Op-Amp circuit with applied external compensation

In summary, the THS4021 op-amp circuit with applied external compensation has a low frequency gain of -R2/R1, which can be determined by analyzing the feedback network and applying the ideal op-amp model. The capacitor at dc will have no current flowing through it, resulting in a stopped charging process and a zero contribution to the gain.
  • #1
topcat123
78
1

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

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  • #2
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 ...".
 
  • #3
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.
 
  • #4
I think you've missed @rude man 's hint. What's the lowest frequency you can think of?
 
  • #5
Fractions above zero?
 
  • #6
topcat123 said:
Fractions above zero?
What's the lower limit?
 
  • #7
0Hz

Here is some more info on the op-amp.
 

Attachments

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  • #8
topcat123 said:
0Hz
Yup. So What's the 0 Hz gain of the circuit?
 
  • #9
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 [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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  • #13
OK, what would that be using the nomenclature in your figure?
 
  • #14
So the gain Av
[tex]A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}[/tex]
 
  • #15
topcat123 said:
So the gain Av
[tex]A_v=\frac{V_{out}}{V_{in}}=\frac{R_2}{R_1}[/tex]
Ooh, almost. Sign?
 
  • #16
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]
 
  • #17
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.
 
  • #18
Thanks rude man and gneill for all your help.
 

1. What is an op-amp circuit with applied external compensation?

An op-amp circuit with applied external compensation is a type of electronic circuit that uses an operational amplifier (op-amp) to amplify and process electrical signals. External compensation is used to improve the performance of the op-amp circuit by adjusting the feedback and compensation components.

2. Why is external compensation necessary for op-amp circuits?

External compensation is necessary for op-amp circuits because it helps to stabilize the amplifier and improve its performance. Op-amps can have issues with stability, bandwidth, and distortion, and external compensation can help to address these problems.

3. What are the different types of external compensation used in op-amp circuits?

There are two main types of external compensation used in op-amp circuits: frequency compensation and feedback compensation. Frequency compensation involves adjusting the gain-bandwidth product of the op-amp, while feedback compensation involves adjusting the feedback and compensation components to improve stability.

4. How does external compensation affect the performance of an op-amp circuit?

External compensation can have a significant impact on the performance of an op-amp circuit. By stabilizing the amplifier, it can improve the bandwidth, reduce distortion, and increase the overall accuracy and precision of the circuit. It can also help to prevent oscillations and improve the op-amp's response time.

5. What are some common applications of op-amp circuits with applied external compensation?

Op-amp circuits with external compensation are commonly used in a variety of electronic devices and systems, such as audio amplifiers, signal processing circuits, and control systems. They are also frequently used in instrumentation and measurement equipment, as well as in communication systems.

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