# KCL Question

1. Aug 26, 2011

### Red_CCF

Hi

I was starting to read up on Op Amps and near the beginning they said that current flowing into an Op Amp equals current going out (output current + 2 input current + 2 source current = 0) on the basis of KCL. I'm confused because I thought KCL only applied for all currents entering and leaving the same node (where there is no voltage change) and not for a component?

Any help is appreciated
Thanks

2. Aug 27, 2011

### tiny-tim

Hi Red_CCF!
That's right …

https://www.physicsforums.com/library.php?do=view_item&itemid=93" current law is the law of conservation of charge

charge in (per time) = charge out (per time) …

it applies to anything

a node, a component, or even a whole network

Last edited by a moderator: Apr 26, 2017
3. Aug 27, 2011

### Studiot

Good morning Tiny Tim

What happened to

Input = Output plus Accumulation

eg a Van Der Graff generator ball?

4. Aug 27, 2011

### tiny-tim

Last edited by a moderator: Apr 26, 2017
5. Aug 27, 2011

### Studiot

When you run a VDG there is certainly accumulation of charge in the ball.

There is also output charge due to leakage etc.

When you charge a capacitor from a direct source there is input, accumulation of charge on the capacitor and possible some leakage output, along with the displacement current.
As charging proceeds the input diminishes and so does the displacement current.
Eventually a steady state is reached where the charge on the capacitor remains fixed (accumulation =0), the input has fallen to (near) zero as has the displacement current.
In that circumstance you can say

input = ouput (leakage only).

KCL is derived on the explicit assumption that there can be no charge accumulation at a point and so can only be applied where this is true.

6. Aug 27, 2011

### tiny-tim

But KCL does apply to a capacitor (which has charge accumulation) if we include electric displacement current?

7. Aug 27, 2011

### Studiot

The ac version, yes.
There is no accumulation at ac.

8. Aug 28, 2011

### Red_CCF

Hi, thanks for the response

Just to confirm, given that an Op Amp does not have any component where charge can accumulate, KCL would apply to an Op Amp?

Just wondering in the link, they made a reference to the ""facing" sides of the two capacitors" for two capacitor in series. I'm confused which sides they're talking about, is it the side of the capacitor that faces the other capacitor?

Thanks

Last edited by a moderator: Apr 26, 2017
9. Aug 28, 2011

### tiny-tim

Hi Red_CCF!
hmm
For two capacitors in series (that is, where the charge between them is "stuck", with nowhere else to go), the charge on the "facing" sides of the two capacitors must be equal and opposite, and so the charge across both capacitors must be equal ($Q_1\,=\,Q_2$).​

… yes it is a bit unclear, isn't it?

let's change it!

what do you think of …
For two capacitors in series (that is, where the charge between them is "stuck", with nowhere else to go), the charge on the "between" sides of the two capacitors must be equal and opposite, and so the charge across both capacitors must be equal ($Q_1\,=\,Q_2$).​

… or do you think it needs to be a bit longer?

thanks for pointing it out!

10. Aug 28, 2011

### Studiot

Good morning Red,

I am not used to the analysis presented in your book for op amps, perhaps you could post more detail, including a circuit of the application analysed?

Here is the version I am more used to which, as you say, considers KCL at two points.

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11. Aug 28, 2011

### Red_CCF

Hi

My question is a bit more general in that it includes the power sources. I attached the circuit for one of them.

Basically they did a "global" KCL equation for the whole Op Amp. They defined all five currents (inverting, non-inverting, output, and two voltage sources) associated with the Op Amp as positive entering the Op Amp and stated:

i_p_ + i_n_ + i_o_ + i_c+_ + i_c-_ = 0; basically sum of currents entering Op Amp is 0.

I was just wondering if KCL is valid for this case since in several books I consulted, they all say KCL is valid for a node but did not mention anything about for a component, but what tiny-tim said also makes sense given that an Op-Amp has no source of charge accumulation so based on charge conservation the currents should add to 0.

Thanks

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12. Aug 28, 2011

### Red_CCF

Hi

I personally think a simple diagram would be much better than a description; I'm still a bit confused on what "between side" is.