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I was trying to understand the non-ideality of op-amps, and specifically how the input bias current is modeled. The image was taken from a nice Youtube tutorial (Mateo Aboy - DC Offsets).

The input bias current (\$ I_{b+} \$ and \$ I_{b-} \$) are the undesired currents generated from within the op-amp (i.e. a base current or a leakage gate current heading to the first transistor). The op-amp has to have these current inside the op-amp internal circuitry. But when \$ I_{b+} \$ and \$ I_{b-} \$ are modeled with superposition, these bias currents flow through the compensation resistor \$ R_C \$ instead, and none into the op-amp as if the op-amp is ideal.

I can't make sense out of this model: the input bias currents are the currents flowing into (or out of) the inverting or non-inverting input, and there should be a (theoretically) measurable current at the input terminal. There should be a \$ I_{b+} \$ entering a non-inverting input, and a bigger current \$ I_{total} - I_{b+} \$ flowing through \$ R_C \$ into ground (i.e. current divider).

How can one assume \$ I_{b+} \$ existed internal within the op-amp but when modeling the circuit assume it all flows through \$ R_C \$?

z5


[EDIT]

Shouldn't the input bias current shown in the above circuit (b) be modeled a bit more like the following circuit? The D1 diode is an idealized component that takes zero voltage drop and functions to restrict the direction of the current to only flow into the non-inverting input (hence, input bias current)? Otherwise, if without the diode, we'll run into the dilemma where we're modeling an input bias current that does not flow into the input - make no sense.

enter image description here

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I think this is just a visualization problem. The current sources should technically be drawn within the opamp triangle and not outside. Refer to the dotted yellow triangle in the top-left corner of your original image, which gives you a better idea of how you're supposed to interpret it. The inner black triangle doesn't really have "inputs", but its [+] and [-] pins are just the control nodes for the dependent voltage source of the ideal opamp model's output. If you substitute the black triangle for what it really is, then you're left with the outer yellow one, as shown in the very quickly slapped together MSPaint image below. The yellow [+] and [-] pins are the actual inputs and where the bias currents flow into/out-from.

enter image description here

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Here's a simple opamp input stage with 2 bipolar transistors (I didn't bother drawing the rest of the opamp). The bases of these transistors are the two inputs of the opamp.

enter image description here

So, bias and offset input current, which is the base current for these transistors, always travels in a loop, like all currents do. Since it's a loop, it has no "start" and "end" points, but it helps to always start with ground. So, the loop is

Ground

-> through caps and power supply into your power supply +/-15V rails

-> into the opamp via the power supply pins

-> through the internal circuitry

-> out of the input transistors bases

-> through Rc

-> and back to ground.

The ideal opamp model doesn't have power supply pins, so it does NOT respect Kirchhoff's circuit laws : the output current (and input bias/offset currents) just magically appear out of nowhere. I think that explains your puzzlement. But in a real opamp, of course the input and output currents loop through the power supplies.

Note that if you use SPICE to simulate a circuit, many opamp models will cheat and reference ground internally, which means the simulated current it draws from the power supplies will be incorrect, among other problems.

there should be a (theoretically) measurable current at the input terminal.

Yes, you can measure it. It's tiny, so you need a high value resistor. A simple way is to set the opamp to a high gain, use two different values for Rc and measure the change in output offset voltage. So you use the opamp to amplify its own input bias current.

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  • \$\begingroup\$ I don't see how that answers my question though which was essentially asking why \$ I_{b+} \$ loops through \$ R_C \$ and no current goes into the non-inverting input in the model. If there's no current flowing into the input, then there is zero input bias current. The model is self-contradictory. \$\endgroup\$
    – KMC
    Apr 22 at 1:19
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    \$\begingroup\$ @KMC I think this is just a visualization problem. The current sources should technically be drawn within the opamp triangle and not outside. See the dotted yellow triangle in your original image. That's how you're supposed to look at it, where the inner black triangle doesn't technically have "inputs", but its (+) and (-) are just the control nodes for the controlled voltage source of the ideal opamp model's output. If you substitute the black triangle for all that it is, then you're left with the outer yellow one, as in this image: i.stack.imgur.com/MkKbY.png \$\endgroup\$
    – Ste Kulov
    Apr 22 at 5:10
  • \$\begingroup\$ @SteKulov ...omg... your comment and image clears everything up for me. While bobflux's has some add in info, your comment answers (or more precisely "fixes") my question. If you can kindly move this comment to a proper answer I'll accept it \$\endgroup\$
    – KMC
    Apr 22 at 5:47
  • \$\begingroup\$ @KMC Ehhh....ok. I'll try to do that later today. \$\endgroup\$
    – Ste Kulov
    Apr 23 at 15:37
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Your questions and reasoning are very interesting and reasonable... but the only way to find the answer to them is to "open" a simple transistor differential amplifier stage (the so-called "long-tailed pair") and see where the currents flow.

I have done it in my question, Where do input bias currents flow and what voltage drops do they create? and in my comprehensive answer to this question.

Here is one of the pictures that shows the paths of the input bias currents and the voltage drops across the (compensating) resistors:

Differential pair with RB1 and RB2

Just to clarify that there are power supplies only on the right. The ones on the left (in pale gray) I drew only for symmetry; they are the same sources on the right.

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