You derived the gain expression by assuming that the amplifier has infinite A and thus the voltage difference across the inputs equals zero. In reality there some voltage across the input terminals
Well this is just a question from my exam, not a circuit design for any purpose. and i get the same value from multisim simulation even without including the open loop gain by using my own calculation previously. The full question is as below:I think the frequency is high so the opamp has an open-loop gain of only 20 instead of 200,000 as it has at low frequencies. Then Rf/Ri is wrong.
The low value for R2 implies an old opamp with a high bias current.
Nope, the gain of the entire circuit is 2.477. Below is the derivation of the closed loop gain of the amplifier, not including the attenuator.The input 47k resistor feeding the 10k resistor forms a 0.175 attenuator.
If the opamp has an open loop gain of more than 1000 then its closed loop gain is 1+ (470k/10k)= 48.
So the gain of the entire circuit is 0.175 x 47= 8.4.
If the opamp has an open loop gain of only 20 then the entire circuit is an attenuator.
He posted this in GEC, but it looks like a homework exercise to me.Opamps work best when their negative feedback is the highest.
I wonder why the input resistors are used which is an attenuator.
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