Op Amp Circuit Help: Solving KCL Eq with Source Transform

In summary, the source transform threw off the KCL equation and it was much easier to just scan the pictures and problem solve than trying to recreate it.
  • #1
wclawson
17
0
It was much easier scanning the pictures and problem than trying to recreate:

circuit1.jpg


circuit2.jpg


I began by doing a source transform, making Vs = is*Rs, and placing Rs in series instead of parallel. I then did KCL at Vn.

I tried to break the KCL eq down to il/is, but just ended up with some un-godly mess that ended up approximating to 0 for part b. Is the source transform throwing me off? Or is there just a much easier way to solve this than what I tried.

Thanks a lot.
 
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  • #2
wclawson said:
It was much easier scanning the pictures and problem than trying to recreate:

circuit1.jpg


circuit2.jpg


I began by doing a source transform, making Vs = is*Rs, and placing Rs in series instead of parallel. I then did KCL at Vn.

I tried to break the KCL eq down to il/is, but just ended up with some un-godly mess that ended up approximating to 0 for part b. Is the source transform throwing me off? Or is there just a much easier way to solve this than what I tried.

Thanks a lot.

Since they want you to calculatet Il/Is, I'd probably leave the source un-transformed. Write the 2 KCL equations (one for each input on the opamp), and see how it shakes out. And yes, the approximation when the gain is infinite helps simplify it a lot.
 
  • #3
I happen to be working on this same problem right now (University of rochester, ECE 111?)

Wouldn't the currents of each input always equal zero?
 
  • #4
danhamilton said:
I happen to be working on this same problem right now (University of rochester, ECE 111?)

Wouldn't the currents of each input always equal zero?

For an ideal opamp, yes. But Figure 4-4 shows some of the more practical (non-ideal) model bits...
 
  • #5
danhamilton said:
I happen to be working on this same problem right now (University of rochester, ECE 111?)

Wouldn't the currents of each input always equal zero?

Haha yes, ECE 111 at U of R.

Thank for the help, I've got it now.
 

FAQ: Op Amp Circuit Help: Solving KCL Eq with Source Transform

What is an op amp circuit?

An op amp circuit is a type of electronic circuit that uses an operational amplifier (op amp) to amplify, filter, or perform mathematical operations on an electrical signal. Op amps are commonly used in a variety of applications, such as audio amplifiers, signal processing, and control systems.

What is KCL and why is it important in op amp circuit analysis?

KCL stands for Kirchhoff's Current Law, which states that the sum of currents entering and exiting a node in a circuit must be equal. In op amp circuit analysis, KCL is important because it allows us to determine the currents flowing through different branches of the circuit and ensure that the circuit is functioning correctly.

How do you solve KCL equations in op amp circuits?

To solve KCL equations in op amp circuits, you can use source transformations to simplify the circuit. This involves replacing voltage sources with equivalent current sources and vice versa. Once the circuit is simplified, you can use KCL to solve for the currents in each branch.

What are some common challenges when solving KCL equations in op amp circuits?

One common challenge when solving KCL equations in op amp circuits is determining the correct direction of current flow. Another challenge is dealing with virtual grounds, which are points in the circuit where the voltage is assumed to be zero. It is important to carefully analyze the circuit and apply KCL correctly to overcome these challenges.

Are there any tips for solving KCL equations in op amp circuits?

One helpful tip for solving KCL equations in op amp circuits is to redraw the circuit in a simpler form using source transformations. This can make it easier to identify the currents and apply KCL correctly. It is also important to carefully label all currents and use the correct sign convention to ensure accurate results.

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