Ian Thompson-Bell wrote:
>
> Patrick Turner wrote:
> >
> > Alex wrote:
> >> "Ian Thompson-Bell" <ruffrecords@[EMAIL PROTECTED]
> wrote in message
> >> news:fukkiv$qch$1@[EMAIL PROTECTED]
> >>> 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
> >> Hello, Ian.
> >> Your equations are perfectly correct.
> >>
> >> To test them use an (imaginery) ridiculous amplifier with very low
gain, say
> >> Ao=0.01. Also let us assume beta=1, i.e. Ri = Rf. Well, because of
> >> ridiculously low gain the feedback is virtually inoperative. What
will the
> >> gain An be?
> >
>
> Patrick replied.
>
> > But ß cannot be 1.0 where R1 = R2 ( Ri = Rf ).
> >
> > ß is the fraction of the output voltage fed back to the input.
> >
> > Consider the input voltage terminal to be grounded.
> >
> > A noise signal within the amp appears at the output, Vn.
> > The fed back Vn at the grid, or inverting input = Vn x R1 / ( R1 + R2
).
> >
> > The Vn at the grid is thus 0.5Vn, and is amplified by the OLG to
oppose
> > its own production.
> >
> > ß = 0.5 when R1 = R2.
> >
> > R1 and R2 operate as a resistance divider.
> >
>
> This is indeed confusing and hopefully only for this particular form of
> feedback topology.
What is confusing?
ß is the fraction of the output signal fed back, period.
In shunt FB, ß = R1 / ( R1 + R2 )
In series FB its also R1 / ( R1 + R2 ),
but where R1 = 100 ohms between the Rk&Ck bias network of an input
triode and 0V,
and R2 is the R from the speaker terminal to the top of the 100 ohms,
often 1k5.
The series network is generally a low resistance type.
>
> A lot depends on your definitions. R1, R2 do indeed act as a pot divider
> and if they are equal you might expect ß to be 0.5 but that would mean
> the closed loop gain would be 2 (1/ß) when we know it is one with R1=R2.
The CLG is what it is as I have calculated.
Only where OLG is enormous that CLG = near enough to unity when R1 = R2.
> Thing is, by the same argument, R1,R2 act as a pot divider for the input
> signal, thus halving its value at the amp input so you could say the
> input is halved then amplified by X2 with an overall unity result.
Not so. The input signal is what it is at the input terminal
which is regarded as a voltage source.
The input signal simply has to
provide the required CURRENT to equal the CURRENT
flow from anode to grid through the R2 feedback resistance.
>
> The problem is the topology of the input because the summing point is
> somewhat isolated from the input by the input resistor. It is
> interesting that RDH takes a different approach with no input resistor
> and instead shows how the overall gain depends of the resistance of the
> signal source which amounts to the same thing.
Some apps of shunt NFB involve the use of a high resistance source which
becomes
R1 of the R1 + R2 network, see pages 333 and 334 of RDH4.
With a pentode driving an output stage, the shunt NFB is most effective,
and see
my website for many applications.
>
> Funny stuff the NFB.
Depends....
Patrick Turner.
>
> Cheers
>
> Ian
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