Help! Solving Nodal Equation for Vx @ Op-Amp Node

V- of the first op-amp is unknown because we are trying to solve for it. So, V- of the second op-amp subtracted from V- of the first op-amp would be V0 - V-. Therefore, the potential difference for the 30k resistor would be (V0 - V-). In summary, the conversation is about filling out the nodal equation for the node of the first op-amp towards the left. The user is having trouble figuring out what to put for the ~~~~ part in the equation Vx - 5 / 70k + ~~~~ = 0. They are also trying to verify the KCL equation for the node by the + polarity of V0
  • #1
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Hi, I am having trouble filling out the nodal equation for the node of the first op-amp towards the left. Let's say that node voltage is Vx. (V-)

I know that, Vx - 5 / 70k + ~~~~ = 0... I don't know what to put for the ~~~~ part. I know the current has to equal the current across teh 70kohm resistor because the current into an ideal op amp is equal to 0. I'm a bit confused as to HOW to model the current leaving that node to the particular branch.

If you can't figure out what I'm talking about, it's toward the left part of the 60k and 30k resistor... that node.


http://img172.imageshack.us/img172/3728/screenshot01sd4.jpg [Broken]

I also want to verify that there is another node by the + polarity of the V0. That node will have voltage V0. So, the KCL equation for that node will be:

V0 / 30k + (V0 - Vx) / 60k = 0, right?
 
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  • #2
Ask yourself this question. What are the voltages (V+,V-) at the op amp terminals. In an ideal op amp V-=V+

Sorry, I was in the middle of editing and trying to answer your questions till the phone rang...

The 70, 30, 60 resisters at the top are connected by the same node. Yes, Vo=V- at the 2nd op amp. Start with the left and work your way to right
 
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  • #3
teknodude said:
Ask yourself this question. What are the voltages (V+,V-) at the op amp terminals. In an ideal op amp V-=V+

Sorry, I was in the middle of editing and trying to answer your questions till the phone rang...

The 70, 30, 60 resisters at the top are connected by the same node. Yes, Vo=V- at the 2nd op amp. Start with the left and work your way to right


When I apply KCL at the left op-amp, I don't know how to model the 30k resistor as a difference of potential divided by (/) a certain resistance. Don't you still have to apply KCL?

For the KCL at the second opamp's positive terminal, I meant to say this:

Assume Vy is the voltage out of the op-amp with respect to GND (at the bottom of the circuit)

KCL Op Amp #2: 0 = V0/30kOhm + V0 - Vy / 140 kOhm
 
  • #4
Your image of the op amp is gone. Can you upload it again?
 
  • #6
ravenprp said:
When I apply KCL at the left op-amp, I don't know how to model the 30k resistor as a difference of potential divided by (/) a certain resistance. Don't you still have to apply KCL?

The potential difference for the 30k resistance is just going to be the V- of the first op amp subtracted from the V- of the 2nd op amp. I'll leave it to you to find what those are.
 
  • #7
teknodude said:
The potential difference for the 30k resistance is just going to be the V- of the first op amp subtracted from the V- of the 2nd op amp. I'll leave it to you to find what those are.

V- of the second op-amp is going to be +V0 because it's the same node.
 

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