resistor getting hot

Thread Starter

stephenstern

Joined Jul 11, 2020
5
Hi all,

Hopefully someone can help me out here.
Ok, so I have:

9 x 1w LED, its 12v, 300ma, 3,4v.
9 x 3w 33R resistor.
Power supply 12v, 200ma (testing with this one but i intend to use larger ma when all 9 leds are in use).

I believed this resistor to be good enough (its a fair size) to cope, but it seems to get very hot..

Using this guide..;
resistor.PNG
 

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Audioguru again

Joined Oct 21, 2019
1,593
Of course the resistor is VERY hot, the calculator program is wrong with 2.2W in a 2W resistor. Use a 4W or 5W resistor.
Lookup the temperature of a resistor that is operating at its maximum allowed power. The horrible calculator program even used an OLD 10% resistor value of 33 ohms instead of a common 5% resistor value of 30 ohms.

The problem is that the supply voltage of 12V is much too high for a 3.4V/300mA LED. If the supply is reduced to 5V then the resistor has 5V - 3.4V= 1.6V across it and with 300mA of current its heating is only 1.6V x 300mA= 0.48W and a 5.6 ohm/1W resistor could be used.
How will the LED be cooled?
 

ErnieM

Joined Apr 24, 2011
8,046
I'm going to assume that "9x1W LED" and such means you will eventually build 9 parallel resistor - LED lights but just testing one for now.

Using I^2*R (current squared time resistance) with .26A and 33 ohms yields 2.2 watts, so 3 watts is OK but I would use 5 watts for long term survival.

Now 2.2 watts isn't a small power, you should feel the resistor getting warm but not hot. Do you have a meter to measure the voltage drops? Even a cheapie Harbor Freight one will help here.
 

Alec_t

Joined Sep 17, 2013
11,402
Could you make 3 strings of LEDs, each string comprising 3 LEDs in series? That way much less power would be wasted in the resistors. I leave you to calculate the necessary current-limiting resistor values.
 

Tonyr1084

Joined Sep 24, 2015
4,710
The big mistake I used to make when powering an LED was to forget about calculating the wattage. Using a 12 volt source with an LED with a 2.2 forward voltage and a target current of 20 mA meant I'd need a resistor value of 490 ohms.

12V - 2.2 Vf = 9.8V
9.8V ÷ 0.02 = 490Ω

OK, great. I can now successfully light my LED. BUT WAIT! How many watts is that?

9.8V x 0.02 = 196 milliwatts.

A 1/8 (0.125) watt resistor can't handle 196 milliwatts. So if I wanted to keep things small AND cool I'd have to use a 1/4 watt resistor. If I wanted to push 300mW through my LED then that's going to be (assuming the Vf of the LED is 3.4 volts)

12V - 3.4 Vf = 8.6V
8.6V ÷ 0.300 = 28.6•••Ω (call it 29Ω)
8.6 x 0.3 = 2.58 watts.

At 2.6 watts, forget about using a quarter watt resistor. Forget about a 1 watt, or even a 2 watt. 3 watt? Feasible. But WILL run hot. 5 watt? Much better. But keep in mind the circuit WILL produce 2.6 watts of heat. Whether you distribute it in a 5 watt or a 10 watt resistor, you still have to manage that heat. Even though a 10 watt resistor will feel cool to the touch, the amount of heat energy will still be the same.
 

Thread Starter

stephenstern

Joined Jul 11, 2020
5
So the real issue here is 12v for 9 x 1w leds ?
I have found a 9v power supply. 2.1a

Can i use this, with the resistor i already have (33ohm)? (it does power it as i tried but if they will just burn out that wont help!). i assume this wont happen as im actually running less through?
 

Audioguru again

Joined Oct 21, 2019
1,593
With the 12V supply, can't you connect three strings of three LEDs in series? Then the LEDs need 3 x 3.4V= 10.2V and the series resistor for each string will be (12V - 10.2V)/300mA= 6 ohms and each resistor will heat with 0.3A x 0.3A x 6 ohms= 0.54W and 1W resistors will be fine.
 

Thread Starter

stephenstern

Joined Jul 11, 2020
5
I cant put the leds in series as each of the 9 are on 5m wire back to a box (which just joins power to the runs) - think decking lighting. so i can only power 1 led in a circuit.

if i use the <10v supply i have found, the math says a 22ohm resistor but will 33 be safe?
 

Audioguru again

Joined Oct 21, 2019
1,593
Did you measure the forward voltage of each LED? Nobody makes a 3.4V LED, it has a range of voltages which might be 3.0V to 3.7V.
If all your LEDs are 3.4V then with a 9V supply the resistor will have 5.6V across it then a 22 ohm resistor will pass a current of 5.6V/22= 255mA and heat with 1.43W. A 33 ohm resistor will pass a current of only 170mA and heat with 0.95W.
If some of your LEDs are 3.0V then you do the math.
 

Thread Starter

stephenstern

Joined Jul 11, 2020
5
Hi Tony. That's all in OP.

Rethinking this. I originally used the psu for the original leds what was in place. Which was 12v. But I don't really get why it was ever made with 12v..

Anyway. I now have a 5v 2a supply. This means I can buy a much smaller resistor and displace little. Meaning no heat.
The 33ohm one I have now limits current to 50~60ma. Which weirdly enough is still super bight on the led (rated for 300ma).. That I can't work out? If anything I'd like to restrict more current and dim the led further
 

Audioguru again

Joined Oct 21, 2019
1,593
Can you see things in very bright sunlight? Can you still see things in very dim moonlight?
Yes, our vision has an extremely large range of brightness so you can reduce the current in the LEDs and they still look bright.
 

Tonyr1084

Joined Sep 24, 2015
4,710
I now have a 5v 2a supply. This means I can buy a much smaller resistor and displace little. Meaning no heat.
The 33ohm one I have now limits current to 50~60ma. Which weirdly enough is still super bight on the led (rated for 300ma).. That I can't work out? If anything I'd like to restrict more current and dim the led further
Smaller resistors? Yes. "No heat"? Not possible. Suppose you have 5V and a 3.4Vf, that leaves 1.6V to be dropped through the resistor. You have a target (something you've chosen) current of 100mA. 1.6 divided by .1 = 16Ω. That resistor is going to have to drop that voltage and dissipate 160mW of heat. That's not a lot of heat, but it still is heat dissipated through a quarter watt (250mW) resistor.

To change the brightness you change the amount of current. Brighter LED's mean higher current AND wattage. Dimmer LED's means lower current and wattage. But using a potentiometer to accomplish that will fail because pots aren't designed to carry that much current. You COULD construct a "Current Mirror" (HERE) and control the brightness that way ( I think ). Would need to look that up and do a little more research; but a current mirror is a pretty basic control circuit.
 
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