Why is the Input Current of Ideal Op-Amps Always Zero?

[Swapnil]

Hi, I was wondering why is it that the input current for an ideal op-amp is always zero? They say that it is zero because of infinite impedence but where exactly is this impedence? Across the op-amp?

 

[FrogPad]

Hi, I was wondering why is it that the input current for an ideal op-amp is always zero? They say that it is because of infinite impedence but where exactly is this impedence? Across the op-amp?

The input currents of a non-ideal op-amp are not exactly zero. They are close to zero though. My book (Sedra/Smith) says, typical input bias current values for general-purpose op amps that use bipolar transistors are I_B= 100nA

Where:
I_B = \frac{I_{B1}+I_{B2}}{2}

IB1 and IB2 are the currents that you see going into the (-) and (+) terminals.

It is helpful to think of the impedence as being zero in a ideal op amp, because this allows you to perform speedy circuit analysis. You can apply the non ideal conditions later.

Imagine you have the following:************R1
*******------/\/\/-----(n2)
*******|
*******|
(n1)-----=----(n3)

Lets say you connect the positve terminal of a 9-volt battery to n1, and the negative terminal to ground. Now let's say you connect a resistor R1 to the n2 terminal, and the other side of the resistor to ground.

Do you see how a current will flow through the battery through the n1 terminal, it will then hit the 90 degree angle (I marked it with a equals sign), and then it will flow up into the resistor marked R1, and then down through the load we connected into R1 and then to ground. Right?

Well what is the current from the equals sign to (n3)?

It would be zero right?
Notice that this would be the same as saying.

************R1
*******------/\/\/-----(n2)
*******|
*******|
(n1)-----=----/\/\/----(n3)----ground
************Ri

Where Ri = infinite resistance

it would be so nice if there was a LaTeX package for drawing circuits...

 

[Swapnil]

Thanks for the comprehensive reply but I am a little confused about the drawing. Do n1, n3 and n2 stand for Non-Inverting Input Current, Inverting Input Current and Output Current respectively?

 

[FrogPad]

woops. I realized I never finished this post.

n1, n2, and n3 just stand for node 1, node 2, node 3.

What I wanted to end up showing you was this,


*******|---------\
*******|********\
*******|*********\
(+ )-----=----(n1)****\
*******************\______()
*******************/
(- )-----=----(n2)****/
*******|*********/
*******|********/
*******|---------/


So n1, n2 stand for node 1 and node 2 again. Now apply the same logic that I showed above to the branch at node 1, and node 2. Think about this branch as being connected to a resistor, and then to ground. If that resistor is has infinite resistance, then it is going to act like an open circuit and you will see zero current in that branch.

Just remember that when you add a little triangle (the circuit symbol for an op amp) and use the non-ideal analysis techniques, you are using a model of a real device. The real device doesn't fit the-non ideal characteristics. However, because of the physics of the device we are able to think of the inputs inside the amp as being open, and thus zero current flows into it. Inside the op-amp are a bunch of transistors that create these properties. Like I said above though, you will see some current being drawn at these branches (and this is specified in the data sheet for the amp), but the amount of current is negligible in most cases. So the non-ideal analysis is usually sufficient.

This is actually what I'm studying right now (undergrad), so I'll do my best to answer. Like always, the more experienced people around here can fill in the gaps.

 

[FrogPad]

PS - The "lovely" ASCII art above is supposed to be an op amp :)