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### Network # Op-Amp Gain ?

#### ElectStudent

Apr 8, 2014
13
In this question i have attached, do i find out the gain by just reading the graph?
It seems a bit too easy to get 6 marks for reading the graph, that's why i'm confused.
However the solutions are (a) 1001 and (b) 100, which are not values obtained from this graph.

Last edited:

#### duke37

Jan 9, 2011
5,364
I agree with a gain of 1001 at low frequency, near enough to 10^3. There is no way that circuit can give a gain of 10^6.

• ElectStudent

#### ElectStudent

Apr 8, 2014
13
In the graph it says gain of amplifier, is there any difference between amplifier gain and actual circuit gain? Since by calculation the gain is (1000+1/1 = 1001)

#### LvW

Apr 12, 2014
604
The closed-loop gain Acl of the circuit (non-inverting) for low frequecies is Acl=1/Hf with Hf (feedback factor)=Ro/(Ro+Rf)=1/1001.
Hence Acl=1001.
This value applies as long as the open-lop gain Aol is much larger than Acl (identical to loop gain LG>>0 dB).
For all frequencies above the cross-over frequency (where the loop gain LG=0dB) the closed-loop gain Acl follows the open-loop gain Aol.
That means: Aol=Acl=100 at f=1MHz.

Comment: The above estimates are valid only, of course, in case of stable operation of the feedback circuit.
However, the open-loop response exhibits a gain slope of -40dB/dec, which means: Instability!
I assume a corresponding error in the plot because, normally, the open-loop characteristic has a slope of -20dB/dec only.

Last edited:
• ElectStudent

#### ElectStudent

Apr 8, 2014
13
oh ok ,thanks. So at large frequencies acl follows the open loop gain.

#### LvW

Apr 12, 2014
604
oh ok ,thanks. So at large frequencies acl follows the open loop gain.
Yes - but tell your teacher that this answer is purely theoretical and has no practical relevance because the shown frequency response is by far not realistic. It leads to instability.

#### Ratch

Mar 10, 2013
1,098
In this question i have attached, do i find out the gain by just reading the graph?
It seems a bit too easy to get 6 marks for reading the graph, that's why i'm confused.
However the solutions are (a) 1001 and (b) 100, which are not values obtained from this graph.

OK, let's figure this out. Vi is the input, Vneg is the voltage on the negative terminal of the op amp, Vout is the output voltage, beta is the open loop gain. Vneg = Vo*(1/1001) and (Vi-Vneg)*beta = Vo . Solving the two equations for Vo/Vi, we get Vo/Vi = (1001*beta)/(1001+beta). Both the solutions you presented are wrong. If beta were 1E50, then the closed loop gain would be 1001. For beta= 1E6, the gain is 1000, and for beta=100, the gain is 90.92.

Ratch

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