Peavey Heritage VTX - Bias Question

[Feldspar]

I picked up a Peavey Heritage VTX, and I was curious about bias adjustment on this design.

Here is a link to the schematics: http://www.schematicheaven.com/newam...ritage_vtx.pdf

Is it possible to modify this amp to allow adjustment of the bias via a potentiometer? If so, what how would you wire it to aciheve the desired result?

Thanks,
Micah Walker

[gingertube]

Quick Notes:

The Output Tubes are cathode driven. See the notes section on the RHS of page 1.
It says V(idle) which is the DC voltage across R155 and R159 should be 40mV max, 20mV min and within 10mV of each other (push pull balance).

Since those resistors are 3 Ohms that translates to:
13.3mA max
6.7mA min
3.3mA maximum difference between sides

That is the idle current for 2 tubes.

U9A, U9B, Q6, Q8, Q7 and Q9 form an AC drive servo for the output tubes with a DC demand superimposed upon it for the idle current.

The AC demand is fed into point "A" (bottom LHS of schematic) and the DC demand is generated off the +15V rail via R147 and CR39,CR40. All those other didoes and the secondary bias network R149, CR42, CR43 all influence the idle currents. The idle currents are extremely low with Tube dissipations around 7 watts total each side (3.5 Watts per tube).

It also talks about a "compression effect" in the power amp which suggests that the idle current must shift with signal level. This compression effect is probably what Q6 and Q8 are doing. Certainly they only conduct on the peaks at high signal levels.
The NOTES section says that to measure maximum power output then a link must be added to TP1 to defeat this compression effect.

Without doing an indepth analysis I would GUESS that R149,CR42,43 set the idle current and R147,CR39,40,R148 set the level of compression BUT they are bound to interact.

I'd be very disinclined to mess with it, and can't see any good reason to do so.

Cheers,
Ian

[Feldspar]

Thanks for the analysis. From what you said, I'm assuming that the bias is dynamically adjusted by the cathodes based on the current across the tubes.

This amp has a high power/low power which allows you to run the amp at 1/4 power or full power (~130 watts).

Can someone explain how the circuit works? I'm new to this, but I would like to understand the logic behind it.

Thanks,
Micah

[gingertube]

The HI/LO Power switch simply switches the high voltage to the output tubes between +525V and +262V respectively.

In simplest terms, the effective "resistance" of the tubes doesn't change so with twice the voltage you get twice the current and since power is volts times current you get 4 times the power.


Cheers,
Ian