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Old 28th July 2013, 08:29 AM   #101
H.Michi is offline H.Michi  Japan
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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
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Old 28th July 2013, 09:59 AM   #102
tcqanh is offline tcqanh  Viet Nam
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Quote:
Originally Posted by H.Michi View Post
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 10:02 AM.
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Old 28th July 2013, 05:35 PM   #103
JLH is offline JLH  United States
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Quote:
Originally Posted by tcqanh View Post
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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Old 29th July 2013, 04:18 AM   #104
tcqanh is offline tcqanh  Viet Nam
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Originally Posted by JLH View Post
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 ?
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Old 29th July 2013, 04:28 AM   #105
tcqanh is offline tcqanh  Viet Nam
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Quote:
Originally Posted by H.Michi View Post
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.
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Old 29th July 2013, 05:44 AM   #106
H.Michi is offline H.Michi  Japan
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Quote:
Originally Posted by tcqanh View Post
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
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Old 29th July 2013, 09:59 AM   #107
JLH is offline JLH  United States
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Quote:
Originally Posted by tcqanh View Post
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.
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