[Amps] Passive grid stability calculation for a tetrode?

Markku Oksanen ww1c at outlook.com
Sat Nov 16 14:08:54 EST 2013


Hi
No problem.  I am also a somewhat seasoned amp builder and so far only one amp was tricky, that was because I had a near  SECOND resonance on the plate choke chain and rf voltages can build up. That was not easy to find, however, component changes fix these things with ease.
I simply wanted some advice on what not to absolutely do, then try something with some stability margin and see what happens in reality.
So, the conclusion is that without neutralization somewhere in the range 150 ohm grid resistor if the plate load resistance is 1.7 kohm. 
Thanks!
Markku
> From: wlfuqu00 at uky.edu
> To: markku.a.oksanen at kolumbus.fi
> Subject: RE: [Amps] Passive grid stability calculation for a tetrode?
> Date: Sat, 16 Nov 2013 16:53:43 +0000
> 
>    Ok, my point is simply that you can't know for sure what impedance the grid or plate are seeing under all conditions.
> Nor, all the possible feedback paths. So, you design with plenty of margin in choosing your swamping resistor. Use 
> the smallest value you can for the drive and hope there is something not lurking in your amplifier to surprise you.
>    Never, have an amplifier biased where the tube may amplify with an unloaded output or input. This is a sure way
> to have an oscillation. 
>    With some luck and good component layout and bypassing/rf filtering of all non-RF wiring things will be fine.
>    I have built a number of amplifiers in the past and have been unfortunate on occasion. 
>    You do the best you can.  
>    I am not complaining about any particular mathematical analysis just that there are often unexpected surprises when 
> prototyping anything.
>    I have seen my share. 
>    Now, all this may worry you when doing your project. If a problem creeps up when you are done, then bring it to the group and we will
> all do what we can to help you resolve it.  
>    Regardless let us know how it turns out.
>    
> ________________________________________
> From: Amps [amps-bounces at contesting.com] on behalf of Markku Oksanen [markku.a.oksanen at kolumbus.fi]
> Sent: Friday, November 15, 2013 5:05 AM
> To: Bill Turner; Amps
> Subject: Re: [Amps] Passive grid stability calculation for a tetrode?
> 
> Hi
> Bill,
> Thanks for the info. I understand that very low values of swamping resistor will probably result in a stable design.However, it looks like the capacitive feedback in a 4CX5000 is so low that just about any value would do, even higher than 50, higher by lots.Still commercial designs (from which these tubes and components were pulled, Marconi from 1967) use neutralization.
> So how do I calculate the stability criteria for this tube?
> Thanks!
> Markku
> 
> > From: dezrat1242 at yahoo.com
> > To: amps at contesting.com
> > Date: Fri, 15 Nov 2013 00:42:59 -0800
> > Subject: Re: [Amps] Passive grid stability calculation for a tetrode?
> >
> > ORIGINAL MESSAGE:          (may be snipped)
> >
> > On Fri, 15 Nov 2013 07:32:39 +0000, WW1C wrote:
> >
> > >
> > >All
> > >I am back on my 4CX5000 project and as it not done yet, I am still contemplating the circuit to be used.Earlier this year I got solid info from this reflector for GG operation and that look like a pretty good solution.However, I was wondering about passive grid, grid driven also.
> > >The question is:  How do a calculate the stability?  If I simply do math for tube voltage gain I get something like 25.Then if I calculate the feedback capacitive voltage divider, Grid to Cathode 120 pf, Plate to Grid 0.7 pf, I get less than 0.01 for the ratio. If the beast needs the voltage gain times feedback gain to be above 1 to oscillate, to me it seems this tube is stable no matter what.  Where did I err if I did?
> > >Thanks!
> > >MarkkuWW1C
> >
> > REPLY:
> >
> > The thing that makes the passive grid design stable is the very low swamping
> > resistance from grid to ground, typically 50 ohms. This, together with the
> > very low anode to G1 capacitance of a tetrode means the feedback gain is
> > much less than 1 and that's why it is stable.
> >
> > Of course it's very important to use a non-inductive resistor of adequate
> > power handling capability. I like the Caddock resistors. Mouser carries then
> > and no doubt others do too.
> >
> > It's a simple, effective design. Go with it.
> >
> > 73, Bill W6WRT
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> 
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