# Different answer using inverting op-amp and KCL?

Discussion in 'Electronics Homework Help' started by Constant, Feb 15, 2015.

1. ### Constant

6
0
Feb 15, 2015
Hi all.

I have a bit of an interesting problem (see attachment),

I've calculated V_a in the first part of the question. Then I've applied KCL at the ground node to try and solve for V_0.

I know that V_0 should be -4.5V due to the inverting op-amp being present but KCL gives me 4.5V?

Thank you!

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2. ### LvW

604
146
Apr 12, 2014
The calculation for V_a is correct: V_a=4.5V.
This voltage is amplifier by the inverting opamp amplifier having a closed-loop gain of Acl=-96/32=3.
Hence: V_0=1.5*(-3)=-4.5V.

3. ### Constant

6
0
Feb 15, 2015
So what is wrong with my KCL calculation which gives V_0 = +4.5V?

4. ### Arouse1973Adam

5,165
1,087
Dec 18, 2013
Because it's an inverting amplifier so the output will go negative for a positive input.
Adam

5. ### Constant

6
0
Feb 15, 2015
I understand that +1.5V fed into the inverting op-amp will give an output of -4.5V. But using KCL at the ground node give me +4.5V, why is this?

6. ### Arouse1973Adam

5,165
1,087
Dec 18, 2013
Because it's inverting. So you have to change the sign to minus. -96/32 = -3 or 96/-32
Adam

7. ### Ratch

1,093
334
Mar 10, 2013
I have a bit of an interesting problem (see attachment),

I've calculated V_a in the first part of the question. Then I've applied KCL at the ground node to try and solve for V_0.

I know that V_0 should be -4.5V due to the inverting op-amp being present but KCL gives me 4.5V?

Thank you![/QUOTE]

You can use a combination of node and loop methods to calculate the voltages. The output voltage vo forces the current vo/96k into the va node, because it cannot pass into the op-amp input pins. Therefore, that voltage caused by current across the 32k resistor, added to the va voltage will be zero. The attachment shows the solution equations. Ask if you have any probs understanding it.

Ratch Last edited: Feb 15, 2015
8. ### Constant

6
0
Feb 15, 2015
If we forget about the inverting op-amp and just consider KCL.

Why does that particular route of analysis give me +4.5V?

9. ### Ratch

1,093
334
Mar 10, 2013
The solution I gave does not take into consideration whether it is an inverting amp or not. The second equation in the solution I submitted is a loop equation, and gives vo = -4.5 . Compare your loop equation with mine, and see where the difference lies.

Ratch

Arouse1973 likes this.
10. ### Ratch

1,093
334
Mar 10, 2013 Here is an analysis using only the node method. The node "vb" is the minus terminal of the op-amp. The term "amp" represents the finite amplification factor of the op-amp. In the solution, setting "amp" to infinity gives the same answer as before.

Ratch

11. ### hevans1944Hop - AC8NS

4,608
2,151
Jun 21, 2012
You can't just "forget about the inverting op-amp". The op-amp, acting as a voltage-controlled voltage source, connected between the op-amp output terminal and the ground node (your diagram does not explicitly indicate this) is providing current to the 96 kΩ load, as well as (it turns out) an identical current through the 96 kΩ feedback resistor. The ratio of the 96 kΩ feedback resistance to the 32 kΩ input resistance sets the op-amp circuit gain to -96/32 = A = -3. Since Vo = A Va, and Va = 1.5 volts, then Vo = -4.5 V.

Your KCL "analysis" at the ground node does not account for all the currents at the ground node because it ignores the implicit ground node current returning to the op-amp represented as a voltage-controlled volage source.

Because you didn't apply KCL correctly to account for ALL the current at the ground node. It was just a fortuitous choice of component values that made it appear that Vo was a function of Va with the wrong sign.

The op-amp can be modeled as a voltage-controlled voltage source, in which case its output is represented by a voltage source, Vo, connected between the op-amp output and the ground node. Note that this voltage source, Vo, is conspicuously missing from your diagram. The current through the 96 kΩ load resistor, as well as the current supplied to the 96 kΩ feedback resistor, must flow back to this voltage-controlled voltage source which has one terminal connected to the ground node. You do not present an equation that accounts for this ground return current.

The output Vo should not be a function of the load resistance. Try substituting 100 Ω, 1 kΩ, and 10 kΩ load resistors in your "KCL ground node calculation" to find out what you get for Vo with those three different values of load resistance.

12. ### Merlin3189

250
69
Aug 4, 2011
Well done Heavans! I hadn't spotted that until you said.
We just think about putting this sort of ground for a voltage reference, but of course current may flow. It's easy to forget.

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