Using KCL on this Op-Amp circuit to solve for V0 and I0

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  • Thread starter bornofflame
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  • #1

Homework Statement:

Find v0 and i0 for the circuit.

Relevant Equations:

KCL, KVL(?), Ohm's Law
ENG220-p6.3-2.png
eng220-p6.3-2soln.jpg

Another classmate and a web search show that the answers for this problem are v0 = -12 V, and i0 = 2.4 mA. This was done using KVL around the circuit. I should be able to reach the same answer using KCL but I haven't and I'm not sure why. Is it because the source isn't connected to ground and so I can't do a KCL using the voltage source?
 

Answers and Replies

  • #2
DaveE
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Start by assuming that the op-amp is an ideal voltage amplifier. Then determine, by inspection, where the current io must flow. That will give you lots of information to solve this.
 
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  • #3
LvW
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I recommend to apply the superposition rule. You can assume that the opamp output resembles an ideal voltage source - hence, you have two voltage sources in the circuit.
 
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  • #4
The Electrician
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Use a source transformation. The 12 volt source in series with 5k can be transformed into a current source of 2.4 mA in parallel with 5k ohms. Having done this, it is immediately apparent that Io is 2.4 mA since no current flows into the inverting input of the opamp. The other voltages are also trivial to obtain.
 
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  • #5
Thank you for the replies! I ran with your recommendations, but I'm afraid I'm still unsure of how to proceed.
Here's what I was able to do so far:

IMG_20200429_001842.jpg


Start by assuming that the op-amp is an ideal voltage amplifier. Then determine, by inspection, where the current io must flow. That will give you lots of information to solve this.
Unfortunately op-amps are the first amplifiers that we've covered. This is the first that I've heard about Ideal voltage amplifiers actually. That said, thinking about where current is flowing, there's nothing at the inputs but Iout is > 0. Is Io considered a dead end, though: current at one end of the amp, but none at the other.

I recommend to apply the superposition rule. You can assume that the opamp output resembles an ideal voltage source - hence, you have two voltage sources in the circuit.
Since there's no difference in the voltage for either input at the op-amp, would I redraw it as a voltage source with the input voltages as the supplied

Use a source transformation. The 12 volt source in series with 5k can be transformed into a current source of 2.4 mA in parallel with 5k ohms. Having done this, it is immediately apparent that Io is 2.4 mA since no current flows into the inverting input of the opamp. The other voltages are also trivial to obtain.
Since the 5k resister is in parallel with the current source and the op-amp wouldn't that mean that some of the 2.4mA branches off and that Io is less than the current source?
 
  • #6
LvW
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"Since there's no difference in the voltage for either input at the op-amp, would I redraw it as a voltage source with the input voltages as the supplied "

For an ideal opamp (as in your case) the voltage vp at the non-inv. opamp input node is identical to the voltage vn at the inv. input node. Therefore, calculate both voltages applying the superposition rule and then equalize both.
But I think the method as proposed by "The Electricia" is simpler and more direct and leads immediately to the wanted quantity.
 
  • #7
The Electrician
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Since the 5k resister is in parallel with the current source and the op-amp wouldn't that mean that some of the 2.4mA branches off and that Io is less than the current source?
The 5k resistor is connected across the opamp inputs, so there is no voltage across it. If there's no voltage across the 5k resistor then there is no current passing through it, and none of the 2.4 mA branches off.

Also, notice that the current in the 1k resistor is also 2.4 mA for the same reasons and therefore the voltage across it is immediately known, and then the voltage at the plus and minus inputs of the opamp.
 
  • #8
DaveE
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Unfortunately op-amps are the first amplifiers that we've covered. This is the first that I've heard about Ideal voltage amplifiers actually. That said, thinking about where current is flowing, there's nothing at the inputs but Iout is > 0. Is Io considered a dead end, though: current at one end of the amp, but none at the other.
The ideal voltage amplifier (voltage input, voltage output) approximations are each of the following:
- Infinite impedance at each input. No current flows into the inputs.
- Zero output impedance. The output acts like an ideal voltage source.
- Infinite gain from input to output. This also implies that when negative feedback is properly applied, the amplifier will output whatever voltage is required to make the inputs equal to each other.
- Completely linear response. No saturation or limitations on voltages or currents into or out of the amplifier.

So, with infinite input impedance the current Io must flow through each circuit component except R. There is no other path available. This tells you, for example, that the voltage at the + input (across the 1K resistor) is -1000*Io.
 

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