I just encountered this circuit and I'm a bit confused by it

enter image description here

I see that it's an NMOS current mirror. At first I thought it's a cascode current mirror due to M3 on the right but it isn't. I have two questions

  1. What is the point of M3 in this circuit? I mean it's going to have some factor times Iref through it?
  2. What is the point of R1? I mean surely I_REF will push that much current through it no matter what so?

If somebody could just give me an insight on how it works, that would be great.

  • 2
    \$\begingroup\$ As for R1 read this wiki.analog.com/university/courses/electronics/text/… And for M3 do you know what channel length modulation is? Basically, M3 eliminates channel length modulation in the M2 transistor. \$\endgroup\$
    – G36
    Commented Nov 8, 2018 at 16:45
  • \$\begingroup\$ Yes, I know channel-length modulation. How exactly is M3 doing that? Also is M3 and M2 together not just a cascode current source? \$\endgroup\$ Commented Nov 8, 2018 at 22:14

2 Answers 2


I believe both the resistor and the cascode are techniques for keeping \$v_{DS}\$ constant to avoid short-channel effects.

Resistor \$R_1\$

A regular diode-connected transistor (\$R_1 = 0\$) will also change \$v_{DS1}\$. Increasing \$I_{REF}\$ will make \$v_{GS1}\$ larger, which in turn also causes \$v_{DS1}\$ to increase as they are short-circuited.

If \$R_1 > 0\$, then as \$v_{GS1}\$ increases with \$I_{REF}\$, \$R_1\$ will push \$v_{DS1}\$ down compared to \$v_{GS1}\$. If you choose \$R_1\$ carefully you can make it compensate the change of \$v_{GS1}\$ such that \$v_{DS1}\$ remains approximately constant.

Cascode M3

Cascodes are a technique that employ negative feedback to keep the current constant.

If we try decreasing \$v_{DS2}\$ by sinking extra current to ground, then \$v_{GS3}\$ increases which makes M3 conduct more current. As M3 injects more current into the node between M2 and M3, it counteracts the decrease of \$v_{DS2}\$ (negative feedback).

\$v_{DS2}\$ variation is reduced by M3, reducing short channel effects.

  • \$\begingroup\$ Very nice explanation. What exactly do you mean by "sinking extra current to ground"? \$\endgroup\$ Commented Nov 9, 2018 at 11:15
  • \$\begingroup\$ Can be anything really, like M2 conducting more current or adding another path through which current can flow out of that node. \$\endgroup\$
    – Sven B
    Commented Nov 9, 2018 at 12:32

Yes. It is just a cascode current mirror that:

  • uses a separate voltage bias source for the cascoding transistor.
  • Reduces the Vds of the bias transistor with a resistor.

Both choices actually make it worse than a standard cascode, as the channel length modulation might not simply match on both sides of the mirror.

It seems to me more of a homework exercise than a practical circuit.


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