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Old 23rd May 2013, 01:00 AM   #1
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Default Input capacitance when tube is not conducting

When a triode is not conducting, does the following formula still applies?

Cin = Cgk + Cgp*(gain+1)

Specifically with regards to EL84 in triode mode. I would like to connect a preamp to multiple power amps. Only one would be active at any one time. I would like to know if there is any detrimental effect by leaving those not turned on connected tot he output of the preamp.

Thank you.
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Old 23rd May 2013, 01:12 AM   #2
FoMoCo is offline FoMoCo  United States
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I'll wager a guess that it is either that formula with gain equal to zero or simply Cgk. The reason I think Cgk is that although Cgp exists, it's in series with Rp.

Hopefully someone else will come along that knows for sure.
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Old 23rd May 2013, 01:15 AM   #3
kevinkr is offline kevinkr  United States
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Since the amp in question is not turned on, the gain is 0 so there is no miller capacitance.. There is no reason I can think of quickly why it won't work, I'd expect the capacitance of the interconnects to be more significant, and do make note of the overall load resistance reflected by those amps in parallel back to the line stage driving them..
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Old 23rd May 2013, 01:32 AM   #4
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I planned to make the input impedance of the power amp highish at 100kΩ, so that when I parallel 2 of them it is still a manageable 50kΩ. But I didn't consider the input capacitance aspect. I suppose I don't have much to worry about.

I use Belden 1694A interconnect which is rated at 16.2pF/ft which is pretty low as for as coax goes. The tube probably add the equivalent of another foot in cable.

Thanks guys.
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Old 23rd May 2013, 01:48 AM   #5
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Kevin makes a good point about the line stage's drive capability. What does the line stage look like?

I can tell you from 1st hand experience that a power MOSFET source follower setup for substantial drain current will be able to drive the load of multiple cables and power amps, in parallel. In this situation, substantial is about 25 mA. Both a low O/P impedance and the current to charge/discharge the high total cable capacitance are needed.
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Old 23rd May 2013, 05:08 AM   #6
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Quote:
Originally Posted by Navyblue View Post
When a triode is not conducting, does the following formula still applies?

Cin = Cgk + Cgp*(gain+1)
No.

Quote:
Specifically with regards to EL84 in triode mode. I would like to connect a preamp to multiple power amps. Only one would be active at any one time. I would like to know if there is any detrimental effect by leaving those not turned on connected tot he output of the preamp.

Thank you.
You have more cable capacitance/stray capacitance than Cgk here. The latter will be insignificant as compared to the former. The only detrimental effect is that the DC grid return resistors of the inactive amps will load down the output of that pre. How many are you paralleling, and what's the drive capability of the pre?
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Old 23rd May 2013, 08:09 AM   #7
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The line stage output stage:

- anode follower 6H30 unbypassed cathode (output impedance around 1kΩ?)

Power amp input stage:

Amp 1: 100kΩ input impedance, 2SJ74/2SK170 JFET source follower (F4 clone)
Amp 2: 100kΩ input impedance, EL84 triode mode anode follower (Can the line stage push this through interconnect?)
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Old 23rd May 2013, 11:45 AM   #8
DF96 is offline DF96  England
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Quote:
Originally Posted by Navyblue
When a triode is not conducting, does the following formula still applies?

Cin = Cgk + Cgp*(gain+1)
No. Cin is just Cgk.

The formula assumes the stage has an output impedance which is small enough compared with the reactances that it can be ignored i.e. that you have a true voltage amplifier. A cold valve has a high output impedance so Cgp effectively disappears.

Incidentally, this also means that the formula should not be blindly applied to a pentode stage or other transconductance amplifier with a very high impedance output. The textbooks rarely mention this. In many cases, of course, a pentode will have a sufficiently low anode load resistor that it can be regarded as a voltage amplifier.

If you deliberately add a Miller cap to a transconductance stage (e.g. the VAS in a typical SS amp) then you can end up with the stage input impedance not being capacitive but resistive: roughly X/gm, where X <= 1 depends on what proportion of output current flows back through the Miller cap. However, the stage still acts as an integrator so the output is 90 deg from the input as expected. Its just that the integration happens at the output instead of the input.
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Old 23rd May 2013, 07:59 PM   #9
tomchr is offline tomchr  United States
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Well..... You'll still have the raw Cgp left. It'll be in series with the plate load, so it may not matter. That really depends on how that load behaves when the supply is turned off. So technically, the input impedance would be Zin(s) = 1/(s*Cgk) || (1/(s*Cgp) + Zplate(s)).

I'm guessing Cgk will dominate at most relevant frequencies, hence, Cin = Cgk.

~Tom
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Old 23rd May 2013, 11:28 PM   #10
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Quote:
Originally Posted by Navyblue View Post
The line stage output stage:

- anode follower 6H30 unbypassed cathode (output impedance around 1kΩ?)

Power amp input stage:

Amp 1: 100kΩ input impedance, 2SJ74/2SK170 JFET source follower (F4 clone)
Amp 2: 100kΩ input impedance, EL84 triode mode anode follower (Can the line stage push this through interconnect?)
The line stage is probably OK driving the paralleled I/P impedances. HOWEVER, I suspect it could be in trouble trying to drive the paralleled cable capacitances. Kevin gave you the "heads up" about cable capacitance being more important than tube inter-electrode capacitance, early in this thread. Remember, cable capacitance is specified in pF/ft.

Please post the line stage schematic. Grafting a DC coupled IRFBC20 source follower onto the line stage rates to be easy. The big stumbling block is the current capability of the B+ supply. An additional (sic) 50 mA. are going to be needed.
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