Yes, Miller effect is due to the current fed back from the output
to the input of the amplifier. Neutralization nulls out this current this
current. Imagine you have a tube amplifier (common cathode) with a feedback
capacitance of 1 pF. And it has a voltage gain of -10. In other words it
inverts the signal and multiplies the voltage by 10. The feedback current
would be the same as 11pf across the input (grid to cathode). Because for
a 1 volt input signal the output of the amplifier would be -10volts (11
volts across the feedback cap.) thus the current thru the 1pF capacitor
from plate to the input would be 11 times as much for 1 pF across the
input. Or the effect as a 11 pF across the input. But, if you have no
feedback capacitance you would have no Miller capacitance.
In grounded grid it is the same due but phase is different. The
value of the reactance from cathode to ground would be equal to the
capacitive reactance multiplied by (voltage gain-1) Since the input is in
phase with the output. The effective input reactance should be (phase wise)
-capacitive. Or sort of inductive but reactance would decrease with an
increase of frequency, assuming voltage gain remains constant.
Miller effect is what made the reactance tube modulators work. They
required a variable mu tube so that the AF into the tube would modulate the
gain thus modulation the Miller capacitance.
I sometimes mess up on math on the computer because I am used to
doing it on the back of envelopes.
73
Bill wa4lav
At 10:05 AM 2/23/2005 -0800, you wrote:
>On Feb 23, 2005, at 8:11 AM, Bill Fuqua wrote:
>
> > I am not referring to input resistance but to the "Miller effect" or
> > coupling from the plate to input circuit that in effect changes it's
> > input reactance.
>
>Does neutralization affect Miller-effect?
> >
> > 73
> > Bill wa4lav
> >
> >
> > At 07:25 AM 2/23/2005 -0800, R.Measures wrote:
> >
> >> On Feb 22, 2005, at 1:40 PM, Bill Fuqua wrote:
> >>
> >> > One minor glitch in the process is that plate current may drop
> >> > due to the impedance at the tube's input changing. This is noticed
> >> at
> >> > times with either grounded grid or neutralized grid driven
> >> amplifiers.
> >> > In fact one common test for neutralization in the older tube type
> >> > transmitters is to see that influence on the grid current is
> >> > symmetrical about the dip in the plate circuit.
> >>
> >> Class AB1 grid-driven amplifiers do not change input Z unless the grid
> >> terminator R is changed to a different value. In fact, the input Z is
> >> the same whether the filament is lit or not.
> >>
> >> > I guess what I am saying if the test does not turn out perfect
> >> it
> >> > is not due to the plate circuit being off resonance but the input
> >> > impedance changing. But other than that glitch, it seems as a
> >> > reasonable test.
> >> > A the primary resonant frequency of a resonant system is the
> >> > frequency at which the stored oscillating energy divided by the
> >> > applied energy per cycle is maximized after the system has reached
> >> > equilibrium. In the case of most impedance matching networks it is
> >> > where Pout/Pin of the network is maximized. But there are cases
> >> were
> >> > there is no RF output from the resonant network and all of the RF
> >> > power goes into heat. Or in the case of an antenna most of Pin is
> >> > radiated as electromagnetic waves at the applied frequency and a
> >> > little in heat (electromagnetic waves of much shorter wavelengths).
> >> >
> >> > 73
> >> > Bill wa4lav
> >> >
> >> >
> >> >
> >> >
> >> >
> >> > At 04:15 PM 2/22/2005 -0500, TexasRF@aol.com wrote:
> >> >> In a message dated 2/21/2005 4:47:23 A.M. Central Standard Time,
> >> >> r@somis.org
> >> >> writes:
> >> >>
> >> >>
> >> >> On Feb 20, 2005, at 6:35 PM, TexasRF@aol.com wrote:
> >> >>
> >> >> >
> >> >> > Hi Rich, no, I said "C1 resonates the network" but no matter, we
> >> >> both
> >> >> > know what the intent was.
> >> >>
> >> >> No capacitor in a L-network or a Pi-network (double L-network)
> >> >> resonates the network.
> >> >>
> >> >>
> >> >>
> >> >> Hi Rich, here is the plan for the bullet proof dip meter and test:
> >> >>
> >> >> The PA has an 8877 tube in it with a 1000 ma plate current meter
> >> >> installed
> >> >> and connected. We can use this meter to observe the resonance
> >> "dip".
> >> >> The Pi
> >> >> network is adjusted for maximum output power with 75 watts of drive
> >> >> power
> >> >> applied. We have to do this with a dummy load so any antenna
> >> related
> >> >> influence in
> >> >> our test is eliminated.
> >> >>
> >> >> At resonance, the plate load impedance is all resistive, no shunt
> >> >> reactance.
> >> >> Off resonance in the higher frequency direction would entail the
> >> >> presence
> >> >> some shunt inductive reactance, which in parallel with the plate
> >> load
> >> >> resistance would cause the load impedance to be lowered. Off
> >> >> resonance in the lower
> >> >> frequency direction would entail presence of some shunt capacitive
> >> >> reactance,
> >> >> also lowering the total load impedance.
> >> >>
> >> >> Since we know from Ohm's law that current equals voltage divided by
> >> >> resistance (or impedance in an ac circuit) we would expect the
> >> plate
> >> >> ma meter to be
> >> >> minimum when the load (network) is at resonance and non minimum
> >> when
> >> >> the load
> >> >> (network) is off resonance.
> >> >>
> >> >> Now comes the dip check: Expecting the network to be non resonant,
> >> as
> >> >> the
> >> >> driver frequency is slowly changed, in the direction of expected
> >> >> resonant
> >> >> frequency, we would expect the plate current to slowly reduce until
> >> >> we reach the
> >> >> actual resonant frequency of the load (network). If we go the wrong
> >> >> way then
> >> >> the plate current will rise. No problem, we just tune the driver
> >> >> frequency in
> >> >> the other direction in this case.
> >> >>
> >> >> Once we find the frequency of minimum plate current (the dip),
> >> bingo!
> >> >> We
> >> >> have found the resonant frequency of the load (network). If it is
> >> >> different than
> >> >> the starting frequency then the idea that Pi networks do not
> >> operate
> >> >> at
> >> >> resonance will be proven. Also, no dip meters have been sacrificed
> >> in
> >> >> the process
> >> >> and any question of what influence is caused by cover removal is
> >> >> avoided.
> >> >>
> >> >> Anyone out there willing to give this test procedure a try? I am at
> >> >> work
> >> >> right now and no access to a big PA to check this out.
> >> >>
> >> >> Standing by for test results!
> >> >>
> >> >> Thanks/73,
> >> >> Gerald K5GW
> >> >>
> >> >> _______________________________________________
> >> >> Amps mailing list
> >> >> Amps@contesting.com
> >> >> http://lists.contesting.com/mailman/listinfo/amps
> >> >
> >> >
> >> >
> >>
> >> Richard L. Measures, AG6K, 805.386.3734. www.somis.org
> >>
> >> _______________________________________________
> >> Amps mailing list
> >> Amps@contesting.com
> >> http://lists.contesting.com/mailman/listinfo/amps
> >
> >
> >
>
>Richard L. Measures, AG6K, 805.386.3734. www.somis.org
>
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