Patrick Turner wrote:
>
> 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.
>
Indeed and by your calculation this is 0.5. This means when the open
loop gain is large enough the closed loop gain tends to 1/ß which in
your case is 2. In practice the gain is unity. That is what is confusing.
Cheers
Ian
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