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 14th November 2003, 06:56 PM #1 e.lectronick   diyAudio Member     Join Date: Feb 2002 Location: Ocean, NJ Dumb Question about Wattage Calc Hi, guys. I'm still just a noob when it comes to DIY, and it's been a while since I've been in an electronics class. Can somebody direct me to a quick and dirty resource or "rule of thumb" calculation for determining the wattage requirement for resistors and heatsinks for a given circuit design? I know that there are certain places where a resistor will be in contact with the power supply and need to be much beefier than a resistor in the signal path, but the exact calculations to determine the required ratings is confusing to me. I have a mental block when it comes to this. Also, what is the quick and dirty rule of thumb for determining the size of heatsink you need for a design? Thanks in advance for answers. Apologies in advance for asking such a DIY 101 question. -Erik. __________________ -Who put the Tribbles in the Quadrotriticale?
 14th November 2003, 07:39 PM #2 BobEllis   diyAudio Member     Join Date: Sep 2003 Location: Upstate NY Power = V*I = I^2 * R = V^2/R, just plug in the values that you know. NP has walked through the heat sink sizing calculations several places in this forum, and in the A75 articles at passdiy.com Pretty straightforward stuff - determine the rise that you can live with, divide by the power dissipated in the output stage and from that you get the required heat sink rating in degrees C/watt dissipated. Apply a fudge factor to account for things like the back of the heat sink being in the case. I think AAVID and Wakefield have tutorials on their sites. Aavid has a calculator that lets you determine the rating of a section longer than the cataloged 3".
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Join Date: Jan 2003
Quote:
 Originally posted by e.lectronick I have a mental block when it comes to this.
Me too, but I will try anyway. It's only resistrors after all...
First I keep Ohm's law figure under my pillow.
Choose yourself a nice one, print it and hang it in front of your design table or anywhere close to your DIY-environment. It helps. A lot...
From there you will find out that the more current will flow through a resistor the more powerful it has to be.
There are a lot of nice JavaScript calculators around as well. Do a google search. Every time I go there, more implementations have been written.
There is no rule of thumb, certainly not for power resistors. You’ll have to do the maths. As for other resistors in the signal path, usually 0.6W is enough. However, when not calculated, you might find some surprises now and then.
I think it’s not a bad idea to double or even triple the found wattage value of a resistor. At least not for DIY.
Commercial products have to cut costs on everything, including resistors.

As for heatsinks, suppose you have to dissipate 25W.
A heatsink is rated 1°C/W. This means your heatsink will be 25°C at full 25W dissipation.
If the same heatsink is rated 0.1°C/W it will only heat up to 2.5°C.
That is above room temp. of course.

/Hugo

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