I'm needing to neutralise the "battery activation" circuit on my Hotpoint Q75 (AWA 653) to protect a B-battery simulator from being fed a charging current that would have been sent to a 90v battery.

I think I can just take out a couple of resistors that couple the 90v and 9v sources to the A and B batteries when the rotary power selector is switched to "Activate".

I notice that the 6X4 rectifier has both plates connected to chassis ground, except that on the "Activate" setting one plate is isolated from ground and instead connected to the negative of the A battery (9v).

In this circuit, the heaters are connected in series to the 9v, giving an average of <2v per filament. The 6X4 heater has its own 6vac tap on the transformer so is on a different circuit.

Never seen this rectifier setup before, with the B circuit getting its DC apparently positive from chassis during AC operation, but negative from chassis during battery operation.

Do I understand correctly, and does this mean I should forget any ideas of charging the batteries in circuit?

See schematic here

Maven

I will print the circuit rather than have it online. But you cannot recharge if those are dry cells albeit there were some where power was applied. But it would not be in reverse.

However, the easiest way to stop current going backwards is to use diode blockers.

If you put a mains, SS regulated PSU in it make very sure that nothing on the output side can touch any form of mains earth.

Marc

With this set the polarity of the filaments comes into play. These are polarised as the filament voltage drop forms part of their bias.

Secondly the grid is negative with respect to cathode (filament in this case), so in order for this to be achieved & for the filament battery to act as "C" as well as "A" it has to be positive to the common ground. The B+ is a separate circuit and remains negative to the common ground.

Marc

On the circuit, I see +90vdc from the B battery connecting directly to the second grid of the 3V4 power amp and via resistors to the second grids of valves 1-4. Are you saying that the various caps to ground and resistors in series reduce those grid connections to less than the very low filament voltages?

When the AC power is in circuit, what does it mean that both plates of the 6X4 rectifier go straight to ground?

Maven

I think you are not showing a familiarity with the valve. There are two elements on valves like 1T4 & 3V4 which are Pentodes, that get B+ 3V4 will have full B+ on its screen and the plate will get B+ from the Output transformer.

The screens of the other Pentodes may have a common rail, perhaps several, that may be less (eg 67.6V) than B+.

With AC on the diode plates the cathode of the rectifier should be positive.

The only batteries that you will recharge in circuit would be rechargeable types & you would need current limiters. I doubt the current transformer would hack it.