'72 Fender Bassman Ten Pot on Pream Heaters

[Heavenamps]

Hello,

I recently picked up a 72 fender bassman ten. I opened it up to service a few issues and discovered something I've never seen before. There is a 100 ohm bias pot connected to the heaters of the preamp tubes (see attached image and schematic). I'm wondering if anyone has ever seen anything like this before. I suppose its some kind of mod to be able to vary the voltage on the heater supply? It doesn't appear in any schematics I've found and I'll probably just remove it, but I'm wondering if anyone can shed some light on this. Thanks!

[sluckey]

https://el34world.com/Forum/index.php?topic=28241.0

You can see it in the Bassman 20 schematic.

[Heavenamps]

Oh cool, thanks!

[acheld]

It's called a humdinger.    See:  https://www.valvewizard.co.uk/heater.html  about halfway down the page.

[WimWalther]

It's called a humdinger.    See:  https://www.valvewizard.co.uk/heater.html  about halfway down the page.

Not quite sure how, when or why "humdinger" took on this meaning, but I've always known these as "hum balance" pots.

They're typically 100R 1W devices. They dissipate about 400mW on 6.3V heaters, so a 1/2W pot is a bit light for that use.

[acheld]

Not quite sure how, when or why "humdinger" took on this meaning, but I've always known these as "hum balance" pots.

LOL.  I expect it's the same as some people saying "soda" when what they mean is "pop", or when we say "tube," but mean "valve."   

I dunno, I learned the term when I read Merlin's book years ago.  It always made sense to me, and while I know that a "hum balance" pot does balance current voltage in the two legs of the heater circuit, a humdinger actually "dings" the hum. 

Edit:  By the way, if you use a 250 to 500R pot, they of course dissipate less power.  When I use them, I'll use a 500R 1W Bourns pot.  That seems to work for my builds.

[sluckey]

They're typically 100R 1W devices. They dissipate about 400mW on 6.3V heaters, so a 1/2W pot is a bit light for that use.

Fender used 100Ω 1/2W CC resistors, usually conveniently installed on the power indicator assembly. This started after the AB763 series.

When you connect two 100Ω resistors in series and apply 6.3V the total power dissipation is 6.3²/200=200mW. So each resistor is only dissipating 100mW. Leo would never have wasted money on a 1 watt resistor when a 1/2 watt resistor was more than sufficient.   

[Williamblake]

The current is the same across the pot. The wiper sets the voltage in the voltage divider.