Continue to Site

# Calculating required Wattage of Resistors

Status
Not open for further replies.

#### Gandledorf

##### New Member
First off, thanks so much to all of you who have helped me with my questions, I hope to be able to share some of what I know with you in time, and what I am learning with the new comers.

I am trying to calculate the wattage required of a resistor used to limit the current flowing to an LED. My circuit is roughly:

V_signal is connected to the output line of a uC, which is pulsing in the KHz range. I'm not 100% sure what my required pulse width will be, but just for starters lets say 50%.

The max current through the diode ought to be around 1 Amp with the current setup (correct?), now given that watts = amps * resistance, this means that where the current to be continuous, I'd need a 5W resistor. I'm guessing though that I only need to meet the average wattage? So only 2.5W?

Further more, if I split up the resistance into two 2.5 ohm resistors, does each only need to be 1.25W, and so on?

I am not certain I understand what your circuit does, specifically, the purpose of the capacitor. In any case, the power dissipation in the resistor is given by P=I*I*R rather than I*R. So if you have 1A flowing through your resistor, you would use a 5W resistor. If for instance your uC locked up, and allowed the base current to continuously flow (ignoring the cap for a moment), you would cook a lower power resistor. Of course, you should design around the maximum you expect to see.

Putting 2 2.5Ohm resistors in series would allow you to "split" the dissipation between the resistors so you would only need a 2.5W resistor.

crust said:
I am not certain I understand what your circuit does, specifically, the purpose of the capacitor.

It's a battery powered system which can only draw 500mA constant current from the regulator. The 470uF capacitor is to bank power in between beams (the LED is in the IR spectrum, and used for communication).

In any case, the power dissipation in the resistor is given by P=I*I*R rather than I*R. So if you have 1A flowing through your resistor, you would use a 5W resistor. If for instance your uC locked up, and allowed the base current to continuously flow (ignoring the cap for a moment), you would cook a lower power resistor. Of course, you should design around the maximum you expect to see.

True, true. I hadn't even thought about that. Burning up the resistor would toast my expensive LED too I imagine. At \$3 a pop, I don't want to do that.

How do you figure that? I'm a bit rusty in my analog, so I've forgotten a lot of the formulas. Tell me if this is right:

The Vf you are referring to is the forward voltage drop of the LED, correct? If so it is about 2.6 at 1 Amp according to the data sheet: https://www.electro-tech-online.com/custompdfs/2004/01/172321.pdf

So your 2.5 is about right. Now where do you get the remaining 0.7V drop? Is this from the transistor? I had forgotten about the voltage drop on the LED. So actually I ought to make the resistor 2.5 Ohms, correct?

Putting 2 2.5Ohm resistors in series would allow you to "split" the dissipation between the resistors so you would only need a 2.5W resistor.

Thanks, that was what I was thinking. I'm trying to avoid buying 5W resistors, as it would require ordering from yet another company, and more shipping.

As already mentioned, the formula for wattage is W=I*I*R, but this is for continuous power - in your application the power should only have a low mark/space ratio and be at fairly widely spaced intervals. This is why you can pulse an IR LED at high currents, because you only do it for small intervals of time.

but with only 10% use, 0.1*0.1*5=0.05W

Somewhat of a difference!.

what kind of led do you have that you can pulse at 1A for 50% duty cicle??? it seems that there is a problem with the datasheet. i can't open it.

bogdanfirst said:
what kind of led do you have that you can pulse at 1A for 50% duty cicle??? it seems that there is a problem with the datasheet. i can't open it.

Data sheet opens fine from my end, so I am not sure what the problem is.

It's an Optek133 IR LED. It can withstain a max of 10A forward current on a 2us pulse width .1% duty cycle. Maybe it can't sustain 1A over the period I am looking for... I don't remember where I came across that figure.

What is the rate of change in forward current to duty cycle? It can sustain 100mA constant forward current, which isn't a linear relationship with 10A on .1% duty cycle.

Would I be safer limiting it to 200mA? Nigel suggests a 10% duty cycle, and after working out some diagrams last night, I can see why. So given this, can I assume 1A forward current?

Gandledorf said:
Would I be safer limiting it to 200mA? Nigel suggests a 10% duty cycle, and after working out some diagrams last night, I can see why. So given this, can I assume 1A forward current?

I didn't suggest 10% as an exact value, just to show how the dissipation works out for the resistor. If the maximum continuous current is only 100mA, I would like to see less than 10% - probably I would aim for no more than 5%.

There are a number of decisions which will affect the duty cycle, firstly the mark/space ratio of the modulation used. This is usually around 38KHz, so for 50/50 we're talking roughly 13uS on, 13uS off - so we're already down to 50% duty cycle.

The actual IR encoding should only use fairly short pulses of this 38KHz, the Sony SIRC's example in my tutorial uses pulses of 2.4mS (start pulse), 1.2mS (1's) and 0.6mS (0's) separated by pauses of 0.6mS. The entire command is sent every 45mS, so for the worst case (all 1's) the command takes 24mS, followed by a pause of 21mS, and for the best case (all 0's) it takes 16.8mS, followed by a pause of 28.2mS.

Assuming the worst case scenario, this means the LED is pulsed for 37% of the time (at 50/50), giving 18.5% duty cycle over 45mS. Under the best case it will only be pulsed 21% of the time, giving 10.5% duty cycle over 45mS.

These figures both exceed the 10% at 1A we are talking about, but if you alter the modulation of the LED to 25/75 this halves the duty cycle again - giving 9.25% worst case, and 5.25% best case.

Also, as Gandledorf has already told me he's not using 12 bit data, only 8 bit data - if you keep the 45mS timeslot it reduces the duty cycle still further.

Sorry for the long post (and for any calculator errors I may have made), but I think it may help the decisions required - it should be noted, that in my tutorial, I use two LED's in series, this give twice the output at no extra current drain - and decreases waste in the series resistor (which has to be a different value of course) - this is common practice in many remote controls!.

Status
Not open for further replies.

Replies
5
Views
948
Replies
12
Views
3K
Replies
14
Views
2K
Replies
8
Views
2K
Replies
15
Views
2K