Instrumentation amplifier input bias currents direction and error

Question

I am trying to understand how the input bias currents flow in op amp and instrumentation amplifier circuits and what errors they create. I came across a circuit that looks like the following:

^{simulate this circuit – Schematic created using CircuitLab}

My questions are:

1. If I assume the bias currents flow into the terminals, which way do the bias currents go? For the positive terminal I would assume that the bias current flows from V1 through R1. But what about the negative terminal? Does the bias current flow from ground to the negative input?
2. What errors do the bias currents create? Is it only the voltage drop across R1 that I need to consider?

Answer 1

If I assume the bias currents flow into the terminals, which way do the bias currents go?

Bias currents will flow like this: -

For the positive terminal I would assume that the bias current flows from V1 through R1.

The bias current will split into both paths; R1 and R2.

But what about the negative terminal? Does the bias current flow from ground to the negative input?

Yes it does.

What errors do the bias currents create? Is it only the voltage drop across R1 that I need to consider?

They can create a differential offset voltage that gets amplified by the amplifier and no, you need to consider the volt drops across R1 and R2 together. To make it easier, you can assume the input voltage is zero then, R1 is in parallel with R2 so, the resistive paths on both inputs are clearly different and the bias current will create a differential offset voltage that produces an error.

Answer 2

1. It depends on the amplifier's input structure: see the data sheet. Bias current refers to the average current into the inputs, so by definition it flows in the same direction for both. The difference is called "offset current".

2. It produces a voltage offset based on the difference between the resistances the two inputs see. In your case the (-) sees 1k, the (+) sees 1k || 10k ~ 9.09k, a difference of about 90 ohms (assuming perfect resistors). For a max +-2.5 nA, that'll give you an offset due to bias of ~230 nV. You'll never notice that.

Answer 3

The full path (loop) of the input bias current in the simplified circuit of an op-amp differential input stage below is: the positive terminal of the negative power supply V- >>> ground >>> input voltage source >>> base-emitter junction >>> emitter current source >>> the negative terminal of V-.

^{simulate this circuit – Schematic created using CircuitLab}

Why did the circuit designers decide to resort to this non-standard solution of closing the biasing current path through the input source? The answer is obvious - in this way they achieved the maximum possible op-amp input resistance because nothing shunts the input.

See more in the links below pointing to related answers of mine:

Why does this op-amp circuit require a resistor to ground?

Why do we need R1 in a non-inverting op-amp?

Need clarification regarding op-amp's ground/ reference and common mode dependence on supply

BJT Differential Amplifier problem

Why put a resistor in series with the positive terminal in a non inverting op-amp circuit?

Designing a differential amplifier

How does the current and voltage stabilize in a differential amplifier?

Un-biasing transistors

See also my question:

Where do input bias currents flow and what voltage drops do they create?

and my answer to it:

Answer 4

In practice for a single supply amplifier, if the input is GND, the current will flown out from the input. However this doesn't affect the design goal - make the effective resistance at each input equal.

Note this can be achieved by just adding 10 kΩ in parallel with R3 on the example.

In some systems, the added resistors affect slightly the AC performance, and a small C may be added to bypass the DC resistor.