On Feb 23, 2005, at 11:01 AM, Bill Fuqua wrote:
> 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.
Bill -- Is not the 1pF of feedback C between the grid and the anode?
> 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
>> >> >>
>> >> >> _______________________________________________
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>> >> >> Amps@contesting.com
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>> >> >
>> >> >
>> >> >
>> >>
>> >> 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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>> Amps@contesting.com
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>
>
>
Richard L. Measures, AG6K, 805.386.3734. www.somis.org
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