# Current source resistor values in differential amplifier

In this circuit How measure input impedance this differential amplifier in LTspice? the OP has used a 4ma current source in a differential amplifier. I'm taking it apart to learn how it works. I see that the diodes will give a pretty reliable 1.4v drop, and the transistor will give a 0.7v drop, which allows us to use a single 180 resistor to set the desired current to approx. 4ma. This makes sense to me and checks out in a simulation. Below is that portion of the circuit. Of course, the collector in my schematic would connect to the emitter resistors in the diff. amp. What I'm wondering about is the choice of a 3.3k resistor at the base. It seems to me - as a novice - a waste of energy, as it sees a rather large current of 4ma. Since the 3.3k resistor will always drop around 13.6v, couldn't its value be increased significantly to reduce how much current it has to handle? As far as I can tell, the base only requires a small fraction of this 4ma, say around 40uA, sending 99% of that current through the diodes. So my question is, how should one size this resistor biasing the base in a current mirror? Is there a minimum required current to get the diodes to drop 0.7v? Is there some compelling reason to set it at a particular value?

• The circuit you linked is one generated by another OP. There is no guarantee at all that what he posted has been honed to perfection so why don't you ask him. I see no reason to have 4mA in the 3k3 resistor but I also see no reason for this question as it stands because you should address your concern to the OP in the other question. Oct 12, 2020 at 8:36
• Thanks andyaka, I'll ask it there also. It was a post from several years ago and I didn't expect the OP to still be active. Oct 12, 2020 at 18:42