by pebe » Tue Jun 03, 2014 8:26 pm
In answer to your last question. No, I don't know any more about the unit than is given in the manual and the circuit. I thought I had explained in earlier posts how I arrived at voltages required, but for the sake of continuity I'll repeat it here.
It is a 30V supply so the absolute minimum voltage required at TS29 emitter is 30V + the c-e drop of the TS29, So probably about 31V min allowing for some ripple on C30/C31.
The absolute minimum voltage on C26 must be (3 x 6.2) for the 3 zeners in series + 4.5V across R27 + about .5V (for TS26 with c-e saturated) = 23.6V. Those are the DC voltages.
Now for the AC voltage requirements. A rectified AC voltage from a bridge feeding into a reservoir capacitor with no load would theoretically charge it to the peak of the AC waveform (1.4 x its RMS value). In practice, the capacitor would draw heavy current from the AC when the diodes conduct during the peaks of the waveform, and so would 'flatten' them. So the peak rectified figure is nearer RMS x 1.25 or x 1.3. Then there is the voltage drop across 2 diodes in series of the bridge rectifier ( = 1.4V which I left out of my earlier calculations), so the formula simplifies to the minimum AC input voltage to rectifier = 1.4V + Peak voltage/1.4
In the two cases it would be:
For S2: 1.3V + 31/1.4 = 23.54VAC
For S3: 1.3V + 23.6/1.4 = 18.26VAC
And these needs are higher than the 17V and 14.9V voltages you read.
I hope that all makes sense.
So you run on an inverter. You have just added another unknown into the puzzle. So that I can get my head around all the possibilities, could you tell me:
1. Is the output waveform of your converter a square wave or a sine wave?
2. What voltage have you set the input taps for?
3. Is the converter the same one you were using when the power supply was working OK, or has it been changed?
4. Does your voltmeter calculate true RMS voltage (as distinct from one that derives a voltage from the average rectified DC level)?