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I have seen current mirrors being used as current source in differential stage of an amplifier design.

The reason given is that the current mirror has a very high output impedance and behaves like an active load. I am not sure what this means.

Why do we need to use a current mirror in differential stage of an op-amp? I understand that it is to provide the same current for collectors of both transistors in the differential stage. However, I do not understand why we do this in the first place, why does current mirror have a high output impedance and what does the current mirror being used as an "active load" means here?

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    \$\begingroup\$ If you read the right columne of this page ("RELATED") you will find related posts (questions/answers) with some more "conventional" explanations. \$\endgroup\$ – LvW Mar 8 '15 at 13:47
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Here is a unconventional explanation (you can find many other conventional explanations on the web)...

The basic idea is to oppose two current sources (transistor collectors); so we have to connect them in the same loop and to vary their currents in opposite directions:

enter image description here

As a result the voltage of the common point between the sources will "move" extremely vigorously (a fully geometrical phenomenon):

enter image description here

A good example of this configuration is the classic CMOS stage where the two transistors are connected with their sources to Vdd and Vss (ground) and their drains are joined:

enter image description here

But in the emitter-coupled (long-tailed) pair, the collectors of both transistors (T1 and T2 below) are turned up. So, we should turn the one collector (the T1's in this example) back down to stand against the other (T2's) collector. For this purpose, we have connected the simple current mirror (T3-T4) in the picture below:

enter image description here

See also this paper and its more sophisticated version.

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