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Input capacitance when tube is not conducting

Navyblue

Member
2012-03-22 8:37 am
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. :)
 

kevinkr

Administrator
Paid Member
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..
 

Navyblue

Member
2012-03-22 8:37 am
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. :)
 
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.
 
When a triode is not conducting, does the following formula still applies?

Cin = Cgk + Cgp*(gain+1)

No.

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?
 

Navyblue

Member
2012-03-22 8:37 am
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?)
 
Navyblue said:
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.
 

tomchr

Member
Paid Member
2009-02-11 12:58 am
Calgary
www.neurochrome.com
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
 
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.
 
When a triode is not conducting, does the following formula still applies?

Cin = Cgk + Cgp*(gain+1)

Yes, it always follows the formula, even when Gain = 0

Cin = Cgk + Cgp*(0+1) = Cgk + Cgp -- however, Cgp has plate load resistance in series, strictly speaking, so the result will be slightly different.
 
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Think of the gain term in the equation as multiplying the effect of the Cgp capacitance because the plate voltage variation with signal is affecting the signal voltage across Cgp and is hence the reason that Cgp appears to be larger that it physically is - so when there is no amplified plate voltage variation (due to valve not being in an operating state where it can amplify) then Cgp just acts like a raw Cgp component in the circuit.
 
Trying to start a car with Lucas electrics on a cold wet November morning was part of the culture of the UK. The dawn chorus of churning starter motors, followed by cursing and a ten minute break to allow the fuel to evaporate from a flooded carburretor before retrying helped wake me up for school. When I was old enough to buy a car I opted for Toyota - they were fine in the cold, but would not start when hot!
 

Navyblue

Member
2012-03-22 8:37 am
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.

It's a standard anode follower, doesn't have much juice, would appreciate it if you can tell me if it would work well. Something like this:

[IMGDEAD]http://members.chello.nl/~m.heijligers/DAChtml/Analogue/Analog-21.gif[/IMGDEAD]

Rg - 560kΩ (paralled with 50kΩ potentiometer}
Rc - 240Ω
Rp - 8.2kΩ
Cp - 2.2µF
Cc - none

6H30 datasheet:

http://www.hifitubes.nl/weblog/wp-content/electro-harmonix-6h30-eh.pdf
 

Navyblue

Member
2012-03-22 8:37 am
A somewhat related question, what do you guys think of this claim:

6BQ5 internal capacitances in triode mode - smoking-amp - Tube DIY Asylum

If the EL84/6BQ5 input capacitance in triode mode is in the 100pF range, it seems to be quite similar to something like 6H30. Does it mean that both can be driven directly from source? In other places I've read that the input capacitance of EL84/6BQ5 is too high to be driven directly from source.