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## 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?

$$ u(s) = \frac{u_o}{1+\frac{s}{w_o}} = \frac{100}{1 + \frac{s}{40}}$$

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?