Author Topic: Calculating power requirement for a resistor  (Read 232 times)

mark2

Calculating power requirement for a resistor
« on: February 02, 2021, 01:18:45 PM »
This may be asking too much, so no worries if so (and I'll simply continue digging until I have a narrower question)...

Can anyone walk me through the considerations and calculations you do to determine the max power through the 220 resistor to ground coming off the sustain pot in the middle of the schematic below?

Or more generally, can anyone point me to resources to better understand how you holistically apply ohms law on a circuit.

For a single component it's easy enough.  I can of course calculate worst case: (9V^2)/(220 ohm) = 0.37W ... but I'm guessing it's not that simple when placed in this larger network. e.g. do the decoupling caps on its + side have any impact?

Thanks!


from http://fuzzcentral.ssguitar.com/foxx.php

antonis

Re: Calculating power requirement for a resistor
« Reply #1 on: February 02, 2021, 01:40:39 PM »
Only AC can flow through 220R resistor, due to 10 μF coupling caps, the amplitude of which is in accordance with Q2 Collector output (+/- one diode forward voltage drop..)

Power dissipated is p = Vrms X Irms = I2rms X R = V2rms / R and for instaneous power P = [V2 / R] X cos(θ) or I2 X R Χ cos(θ)

P.S.1
In case you worry for items power rating this should stand for 50k pot rather 220R..  :icon_wink:
(same or heavier series current for 220R and pot resistance, depending on wiper setting, but much higher value than 220R)

P.S.2
>can anyone point me to resources to better understand how you holistically apply ohms law on a circuit<

Simply by Kirchoff laws (KCL & KVL) implementation..
« Last Edit: February 02, 2021, 05:59:38 PM by antonis »
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

mark2

Re: Calculating power requirement for a resistor
« Reply #2 on: February 02, 2021, 01:49:18 PM »
which is in accordance with Q2 Collector output (+/- one diode forward voltage drop..)
So on the order of 0.001W?  (1.4V^2)/(220ohm)

If it's in that order of magnitude I should be able to safely use 1/10W 0603 resistors.

Quote
P.S.1
In case you should worry for items power rating this should stand for 50k pot rather 220R..  :icon_wink:
(same or heavier series current for 220R and pot resistance, depending on wiper setting, but much higher value than 220R)
I'm sorry, I don't understand.  Are you saying the pot's low power rating makes it more of a concern than resistors in most cases?

And thanks for all the info!

antonis

Re: Calculating power requirement for a resistor
« Reply #3 on: February 02, 2021, 01:58:43 PM »
So on the order of 0.001W?  (1.4V^2)/(220ohm)

More close to 0.009mW.. :icon_wink:

Are you saying the pot's low power rating makes it more of a concern than resistors in most cases?

I'm saying that for the same current, power rating is directly proportional to resistance vaue..  :icon_wink:

e.g. for 1mA current say, 220R must be rated at 220μW (0.00022mW) and 50k at 50mW..!!
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

mark2

Re: Calculating power requirement for a resistor
« Reply #4 on: February 02, 2021, 02:01:16 PM »
Whoops, yes, i added an extra zero (was trying to round up to 0.01).

Thank you. I understand now. I appreciate the help, tips, and consideration!

Rob Strand

Re: Calculating power requirement for a resistor
« Reply #5 on: February 02, 2021, 05:52:59 PM »
Because a common current I must flow through the 100k's and the 50k pot and the fact power dissipated is P = I^2 R the higher valued resistors will be the ones dissipating more power; you can see that from antonis's calcs.   Not to mention all the other resistors in the circuit which are dissipating more than the group around the 220R.   The worst ones will be the 4.7k's, roughly 3V ^2 / 4.7k = 1.9mW.
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