Understanding Op Amps and Kirchoff's Current Law

In summary: Yes, KCL is valid for a node. However, it is not valid for an Op Amp. :)In summary, KCL is valid for a node but not for an Op Amp.
  • #1
Red_CCF
532
0
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
 
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  • #2
Hi Red_CCF! :smile:
Red_CCF said:
… current flowing into an Op Amp equals current going out …

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 :wink:
 
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  • #3
Good morning Tiny Tim

What happened to

Input = Output plus Accumulation

eg a Van Der Graff generator ball?
 
  • #5
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
Studiot said:
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.

But KCL does apply to a capacitor (which has charge accumulation) if we include electric displacement current? :confused:
 
  • #7
But KCL does apply to a capacitor

The ac version, yes.
There is no accumulation at ac.
 
  • #8
Hi, thanks for the response

Studiot said:
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.

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


tiny-tim said:

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
 
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  • #9
Hi Red_CCF! :smile:
Red_CCF said:
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

hmm :confused:
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 ([itex]Q_1\,=\,Q_2[/itex]).​

… yes it is a bit unclear, isn't it? :redface:

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 ([itex]Q_1\,=\,Q_2[/itex]).​

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

thanks for pointing it out! :smile:
 
  • #10
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
Studiot said:
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.

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
tiny-tim said:
Hi Red_CCF! :smile:


hmm :confused:
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 ([itex]Q_1\,=\,Q_2[/itex]).​

… yes it is a bit unclear, isn't it? :redface:

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 ([itex]Q_1\,=\,Q_2[/itex]).​

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

thanks for pointing it out! :smile:

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.
 

1. What is an op amp and how does it work?

An op amp, short for operational amplifier, is an electronic device that amplifies a voltage signal. It has two inputs, an inverting (-) and a non-inverting (+) input, and one output. The output is the difference between the two inputs, multiplied by a very large gain. This allows the op amp to amplify even small input signals to a desired level.

2. How do I apply Kirchoff's Current Law to analyze op amp circuits?

Kirchoff's Current Law states that the sum of currents entering a node in a circuit must equal the sum of currents leaving that node. To apply this to op amp circuits, you must first identify all the nodes in the circuit and write down the equations for each node. Then, use Kirchoff's Current Law to solve for any unknown currents.

3. What are the common applications of op amps?

Op amps have a wide range of applications in electronic circuits. Some common uses include signal amplification, filter circuits, oscillators, and comparators. They are also used in audio and video equipment, sensors, and control systems.

4. What are some important considerations when choosing an op amp for a specific application?

When selecting an op amp, it is important to consider factors such as gain, bandwidth, input and output voltage ranges, and power supply requirements. Other important considerations may include noise levels, input impedance, and output impedance. It is also important to choose an op amp with the appropriate package and pin configuration for your circuit.

5. How can I troubleshoot and troubleshoot common problems in op amp circuits?

When troubleshooting op amp circuits, it is important to first check for any loose connections or incorrect wiring. If the circuit is properly connected, you can use a multimeter to check for proper voltage levels at various points in the circuit. Common problems in op amp circuits include oscillation, saturation, and noise. Referencing the op amp's datasheet and experimenting with different component values can help resolve these issues.

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