Voltage Measurement Question

[JoeTele]

I'm finishing up a home brew tube tester I've been working on for awhile. The "gas"/grid current check will place resistance in series with the grid circuit and look at effect on plate current, which I understand is a standard way to do this. In testing this concept, I soldered a 1 meg resistor to the negative bias voltage terminal and either let the other side flap in the breeze or soldered it to a cap which was then grounded. In both cases, with about -25 volts, I was getting about a 3 volt difference when I measured either side of the resistor to ground despite the fact that there was no drop across the resistor itself (well, technically, there was maybe .2 millivolts). Can someone tell me what would account for this? If this was a real life scenario, the resistor would switched in and out with the tube running under quiescent conditions and the meter connected with the hot side between the grid and the resistor and the other to ground/cathode. Would the grid "see" this voltage difference to ground? If I set the desired grid voltage without the resistor, can I basically ignore a change on my meter when the resistor is switched provided plate current stays the same?

Thanks as always!

Joe

[Malcolm Irving]

Not sure if I understand the circuit configuration you are describing. But I know anomalous readings are often obtained when you try to measure voltage at a tube grid.
A ‘perfect’ voltmeter would have infinite resistance, but real ones usually have 10M or so. This means that current through the voltmeter can change the voltage that you are trying to measure. 10M is not huge compared to the typical 1M grid leak resistor. At the same time, 10M can be small compared to the grid-to-cathode resistance of a normally biased tube.

[martin manning]

I agree it'd be good to see your schematic to help understand what you are building and the context of this measurement.

[JoeTele]

Here is a sketch. It's a negative bias supply. Measuring from ground to point 1 is about -25v and ground to point 2 is about -23. I think I understand what's going on, i.e. the meter pulling current through the resistor as with the meter connected to point 2 there is 2 volts across the resistor as measured with a second meter and no voltage across it otherwise (obviously, as it's connected to nothing on the other side). Is this burden voltage?
[Edit, add on]

In my test setup, point 2 would be connected to a tube grid to supply the negative bias with the cathode grounded. The 1 meg, which is actually a potentiometer, would get cranked up from it's lowest setting to see if plate current changes (i.e. there is excessive grid current creating voltage drop). From what I've observed though, if the meter was also at point 2 monitoring the bias voltage on the grid, cranking up the pot would change the grid voltage due to drop created by the meter, so I guess I would just want to either disconnect the meter after the bias voltage settles or be content to have it at point 1, where it would not be drawing the current through the 1 meg resistance.

[Malcolm Irving]

Yes, I think a meter resistance of 10M explains the measurements you are getting. 25V across the series combination of the 1M and the meter (10M) would produce 25/11M = 2.27uA which would give a voltage drop of 2.27V across the 1M resistor.

‘Burden Voltage’ is a small drop in voltage across a current measuring device. Here we have a small current into a voltage measuring device, which perhaps should be called ‘Burden Current’, although an internet search on that term did not bring anything up.

What is the circle with 50k in it?