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I was going through data sheets of op amps and saw that bias currents usually add unwanted voltages due to voltages generated from the bias current flowing through the source impedances.

What exactly does the source impedance mean ? I understand what impedance is, but since there could be a lot of circuitry before the input to the op amp, do we just use normal laws to calculate the total resistance/impedance before it(input) ?

There also might just be a power supply providing constant DC voltage , how are we supposed to calculate the impedance of that ?

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The device that supplies a voltage signal or a dc power voltage is the source. The source impedance is the internal resistor in series with that otherwise perfect source voltage.

A battery for instance may have a source impedance of about 0.1 ohms. See this link to Duracell's website and click on the pdf for an AA battery: -

enter image description here

It's the same for linear voltage regulators, microphones and anything that can generate an AC or DC voltage - there will be some internal resistance that may have to be taken into account when the device is connected to a load.

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I was going through data sheets of op amps and saw that bias currents usually add unwanted voltages due to voltages generated from the bias current flowing through the source impedances.

Bias currents are the dc currents for biasing the input differential stage of the opamp. Because these currents are very small (<1µA) these currents, in many cases, are neglected during calculation of the feedback network. However, there are some cases requiring rather large resistors (some hundreds of kOhms) which cause a dc voltage drop that might be unacceptable. (It acts like a dc offset and causes a dc shift of the output operational point).

In those cases, we can make something like balancing/compensating these unwanted dc voltages at the input. That means: We try to make the dc voltage drops equal at both opamp input terminals.

Example: Inverting amplifier with a feedback resistive chain 500k-1k (gain of -500). In this case, we could ground the non-inv. opamp terminal via an additional compensating resistor Rc of app. (1k||500k~1k).

In this example, it was assumed that the signal source has an internal source resistance Rs=0. If this is not the case, this resistance Rs has to be taken into account (in series to the 1k resistor) - for computing the gain value as well as for finding the proper compensation resistor Rc.

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You can calculate the approximate impedance of the source if not exact. Here you go. Connect a load across the battery.Measure the drop across the load. Get the difference between the defined input voltage and the actual voltage across the load. Divide the difference by current. That is your source impedance.

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