Question about bias supply and negative voltage on a PT with no CT

[The Jonald]

Hello all. I'm having trouble figuring something out on my own. I'm hoping you excellent people will be able to set me straight.

1) In the circuit shown, if I change the value of the 390k resistor for the bias supply, does that affect the B+ voltage? If so, how and why?

2) What about the 47k resistor?

Thank you!

[The Dude]

Nope. Changing those resistors shouldn't have any effect on B+. It's a separate supply and bias voltage is primarily a reference voltage. There is very little current draw on a bias supply.

[The Jonald]

Thank you! I've been scratching my head about this. I had to add a pot in line with the 47k resistor to ground, in order to make the bias adjustable and bring it to a reasonable level, but when I adjust the bias, the plate voltage varies too. I know varying the bias will affect current draw through the tubes and cause the plate voltage to change, but I wasn't expecting the effect to be as large as I found, so I wondered if there was something else happening in the power supply that I didn't understand. I appreciate you clearing this up for me.

[Enzo]

Plate current does have a large effect on B+. The bias affecting the tube is doing it. The bias supply itself has no effect on the B+ supply.

[Mick Bailey]

A transformer has an open-circuit (no-load) voltage, and a loaded voltage. A simplified model is to think of the transformer as having an internal series resistance. According to Ohm's law, the more current you pull the more the voltage drops across this resistance. If the transformer was massively proportioned its resistance would be low and the ability to supply current greatly increased. There are other factors such as core size and lamination material that also affect current delivery. So, as your bias voltage approaches zero, the tubes pull more current and the voltage drops. How much a transformer can maintain a steady voltage is referred to as regulation - small transformers have poor regulation so their voltage varies considerably with load.

[pdf64]

I don't see how the thread title applies to the circuit in question, which seems to have a CT on the HT winding of its PT

[Enzo]

I thought about that too, and I decided he might have meant the half winding for the bias supply effectively has no center tap.

[The Jonald]

Enzo, you're right. I know there is a CT in the circuit I posted, but it's not being used to generate the bias supply, so I ignored it.

[pdf64]

Enzo, you're right. I know there is a CT in the circuit I posted, but it's not being used to generate the bias supply, so I ignored it.

I think you may be getting the 'centre tap' (CT) mixed up with a 'bias tap'?
In your amp, neither the HT or the bias supply would operate as intended, without the CT of the HT winding being connected to 0V.
See the section 'Two Phase Rectifier' http://www.valvewizard.co.uk/bridge.html

[The Jonald]

I know that. I mean it's not a center tap negative bias. Something like the following.

[pdf64]

I know that. I mean it's not a center tap negative bias. Something like the following.

Sorry, I can't make sense of that circuit
It's vaguely reminiscent of a 'back bias' arrangement, but that's extremely unusual, I can't think of any regular guitar amps with it.
What amps are you thinking of?

[Helmholtz]

The HT winding needs to have some ground reference. If that's not shown it's not possible to see how the bias voltage develops. Any current needs a return path to its source (e.g. transformer winding). Currents only flow in closed loops. Without a current no voltage across resistors.

Anyway, in your Kalamazoo schematic the CT is grounded and the bias circuit uses one half of the HT winding. A very straight forward circuit.

[The Jonald]

Sorry, I can't make sense of that circuit
It's vaguely reminiscent of a 'back bias' arrangement, but that's extremely unusual, I can't think of any regular guitar amps with it.
What amps are you thinking of?

Yes, a back-bias. Unusual for guitar amps. Used by some older Hammond amps apparently.

[tubeswell]

...

1) In the circuit shown, if I change the value of the 390k resistor for the bias supply, does that affect the B+ voltage? If so, how and why?

2) What about the 47k resistor?

Thank you!

The bias supply voltage divider is split into 3 resistors. The 390k is seeing VAC (but nevertheless still forms part of the voltage divider.) The output of the divider is between the 4k7 and the 47k. If you decrease the 47k (but leave the other resistors the same), the -ve bias voltage output will be decreased (i.e. smaller -ve voltage)

If you decrease the 390k to some lesser resistance, there will be less VAC dropped across this resistor, and so the -ve current pulses 'seen' after the diode will be bigger - meaning that the VDC dropped across the other 2 resistors will be bigger - and so the bias supply output's -ve voltage will be increased slightly (i.e. more -ve voltage). Make sure your diode and 20uF caps are rated to handle the higher VAC it will be seeing (in all likelihood, it probably will be rated adequately, if you keep the changes to the resistances incremental*)

The fundamentals of voltage dividers apply. Current is the 'same' through the total circuit, and so will be the same through each resistor (considering that there is 'RMS current' in the 390k vs DC in the other 2 resistors). Ohms Law applies. (current x resistance = voltage) The amount of -ve VDC at each point in the voltage divider will be measurable with your VDC meter at each 20uF cap node. (So clip your meter on and see where you land)

* P.S. don't make huge reductions to the 390k, otherwise you risk slamming the diode or bias supply filter caps with too-high-a-voltage.





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[The Jonald]

tubeswell, thank you for your clear explanation. However, my question was about how those resistors affect B+, not the bias voltage.