# Help with smoking resistors

1. Sep 22, 2008

### mgibson

Hello,

I am working on a LED project that involves the use of three 3W LEDS. These LEDs are being driven at about 700mA (I believe). My source is 12V (8 AA's in Series) and I am using 16 Ohm resistors (one for each LED) capable of handling at least 5.4 watts ( ~15 Ohm resistor in series with a .47 Ohm, 5W resistor). When I turn the LEDs on, the .47 Ohm, 5 Watt resistor begins to slowly smoke after about 15 seconds. Why is this? Is this normal for the first few times the circuit is turned on? I thought the reason for the 5.5 Watt resistor was to handle this heat load. How do the watt ratings of resistors add in series and parallel?

If anyone can help me figure out what is wrong, I would greatly appreciate it. Thanks!

2. Sep 22, 2008

### Pumblechook

Do you mean each LED is in the series with two resistors and then this combination forms 3 parallel circuits?

It seems a very wasteful way of doing it. Small LEDs run with around 1.6 volts across them. Higher power ones presumably have a higher voltage drop but much less than 12 Violts ??

I would think the 15 Ohms are the ones smoking.

If you are running 700 mA through a 15 Watt resistor it is disippating 7.35 Watts (0.7 x 0.7 x 15). A 0.47 Ohm will dissipate 0.23 Watts (0.7 x 0.7 x 0.47).

Better to put the LEDs in series and have a smaller value dropper resistors and maybe not so many cells.

You can get LED drivers (integrated circuits) for this purpose which may work on an efficient switched mode principle rather than crude dropper resistors.

Can you measure what the voltage across the LEDs is with 700 mA current flowing through them?

3. Sep 22, 2008

### Pumblechook

4. Sep 22, 2008

### mgibson

The three LED circuits are identical so I will describe one of them.

The LED is in series with 2 resistors (I thought the 15 Ohm one drops the current while the .47, 5 Watt one should dissipate the heat). The voltage drop across the LED is 3.6-4.2V. When I calculated the resistor needed for this set-up (Source: 12V - Vdrop of LED: 4V - desired current: ~700mA), I got a ~16 Ohm, 5.4 Watt resistor. I thought that I could put the two resistors in series to get a cumulative (15+.47) 16 Ohm, (.5+5) 5.5 Watt resistor. This must not be the case as you have shown with your power calculations.

It may be the ~15 Ohm (which is actually 2 10 ohm resistors (Parallel) in series with 2 20 ohm resistors (parallel)) that are smoking. Its hard to tell since they are so close to each other.

I cannot alter the circuit much, just the resistors. Is their a way to make a ~16 Ohm, ~5.5 Watt resistor by combining easily available (radioshack) resistors? Would 6 100 ohm, 1 watt resistors in parallel (making 16.6 ohm resistor) work? Would the power be ~8W (.7x.7x16.6)? with each resistor dissipating 1.36 watts ((.7/6)x(.7/6)x100)? for a cumulative 8.16 watts (1.36x6)?

Thanks so much for your help. I know their are better ways to design the circuit but because of the design layout and purpose, this set-up was ideal.

5. Sep 22, 2008

Well, there's an oversight here. The resistor is going to see 8V and 700mA, which works out to 5.6 Watts of power that is needs to dissipate. However, you need to leave some margin in the power ratings of the resistors. A 5W resistor is not going to handle 5.6W for any length of time, unless maybe you've got a fan blowing cold air directly onto it or something. And note that as the resistors overheat, they may well lose resistance, which will result in higher currents through them, and so heat them even more. You should be using resistors rated for at least 7W, or even more if this circuit is going to be housed somewhere enclosed, particularly with lots of other circuitry or heat sources.

Whichever circuit configuration you go with, you need to calculate how much power will be dissipated across each resistor, and then add some margin to that (maybe even double it) to get the power ratings that will result in reliable operation.

That's all correct, except that 1W resistors will not safely dissipate 1.36W. Instead, they will overheat and start smoking. You need resistors capable of at least 2W for such a circuit. Instead, you might consider using 12 200 Ohm, 1W resistors, although 24 400 ohm, 1W resistors would be better.

6. Sep 22, 2008

### mgibson

Thanks a lot. I think I have a good understanding of the mistake I made and how to fix it. My only problem now is going to be finding resistors for this at radioshack so I can finish this project on time.

Thanks again for all your help

7. Sep 22, 2008

### Pumblechook

If you are dropping 8 volts with a 700 mA current.. That is 11.4 Ohms...5.6 Watts.

So around 9 Watts are going to the LEDs and 16.8 Watts are being wasted in resistors.

You might get away with all three LEDs in series and just a small value resistor if you can get away with a lower current and therefore lower LED forward voltage. Or you might need one or two extra cells to get the total available voltage up but you will taking 700 mA from the cells and not 2100 mA. AA's will last minutes at that high current and not too long at 700 mA for that matter.

I would use a power transistor not resistors and bias it for 700 mA collector current.

What is the purpose for these LEDs ?

8. Sep 22, 2008

### mgibson

The purpose is for a light fixture I designed in such a way where it would be very inconvenient to redesign the circuit. So I am going to use either:

2 8ohm, 20W resistors in series, giving me 16ohms with up to 20W, which is more than double the 7.84 (.7x.7x16) watt requirement.

or

3 50ohm, 10W resistors in parallel, giving me 16.66ohms with up to 10W per resistor, which is more than triple the 2.72 (.7/3x.7/3x16.66) watt requirement per resistor.

Thanks everyone for all your help, I greatly appreciate it.

9. Sep 23, 2008

Either of those configurations should work without overheating resistors. But, I think Pumblechook's advice is all very pertinent if you plan to use AA batteries for this. If nothing else, why not just reduce the number of batteries down to closer to the bias voltage of the LEDs?

Suppose you use $N$ AA batteries in series, with an LED drop of 4V, drive current of 0.7A and a resistor of $R$ Ohms. Suppose that the desired maximum power dissipation in the resistor is $X$W. Ohm's law for the resistor gives $1.5N - 4 = 0.7R$, while the power constraint is $(0.7)^2R<X$. Solving Ohm's law for $R$ and substituting into the power constraint gives $0.7(1.5N-4)<X$. Suppose you want to buy 5W resistors, and so set $X=2.5$. Then, you have

$$N < (\frac{2.5}{0.7}+4)\frac{2}{3} \approx 5.05$$

So, if you want to use resistors in the 5W or lower rating, you need to use no more than 5 batteries. Using 3 or 4 batteries will result in even lower waste heat, but the required resistance starts to get pretty low. For $N=5$, Ohm's law gives $R=5$ Ohms, $N=4$ gives $R=2.85$ Ohms and $N=3$ gives $R=0.7$ Ohms.

Last edited by a moderator: Sep 23, 2008
10. Sep 23, 2008

### Oberst Villa

No, you definitely don't want to see any smoke when you switch on any electric circuit. There is a rule-of-thumb: You should be able to touch a component for a few seconds with your finger. If you go "Ouch !" immediately, then it's probably too hot. (Warning: Before you actually touch it, make sure that there is no voltage present that is high enough to harm you. Also take care that you don't kill the circuit with static discharge...)

Furthermore, if you want to be on the safe side you shouldn't use the (partially) burnt resistors again. As quadraphonics said in #5, their values might have changed. Or the thermal stress might have resulted in small cracks that might have no result right now, but might make them fail later. The point is, any componment can only fullfill its specifications as long as you treat it kindly. Once you have tortured it so much that it started sending smoke signals all bets are off.

11. Sep 23, 2008

### mgibson

The reason I am using 8 AA's in series is because another part of the light fixture involves 24 basic LED's which came with free resistors for a 12V source. Also, Radioshack had convenient battery holders for 8 AA's. It was either 8 or 4 and 8 seemed most appropriate since I wouldn't have to buy new resistors and it would be sufficient in supplying plenty of current to the 24 basic LEDS and 3W LED with okay battery life. (I think)

Oberst Villa, Thanks for your input. My resistors are touchable and thus not very hot. I threw out the old burnt resistors.

Thanks everyone for your input and help.

12. Sep 23, 2008

Well, what's okay depends on your expectations, but you're talking something in the range of a few hours, supposing you buy expensive, long-life batteries. The current design is not so much an LED lamp that produces waste heat, as much as a heater that happens to produce light with a small portion of the input power.

It shouldn't be too difficult to tap out the battery holder at whatever position you need, should it? Then you could leave the low-power LEDs on the regular 12V source, and run the high-power LEDs off of lower voltage. Doing so would roughly double your battery life.

13. Sep 23, 2008

### mgibson

Okay, so I think I should re-think the source voltage. I didn't realize so much power was being wasted in heat. Is running the 24 low-power LEDs off of a 12V source (8 AAs) okay? Or is it wasting tons of energy, enough to make it worth re-wiring every resistor/LED pair? For the 3 3W LEDs, could I run them off of a 6V source (4 AA's) and still get 700mA through each of the three branches, for a cumulative of 2.1 Amps? Would this increase my battery life and efficiency significantly?

What supply would you suggest running 3 3W LEDs off of to get good battery life and efficiency? I can change the supply and the resistors without much hassle.

Thanks for helping me turn my wasteful, inefficient circuit into an effective one. I appreciate your patience.

14. Sep 23, 2008

The amount of power wasted by the low-power LED/resistors shouldn't be a big deal, even though the source voltage is much higher than the forward bias voltage of the LED. The current should still be low enough that we're not talking big power here, at least compared to the other parts of the circuit. So, I wouldn't change them for their own sake; if you decide to change the power supply because of the high-power LEDs then you will of course have to change the low-power resistors so that the low-power LEDs still work.

Running the 3W LEDs off of a 6V source would work (as described in my previous post), provided you can find a 2.5 Ohm, 10W resistor. Doing so will reduce wasted power from 5.6W to 1.4W, which is then less than a majority of the power draw. However, making this change won't affect battery life, because you're still drawing the same amount of current in either case. It just cuts down the number of batteries that you're wearing out at any given time, as well as the heat produced.

However, you're still not looking at very much battery life if you try to power three such circuits simultaneously. If you're drawing 2.1A, you'll run out even heavy-duty AA batteries in around an hour. Moreover, drawing that much current may not be possible (or, if it is, advisable) with AA batteries, which are designed more with low-draw applications in mind. I would consider using something beefier, like C or D cells, or the batteries used in heavy-duty electric lanterns.

15. Sep 23, 2008

### mgibson

For the 3 3W LEDs:

I am changing my source to 4 D cells creating 6V. Hopefully this will meet my current requirements and improve battery life. I am also changing my resistors to 3 ohms with up to a 10W power rating.

Thanks for your help and patience. I have learned a lot in the process.

16. Sep 24, 2008

### Pumblechook

If you can afford it buy a bench power supply unit which has constant voltage and constant currents modes and independent digital metering of both voltage and current .

You can play around with LEDs and all sorts of circuits at different voltage and current settings.

I aslo charge batteries with mine. You can see at all times the voltage and current.

I have 2 Thurlby-Thandar PL310 0-32 V.. 1.1 Amp PSUs.

http://www.testequipmenthq.com/SelectProduct.asp?P=Thandar_PL310&Code=PSD&ProductCode=PSDJ797

17. Sep 24, 2008

### mgibson

Great advice. I agree a bench power supply is a great thing to have when experimenting with circuits. I tried to buy one at the beginning of the year but it was back ordered too long. So I cancelled the order. Hopefully sometime this fall I will re-order it. Thanks