# Op-Amp circuit with applied external compensation

1. Sep 16, 2016

### topcat123

1. The problem statement, all variables and given/known data
FIGURE 3(b) shows the THS4021 in a practical op-amp circuit with applied external compensation. Determine an expression for the low frequency gain of the circuit.

2. Relevant equations

3. The attempt at a solution
I really don't know where to start.
Is there a virtual earth at the inverting input?

#### Attached Files:

• ###### Fig 3.PNG
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2. Sep 16, 2016

### rude man

Big hint: " ... low-frequency gain ...".

3. Sep 16, 2016

### topcat123

Thanks Rude Man
I did see that in the question.
$$f_c=\frac{1}{2πRC}$$
This is the only equation I can find for low pass RC circuit.

4. Sep 16, 2016

### Staff: Mentor

I think you've missed @rude man 's hint. What's the lowest frequency you can think of?

5. Sep 19, 2016

### topcat123

Fractions above zero?

6. Sep 19, 2016

### Staff: Mentor

What's the lower limit?

7. Sep 19, 2016

### topcat123

0Hz

#### Attached Files:

• ###### fig.2.PNG
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8. Sep 19, 2016

### Staff: Mentor

Yup. So What's the 0 Hz gain of the circuit?

9. Sep 20, 2016

### topcat123

From the op-amps characteristic graph we can see 90 dB?
so how do I "Determine an expression for the low frequency gain of the circuit"

10. Sep 20, 2016

### Staff: Mentor

The op-amp is surrounded by a feedback network. Start with the ideal op-amp model and find the gain. If the result is much smaller than 90 dB then you can take the result as accurate enough. Otherwise, if what you get is a large fraction of 90 dB you'll have to introduce a more realistic model for the op-amp and go through the work of analyzing it. But that's unlikely to happen here, since I can see by inspection of the circuit that the gain won't be large...

11. Sep 20, 2016

### rude man

Hint: #1: you don't need any "op amp characteristics" here.
Hint #2: what can you say about a capacitor at dc after a "long time" has passed?

12. Sep 20, 2016

### topcat123

Ah so when the cap has charged current through it stops.

and the general op-amp inverting input gain equation is $${V_{out}}=-\frac{R_f}{R_1}V_{in}$$.
or $$A_v=\frac{V_{out}}{V_{in}}=\frac{Rf}{R_1}$$

Last edited by a moderator: Sep 21, 2016
13. Sep 20, 2016

### rude man

OK, what would that be using the nomenclature in your figure?

14. Sep 20, 2016

### topcat123

So the gain Av
$$A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}$$

15. Sep 20, 2016

### rude man

Ooh, almost. Sign?

16. Sep 21, 2016

### topcat123

I put it in the original equation then missed it out on the second.
Because it is inverting
$$A_v=\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1}$$

17. Sep 21, 2016

### rude man

Right.

18. Sep 21, 2016

### topcat123

Thanks rude man and gneill for all your help.