Engineering Op-Amp circuit with applied external compensation

AI Thread Summary
The discussion focuses on determining the low-frequency gain of a practical op-amp circuit featuring external compensation. Participants clarify that the low-frequency gain can be derived using the inverting amplifier formula, A_v = -R_2/R_1, while noting the importance of considering the op-amp's characteristics. The conversation emphasizes the behavior of capacitors at DC and the concept of virtual ground at the inverting input. Acknowledgment of the op-amp's gain being around 90 dB is mentioned, but participants agree that a more realistic model may not be necessary for this analysis. The thread concludes with participants expressing gratitude for the guidance received.
topcat123
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Homework Statement


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.

Homework Equations

The Attempt at a Solution


I really don't know where to start.
Is there a virtual Earth at the inverting input?
 

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topcat123 said:

Homework Statement


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.

Homework Equations

The Attempt at a Solution


I really don't know where to start.
Is there a virtual Earth at the inverting input?
Big hint: " ... low-frequency gain ...".
 
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.
 
I think you've missed @rude man 's hint. What's the lowest frequency you can think of?
 
Fractions above zero?
 
topcat123 said:
Fractions above zero?
What's the lower limit?
 
0Hz

Here is some more info on the op-amp.
 

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topcat123 said:
0Hz
Yup. So What's the 0 Hz gain of the circuit?
 
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
topcat123 said:
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"
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
topcat123 said:
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"
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
rude man said:
Hint #2: what can you say about a capacitor at dc after a "long time" has passed?
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}
 
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  • #13
OK, what would that be using the nomenclature in your figure?
 
  • #14
So the gain Av
A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}
 
  • #15
topcat123 said:
So the gain Av
A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}
Ooh, almost. Sign?
 
  • #16
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
topcat123 said:
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}
Right.
 
  • #18
Thanks rude man and gneill for all your help.
 
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