# How can I find the open loop voltage gain of this opamp?

## Homework Statement:

Given a non-ideal op amp model, find the open loop voltage gain in the given format.

## Relevant Equations:

Below
The open loop voltage gain is given as :
$$u(s) = \frac{u_o}{1+\frac{s}{w_o}} = \frac{100}{1 + \frac{s}{40}}$$
Where u_o is the d.c. voltage gain and w_o is the pole.
The op amp that is given is: And I am told to use the non ideal op amp model as follows: Well my guess is that I can find the D.C voltage gain by substituting this model into the circuit shown in figure 1. I would not remove the feedback in order to find the D.C. voltage gain. If i find the voltage across r_o and then using it in the following formula:
$$u_o = \frac{V_{in}}{V_{out}} = \frac{V_{ro}}{V_{out}}$$

then i should be able to plug this into the equation for u(s) but then I would be missing the value for w_o. Am I thinking this correctly? What am I doing wrong or in what direction should I take this?

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LvW
So you want to find the "open-loop voltage gain" ....which is then written down by you in the next line...?
Sounds a bit confusing....

NascentOxygen
Staff Emeritus
Draw the resistor network in Fig 15a, but without the op-amp. Between Rs and R1 add the 1k resistor, and feeding the junction of R2 and RL add the 500Ω resistor, and to the free end of that 500Ω draw a voltage-controlled voltage source, A.vΔ. Now analyse that circuit to derive the transfer function, Vo/Vin. There will be a denominator term of the form (1 + s/ωf) to tell you the bandwidth of that circuit with a finite-gain amplifier.

So you want to find the "open-loop voltage gain" ....which is then written down by you in the next line...?
Sounds a bit confusing....
Sorry I am trying to derive it.

Draw the resistor network in Fig 15a, but without the op-amp. Between Rs and R1 add the 1k resistor, and feeding the junction of R2 and RL add the 500Ω resistor, and to the free end of that 500Ω draw a voltage-controlled voltage source, A.vΔ. Now analyse that circuit to derive the transfer function, Vo/Vin. There will be a denominator term of the form (1 + s/ωf) to tell you the bandwidth of that circuit with a finite-gain amplifier.
Alright I will try this method but I do have one concern: usually inductors and capacitors have the s variable when they are transformed into the frequency domain, where will I get the s variable here?

NascentOxygen
Staff Emeritus
Alright I will try this method but I do have one concern: usually inductors and capacitors have the s variable when they are transformed into the frequency domain, where will I get the s variable here?
The only component in the given circuit with an "s" in its relationship is the first-order falloff characteristic of the op-amp, μ(s).

BTW, I assume you are wanting to find the closed-loop response of this amplifier?

LvW
Sorry I am trying to derive it.
.....to derive it.... WHAT do you want to derive???
Please try to formulate a clear question .

For my opinion, only two alternatives do exist:

1) What is the closed-loop gain when an open-loop gain expression is given?
2.) What is the open-loop gain when an equivalent circuit diagram is given?

So - what do you need?

• BvU
The only component in the given circuit with an "s" in its relationship is the first-order falloff characteristic of the op-amp, μ(s).

BTW, I assume you are wanting to find the closed-loop response of this amplifier?
Well I am given the open loop voltage gain which I show as the first equation. I want to find out how they got that.

NascentOxygen
Staff Emeritus
Well I am given the open loop voltage gain which I show as the first equation. I want to find out how they got that.
I would think they constructed an amplifier on a silicon chip, then measured its output versus input for sinusoids and approximated that plot to a first-order system.

The wording in the instructions doesn't seem right, that's why I'm talking about finding closed loop response when told the open-loop response. It would be a good idea to check exactly what the task is that you have been set.

Tom.G