Quote:
Originally Posted by 6SN7WGTB
Thanks Lawrence for taking the time to reply.
I’ve now got the full schematic which makes it clear.
In the two HV AC ranges (500 and 1500) the circuit is as follows:
1. 0u1 input cap
2. A ladder of its own amounting to 388k total going to common
3. A takeoff from between 0u1 and 388k going to the rectifier tube going via a 47n
4. A full ladder ‘below’ the tube of about 30M
So, Xc at say 50Hz is 32k (btw that calculator gave me milliohms instead of kilohms).
The rectifier and 30M ladder has little parallel effect on the 388k, so the Z of the 0u1 and 388k is root sum squares = 389k. Current at 1500V thus ca. 4mA.
Voltage across cap thus ca. 32/388 x 1500 = 141V or ca. 200 WVDC.
However, I assume the theoretical edge case is if you apply DC to the AC input, I.e. frequency = zero Hz. Surely at this point the voltage drop approaches 1500V DC?
So I’m still unconvinced about my cap being rated 1000 WV DC…
(Also BTW doesn’t the V7A have a dedicated HV ladder up front of the rectifier, so does that undo your calculations? Meaning that they should look more like mine?)

The AC input circuit on the highest AC voltage range (1,500V) in the Heathkit V7A primarily consists of a series circuit comprising of a 0.01uF capacitor in series with a 900k resistor, a 320k resistor and a 150k resistor, the feed to the diode rectifier is taken from across the 150k resistor and as I said in my 1st post the effect of the diode circuit was ignored, the reason being that it wouldn't have greatly affected the outcome of the calculation, the calculations I did seem to be correct to me, not sure about your reactance being given in milliohms, here's a screen shot showing the calculation from my figures.
Lawrence.