Adjustable NFB pot wattage

[Madditch]

I wanted to add a pot in line with the NFB resistor and I have a 100K 16mm Alpha that I wanted to use. I believe its only rated at 1/4 watt. What is the lowest wattage pot I can safely use for that application?

[SDM]

Using Ohms law in the form of P(watts)=V^2/R or rearranged as V=sgrt(P*R):

Say your amp actually puts out 100 Watts under full clipping, a quite safe overestimate taking about a JTM 45. Say you also leave a 27K NB resistor on the board to set the maximum NFB you'll allow (never want way too much NFB), and say you are using the 16 ohm tap which also has the highest voltage of any output tap. The added NFB pot then goes from 16 ohm tap to the 27K which then hits a 5K to ground, forming a voltage divider. 100 watts into 16 ohms = sqrt(100*16)= 40 volts on that 16 ohm tap.

To see how the formula works, can just look at what kind of wattage the 100K pot will need to dissipate set off and cranked for starters. The pot forms a divider with the rest of the NFB ciruit, which looks like a 27K+5K=32K resistor to ground. The divider starts with 40V on it, has the pot setting forming the first part, 32K forming the second part to ground.

So, if the pot is set at 0/off, easy there as it's basically shorted out. For the heck of it though, say the pot only dials down to 1 Ohm at its off/lowest setting. Voltage the pot drops is then calculated by summing the voltage divider parts, here 1ohm+32Kohms = 32,001ohms. Now divide the pot's resistance by the total to see the share of the voltage it drops= 1/32,0001=.00003125. Have 40V at the start of the divider here, so the pot drops .000031252*40V=.00125 volts. The final equation now has: V=.00125, R=1 (since pot is set to 1 Ohm), and the power the pot dissipates here is P=.00125^2/1=.000001563 Watts or 1.56uW, that's way way too low to even begin to care about.

Say pot is set at 100%, fully cranked now to 100K. Its resistance is now 100K, and going through the above, 100K+32K=132K. 100K/132K=.7576, 40*.7576=30.304 volts. P=30.304^2/100,000= .00918 watts. That's only 9.18 milliWatts, no problem there.

May think Ok, in the clear here as max power dissipation for the pot would happen with the pot at maximum resistance, but that would be incorrect. It'll instead happen when the pot is set = to the second half of the divider, set to 32K. So lets see what it dissipates when set to 32K, most power it'll ever need to dissipate: 32K+32K=64K, 32K/64K=.5, .5*40=20volts, P=20^2/32,000=.0125Watts = 12.5milliWatts. 12.5mW is way below the 1/4W (or 250mW) rating of the pot, so no problem there.

Thus the 1/4 watt pot will work just fine in this example, no real stress at all as the most power it'll ever see is 12.5mW. Things are overall simplified a bit here, but this basic method of calculating is more than close enough, especially as the amp's output power was overestimated to start with and so little power is involved here.

Hope that made sense.

[dirtybogner]

great post... from newbie