Hi I've attached this diff. amp. circuit schematic. Normally there would be an ideal current source connected between the Re terminals for biasing purpose. I think that this current source is specifically neglected for this case because an ideal current source's output resistance would've be infinite and so since the subject is related to finding Rid it is neglected. Is this the true reason of why it is neglected? What I dont understand is that why an ideal current source's output resistance would need to be so high? Well as far as I know it is high so that the the voltage change at the current source won't cause a huge change on the induced current, but why?

 

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The circuit should show the Rc resistors going to Vcc, not ground if you are using npn transistors as shown.
At the junction of the two Re resistors there would normally be either a resistor (call it Rcc) or a constant current source.
The CMRR of this circuit is approximately (Re+re)/Rcc. Small values of Rcc make for poor CMRR while an infinite resistance value value, like in an ideal current source would give you pretty good CMRR.

 

the source are neglected ie voltage source short current source open. Actually the topic that I'am having trouble with is the current source. Why would a current source have an infinite output resistance?

 

Why would a current source have an infinite output resistance?

The answer is very simple.
When we have ac signal there is a change in Voltage but there is no corresponding change in current because this is a ideal current source.
Here you have the example

 

The figure in the OP's first post looks like a scan from a textbook showing an AC model of a differential pair. In AC models DC voltage sources are shorted to ground, and DC current sources are opened.

That's why the figure correctly shows current flowing into the emitter and out the base of Q2. In the full circuit there is a current sink at the junction of the resistors so for every mA of increased current in Q1 there is a corrosponding decrease of current in Q2.

 

The answer is very simple.
When we have ac signal there is a change in Voltage but there is no corresponding change in current because this is a ideal current source.
Here you have the example

well it is an interesting example. firstly thnx for it. secondly looking at the parallel connected resistance, would it represent the output impedance of the current source? Well if this resistance in your example would be, say couple of Mega ohms, the current I1 would be stable. So is this what they mean about that the ideal current source would have very large output impedance?

 

looking at the parallel connected resistance, would it represent the output impedance of the current source?

Yes, R1 represent the output impedance of the current source.

Well if this resistance in your example would be, say couple of Mega ohms, the current I1 would be stable. So is this what they mean about that the ideal current source would have very large output impedance?

Yes, But the first example show ideal case.
When the output impedance of the current source is equal to infinity.

 

Well than looking to the image that I've attached, all dc sources are turned to 0. So voltage source short and the current source needed to be open. But here the although its current source is left open, its output impedance is also left where it was, i.e not neglected. Firstly in differential amplifier why finite output impedances of current sources started to be considered instead of takinkg them infinite and simply negelct them? Secondly we left open the current source, why not also cutting of its output impedance (the resistor REE) also, instead in the schematic that I've posted it still floats at the emitter terminal where the current source was located before the dc sources were turned to 0.

 

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We take into account output impedance of the current source, if we want more real life example.