Op-Amp circuit with applied external compensation

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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.
[tex]f_c=\frac{1}{2πRC}[/tex]
This is the only equation I can find for low pass RC circuit.
 
Fractions above zero?
 
0Hz

Here is some more info on the op-amp.
 

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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"
 
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...
 
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?
 
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 [tex]{V_{out}}=-\frac{R_f}{R_1}V_{in}[/tex].
or [tex]A_v=\frac{V_{out}}{V_{in}}=\frac{Rf}{R_1}[/tex]
 
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So the gain Av
[tex]A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}[/tex]
 
I put it in the original equation then missed it out on the second.
Because it is inverting
[tex]A_v=\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1}[/tex]
 
Thanks rude man and gneill for all your help.