How can I analyze this transimpedance amplifier?

  • #1

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...ier-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 dont think this is the way to solve the problem. I dont really know the voltage accross 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?
 

Answers and Replies

  • #2
1,956
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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 dont think this is the way to solve the problem. I dont really know the voltage accross 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.
 
  • #3
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 accross Jg (which is what i need if I were to use KCL.
 
  • #4
gneill
Mentor
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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

  • #5
1,956
252
You mean the voltage? I am given the current of Jg to be 1mA but not the voltage accross 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?
 
  • #6
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 dont know? A current source in series with a resistor can be transformed to a voltage sorce with a resistance in parallel.
 
  • #7
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?
 
  • #8
gneill
Mentor
20,803
2,784
I dont 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...
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?
 
  • #9
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
gneill
Mentor
20,803
2,784
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
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
gneill
Mentor
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That looks better, yes.
 
  • #13
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.
 
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