with LED questions *aren't* using IR LEDs. I haven't used IR LEDs since the 1970's, but recall the forward voltage being lower than visible LEDs.
ok so what is power rating each resistor and all 4?
Your questions are still unclear. You show 2 parallel strings of 2 series LEDs, but you talk about "all 4" resistors.

First, power dissipation in the resistors are independent of each other.

In the 2 parallel strings of 2 series LEDs configuration, each resistor will dissipate:
$$\large P = I^2R = 16.8mA^2*150\Omega = 42.3mW$$
So each resistor can be 1/8W.

In solution #1, assuming you want the same current you'd use 220 ohm resistors.
$$\large P = \frac{V^2}{R} = \frac{3.75V^2}{220\Omega} = 63.9mW$$
So each resistor can be 1/8W.
total dissipation = .0423w * 4 = 1.69w
Your arithmetic is wrong. 4 times 42.3mW is 169.2mW

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#### WBahn

Joined Mar 31, 2012
29,519
Am using IR Leds
I = 16.8 ma

ok so what is power rating each resistor and all 4?
P = I^2 * R
= (.0168A)^2 * 150 ohms(above schematic)
= .0423w per 150ohm/resistor
total dissipation = .0423w * 4 = 1.69w

Is the power rating the highest wattage of the individual resistors or the total circuit? The 100ma in schematic above is total I think.
You total dissipation number is wrong -- does it make since to multiply 0.04 by 4 and get well over 1.5? Always ask if the answer makes sense. Your total power dissipation is 169 mW.

The power rating that you need for a resistor is based on the power that THAT resistor will be dissipating (possibly taking into account other factors, which I'll mention in a moment).

So each of your resistors needs to dissipate about 42 mW -- call it 50 mW. Common practice is to double that, so 100 mW. Now use a resistor with a power rating at least that large, which would be 1/8 W. You can use 1/4 W if that's what you have handy (or you can use 5 W if you want) -- you just what a resistor that can handle at least 100 mW.

Now, the power rating on a resistor is based on it being used in a particular type of environment (the data sheet will spell what this happens to be). For the kind of resistors you are looking at using, that usually means that they are mounted on a circuit board in free air (so no forced air cooling), at room temperature, and away from other significant heat sources. So if you had a bunch of resistors packed into a tight space with no ability for air to move, you would need to do better calculations based on the actual thermal environment. But as long as each of your resistors has a bit (doesn't take much when you aren't pushing the power limits) of space between it and it's neighbors, you should be fine.