In The Art of Electronics by Horowitz & Hill, there’s a section about current mirrors with multiple outputs. We're then given the following circuits:

enter image description here

It's then described how the circuit shown in Figure 2.49 doesn't compensate for current hogging, since if load 1 was removed, the corresponding transistor will saturate and pull more current than the other transistors connected to the common reference base.

The solution to this is then given via Figure 2.50, where compensation for current hogging is provided via an extra transistor connected at the collector of the "programming transistor" on the left. This extra transistor effectively pulls the voltage of the programming transistor's collector down to 2 diode drops below \$V_{CC}\$, moving it further into the forward active region of operation.

What confuses me is this: if you remove load 1 from Figure 2.50, the corresponding transistor will still be saturated there, just as in Figure 2.49. With that said, how does the extra transistor resolve this?


1 Answer 1


The fact that the current defining transistor's Collector is moved out of saturation is irrelevant.
Think of the added transistor as an Emitter follower increasing the current capability of the reference rail at the given Vbe voltage (It's now a voltage reference rail).
You might find this excellent reference a valuable read.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.