question about power rating on a potentiometer

[mort]

ok, so I'm planning out an 18W marshall style build with a VVR bjuilt in. The plans I have call for a 1/2W 1m linear switched(0ff at 0) pot for the VVR. I'm looking at a pot labelled as 2A, 125V, .25W. Does that mean that 125v should not be exceeded through this pot? Or can we assume if the voltage is doubled then the amp rating cuts in half? The secondary of the VVR will have 2 el84's and a 12ax7 and will be fed by a 302/0/302 PT through an EZ81. Let's say the loaded voltage is ~350. 350v/125v= a 2.8 multiplier---- and 2A/2.8=714mA which would be plenty more than enough rating to handle the VVR load, correct? Am I doing this correctly?

[g1]

Can you link the exact pot?
I'll guess that the switch contacts are separate and that is what the 2A 125V refers to?

[mort]

ah you're probably right..

here's the pot: https://www.cedist.com/products/R-VS-1MA

[Tom Phillips]

The person who wrote the specs for that pot catalog listing included quite a mish mash of specs.
In one place it lists "2 amp, 125VAC, 0.25W, single pole, single throw switched potentiometer."
Further down it lists "Max Rating: 0.25W, 250VDC (Audio)"

As already pointed out,2 amp, 125VAC is the switch rating but the 0.25W is the dissipation rating of the pot element so it's deranged that it should be included in the middle of that first line.

0.25W is that maximum continuous dissipation rating of the pot element (when you are dissipating across the whole element) and 250VDC is the max voltage that can be safely applied anywhere to the pot track.

I assume that it is an audio taper pot but why they wrote "250VDC (Audio)" is a very strange presentation. Maybe they were giving the shipping clerk or the stock room person an opportunity to write catalog copy. Or...maybe this pot is meant for use in powerful amps that can be played a bedroom volume without disturbing the people in the rest of the house. 250VDC (Audio) would be very quiet. At least until the speaker caught on fire.

[J M Fahey]

I bet the person using a 250V DC amplifier will soon write here complaining about the LOUD !! turn on thump

And his personal bedroom sized:

[Justin Thomas]

Always fun times when Juan uses Mega-Font and his nuke-bomb picture...

Justin

[mort]

So based on the load that the pot will be under, I need to find a different pot, don't I?

I've been hunting for a higher powered one with a switch but am having no luck.

[Leo_Gnardo]

So based on the load that the pot will be under, I need to find a different pot, don't I?

I've been hunting for a higher powered one with a switch but am having no luck.

Enough faffing around: the pot you had in mind in the first place will work just fine. 2 amp 125V refers to the switch. Can't say I've ever seen a 1 Meg pot roasted from over powering: you'd really have to work hard to wreck one, not gonna happen in any normal amp.

[mort]

Enough faffing around: the pot you had in mind in the first place will work just fine. 2 amp 125V refers to the switch. Can't say I've ever seen a 1 Meg pot roasted from over powering: you'd really have to work hard to wreck one, not gonna happen in any normal amp.

Thanks for that.

Just for my own self sufficiency, where am I missing the math? ~350vdc x .12A (the sum current of 2 el84's and 1 12ax7 that this pot will be feeding) = 42w but that seems very wrong for some reason.

[g1]

How about showing us what the pot & switch are doing in the circuit.

[Tom Phillips]

How about showing us what the pot & switch are doing in the circuit.

+1 Yes. Let's see the schematic.

...where am I missing the math? ~350vdc x .12A (the sum current of 2 el84's and 1 12ax7 that this pot will be feeding) = 42w but that seems very wrong for some reason.

Yes it's very wrong for the application. Your math equation is correct but your assumptions are wrong so you are calculating the wrong thing. The schematic will clear the haze and allow people the understand your planned design.

[mort]

ok the schematic for the amp is the first attachment. It's a 2 channel 18W Marshall style amp.

The second attachment is for referencing the VVR itself, which is where the pot in question is drawn.

oh and the switch is the main power switch for the whole amp.

[Leo_Gnardo]

The second attachment is for referencing the VVR itself, which is where the pot in question is drawn. oh and the switch is the main power switch for the whole amp.

The pot is providing a "programming" voltage to your FET, and the FET is passing current to your tubes. Back of the envelope calc gives me about 1/10 watt dissipation in the pot. For safety I would avoid mounting that pot on chassis metal, put it on a piece of non conductive plastic instead. Plexiglas, Lexan, something like that. Keep in mind there's not going to be much space inside the pot between its shell and the track & electrodes. Breakdown voltage of air is 8000V per inch. The 350V in your circuit could hop across a millimeter gap. I'd place the pot inside the chassis, with a screwdriver slot on end of shaft, on the other side of a hole with a plastic insulator grommet on it. I can see a knob on this sort of thing would be an invitation to a bad shock.

I'd use a separate switch for the amp power. What sorta genius thought to put the power switch on the back of a pot doing this function...

Also with the switch separate, you may be able to find a pot that has a higher power rating. In the hi-vacuum surface studies lab we'd sometimes have to rig up something like this. Those tough ol' Allen-Bradley or Ohmite pots were rated 2 watts, some came with pre cut screwdriver-adjust shafts. Snoop the surplus sources like ApexJr.

[Jazz P Bass]

Snicker.

NTE2377.

[mort]

Snicker.

NTE2377.

expensive as hell. I did some digging around and found someone from this site using a IRF830 in amps under 20W so that's what I ordered.