Lifting cathode of power tube

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Matador

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I am retrofitting a tube power amp with 1 ohm cathode resistors to monitor plate (plus screen) current. I am adapting a differential sense amplifier to sense the voltage drop across these new resistors (there will be 4 amplifiers in this circuit), and output a voltage proportional to the current so I can use it to calculate the static power dissipation (and thus use it to set the negative fixed bias voltage).

However this particular power amp has a quad of 6L6 tubes, and the inner pair can have their cathodes lifted to reduce the output power by half. With the 1 ohm resistors, this means the switch lifts the ground connection of the inner two 1 ohm resistors to disable the inner tube pair. This is done with a single SPST switch which removes the ground reference to the cathode simultaneously for both tubes.

My question: what happens to the voltage across the sense resistor when the ground reference is lifted? The maximum input voltage on the differential amplifier's input is 26V, which means if the sense resistor floats up past this it could destroy the sense amplifier.
 

john12ax7

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The current will go to 0 when the switch is opened, so the voltage across the resistor will be 0V, as both ends will be the same. But assuming you really want to know what voltage the cathode goes to, you will you need to measure that. It's a bit of an indeterminate, but I've also seen tube terminals go to unexpected voltages when one terminal is not connected.

Do you really need to put sense amplifiers inside the amp? Common practice is to simply measure across the 1 ohm resistors directly with a DMM.
 

abbey road d enfer

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I am retrofitting a tube power amp with 1 ohm cathode resistors to monitor plate (plus screen) current. I am adapting a differential sense amplifier to sense the voltage drop across these new resistors (there will be 4 amplifiers in this circuit), and output a voltage proportional to the current so I can use it to calculate the static power dissipation (and thus use it to set the negative fixed bias voltage).

However this particular power amp has a quad of 6L6 tubes, and the inner pair can have their cathodes lifted to reduce the output power by half. With the 1 ohm resistors, this means the switch lifts the ground connection of the inner two 1 ohm resistors to disable the inner tube pair. This is done with a single SPST switch which removes the ground reference to the cathode simultaneously for both tubes.

My question: what happens to the voltage across the sense resistor when the ground reference is lifted? The maximum input voltage on the differential amplifier's input is 26V, which means if the sense resistor floats up past this it could destroy the sense amplifier.
You just need to put the sensing resistor between the ground and switch. If you want to sense each tube separately, you need a DPDT.
 

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trobbins

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I think the OP wants to keep separate 1R resistors for each switched cathode (ie. not use a common 1R for two cathodes).

Given fixed biasing, you could limit each floating cathode to +24V with a zener - depending on the bias voltage, that may push those floating pentodes in to a deep enough cut-off to be functionally ok (ie. off). You probably want to also add a 24V zener to the other unswitched cathodes in case a valve arcs to cathode and you take out a 1ohm sense resistor (as a poor mans fuse).
 

Matador

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I think the OP wants to keep separate 1R resistors for each switched cathode (ie. not use a common 1R for two cathodes).

Indeed (I should have posted a schematic to make this clear).

I had originally thought of moving the 1R's to the ground side of the switch (per Abbey's recommendation), however this seems to only move the problem.

If I move the switch between the sense resistors and the cathode: normally with the switch closed, the differential voltage across the sense R is small (perhaps 100mV or less, depending on plate+screeen current), but the common mode voltage is still around the analog 0V level. Stated another way: one side of the resistor see's 0V, and the other rises in proportion to the plate+screen current. When the switch is opened, both sides fall to 0V (and the amp correctly interprets this as 0 current).

However when the switch is again closed, the top sense signal is going to want to rise to wherever the cathode was previously sitting. Obviously a 1R resistor (with one side bonded to 0V) is going to be strong enough to override the cathode, but the question is: how quickly? If the cathode floated up to 100's of volts that would be enough to punch through a tiny oxide on a SMD amplifier, even if short-lived.

7581_power_amp.png


However my gut tells me that it's not going to be sitting at 100's of volts: if the cathode is lifted, and the plate/screen remains at say, 400V, then with no current flowing the tube is operating along the X axis. Looking at the grid curves, this would occur when the grid is around -65V to -70V?

If the negative bias is around -40V, this means the cathode would float to approximately 25 to 30V, which isn't very much, but it's right in the middle of the maximum common mode input voltage for the amplifier.

Sounds like a 20V, 5W Zener attached in parallel with each sense resistor would be a good measure regardless.
 

Tubetec

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I wired jacks into one amp I had so I could break into the circuit and measure directly in mA with the meter across each cathode ,
There is quite a high voltage at the cathodes when they are lifted , certainly enough to give you a shock , treat the lifted cathode as a source of high DC voltage so as not to get a jolt by surprise . Shrouded banana plugs on a bias probe cable makes a lot of sense .

I remember measuring at the cathodes with them floated from ground , hundreds of dc volts there , across the 1M input of the meter .
 

Tubetec

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If you look at the later model Ampeg SVT II it incorporated a couple of extra op amps to monitor standing current in the tubes , and allowed the biasing to be set up on backpanel leds .
 

abbey road d enfer

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one side of the resistor see's 0V, and the other rises in proportion to the plate+screen current. When the switch is opened, both sides fall to 0V (and the amp correctly interprets this as 0 current).
Correct
However when the switch is again closed, the top sense signal is going to want to rise to wherever the cathode was previously sitting. Obviously a 1R resistor (with one side bonded to 0V) is going to be strong enough to override the cathode, but the question is: how quickly? If the cathode floated up to 100's of volts that would be enough to punch through a tiny oxide on a SMD amplifier, even if short-lived.
Don't you think the 1R resistors would just limit the voltage to a fraction of volt? In order to do this, the tube would have to deliver 1A, which is a lot, and would last for only a fraction of second.
Then what about the possibility of frying the tubes?
Didn't you say that the amp already has this switch? My conclusion is that the manufacturer has already evaluated that risk.
However my gut tells me that it's not going to be sitting at 100's of volts: if the cathode is lifted, and the plate/screen remains at say, 400V, then with no current flowing the tube is operating along the X axis. Looking at the grid curves, this would occur when the grid is around -65V to -70V?
Why don't you measure it? And particularly how much current it is capable of delivering at this high voltage.
If the negative bias is around -40V, this means the cathode would float to approximately 25 to 30V, which isn't very much, but it's right in the middle of the maximum common mode input voltage for the amplifier.
But the amplifier will not see it, being connected to the 1R resistor! Tubes are current sources, they give up when meeting loads that exceed their capability.
 

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trobbins

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As indicated, a grounded 1R sense resistor can't instantaneously have a high voltage across it unless it has substantial inductance in series with it - sort of like switching a floating cathode to gnd whereby the ground doesn't glitch high in voltage.

The disconnected cathode voltage does ride high, as there is a cathode-to-heater leakage resistance that acts like a cathode bias resistor. Another way to look at is if you had a variable cathode auto-bias resistance that you could just keep increasing, the cathode voltage would keep increasing until something started limiting it. One limiting mechanism is when the cathode-heater voltage exceeds its limit - which is why something like a 150V 5W zener is a good idea to put across a 1R sense resistor or a cathode fuse (either of which could open-circuit under a fault for an output stage valve fault if the resistor is say a 0.4W part) - in which case the 150V zener minimises the chance of damage to the valve (if the valve wasn't the cause of the fault, or multiple valves in parallel were in circuit and only one valve had failed).
 

Matador

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Thanks for the input. I should have drawn this out, but was too lazy.

Stock configuration:

stock.png

My typical addition for measuring plate+screen:

radded.png

Then after adding sense amplifier:

problem.png
This has the problem in that if the cathodes float up to 100+V they will take out the amplifier inputs with them.

I think this is Abbey's recommendation:

abby.png

Which means I need to switch (pun intended) from SPST to DPDT. The only source of potential problem would be when this switch is opened, if there was enough parasitic inductance in the sense resistors to cause the + terminal to spike up above 24V for enough time to damage the amplifier inputs (seems unlikely, given the L in the resistor is small, there isn't a lot of parasitic wiring inductance, and even di/dt is likely small as well).

My alternate proposal is this:

alternate.png

This retains the SPST, and clamps the cathodes at about 20V. With the cathodes at around 20V, and the negative bias in the grids, there shouldn't be much leakage current through the diodes (enough to consider them 'on'), and should still disable the tubes when the switch is opened. The Zeners will hold the voltages below the level which could damage the amplifier inputs.

I guess it just comes down to whether the two diodes are cheaper than the switch. :D
 

moamps

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If you are going to use your alternative proposal, you will need to use differential DC amplifiers to determine the cathode's currents. In other case it will not be necessary, IMO.

And you will need to shut down your auto-bias circuitry for these tubes when their cathodes are disconnected, because the auto-bias will otherwise try to increase the cathode currents by reducing the negative grid biases.
 
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Matador

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If you are going to use your alternative proposal, you will need to use differential DC amplifiers to determine the cathode's currents.
I am adapting a differential sense amplifier to sense the voltage drop across these new resistors

In terms of auto-bias, my thinking is this won't be dynamic, as in continually monitoring and adjusting bias while the tube amp is running/being played. The idea is that it behaves much like a tech does: you set the master volume to zero, then hold a button for 5 seconds. Afterwards, the plate voltage for each tube is sampled, the plate+screen current is measured, and the negative bias is adjusted by the circuit is until the static dissipation is set to the correct limit.
 

cxo

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As long as you have adjustable bias supplies, why not just adjust the "disconnected" tubes to a cutoff value and eschew the switching.

If flashover is a worry, a 1N400x diode across each sense resistor will provide protection.
 

moamps

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The idea is that it behaves much like a tech does: you set the master volume to zero, then hold a button for 5 seconds. Afterwards, the plate voltage for each tube is sampled, the plate+screen current is measured, and the negative bias is adjusted by the circuit is until the static dissipation is set to the correct limit.
Sounds pretty complicated. Does that mean you will have some memory for the value of the negative grid bias? Usually, if the HT voltage is not regulated, then the negative grid bias must not be regulated either, so that changing the mains voltage would not significantly change the operating point of the output vacuum tubes. How do you plan to solve this?
 

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