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?
 
Attachments
  • Circuit schematic of an op-amp amplifier labelled Figure 3(b): R1 1K from V1 to the inverting input, R2 1K feedback from output to inverting input, R3 100 ohm in series with C1 22p from the inverting input to ground, and the non-inverting input grounded; op-amp labelled THS4021 X1, output V2.
    Fig 3.PNG
    3.5 KB · Views: 561
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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.
 
0Hz

Here is some more info on the op-amp.
 
Attachments
  • Bode plot with two stacked graphs against frequency in hertz from 100 to 10G: gain in dB (green, top) falling from about 90 dB to about -50 dB, and phase in degrees (red, bottom) falling from 0 to about -180.
    fig.2.PNG
    3.3 KB · Views: 517
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]
 
topcat123 said:
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]
Right.