# non-inverting Op-Amp calculation

Discussion in 'General Electronics Chat' started by reno_0405, Oct 30, 2008.

1. ### reno_0405 Thread Starter New Member

Oct 24, 2008
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Sorry for the question below~
for a 150mV signal to the input, the peak to peak output voltage should be (R2/(R1+R2))*(1+Rf/Ri)*Vin*1.414*2 = 3.57V, but how does the AV = 20 of the op-amp affect the actual gain?

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
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 because A is finite, so you need this small voltage across the inputs to get an output voltage

Vo=A*(V1-V2)

for A=infinite V1-V2=0
A=finite value V1-V2=small value

If you derive the gain expression again by taking into account this voltage across the input terminals then you will see how this affects the output voltage because in the expression the term A will be included.

3. ### Audioguru New Member

Dec 20, 2007
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896
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.

4. ### reno_0405 Thread Starter New Member

Oct 24, 2008
5
0
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:

and i just found an equation from a reference book which is never exist even from my study text book and other reference books, it says that the finite open loop gain might affect the total gain of the op amp, the total voltage gain shud be:
Av = (1+Rf/Ri)/(1+1/Aod(1+Rf/Ri)), which Aod is finite open loop gain, i am not sure this is right because this equation are only exist in 1 book.

if my calculation was wrong, can you correct me?

Last edited: Oct 31, 2008
5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Well, i am not sure if the equation is correct but it seems correct because it includes the Aol term. If you want you can check it by analyzing the circuit as i told you.

6. ### Audioguru New Member

Dec 20, 2007
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896
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.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
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.

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8. ### Audioguru New Member

Dec 20, 2007
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Opamps work best when their negative feedback is the highest.
I wonder why the input resistors are used which is an attenuator.

9. ### Ron H AAC Fanatic!

Apr 14, 2005
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He posted this in GEC, but it looks like a homework exercise to me.
Either that, or it was designed by Space Vomit (inside joke).