Sanity Check

[evan]

Do the same for ...

I didn't know quite what you meant by this since it basically rehashes my arguments. Then I read the thread again and see that you claim there is no HV across the open standby switch on the CT.

Well, heck with thought experiments, I tried it on a real amp.

What I get is close to what I expected. This particular amp runs a B+ of about 250 V under load. With CT open, I measure -275 VDC w.r.t. ground. I even put it on a scope and it's a typical sawtooth ripple. The only thing that surprised me was that the ripple was quite small. I expected to see rectified AC. Probably the reason for that is that this particular amp has "anti-fizz" caps strapped across the solid state rectifier diodes. Anyhow, point proven: the switch on the CT sees just as much open circuit voltage as it would between rectifier and first filter cap.

[VacuumVoodoo]

Your test instrument is affecting the test object. So does mine and everybody elses.

Think current instead of voltage.
When you connect a voltage meter between CT and ground with switch open you effectively put meters internal resistance there. Without the meter the CT is "floating in the air at undefined potential", meters internal resistance provides a current path to ground - that's why you get a reading.
For the CT power supply to work the CT must have a direct low resistance current path to ground. A Voltmeter has high internal resistance that provides current path from CT to ground. Know your meters resistance and calculate the current - that's the most current the switch will have to switch when closing. Pretty low value. That's why you saw very low ripple on the scope.
When switching back to stdby mode you'll be interrupting few tens of milliamps ground current. Maybe 200-300mA if you do it while playing pretty loudly.

If you worry about undefined high potential on the switch causing arcing you can always put a RC snubber across it.

My reason for using this stdby method is I'd rather have the switch at ground potential when the amp is in play mode and very high impedance (air) between switch elements and anything with voltage on it. CT->fuse->switch->ground

[evan]

Without the meter the CT is "floating in the air at undefined potential", meters internal resistance provides a current path to ground - that's why you get a reading.

On the contrary -- the secondaries are not floating. The reference point is through the rectifier and bleed resistor and/or tubes that are in conduction.


-Evan

[VacuumVoodoo]

Take a 9V battery and a capacitor. Ground one end of capacitor connect battery (+) to ungrounded end of capacitor. Leave battery (-) unconnected. Please explain how this can charge up the capacitor without ground return path to battery (-).

[evan]

Your latest example isn't relevant.

Maybe try a sim. You'll see.

Best regards,

Evan

[evan]

Take a 9V battery and a capacitor. Ground one end of capacitor connect battery (+) to ungrounded end of capacitor. Leave battery (-) unconnected. Please explain how this can charge up the capacitor without ground return path to battery (-).

What is relevant is what you've left out of your example. Namely, the resistance to ground across the capacitor, which is a bleed resistor or tube. The capacitor does not charge (and neither do the power supply caps in the amp) but the resistive path to ground does provide a ground reference, so you do have potential w.r.t. ground on the minus terminal of the battery... or the CT of your PT secondary.

If you had an amp with no drain resistor and no tubes, then you would have floating secondaries and no defined potential on the open CT.

[VacuumVoodoo]

So put a resistor across the cap in my battery example and for good measure replace battery with transformer winding and one diode. One end of winding hanging in the air. Which way will any current through the diode return back to source i.e. the free hanging end of the winding? Can you charge the cap or will the resistor only make sure the cap stays drained ? It's basic circuit theory.





















(hint: there will be no current through the diode)

[evan]

No offense... but I'm about to give up. Reality is as I've stated.

I cross checked my "thought experiment" in a real amp, and then in an LT SPice sim just for good measure. The sim and the reality bear me out. Why don't you try it yourself?

-Evan

[VacuumVoodoo]

No offense... but I'm about to give up. Reality is as I've stated.

I cross checked my "thought experiment" in a real amp, and then in an LT SPice sim just for good measure. The sim and the reality bear me out. Why don't you try it yourself?

-Evan

me too. You say you measured (-)275V on B+ with CT lifted. Didn't it make you think why you measured a negative potential on B+?

Spice simulations assume a finite (very large, something in Tera ohms) resistance as open circuit. I do all my sims in MicroCap. I also do measurements with laboratory standard equipment. I have 0V on B+ caps with CT lifted. Leakage current due to reverse leakage in diodes and air is in the nanoamp range both sim and measured. Ok, enough for me too.

[evan]

Yes, I agree, the caps are discharged. Zero volts on B+. No current flows.

And minus B+ on the open CT. Better instruments or sims would change nothing. There's nothing particularly sensitive about this measurement.

[VacuumVoodoo]

Yes, I agree, the caps are discharged. Zero volts on B+. No current flows.

And minus B+ on the open CT.

Aha! Here's our misunderstanding - I misread and thought you were getting negative potential reading on the B+ cap. When you connect a voltmeter between CT and ground you close the circuit with voltmeters internal resistance which is high. You get negative potential reading because in this situation the current flowing in the circuit is the diode very small reverse bias current. The higher internal resistance the more negative potential. You could even get into kVolts. Open switch will have much higher resistance, infinite for practical purposes, that's what defines floating CT: undefined potential and no current flow. It's all alright. Thanks for a good discussion.

[David Root]

I only use this method on cathode biased power section, because of all the problems that come up in fixed bias circuits.