The picure below is a improved current mirror. There are two equal input currents Iin and an output Iout. M1 converts Iin into BIASN2 to bias transistor M4B and M3A converts Iin into BIASN to bias transistor M4A.

So what is the role of transistor M3B here? It is a cascoded transistor and by adding it, the output impedance increases. However, I don't see why we want to increase output impedance at this branch.

enter image description here

  • \$\begingroup\$ What is the output impedance of an ideal current source (or sink as in this case)? \$\endgroup\$ Oct 7, 2016 at 5:56
  • \$\begingroup\$ It is not clear but I assumed it is infinity. \$\endgroup\$
    – emnha
    Oct 7, 2016 at 5:58
  • \$\begingroup\$ So the higher the impedance the closer to ideal, no? \$\endgroup\$ Oct 7, 2016 at 5:59
  • \$\begingroup\$ Oh, I misunderstood your question. I thought that you asked for the output impedance of input current Iin. For your question above, an ideal current source has infinite output resistance. However, what I don't understand is the branch with transistor M3B. This branch is not a current source. It is only to create bias voltages for the current source on the right. So what is the role of transistor M3B here? \$\endgroup\$
    – emnha
    Oct 7, 2016 at 6:08
  • \$\begingroup\$ I think to generate BIASN2. \$\endgroup\$ Oct 7, 2016 at 7:26

1 Answer 1


The transistor M1 will generate the bias voltage for the cascoded current source and has no other function.

The following schematic illustrates the point:


simulate this circuit – Schematic created using CircuitLab

In order to have accurate current mirroring, you would want to have:

  1. The same \$V_{GS}\$, as is the case with all current mirrors
  2. The same \$V_{DS}\$, which is only the case on the right
  3. Matching transistors (layout level)

Adding M3B will improve point 2, by also working on the drain of M5 (M3A in your reference image).


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