How can I analyze this transimpedance amplifier?

Click For Summary

Discussion Overview

The discussion revolves around analyzing a transimpedance amplifier circuit involving an operational amplifier (op-amp) with both negative and positive feedback. Participants are attempting to calculate the output voltage given a specific input current, while exploring the application of Kirchhoff's laws and the behavior of current sources in the circuit.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the initial problem and equations derived from Kirchhoff's Current Law (KCL) but expresses uncertainty about the voltage across the current source.
  • Another participant suggests that finding the current through the current source is simpler than it seems, indicating a potential misunderstanding of the problem.
  • Discussion includes the relationship between the currents through various resistors and the op-amp inputs, with some participants questioning the necessity of the voltage across the current source.
  • There is a clarification regarding the transformation of current and voltage sources, with participants discussing the implications of series and parallel resistances.
  • Participants refine their equations based on KCL, with one participant correcting their understanding of the current through a resistor in relation to the current source.
  • One participant expresses confidence in their approach after receiving feedback, indicating a potential resolution to their initial confusion.

Areas of Agreement / Disagreement

There is no clear consensus on the best approach to solve the problem, as participants express differing levels of understanding and various interpretations of the circuit behavior. Some participants agree on the application of KCL, while others challenge assumptions and clarify concepts related to current sources.

Contextual Notes

Participants are working with an ideal op-amp model and are navigating the complexities of circuit analysis, including the need for specific voltage values that are not provided in the problem statement. The discussion reflects various assumptions about circuit behavior and the application of circuit laws.

Who May Find This Useful

This discussion may be useful for students studying circuit analysis, particularly those focusing on operational amplifiers and transimpedance amplifiers, as well as those seeking to understand the application of Kirchhoff's laws in complex circuits.

Boltzman Oscillation
Messages
233
Reaction score
26

Homework Statement



The opAmp network shown in figure 68 is a transimpedance amplifer that employs both negative and positive feedback. Assuming that the op-amp is ideal, calculate the value for the output voltage for Jg is 1mA.The circuit and question are this:

https://www.chegg.com/homework-help/questions-and-answers/84-op-amp-network-shown-figure-84-transimpedance-amplifier-employs-negative-positive-feedb-q31684479

Homework Equations



Vp = Vn
In = Ip = 0
KCL
KVL

The Attempt at a Solution



I tried using KCL to make the following

(Vn - Vj)/R1 + (Vn-Vo)/R2 = 0

(Vn - Vj)/ R3 + Vn/R5 + (Vn - Vo)/R4 = 0

but I really don't think this is the way to solve the problem. I don't really know the voltage across the current source so it's pretty useless for me to use it here. Can anyone guide me to the right direction? Does anyone know a good resource to read on op amps?
 
Physics news on Phys.org
Boltzmann Oscillation said:
I tried using KCL to make the following

(Vn - Vj)/R1 + (Vn-Vo)/R2 = 0

(Vn - Vj)/ R3 + Vn/R5 + (Vn - Vo)/R4 = 0

but I really don't think this is the way to solve the problem. I don't really know the voltage across the current source so it's pretty useless for me to use it here. Can anyone guide me to the right direction? Does anyone know a good resource to read on op amps?
I think you used KCL correctly.. Finding the current through Jg is easier than you think it is.
 
willem2 said:
I think you used KCL correctly.. Finding the current through Jg is easier than you think it is.
You mean the voltage? I am given the current of Jg to be 1mA but not the voltage across Jg (which is what i need if I were to use KCL.
 
What can you say about the current through ##R_2##? Can you find an expression relating ##V_o## and the potential at the op-amp inputs?

upload_2018-11-25_10-41-34.png
 

Attachments

  • upload_2018-11-25_10-41-34.png
    upload_2018-11-25_10-41-34.png
    11 KB · Views: 999
Boltzmann Oscillation said:
You mean the voltage? I am given the current of Jg to be 1mA but not the voltage across Jg (which is what i need if I were to use KCL.

It certainly appears you are trying to use the fact that the total current into the node at the negative input of the op-amp must be 0 when you write
(Vn - Vj)/R1 + (Vn-Vo)/R2 = 0. It's Kirchhof's Current law, you know what the current is through R1.

Trick Question. What is the difference between a current source, and the same current source in series with a 5k resistor?
 
willem2 said:
It certainly appears you are trying to use the fact that the total current into the node at the negative input of the op-amp must be 0 when you write
(Vn - Vj)/R1 + (Vn-Vo)/R2 = 0. It's Kirchhof's Current law, you know what the current is through R1.

Trick Question. What is the difference between a current source, and the same current source in series with a 5k resistor?

I don't know? A current source in series with a resistor can be transformed to a voltage sorce with a resistance in parallel.
 
gneill said:
What can you say about the current through ##R_2##? Can you find an expression relating ##V_o## and the potential at the op-amp inputs?

View attachment 234634
Yes i can do that and I've actually done it in my notes but i will still end up with a variable for the voltage of Jg in my equation. Do i not have to take that out somehow?
 
Boltzmann Oscillation said:
I don't know? A current source in series with a resistor can be transformed to a voltage sorce with a resistance in parallel.
Nope. You can transform a voltage source with a series resistance into a current source with a parallel resistance. And you can transform a current source with a parallel resistance into a voltage source with a series resistance. But you cannot transorm a current source with a series resistance into anything that doesn't look like a current source. Same goes for a voltage source in parallel with a resistor.

Think about it. If you place a resistor across a voltage source V, what voltage will you measure across the source? Can any resistor value change it?

Now think about the current source with a series resistor...
Boltzmann Oscillation said:
Yes i can do that and I've actually done it in my notes but i will still end up with a variable for the voltage of Jg in my equation. Do i not have to take that out somehow?
You'r doing KCL and so summing currents. What current is coming from Jg?
 
gneill said:
Nope. You can transform a voltage source with a series resistance into a current source with a parallel resistance. And you can transform a current source with a parallel resistance into a voltage source with a series resistance. But you cannot transorm a current source with a series resistance into anything that doesn't look like a current source. Same goes for a voltage source in parallel with a resistor.

Think about it. If you place a resistor across a voltage source V, what voltage will you measure across the source? Can any resistor value change it?

Now think about the current source with a series resistor...

You'r doing KCL and so summing currents. What current is coming from Jg?
Oh so one of my equations should be:

-Jg + In + Ir2 = 0

this is because the current through R1 is the same as the current source? This would give me:

-Jg + (Vn - Vo)/R2 = 0 (1)

(Vn - V3)/R3 + Vn/R5 + (Vn - Vo)/R4 (2)

I can then transform equation one to:

-Jg/R2 + Vn/R2 = Vo (3)

Which can be plugged into (2)? Am I doing this right?
 
  • #10
Boltzmann Oscillation said:
Oh so one of my equations should be:

-Jg + In + Ir2 = 0

this is because the current through R1 is the same as the current source?
Yes. No matter what resistance is placed in the path of an ideal current source, the same current flows.
This would give me:

-Jg + (Vn - Vo)/R2 = 0 (1)

(Vn - V3)/R3 + Vn/R5 + (Vn - Vo)/R4 (2)

I can then transform equation one to:

-Jg/R2 + Vn/R2 = Vo (3)

Which can be plugged into (2)? Am I doing this right?
I don't see where V3 comes from. When you do KCL at the bottom node, what current is Jg pulling from that node?
 
  • #11
gneill said:
Yes. No matter what resistance is placed in the path of an ideal current source, the same current flows.

I don't see where V3 comes from. When you do KCL at the bottom node, what current is Jg pulling from that node?
Sorry I misread my solution.

I believe the current across R3 should be Jg right? So equation 2 should become:

Jg + Vn/R5 + (Vn -Vo)/R4 = 0
 
  • #12
That looks better, yes.
 
  • #13
gneill said:
Nope. You can transform a voltage source with a series resistance into a current source with a parallel resistance. And you can transform a current source with a parallel resistance into a voltage source with a series resistance. But you cannot transorm a current source with a series resistance into anything that doesn't look like a current source. Same goes for a voltage source in parallel with a resistor.

Think about it. If you place a resistor across a voltage source V, what voltage will you measure across the source? Can any resistor value change it?

Now think about the current source with a series resistor...

You'r doing KCL and so summing currents. What current is coming from Jg?
I understand now, a resistance in series with a current source will have a current value of that current source. I can look at this by visualizing the ideal current source.
 
  • #14
Okay i think I have the right tools to complete this question. Thank you all.
 
  • Like
Likes   Reactions: gneill

Similar threads

What is the gain error for this opamp driven by a voltage
  • · Replies 18 ·
Replies
18
Views
3K
Engineering Opamp Circuit Q: Find Vo in Terms of Vs | Chegg.com
  • · Replies 1 ·
Replies
1
Views
2K
Engineering Why is the voltage source high potential into node 1 at -12 V in this op amp problem?
  • · Replies 15 ·
Replies
15
Views
3K
Engineering Derive expressions for the voltage gain of this opamp circuit
  • · Replies 11 ·
Replies
11
Views
4K
Solving for Output Voltage of Op Amp
  • · Replies 4 ·
Replies
4
Views
3K
Solve Op Amp Saturation: Find Rf & Output Current
  • · Replies 7 ·
Replies
7
Views
5K
Engineering Analyzing the internal circuitry of IC 741 Op-Amp step by step
  • · Replies 4 ·
Replies
4
Views
6K
Finding the output current in an op-amp.
  • · Replies 4 ·
Replies
4
Views
31K
Engineering Op Amp : Design the circuit such that
  • · Replies 5 ·
Replies
5
Views
3K
Solving voltage gain in an non-ideal op amp.
  • · Replies 7 ·
Replies
7
Views
4K