Ian Thompson-Bell wrote:
>
> I have just been working through the math for shunt derived shunt
> applied NFB around an amp and Ican't get the expected result.
>
> Imagine an inverting amp with an open loop gain of -Ao with a feedback
> resistor from output to input of Rf and an input resistor from signal
> source to the input of Ri. I get a closed loop gain of:
>
> An = Ao/(1+(Ri/Rf)+(Ri.Ao/Rf)) when I expected some thing of the form:
>
> An = Ao/(1+ß.Ao) where ß = Ri/Rf
>
> but instead I get Ao/(1+ß+ß.Ao)
>
> I have checked the math several times and cannot see where I have gone
> wrong. Anyone throw any light on this?
>
> Cheers
>
> Ian
I think I can explain all, because I ran into the same problem you have
about
10 years ago when I tried to teach myself, and I discovered silly text
books
limited to basic oppamp FB equations just didn't work with simple low
gain triodes.
But let's consider the terms to be used and let input to grid R = R1,
and anode output to grid R = R2, bcause its the most common way I have
seen the R shown
in a shunt FB loop.
Also, to be simple, consider the inverting open loop gain be a positive
number, even though
it really is negative.
And ß = R1 / (R1 + R2) , OK? and not R1 / R2 !
Let us consider an example using an EL34 with R load of 2k2 ohms, in
pentode mode,
and with fully bypassed cathode.
OLG A = 20, ie, 10Vac input gives 200Vac output. Phase is inverted.
Let R1 = 20k, and R2 = 420k.
ß = 20 / ( 20 + 420 ) = 0.04545.
CLG, closed loop gain A' = A / ( 1 + { [ A + 2 ] x ß } ).
In this case, A'= 20 / ( 1 + { [ 20 + 2 ] x 0.04545 } ),
= 10.0, so my notes from my book of tested formulas says.
The effective anode resistance Ra' with NFB = Ra / ( 1 + [ µ x ß ] )
So if Ra = 12k, and µ = 130,
then Ra' with ß = 0.04545
= 12,000 / ( 1 + [ 130 x 0.04545 ] )
= 1,737 ohms.
Since gain is halved with a 2k2 load with the above amount of NFB,
the amount of applied FB = 6dB, but only when RL = 2k2.
If RL was say 4k7, A is maybe 40, and the gain reduction is greater so
the amount of NFB applied is greater.
With a CCS load, but with the above R2 = 420k, the load is approximately
A/( A + 1 ) x 420k.
This is because the R2 has output voltage PLUS the grid voltage across
it.
Notice that Ra' is independant of A, A', or RL;
Ra' is the effective Ra after NFB is applied.
it is the output resistance of the stage, Rout.
If the RL of 2k2 is used as a permanent dc Ia current feeding load,
then Ra' would be the same, but Rout = Ra' // RL.
Notice also that it doesn't take much NFB to dramatically
change the effective Ra' from a uselessly high Ra = 12k down to 1k7.
Obviously, since its not too hard to produce a fairly clean 40Vrms
of drive signal, we could make R2 considerably lower R without loading
down the tube too much
and get a much larger reduction of Ra to below triode connection values,
and with less THD than a triode.
When calculating A with shunt FB applied, the load on the tube = the dc
carrying RL
in parallel with any following load such as a gain pot, AND ALSO
in parallel with the effective value of R2.
To confirm what you measure with what you calculate
you must set up the circuit with FB connected to find out what A is.
Just using the data figures for µ, gm, and Ra is a bull**** practice
with zero merit
for accuracy in practice because these 3 paramaters vary with Ia and
between various makes and versions of tubes.
So its another reason I never simulate.
The data figures are only ever going to give you an APPROXIMATE answer,
but
still useful though if being within 10% accurate is good enough for you.
A with NFB is simply Va / Vg, and A' = Va / Vin at input terminal.
The only valid thing is what you actually have in front of you.
In the example above, R2 = 420k, and is in parallel with 2k2,
and there is no cap coupled load following, thus RL really = 2k2//420k
which I think = 2,181 ohms, which BTW is in my figuring in my notebook.
But set up an EL34 to really find outabout.
Patrick Turner.
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