How can I analyze this transimpedance amplifier?

In summary, the conversation discusses a problem involving an op-amp network with both negative and positive feedback. The question asks for the value of the output voltage for Jg at 1mA, assuming ideal op-amp. The participants try to solve the problem using KCL and discuss the difference between a current source and a current source in series with a resistor. Ultimately, the conversation focuses on finding an expression relating Vo and the potential at the op-amp inputs, and the importance of understanding the nature of a current source with a series resistance.
  • #1
Boltzman Oscillation
233
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?
 
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  • #2
Boltzman 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.
 
  • #3
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.
 
  • #4
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
 

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  • #5
Boltzman 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?
 
  • #6
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.
 
  • #7
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?
 
  • #8
Boltzman 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...
Boltzman 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?
 
  • #9
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
Boltzman 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 gneill

Related to How can I analyze this transimpedance amplifier?

1. What is a transimpedance amplifier?

A transimpedance amplifier is an electronic circuit that converts an input current into an output voltage. It is commonly used to measure small current signals, such as those produced by photodiodes or other sensors.

2. How does a transimpedance amplifier work?

A transimpedance amplifier uses a feedback loop to convert the input current into an output voltage. The input current is passed through a resistor, and the resulting voltage is amplified and fed back to the input. This stabilizes the output voltage and allows for accurate measurement of the input current.

3. What factors should be considered when analyzing a transimpedance amplifier?

Some important factors to consider when analyzing a transimpedance amplifier include the input and output impedance, gain, bandwidth, noise, and stability. It is also important to consider the specific application and the requirements for accuracy and sensitivity.

4. How can I design a transimpedance amplifier?

Designing a transimpedance amplifier involves selecting appropriate components, such as resistors and operational amplifiers, and calculating their values based on the desired input and output characteristics. It is also important to consider the circuit layout and potential sources of noise or interference.

5. How can I troubleshoot issues with a transimpedance amplifier?

If a transimpedance amplifier is not functioning as expected, it is important to check for loose connections, damaged components, or incorrect values of resistors or capacitors. It may also be helpful to simulate the circuit using software or test the amplifier with different input signals to identify and address any issues.

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