Power tube grid bias and Tremolo

[J Rindt]

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[sluckey]

The intensity pot does not change the negative dc bias voltage at all.

[J Rindt]

How does the pot change resistance for the trem circuit but not the bias?
 Aren't they both connected to opposite ends of the same pot.?
As it decreases for the Trem, doesn't it also increase for the bias circuit?

Thank You

[sluckey]

It's a magical mystery.      Not really.

Let's first examine why the INT pot has no effect on the negative DC voltage. The DC voltage is applied to the right side of the pot. But the left side of the pot is connected to a capacitor and we all know that a cap blocks DC voltage, right? Therefore, no current can flow from the right side to the left side of the pot. No current flow means there can be no voltage drop. IOW, the left side has the same voltage as the right side. Turning the INT wiper from one end to the other sees the same DC voltage at each end and all points in between. Makes sense, right?

So, why does it work for the trem? First, understand that the tremolo signal is just a low frequency sine wave AC signal. So that trem signal passes through the cap and appears on the left side of the pot. And the right side of the pot is held to AC ground due to the 25µF bias filter cap. Let's assume the trem signal is 50VAC (that's realistic) on the left side of the pot and since the right side is at AC ground, there will be zero volts AC on the right side. Now the AC voltage on the wiper can vary anywhere from zero volts to 50VAC. Makes sense, right?

The AC trem signal rides on top of the negative DC bias voltage, causing the net voltage sent to the tube grids to vary above and below the DC voltage. This is how the bias vary tremolo works.

[J Rindt]

Oh Man........ as soon as you said capacitor. 
I did not even think about that, or look past the pot on the Trem side.
At any rate, your explanation was textbook clear.  I really appreciate it.
Awesome.....  Thanks Again