# Why do different resistors of the same power rating burn out / not burn out?

I have a circuit board for a small LED light, which comes with an annoying strobe mode. To disable this I have found that I can wire a resistor from a pin on a chip to ground and the LED will stay in the always-on mode that I want. I'm trying to find the lowest resistance that the circuit board will tolerate (if it is too low, the LED turns blue and dims, and chips begin to smoke)

I started with a 1K 1/4W resistor and that worked but was much dimmer than I wanted. I tried a bunch of other resistors (all 1/4W) down to < 50 ohms.

Some resistors, or combinations of resistors, smoke and turn black (although they keep working) such as:

• 1x 47 ohms
• 1x 22 ohms
• 1x 22 ohms, 1x 10 ohms (wired in series)

However, some resistor combinations work fine e.g. 2x 22 Ohms. I don't understand this as I thought the reason the resistors are burning out is due to their power rating being too low, but if that were the case any combination of resistor would burn out as they are all 1/4W.

So I measured the voltage from the pin to ground at 7V (makes sense as the power source is 11-12V) and I measured the current at 2A (with my multimeter in 10A mode and my probe in the 10A port). But that comes out to 14 watts which should instantly blow any resistor I wired in.

It is of note here that, when I measured the current using my multimeter, the circuit board didn't tolerate it (as described at the beginning) presumably because the multimeter offered little resistance in current measuring mode.

Could someone please clarify what I'm missing here as my observations don't seem to match my measurements at all. Sorry for any obvious or stupid oversights as I do not yet understand how the different mechanisms of electricity work.

Here is a picture of the board I am working on:

The resistor in that picture is 47 ohms (though it measures 1.1K when the board is powered). The voltage across the resistor is 7V giving me a calculated current of 140mA and power of just under 1 watt.

• Ah the suck it and see approach - works fine for software (if you know what you are doing and use your head to think why afterwards) but it's a pointless exercise for hardware with no knowledge of the chip. You are wasting your time. – Andy aka Oct 27 '18 at 16:23
• "strobe mode" is not a thing. That is the power supply crowbar-ing from being overloaded. Or because it is defective. Don't run power supplies anywhere near that hard. – Harper Oct 27 '18 at 21:20
• @Harper The light purposefully has three modes: high-beam, low-beam, strobe. These are cycled with each on/off. Only, I don't want low-beam or strobe. – Frayt Oct 28 '18 at 8:56
• Oh; it's a commercial product you are hacking. I did not realize that because you did not say that. The answers also seem to be written on the premise that your "LED light" is a component. – Harper Oct 28 '18 at 15:12

## 2 Answers

To determine the actual power dissipation you should measure the voltage across the resistor when powered on.

The power dissipation is then $$\V^2/R\$$. The most voltage that a 47 ohm resistor can have and still be within the 1/4-W rating is 3.4V.

It does not matter what the current capability is of the supply (and you should never put an ammeter across a voltage source because you could damage the source or your meter), what matters is what the voltage is under load and the resistance.

If you have multiple resistors in series or in parallel, each will have it's own dissipation depending on the resistance and the voltage across it.

When you use (or abuse) the LED the voltage across it will change so the voltage across the resistor(s) changes from open-circuit, and of course you are also fiddling with the resistance.

• When you say "measure the voltage across the resistor" do you mean one probe before the resistor body and one probe after? And how should I correctly measure the amperage without putting across a voltage source? Thanks for the answer. – Frayt Oct 27 '18 at 16:39
• On voltage range, probes in voltage sockets, one probe on one end of the resistor and one on the other. The amperage is known if the voltage and resistance are known. I = V/R. Ohm's law. So if there is 3V and 47$\Omega$ then the current is 64mA. – Spehro Pefhany Oct 27 '18 at 18:11

## UPDATE TWO

I have included a picture with some updated readings

Forget about the resistor.
This board has a constant current driver.

U1 is the driver chip.
If you can read the part number of U1 that would help confirm my answer.
What is needed from the datasheet is the Rsense reference voltage.

I suspect R4 (marked R180) is the current sense resistor with a value of 0.18 Ω.

There are not many high current LED driver chips in a 6 pin package.
Very likely the circuit below is similar to this one.
Rs would be your R4.
All the current flowing through the LED flows through R4.
The driver chip measures the voltage across R4 to know whether to increase or decrease the current.
The internal reference voltage is compared to the voltage between Vin and the SENSE pin.
If the measured voltage is lower than the reference voltage the current is increased until the voltage match (and vice versa if higher).

Replace all the components back to their original values.
Measure the voltage across R4.

The R4 voltage divided by 0.18 is the LED current.

The current will change linearly with changes in the value of R4.

The LED is a very high power LED. Likely a max current of 2A or more.
It appears to have 6 dice.
The arrows point to the spaces between the dice.

This is high lumen density LED used for stadium lighting and street lights. Probably between 1000 and 2000 lumens at 1 Amp. The forward voltage is likely to be 9V or 11-12V. Just measure the voltage across the + and - terminals of the LED, if you want to know, but it does not matter.

To increase the current, lower the value of R4.

EXAMPLE:

Let's say the reference voltage is 0.15V.

The current would be 0.15 ÷ R
0.15 / 1.8Ω = 833 mA

Given the above criteria, if you wanted to increase to 1 A.
The resistor value would be 0.15 Ω (0.15V/ 1 A)

R6 and R7

R6 and R6 may be a voltage divider to apply an external voltage to the ADJ pin. This could be used to fine tune the current to a lower current set by R4. R4 sets the maximum current and a voltage applied to ADJ can reduce the LED current from that set by R4.

R4 could also fix the switching frequency as well as the max current. Then the voltage on the ADJ pin would create an internal PWM signal to reduce the LED current.

END OF UPDATE TWO

UPDATE ONE

Using a 47Ω results in the LED being dimmer than it would be on an unmodified circuit board (I have an identical light which I can use for comparisons). 100Ω and above do not burn out at all. Also, to clarify, the LED is normally white, not blue, and I understand the colour change to blue is because of excess current.

White LEDs are often a blue LED with phosphor wavelength converters to convert the blue to white. Both blue and white are about 3V. It sounds like you have a high power white LED where a safe current would be about 350 mA.

Is it correct to assume the 22 Ω resistor gave you sufficient brightness?

For safety sake let's assume the supply is 12V rather than 7V.
For 350 mA at 12V you will need a 6W 25Ω resistor.
For 350 mA at 7V you will need a 3W 11Ω resistor.

What is the purpose of the LED?
Does it look like this (3 mm x 3 mm)?

Or this?

Post a picture of the LED.

END OF UPDATE

I measured the voltage from the pin to ground at 7V
the LED turns blue and dims
the power source is 11-12V

I am assuming the forward voltage of the LED is about 3V based on the blue.

If it's 7V supply to the LED and resistor, try a 200 Ω resistor.
200 Ω will give you 20 mA and 80 mW for the resistor, which is a very safe place to start.

If 200 Ω is too dim try 100 Ω
If 200 Ω gets too hot, try 500 Ω (supply voltage may be higher than 7V)
The 500 Ω would give you about 20 mA if the supply voltage is 12V.

What you did not say about the 47 Ω was how bright was the LED.
If the supply voltage is 7V 47 Ω would give you about 80 mA and 325 mW from the resistor.

A 22Ω would give you about 180 mA and 720 mW from the resistor with a 7V supply.

A 22Ω would give you about 400 mA and 3600 mW from the resistor with a 12V supply.

how should I correctly measure the amperage without putting across a voltage source?

You measure the voltage across the resistor then divided the voltage by the resistor value. If 5V across a 47 Ω resistor the current is 106 mA (5V/47Ω).

The voltage across the resistor will give you the supply voltage too.
If the resistor voltage is about 5V then the supply is about 7V.
If the resistor voltage is about 9 then the supply is about 12V.

• Using a 47Ω results in the LED being dimmer than it would be on an unmodified circuit board (I have an identical light which I can use for comparisons). 100Ω and above do not burn out at all. Also, to clarify, the LED is normally white, not blue, and I understand the colour change to blue is because of excess current. – Frayt Oct 27 '18 at 17:11
• See update to my answer. – Misunderstood Oct 27 '18 at 17:26
• Thanks. To answer quickly: The LED is of the first type and under a magnifying glass you can see three individual "elements" within the LED. The 22 ohm resistor was sufficient but as bright as an unmodified circuit board. I'll get back to you with some photos tomorrow. – Frayt Oct 27 '18 at 18:38
• I have included a picture with some updated readings – Frayt Oct 28 '18 at 16:52
• A picture is worth a thousand words. See update two. – Misunderstood Oct 28 '18 at 18:20