Re: NFB101 Crisis!
- From: "Alex" <apogosso@xxxxxxxxxxx>
- Date: Wed, 23 Apr 2008 12:14:33 GMT
"Ian Thompson-Bell" <ruffrecords@xxxxxxxxxxx> wrote in message
news:fukkiv$qch$1@xxxxxxxxxxxxxxxxxxxx
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
.
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