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I am not able to understand why the connection between gate and drain of the first MOS is required.

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A similar question to this is asked here but it is BJT based current mirror and the answer is there would be no path for base current: Why use diode connected BJT (and not just connect two BJT back-to-back) for current mirror circuit?

However, in MOS based mirror there is no base current. Then why does one need that connection? Shouldn't the MOSFET be able to decide its VGS based on input current?

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  • \$\begingroup\$ The equations work both ways. For a BJT, the Shockley equation relates collector current to base-emitter voltage. You can supply the collector current and get an emitter-base voltage. Wiring it up as a diode arranges things so that the current flooding the device will be only partially diverted to the base, with most of it as collector current. But the base-emitter voltage (with collector tracking) will set up whatever is needed. This voltage is then applied to the adjacent BJT, therefore controlling its collector current. (Looking at the Shockley eq. the other way.) \$\endgroup\$ Commented Oct 7, 2023 at 0:59

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  1. if you want to copy a current you should have a reference (diode connected MOS) enough robust to spread a good copy to other Mos

  2. the diode connected MOS should be always in saturation (active region) to maintain a fixed current (the diode connection ensure that behavior)

  3. small variation of Vds will not affect the diode connected MOS so the replica won’t be affected.

In summary, the diode-connected configuration in a current mirror is essential for maintaining a stable current and ensuring current matching.

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Especially with Ig = 0, the symmetry is even more apparent: if M1 is biased to some Id, and M1 has the same Vgs, and Vds > Id Rds(on) is true for both, then there you go, that's all you need, you've got it.

Comparing to the other answer, replace "if there weren't any path for base current into either transistor's base, then you wouldn't get any current in either the input or the output side" [paraphrased], with: a path to set gate voltage.

Which, as the answer goes on to discuss to varying extent, is what the BJT is doing. Base current is merely an artifact, more or less, while Ebers-Moll (or further more accurate) equation(s) applies much more strongly (i.e. consistent over decades of range, and within percent between matching transistors e.g. on die), and so setting Vbe is the important part.

The main difference is, because there is base current, disconnecting the C-B link means it stops in its tracks; the MOS version would just sit there where it was left off at (give or take capacitances, and leakage eventually).

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The current through the MOSFETs is the saturation current for a given gain/source voltage. The whole deal of the current mirror is to ensure that

a) the current through the first MOSFET is the desired current. That is achieved via IREF.

b) gate/source voltage of both MOSFETs is identical. That is achieved by connecting them.

c) both MOSFETs are saturated, admitting as much current as possible giving the gate voltage. For the left MOSFET, that is simply achieved by connecting gate and drain: if the gate/source voltage is too small to admit the current, the gate will rise in voltage with the drain until it isn't. If it is too large for saturation at the given current, voltage will fall until the channel is pinched off to saturation.

The gate would not need to be connected directly to the drain: that is just the most straightforward manner to regulate the MOSFET into saturation. And you need some kind of regulation.

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