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.
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.
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.
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.
Funny stuff the NFB.
Cheers
Ian
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