Why a source which has high source impedance modeled in general as current source?
Suppose you have an idealized voltage source with a resistance in series with the output (we'll call this the output resistance).

Now if you place a wide variety of load resistances across this source+output resistance, then the output resistance will form a divider with the load resistance. What happens if...

...the output resistance is very high compared to all the load resistances?

The total resistance in the circuit doesn't change very much no matter what load resistance you use since it is negligible compared to the output resistance. That means the current doesn't change very much with varying loads so it is more convenient to model as a current source.

On the other hand, the voltage across the load resistance changes a moderate amount. Double the load resistance and you can double the load voltage since the current stays fairly constant.

...if the output resistance is very low compared to all the load resistances?

The total resistance in the circuit now almost entirely depends on the load resistance since the output resistance is now negligible. Unlike before, now it is the voltage across the load that remains fairly constant. So in this case it is more convenient to model it as a voltage source.

And unlike before, the current now has moderate changes with load. Double the load resistance and you get about half the current.



So you can see here that we even though we were actually using an idealized voltage source, it was sometimes more convenient to model it as a current source. In the same way, it is sometimes more convenient to model a current source as a voltage source. In the current source's case the output resistance is in parallel rather than in series.

So let's start all over again with an ideal current source with a parallel output resistance and parallel load resistance.

Output Resistance >> Load Resistance
The load resistance dominates since the much smaller resistance always dominates for parallel resistors. Very little current bleeds through the output resistance. Most of the current flows through the load resistance. Therefore, the current through the load resistance is kept relatively constant by the source while the voltage across it will vary moderately as the load varies. Therefore, tt is more convenient to model it as a current source.

Output Resistance &lt;&lt; Load Resistance
The output resistance dominates. Changes in the load resistance will have little effect on the overall current output of the source. Most of the current flows through the output resistance so the output resistance determines most of the output voltage. In this case, it is more convenient to model it as a voltage source.



So you can see here it didn't really matter if it was actually a voltage source or a current source. What really determined the more convenient model was the relative magnitude of the load resistance and source output resistance.



In discrete transistor amplifiers, that's why they use high value resistors in series with the a voltage source. It's meant to approximate a current source since the load presented by the transistor is supposed to be much less than that presented by the high value resistor (R1).



Taken from: https://www.electronics-notes.com/articles/analogue_circuits/transistor/long-tailed-pair-circuit.php

In integrated circuits they just build an actual current source out of more transistors.


Taken from: https://www.researchgate.net/post/Can_we_present_the_long-tailed_pair_as_a_bridge_circuit