by "Alex" <apogosso@[EMAIL PROTECTED]
>
Apr 23, 2008 at 12:14 PM
"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?
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
