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 28th July 2013, 09:29 AM #101 H.Michi   diyAudio Member   Join Date: Jan 2009 Hello Tcqanh San, Thank you for your comment. Z out of 4P1L: 1/GM=200ohm Z in of 811A: is supposedly around 2~3 K ohm, depending on input voltage. Hiro
tcqanh
diyAudio Member

Join Date: Apr 2011
Location: HoChiMinh city
Quote:
 Originally Posted by H.Michi Hello Tcqanh San, Thank you for your comment. Z out of 4P1L: 1/GM=200ohm Z in of 811A: is supposedly around 2~3 K ohm, depending on input voltage. Hiro
I remember that Zin = resistance of G1 in operation point. Is it right ? Can I take Ohms' law in this case for calculation?

Last edited by tcqanh; 28th July 2013 at 11:02 AM.

JLH
diyAudio Member

Join Date: Apr 2005
Location: Indianapolis, IN
Quote:
 Originally Posted by tcqanh I remember that Zin = resistance of G1 in operation point. Is it right ? Can I take Ohms' law in this case for calculation?
No. The worse case grid impedance happens when the grid is driven to its most positive voltage. At that point you can use Ohm's Law to see what the worse case is.
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Ah, how beautifully the orchestra sounds before a rain! In a dry sunny day there is no way for the instruments to sound this way!

tcqanh
diyAudio Member

Join Date: Apr 2011
Location: HoChiMinh city
Quote:
 Originally Posted by JLH No. The worse case grid impedance happens when the grid is driven to its most positive voltage. At that point you can use Ohm's Law to see what the worse case is.
I'm saying 811A in this case with Vg1= 17.5V, Ig1 = 12mA. So, Rg1 ~ input impedance = 1458 ohms. Any wrong ?

tcqanh
diyAudio Member

Join Date: Apr 2011
Location: HoChiMinh city
Quote:
 Originally Posted by H.Michi Hello Tcqanh San, Thank you for your comment. Z out of 4P1L: 1/GM=200ohm Z in of 811A: is supposedly around 2~3 K ohm, depending on input voltage. Hiro
Hi Michi,

There are many ways to keep down Zout of driver stage, especially with high Gm tube. Why do you choice 4P1L ? Because it make more complicated on filament circuit.

If you can, pls use 6BX7/6BL7 or EL34/6L6 in driver stage (these can run higher voltage much more than 4P1L). I'm waiting for your experiment . You can get one amplification stage with high mu, so your design become more simple.

H.Michi
diyAudio Member

Join Date: Jan 2009
Quote:
 Originally Posted by tcqanh I'm saying 811A in this case with Vg1= 17.5V, Ig1 = 12mA. So, Rg1 ~ input impedance = 1458 ohms. Any wrong ?
I think the resistance of vacuum tube does not apply ohm`s low.

I made following experiment with my 811A amp:
1. Connected 10 ohm resister in between Cathode of 4P1L and grid of 811A
2. Measured ac voltage across the resister at 18W output. That was 0.142V ac.
3. Ac current should be 0.142/10=0.0142A
4. Z=V/I 43.2V/0.0142A=3042 ohm
5. Measured ac voltages across the resister at 1W output. That was 0.033Vac.
6. Same calculation: Z=V/I 10.25V/0.0033A=3106 ohm

Frankly I am not sure above measurement is right one or not.

Hiro

JLH
diyAudio Member

Join Date: Apr 2005
Location: Indianapolis, IN
Quote:
 Originally Posted by tcqanh I'm saying 811A in this case with Vg1= 17.5V, Ig1 = 12mA. So, Rg1 ~ input impedance = 1458 ohms. Any wrong ?
That is wrong. For example, if the most positive the grid is driven is +40V and the grid current at +40V is 50mA, then the worse case is the grid impedance is 800 ohms. You should design your driver stage to handle the worse case. If you only design for the idle condition your driver will not be up to the task of driving the tube to full power.
__________________
Ah, how beautifully the orchestra sounds before a rain! In a dry sunny day there is no way for the instruments to sound this way!

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