The goal of this exercise is to study the effects of a modified differential amplifier that reduces the bias current that's shown in the attached image. The first part of the exercise is to calculate by how much the bias current is reduced with the addition of the Q7-Q8 pair. I found this current to be

\(I_B = \frac{2\beta+1}{(\beta+1)^3}\.\times\.\frac{I}{2}\)

which results in a reduction of

\( \frac{2\beta+1}{(\beta+1)^2}\approx\frac{2}{\beta}\)

Next I'm asked to calculate by how much the input resistance changed, which I expected to be by a factor of \(\frac{\beta}{2}\), but looking at the circuit, it seems that the input resistance \(R_{ib}=2r_{\pi}\) is in parallel with the output resistance of the current mirror, which is much larger. Thus, I concluded that the input resistance was unchanged, so \(R_{ib}\approx2r_{\pi}\). This contradicted what I expected, so I decided to do some quick small signal analysis and found

\(i_i = \frac{\beta^3+2\beta^2+\beta^3+1}{\beta^3+2\beta^2+\beta^3+2}i_{b1}\approx i_{b1}\)

Which leads to the same conclusion regarding the input resistance. So my question is, are these two approaches wrong, or am I wrong in assuming that the input resistance should increase just because it draws less current.

I did some simulations and the input resistance did increase, but since I'm not very good at using SPICE, I might have screwed something up.

 

Sorry for the double-post, but I can't edit the original one anymore! The reason why I think my assumption is wrong is because the bias current is a DC value, while the input resistance is dependent on AC ones. If this is true, then I ask what are the advantages of this circuit, except for drawing less current, and thus power, from its input stage

 

The input impedance will not change, if we ignore Q8 output resistance. This circuit only reduces input DC bias current. Which is always good, because the ideal op amp input bias current is 0A.

 

Aha, didn't think about this as the input stage to an op amp, but it makes sense and now I see the advantage. Thanks for clearing this up!

 

We never use this "trick" in discrete BJT circuits due to a larger mismatch between individual BJT's.

 

Seems sensible, but I guess that goes beyond the scope of the exercise itself, but still, that's good information to know. If there's a thing theoretical exercises often lack it's how to deal with non-ideal behaviours, but like I said, it's beyond the point.