power tube acting as a capacitor?

[apa]

In rather idle investigation of a failing power tube that I suspected
might be drawing grid current from the inverter, I pulled the tube and
checked the resistance from the grid to ground. The meter showed a
negative resistance which rose to zero and then eventually up to a few
meg and eventually read as open after a few minutes. So I figured there
must be some voltage stored in the tube that was throwing off the meter
reading (given the negative resistance). I put the tube back in circuit
for a few minutes, pulled it again and put a DVM from G1 to the
cathode. To my suprise it measured just over half a volt. The voltage
reading dropped to zero after a minute or so. Is this some sort of
meter anomolie? Or has the tube developed enough capacitance to still
have a half volt of potential in the time it's taken to pull it from
the circuit and get the leads on it? Or is interelectrode capacitance
enough to account for this?

[TimPerry]

apa wrote:
> In rather idle investigation of a failing power tube that I suspected
> might be drawing grid current from the inverter,

phase splitter?

I pulled the tube and
> checked the resistance from the grid to ground. The meter showed a
> negative resistance which rose to zero and then eventually up to a few
> meg and eventually read as open after a few minutes. So I figured
> there must be some voltage stored in the tube that was throwing off
> the meter reading (given the negative resistance). I put the tube
> back in circuit for a few minutes, pulled it again and put a DVM from
> G1 to the cathode. To my suprise it measured just over half a volt.
> The voltage reading dropped to zero after a minute or so. Is this
> some sort of meter anomolie? Or has the tube developed enough
> capacitance to still have a half volt of potential in the time it's
> taken to pull it from the circuit and get the leads on it? Or is
> interelectrode capacitance enough to account for this?

although you neglected to mention the tube number, the grid capacitance if
the tube that i guess you had will be from 10 to 25 pF. that will not store
much of a charge.

most technicians do not check tubes in this manner (except sometimes 3,000
Watt and higher).
a tube checker is more informative.

most technitions allow power tubes time to cool down after turning power off
to avoid burns when removing them. normally "bleader" resistors drain the
voltages close to zero in this time.

try placing a 25 pF cpacitor across a 1.5 volt cell then measure it with the
same meter.

[Meindert Sprang]

"apa" > wrote in message
ps.com...
> In rather idle investigation of a failing power tube that I suspected
> might be drawing grid current from the inverter, I pulled the tube and
> checked the resistance from the grid to ground. The meter showed a
> negative resistance which rose to zero and then eventually up to a few
> meg and eventually read as open after a few minutes. So I figured there
> must be some voltage stored in the tube that was throwing off the meter
> reading (given the negative resistance). I put the tube back in circuit
> for a few minutes, pulled it again and put a DVM from G1 to the
> cathode. To my suprise it measured just over half a volt. The voltage
> reading dropped to zero after a minute or so. Is this some sort of
> meter anomolie? Or has the tube developed enough capacitance to still
> have a half volt of potential in the time it's taken to pull it from
> the circuit and get the leads on it? Or is interelectrode capacitance
> enough to account for this?

None of the above. The cathode is still hot and emitting electrons. This
produces a voltage across the cathode and the next electrode, the grid.

If you applied a voltage to the filament only, you would see the voltage
build up again.

Meindert

[Scott Dorsey]

apa > wrote:
>In rather idle investigation of a failing power tube that I suspected
>might be drawing grid current from the inverter, I pulled the tube and
>checked the resistance from the grid to ground. The meter showed a
>negative resistance which rose to zero and then eventually up to a few
>meg and eventually read as open after a few minutes. So I figured there
>must be some voltage stored in the tube that was throwing off the meter
>reading (given the negative resistance). I put the tube back in circuit
>for a few minutes, pulled it again and put a DVM from G1 to the
>cathode. To my suprise it measured just over half a volt. The voltage
>reading dropped to zero after a minute or so. Is this some sort of
>meter anomolie? Or has the tube developed enough capacitance to still
>have a half volt of potential in the time it's taken to pull it from
>the circuit and get the leads on it? Or is interelectrode capacitance
>enough to account for this?

Depends. What's the tube, and what is the input Z of the DVM?

With a Simpson 260 you won't see any of this stuff so you won't have
anything to worry about. But with a DVM, the input Z is so high that
you can read charge across parasitic capacitors all the time.
--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Richard Crowley]

There is also the "Leyden jar" effect.
You can store quite a jolt in an old mayonnaise jar. :-)

[apa]

Scott Dorsey wrote:
> apa > wrote:
> >In rather idle investigation of a failing power tube that I suspected
> >might be drawing grid current from the inverter, I pulled the tube and
> >checked the resistance from the grid to ground. The meter showed a
> >negative resistance which rose to zero and then eventually up to a few
> >meg and eventually read as open after a few minutes. So I figured there
> >must be some voltage stored in the tube that was throwing off the meter
> >reading (given the negative resistance). I put the tube back in circuit
> >for a few minutes, pulled it again and put a DVM from G1 to the
> >cathode. To my suprise it measured just over half a volt. The voltage
> >reading dropped to zero after a minute or so. Is this some sort of
> >meter anomolie? Or has the tube developed enough capacitance to still
> >have a half volt of potential in the time it's taken to pull it from
> >the circuit and get the leads on it? Or is interelectrode capacitance
> >enough to account for this?
>
> Depends. What's the tube, and what is the input Z of the DVM?

Tube is an EL34 and the meter is a Fluke 112 - input 10 meg.

> With a Simpson 260 you won't see any of this stuff so you won't have
> anything to worry about. But with a DVM, the input Z is so high that
> you can read charge across parasitic capacitors all the time.
> --scott
>
> --
> "C'est un Nagra. C'est suisse, et tres, tres precis."

[Scott Dorsey]

apa > wrote:
>> Depends. What's the tube, and what is the input Z of the DVM?
>
>Tube is an EL34 and the meter is a Fluke 112 - input 10 meg.

With the tube cold, charge up the cathode and plate with a 9V battery,
then try and measure voltage across it with the Fluke. I bet you see
something there after a couple seconds.

If you measure between grids you might get even more.
--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Arny Krueger]

"Scott Dorsey" > wrote in message

> apa > wrote:
>>> Depends. What's the tube, and what is the input Z of
>>> the DVM?
>>
>> Tube is an EL34 and the meter is a Fluke 112 - input 10
>> meg.
>
> With the tube cold, charge up the cathode and plate with
> a 9V battery, then try and measure voltage across it with
> the Fluke. I bet you see something there after a couple
> seconds.
>
> If you measure between grids you might get even more.

The time constant of 35 pf and 11 megohms is 0.385 milliseconds.

If one actually sees the effects of a DVM or VTVM discharging a cap this
size, the initial voltage must have been very high, and the meter set to
measure low volts.

[TimPerry]

Arny Krueger wrote:
> "Scott Dorsey" > wrote in message
>
>> apa > wrote:
>>>> Depends. What's the tube, and what is the input Z of
>>>> the DVM?
>>>
>>> Tube is an EL34 and the meter is a Fluke 112 - input 10
>>> meg.
>>
>> With the tube cold, charge up the cathode and plate with
>> a 9V battery, then try and measure voltage across it with
>> the Fluke. I bet you see something there after a couple
>> seconds.
>>
>> If you measure between grids you might get even more.
>
> The time constant of 35 pf and 11 megohms is 0.385 milliseconds.
>
> If one actually sees the effects of a DVM or VTVM discharging a cap
> this size, the initial voltage must have been very high, and the
> meter set to measure low volts.

my reference shows that 6AC7 has 11 pF input capacitance.

output capacitance is 5 pF. max plate volts is 300.

if your bleeders are open and/or you pulled the tube without switching off
you might get something briefly visible on a DVM from the plate. i figure
its about .0055 mS per time constant.

in any event i can not think of a meaningful test that can be performed
with a DVM (DMM) on a tube that is out of circuit except filament
continuity.

i would use the DVM (or better a plastic screwdriver handle) to tap lightly
on the tube while it is in circuit to check for intermittent operation.

[apa]

Scott Dorsey wrote:
> apa > wrote:
> >> Depends. What's the tube, and what is the input Z of the DVM?
> >
> >Tube is an EL34 and the meter is a Fluke 112 - input 10 meg.
>
> With the tube cold, charge up the cathode and plate with a 9V battery,
> then try and measure voltage across it with the Fluke. I bet you see
> something there after a couple seconds.
>
> If you measure between grids you might get even more.
> --scott
>
> --
> "C'est un Nagra. C'est suisse, et tres, tres precis."

I do in fact. Thanks Scott.

[apa]

Meindert Sprang wrote:
> "apa" > wrote in message
> ps.com...
> > In rather idle investigation of a failing power tube that I suspected
> > might be drawing grid current from the inverter, I pulled the tube and
> > checked the resistance from the grid to ground. The meter showed a
> > negative resistance which rose to zero and then eventually up to a few
> > meg and eventually read as open after a few minutes. So I figured there
> > must be some voltage stored in the tube that was throwing off the meter
> > reading (given the negative resistance). I put the tube back in circuit
> > for a few minutes, pulled it again and put a DVM from G1 to the
> > cathode. To my suprise it measured just over half a volt. The voltage
> > reading dropped to zero after a minute or so. Is this some sort of
> > meter anomolie? Or has the tube developed enough capacitance to still
> > have a half volt of potential in the time it's taken to pull it from
> > the circuit and get the leads on it? Or is interelectrode capacitance
> > enough to account for this?
>
> None of the above. The cathode is still hot and emitting electrons. This
> produces a voltage across the cathode and the next electrode, the grid.
>
> If you applied a voltage to the filament only, you would see the voltage
> build up again.
>
> Meindert

That was it. Thanks.

[Mark]

apa wrote:
> In rather idle investigation of a failing power tube that I suspected
> might be drawing grid current from the inverter, I pulled the tube and
> checked the resistance from the grid to ground.


If a power tube becomes "gassy" it may well draw grid current, the grid
may tend to go positive (or less nagative) causing excessive plate
current and overheating. I have never heard of this being tested with
a DVM out of the circuit but I suppose if you pull the tube and test
quickly while the cathode is still warm, you might see this effect.
Compare it to a known good tube.

The effect is not due the capacitance of the grid.



Mark