Electronic – How does the compensation resistor in an inverting amplifier compensate for the input bias current


I can't see how a resistor placed on a whole other input terminal has anything to do with compensating this input bias. i.e. the input stages are just the gates of transistors of differential stage. In this circuit, one of them is completely connected to ground and has no current to flow FROM or TO the op-amp.

This is saying that the non-inverting transistor has a current flowing from its collector to its gate. (At least the way I understand it.)

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The most common compensation involves adding a resistor [R3] to the standard inverting amplifier to cancel out bias currents [output offset].
The compensation resistor [R3] causes a current, on the positive terminal, equal and opposite to current flowing into the negative terminal.
So any DC output offset caused by the inverting input is cancelled by the non-inverting input.
The value of R3 should be equal to the parallel combination of R1 [Ri] and R2 [Rf].

Best Answer

Sample Bipolar Schematic and Behavioral Description

Let's look at the LM324. It's a bipolar opamp and it is also a lot easier to follow that some. But it is still fairly representative of the basic ideas related to your question:


simulate this circuit – Schematic created using CircuitLab

You asked about the diff-pair being either NPN or PNP. In this case, it's PNP. It uses a Darlington PNP arrangement, more specifically, with four transistors; \$Q_1\to Q_4\$. The current source, \$I_1\$, is nominally split evenly between the two tail currents (the collectors of \$Q_2\$ and \$Q_3\$, specifically.) So, if the inputs are of equal voltage, we'd expect that there would be \$3\:\mu\text{A}\$ in each tail.

The tail currents are driven into a current mirror, formed from \$Q_8\$ and \$Q_9\$, which means that any current difference will be either be driven outward or sucked inward via the path to the base of \$Q_{10}\$. If the (-) input is lower than the (+) input, then more current is pulled towards the left tail and less current towards the right tail. The diff-pair and current-mirror sections respond to this by sinking the difference as base current via \$Q_{10}\$. This is a very high gain operation and it results in \$Q_{10}\$ pulling its emitter closer to its collector (which is at ground.) That pulls down on \$Q_{11}\$ and therefore also \$Q_{12}\$, causing \$Q_{12}\$ to release its collector a fair bit, allowing the bases of \$Q_5\$ and \$Q_{13}\$ to rise upwards. \$Q_5\$ will soak up current from \$I_3\$ so that the Darlington pair of \$Q_5\$ and \$Q_6\$ will pull their emitters higher, thus raising \$V_\text{OUT}\$.

The overall effect of this is that when the (+) input rises upward with respect to the (-) input, the output rises in response. Which is exactly the desired response.

There is up to about \$100\:\mu\text{A}\$ available in \$I_3\$, of which about half or \$50\:\mu\text{A}\$, is sunk via \$I_4\$. So there will be at most about \$50\:\mu\text{A}\$ available at the base of \$Q_5\$. Given the usual worst-case \$\beta\$ estimates, say \$\beta=40\$ or so, this suggests perhaps a maximum sourcing capability of \$40^2\cdot 50\:\mu\text{A}\approx 80\:\text{mA}\$. The specification says that it is at least \$20\:\text{mA}\$ and typically \$40\:\text{mA}\$, without stating a maximum, which is well-reasoned I think as specifications go.

Some base recombination current is required by \$Q_1\$ and \$Q_4\$. It's modest, because \$I_1\$ isn't a large current. So, nominally, only \$3\:\mu\text{A}\$ is flowing in each tail. Given the Darlington arrangement, the base currents will be on the order of \$1600\times\$ smaller (though we may suggest as little as \$400\times\$ smaller as a conservative limit.) From this, we might suggest at worst, base currents of about \$10\:\text{nA}\$. The specification sheet says that the worst cases are a bit more. But not much more. The reason for this is that they want to deal with cases where the voltage differences are somewhat larger than normal, where one side or the other is moving into saturation mode. So this also is perfectly reasonable.

There is a side-bar worthy of note. Since \$Q_{12}\$'s emitter is at ground, the base of \$Q_{11}\$ is about two \$V_\text{BE}\$'s above ground. That means that the base of \$Q_{10}\$ is about one \$V_\text{BE}\$ above ground. That means the collector of \$Q_{9}\$ is at the same place as the collector of \$Q_{8}\$. And this helps nullify the Early effect that might otherwise be a problem in \$Q_{10}\$. Another good design decision in this circuit. (\$C_\text{C}\$ is a Miller capacitance arranged to set a dominate pole position. Beyond the scope here.)

All of this is just a few very basic circuit concepts and you should make sure, in your own mind, that all of this makes good sense.

Base Currents of \$Q_1\$ and \$Q_4\$

So now we are here. All you have to do is realize that from the circuits there needs to be at least some small base currents in \$Q_1\$ and \$Q_4\$ that are sunk externally towards ground. If you tie one of the bases to ground with a \$10\:\text{k}\Omega\$ resistor and the other input to ground with a \$100\:\text{k}\Omega\$ resistor, then you must realize that there will be a similar, needed bias current so that the diff-pair BJTs can remain in active mode (where they need to be.)

Nominally, in this case, with the base currents about the same but where those currents must be sunk through resistors with values that are an order of magnitude different, it must be the case that there is a small voltage difference at the bases of \$Q_1\$ and \$Q_4\$. Since the emitters of \$Q_2\$ and \$Q_3\$ are tied together, this will mean that the voltage difference results in an exponential difference in tail currents. And that will translate into an output voltage that is offset from nominal by some rather high trans-impedance gain. Feedback can help correct that error, of course. But it's an avoidable problem. So you should avoid it.