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I was looking at a certain circuit in which there was the necessity of generating 3 DC current sources (10uA, 50uA, 250uA), starting from a given current source of 10uA. Let's consider firstly the circuit which has to generate 10uA (let's call it A).

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1st question: can you explain me how does it works, in an intuitive way? Why is the current I4 copied to the transistor M10 (i.e. M10 is a current mirror 1:1)?

Now let's consider the circuits which have to generate 50uA and 100uA (let's call them B and C):

enter image description here

They are connected to I4 through the label VBN, as for circuit A. So B is a current mirror 1:5, C is a current mirror 1: 25. This result is reached by inserting 5 transistors in the first one (molteplicity = 5 so there are 5 transistors in series) and 25 transistors in the second one (molteplicity = 25).

2nd question: can you explain me intuitively why the number of transistors is equal to the multiplication factor of the current?

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    \$\begingroup\$ FYI, the English word is "multiplicity". \$\endgroup\$ – The Photon Nov 12 '19 at 20:58
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Why is the current I4 copied to the transistor M10 (i.e. M10 is a current mirror 1:1)?

Because the \$V_{gs}\$ of M10 is equal to the \$V_{gs}\$ of M9. And if you are operating in the saturation region of the MOSFET, the \$I_d\$ mostly depends on \$V_{gs}\$ and only weakly (due to the channel length modulation effect) on \$V_{ds}\$.

can you explain me intuitively why the number of transistors is equal to the multiplication factor of the current?

Because the drain-source channels of the N FETs are connected in parallel. And the total current flowing through a parallel combination of parts is equal the sum of the currents through the individual parts.

In comments you added,

Surely M9 works in saturation because its gate is connected to its drain, but how can we know if also M10 work in saturation? It depends not only on Vgs, but also on Vds, and Vd of M10 depends on the rest of the circuit in A.

Yes, it depends on \$V_{ds}\$. If \$V_{ds}\$ is too low, M10 will enter the triode region and the output current will no longer track the input current.

And small variations in \$V_{ds}\$ will have small effects on the output current, due to the channel length modulation effect.

This is expected and common to many kinds of current sources. They have a compliance voltage range (a range of output voltages over which they can work properly) and a non-zero output conductance (some variation in output current when output voltage changes).

Moreover, why don't we connect also M10 as M9 (gate to drain)?

If we connected M10's drain to its gate, we' also be connecting it to M9's drain (or rather, M11's drain in this case). We'd be shorting out our control signal to our output current. We'd no longer have isolation between input and output of the controlled current source.

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  • \$\begingroup\$ Very excellent answer, thank you very much. Last question: which is the role of M11? \$\endgroup\$ – Kinka-Byo Nov 12 '19 at 21:31
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    \$\begingroup\$ M11 is in common gate configuration. It's making sure M9 see's a low impedance looking "up" rather than the high impedance that the current source (I4) would present. (I'm not sure how you deal with the fact that connecting VBN on the drain side of M11 sets a lower limit on \$V_{gs}\$ of M9 --- that's a question for a real IC designer) \$\endgroup\$ – The Photon Nov 12 '19 at 21:42
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    \$\begingroup\$ @Kinka-Byo IC designer note, short version, that M11 is used to increase the resistance to help with power supply rejection on M9. If you look for "cascode", you'll find how we used these to minimize DIBL and channel length modulation effects. You can use DIBL to get match between legs on mirrors to microvolts if you are careful. If you assume that you get 100mv/decade in subvt, you can see that you'd want to minimize the effects of the primary bias against noise. Cascodes are an excellent way to achieve this. Shichi Lui's book and Carver Meads Analog VLSI books are excellent sources. \$\endgroup\$ – b degnan Nov 13 '19 at 0:49
  • \$\begingroup\$ Excuse me, I have another question about this circuit. Surely M9 works in saturation because its gate is connected to its drain, but how can we know if also M10 work in saturation? It depends not only on Vgs, but also on Vds, and Vd of M10 depends on the rest of the circuit in A. Moreover, why don't we connect also M10 as M9 (gate to drain)? \$\endgroup\$ – Kinka-Byo Feb 3 at 9:13
  • \$\begingroup\$ M9 and M11 look like a wide swing cascode structure, look it up. \$\endgroup\$ – Mike Feb 3 at 23:21

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