help me understand my readings?

[cooldude666]

I've begun building my first self designed amp. 70 watts from 2 6550s, fixed bias, reverb, clean preamp with a james tone stack.

I'm building the power supply now but I don't understand why I get such varying B+ readings. So first I took it straight off the rectifier and was seeing 600v, but when I connect my reservoir cap (50uf) and the paralleled 220k resistor that drops to 480. What's causing the change? My guess is it's got something to do with loading, but I'm lacking and understanding of how that causes the change. Anyone got an idea? Thanks!

[Steve Conner]

Well, what B+ voltage did you design for? What is the working voltage of your reservoir cap? What is the rated AC voltage out of the PT secondary?

[cooldude666]

They're salvaged transformers so I actually have no idea what the specs are. I measured about 340 off of the secondary so I had designed for what I'm getting, I was really surprised to see 600v when I took that reading. My cap is rated for 500v.

[Steve Conner]

Well, the DC output of a rectifier is 1.4x the AC input. 1.4*340 = 476, so that agrees with the 480V you measured. If it manages to stay above 400V under load you should get your 70 watts.

If you have a cheaper or older DVM, they can sometimes give weird readings with a combination of AC and DC, like you get from a rectifier with no smoothing capacitor.

[cooldude666]

I have a pretty nice DVM and I checked with a backup. Do you really think its just an anomalous reading from the unsmoothed DC?

[Steve Conner]

Could be some sort of ghost voltage coming through stray capacitance, in that case. Try with just the 220k resistor, no capacitor. If everything is working, you should read about 340*(1/1.11) = 306V DC.

If the meter is true RMS then with the smoothing cap removed, you should also see 340V AC on the rectifier output. (Rectifying the waveform doesn't change its RMS value.)

[cooldude666]

Ok here's another. I really just think I don't understand the effect of loading on a circuit.

I don't understand why my bias voltage changes when I add a voltage divider after the bias supply diode. The B+ seems be fluctuating a good amount too. I'm not sure what that's about, but they have the same AC voltage supply. The only changes between the two are time, and the voltage divider.

[cooldude666]

I don't know why it doesn't show in the schematic but the pot is a 47k

[cooldude666]

Does this have something to do with ohms law? I feel like by providing a path to ground here the full potential is diminished, and that's why I'm seeing a lower voltage.

[nickb]

Does this have something to do with ohms law? I feel like by providing a path to ground here the full potential is diminished, and that's why I'm seeing a lower voltage.

This circuit is really quite complex when you get down to the details, but essentially the 50nF capacitor has a significant impedance at the mains frequency and therefore has a voltage drop when current flows. Now since the drop is proportional to the current and the current is dependent on the load then increasing the load increases the drop and you get less volts out.

[Steve Conner]

You never mentioned that you were using the Marshall style bias derived from the HT winding through a capacitor.

I think that if the bias supply is loaded heavier than the main supply, it will act as a voltage multiplier and pump the main supply up, possibly to a theoretical maximum of twice the unloaded output. I really don't like this circuit and try to avoid it. :O

You can make a balanced version with two capacitors feeding a bridge rectifier, instead of one capacitor feeding a single diode. It shouldn't pump the main supply up because the AC current draw is balanced, there is no DC component trying to find its way back to earth through the main supply.

[cooldude666]

do you have a schematic for this improved version? sounds interesting

[g1]

Does this have something to do with ohms law? I feel like by providing a path to ground here the full potential is diminished, and that's why I'm seeing a lower voltage.

Maybe you are confusing unloaded voltage with "full potential"? No load voltage readings are fairly meaningless, as most circuits end up working into loads.
"Full potential" is usually a voltage specified into a load, such as a high voltage winding spec'd at "340V @ 100mA", for example.
No load voltages are sort of like engine horsepower measured at the crankshaft. Once you connect the engine through transmission, differential, and measure horsepower at the wheels, a fair bit of the "unloaded" horsepower is gone.

[cooldude666]

Yes, this. Thank you.