The AM Forum

Hello all,
I am trying to find out what the grid impedance of two 4-400's in parallel would be.
The spec only shows the input capacitance.
I have searched the internet and am not finding any info.
I am trying to redesign an tuned grid input matching circuit and want to simulate the tubes.
Thanks,
Don W9BHI

Whatever resistance you terminate the grid(s) with.

I'm NOT a fan of unterminated grids.  I prefer to wind a transformer to give me the correct voltage needed to drive or excite the grid and terminate the grid in the correct resistance to give a good match.

--Shane
KD6VXI

So if I use a 9.1K grid leak resistor in actual operation, that would be the same value termination that I would use for checking the input network without the tubes?
Don W9BHI

The input impedance would be 9.2k - j1.82k@3.5MHz.

I would not use grid leak to bias a tetrode power tube in this power range.

I would use some protective bias as recommended by the Elmac data sheet and as in the schematic below:

Hi DMOD,

In your circuit, what sets the 9.1 k?

I see the reactance as from the Cin of 12.5 pF/tube. Ok, 2 tubes at 3.5 MHz yields the -j1.82 k value. However, this Cin is in shunt. I did not check, is the parallel of all this when converted to a series form give you the Zin you stated?

This pdf below is interesting read and in the end he uses the Required Drive Power divided by the nominal grid 1 current. That works out to a real value that you can forget about. So yes, seems like forcing a shunt R value for an input Q of say 10 or 12 makes sense. And the rest is just the remaining Cin of the tube.

http://rfcec.com/RFCEC/Section-3%20-%20Fundamentals%20of%20RF%20Communication-Electronics/04%20-%20AMPLIFIERS%20-%20RF%20POWER%20AMPLIFIER%20BASICS/RFPA%20-%20Tetrode%20and%20Pentode%20RF%20Power%20Amplifier%20Information%20(By%20Larry%20E.%20Gugle%20K4RFE).pdf

The 63 edition of the Handbook has a single 4-400A with a link coupled input. I went through the transformer calculations with various assumptions for a single tube. The real part of Z value was all over the place. What Shane says makes sense!
Alan

the real part of the input impedance changes as the grid draws grid current .... do your calculation for peak grid current which will be Zin minimum .... you can terminate at more or less resistance depending on how much drive power is available .... maybe consider some negative feedback for less distortion

My two 4-400 tube amp (grounded cathode, parallel) does use 2 toroid core 1:4 transformers  to step-up the voltage to the required drive voltage. The cores are loaded with resistors to give a nice match. Should be in the order of 16 x 50 Ohms=800 Ohms. But as W4BFS says, the impedance will change with grid current and also with non-ideal neutrodynisation. I ended up with 1.5 kOhm and half way (at the interconnection between the transformers), an other 1,5 kOhm. with a drive of 45 Watts I get approx 1.2 kW at 4 kV anode voltage