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?
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.
Suppose you have a tube with OLG, ( open loop gain) = 20, and ß = 0.5
Suppose you measure noise = +Vn at output, with NFB connected.
+0.5Vn appears at the grid, and is amplified to become -10Vn, the
correction signal.
But you measured +1Vn, so Vn without NFB must have been 11Vn without NFB
connected,
and becomes 1Vn because you have 11Vn without NFB and the correction
signal reduces it to 1Vn.
So it is with distortions. So the amp will have CLG THD = 1/11 of the
OLG THD.
If +20V appears at the anode, -1V is needed at grid.
R1 = R2, so there MUST be 21V across each of R1 and R2,
so input voltage MUST be 22V.
Gain with NFB = 20 / 22. it is a negative number actually
because of the inverting amp.
Notice this A' is less than unity.
For opamps and other ams with OLG A = 50,000,
when R1 = R2, gain is very close to 1.0, always below 1.0 though.
The amp input terminal has such a tiny voltage upon it that its
regarded as a virtual earth, and CLG, A' = R2 / R1, a close enough
approximation.
You could have R1 much larger than R2, then ß approaches unity
and the amount of NFB applied approaches the maximum possible.
But its far more easily done using series NFB where ALL the output
voltage may
be fed back to a second terminal as in the case of a simple cathode
follower
where the output terminal, the cathode, is also a second input terminal.
The CF is THE classic example of a simple series voltage NFB
application.
Shunt NFB gives slightly different figures than series NFB.
Patrick Turner.
>
> Compare two cases:
> 1. Classic case of a non-inverting amplifier with voltage series
feedback.
> In this case
> An = Ao = 0.01 (approx.)
>
> 2. Your case of the inverting amplifier: An = 0.005 -- another half of
the
> gain is lost!
> This is absolutely correct since the signal is divided by two by Ri/Rf
> divider before being applied to the amp input.
>
> Your formulae give exactly that -- correct and consistent with reality!
>
> Regards,
> Alex
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