# Current in OP-cicuit

Discussion in 'Homework Help' started by notthisyear, Apr 2, 2013.

1. ### notthisyear Thread Starter New Member

Mar 30, 2013
2
0
Hello!

I have a question regarding the circuit below. As you can see I've marked two currents, I1 and I2. As it turns out - they are equal! Sadly, I'm not completely sure of why.

The operational amplifier is considered ideal, so that the currents to the inputs are zero seems fine. What I don't really get is the output current. In order for I1 and I2 to be equal, we would require the current from the output of the operational amplifier to be zero, right? Why is it zero? I don't really see the reason for this...

Thank you in advance, sorry for the MS Paint-circuit!

Best regards,
notthisyear

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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To match the circuit conditions as denoted in the schematic the op-amp output current would be -I1 amps.

3. ### WBahn Moderator

Mar 31, 2012
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Q1) Why do you believe that I1 and I2 are equal?

Q2) Even if they are, why does that mean that the opamp output current is zero?

Keep in mind ALL of the paths that EITHER current could be flowing through.

4. ### screen1988 Member

Mar 7, 2013
310
4
I think what OP meant is that the output of OP-AMP is connected to a load, for example, a resistor. And can I consider the OP-AMP is a two port network? If so I1 must be equal to I2?
I1=I2 => the opamp output current is zero according to KCL at output node.

5. ### WBahn Moderator

Mar 31, 2012
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No. Keep in mind ALL the paths.

6. ### notthisyear Thread Starter New Member

Mar 30, 2013
2
0

To WBahn's first question: the only reason for me to think that I1 and I2 are equal is that the solution manual says so...

Here is how I'm trying to solve my problem, but there is apparently something that is not correct...

I2 = outwards current from ground node = (0 - (-V1)) / (100k + 9k) = V1/109k (1)

Then I do KVL in the rightmost loop:

V1 - (I1 * 11k) - (I2 * 9k) = 0

and substitute for I2 the expression (1) and simplify, which leads me to:

I1 = V1/11990

...which obviously is not equal to I2 in expression (1).

Where else can the current go? The inputs to the OPamp are zero, whether or not there is an output current from the OPamp or a load connected through which the current flows I don't know. I'm assuming that the ground node might come in to play, but I not at all sure of how. What am I missing?

Again, thank you for your help!

Regards,
notthisyear

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
I've redrawn the schematic to highlight some important matters. Particularly V1 is shown as a source. Hopefully you will at least understand why your first equation is in error when you consider the circuit topology with V1 as a series element.

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8. ### t06afre AAC Fanatic!

May 11, 2009
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The golden rules are idealizations of op-amp behavior, but are nevertheless very useful for describing overall performance. These rules consist of the following two statements:​
1. The voltage dierence between the inputs, V+ andV-, is zero.
2. The inputs draw no current. ( This is true in the approximation that the Zin of the op-amp is much larger than any other current path available to the inputs.) When we assume ideal op-amp behavior, it means that we consider the golden rules to be exact.

9. ### WBahn Moderator

Mar 31, 2012
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7,382
Not a good way to base your solutions. The solutions manual can be a great resource for learning, but only if you impose on yourself the requirement that you use it to understand WHY they got what they got or did what they did. The real world doesn't come with a solutions manual.

Thanks for finally showing your work. This gives us the starting point we need since now we can see that you have a major misconception right in your very first step.

V1 is the voltage at the top left node relative to the bottom left node. In your equation above, you are saying that voltage on the bottom left node is -V1. Not the case. It is whatever it is, call it V42, and the voltage on the top left node is V42+V1.

I think where the solutions manual is getting the notion that I1 is equal to I2 is if a single component, such as a voltage source, is placed across the inputs. But this is an unreasonable assumption. Another way they might be making the claim is if they think this is a two port network. But it isn't. You can't put this thing in a box with just two ports. There's the power for the opamp and they show a ground connection. Those allow for all kinds of interactions.

I think that V2 is going to be dependent on both the differential input, V1, and also the common mode voltage at the inputs.

If there is no load, then regardless of whether I1 is equal to I2, the ONLY place for the I1 current to go is into the output of the opamp. There is no other path for it! Similarly, without a load, whatever I2 is, the ONLY place for it to originate is out of the ground node.