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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. enter image description here 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?

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  • \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Oct 12, 2020 at 8:36
  • \$\begingroup\$ 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. \$\endgroup\$
    – nuggethead
    Oct 12, 2020 at 18:42

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Often I may (at least initially) overdesign biasing circuit, so they become DONT CARE parts of the overall behavior.

Then, later, if power is a big deal, that bias generator may be optimized, while acknowledging I now have a Beta_dependent (and thus randomly_selected device dependent) region in the circuit.

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  • \$\begingroup\$ So then, analogsystemsrf, there really isn't a compelling reason to have so much current available? \$\endgroup\$
    – nuggethead
    Oct 12, 2020 at 9:49
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This is about making the voltage reference stiff i.e. not susceptible to changes in factors such as supply voltage, temperature etc. The designer wants sufficient current to flow through the diodes to ensure a stable voltage across the diode in varying conditions (supply voltage, temperature, beta, ...) as this voltage is used to program the current mirror.

As with all designs there is a compromise, in this case the power lost maintaining a stiff reference. 4mA through the diode does seem high to me but 40uA seems way to low even though it implies a reasonable sounding beta of 100. If I were building this I would look at something like 1/10 of the mirror current or about 0.5mA roughly corresponding to a 27k resistor in the diode branch.

You might find it interesting to calculate the variations in diode voltage for 5% variation in current at 40uA, 0.5mA and 4mA as an exercise.

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