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I have been reading this current mirror/constant current circuit using MOSFET. MOSFET Constant-Current Source

How does the current not change through Q2 when the voltage at Q2 increases? Ref.Fig1 Where does the excess voltage go when Vcs is increased? Can someone explain the complete working in simpler terms?

Please thanks.

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    \$\begingroup\$ There are several circuits in that article, include the actual circuit you ask about in a picture or drawing. You can also re-draw that circuit using the schematic entry tool. That article is already fairly basic so explaining it simpler isn't easily done. There are also plenty of books and articles describing the NMOS current mirror, just Google for NMOS current mirror to find more. For example: courses.e-ce.uth.gr/CE433/tutorials/… \$\endgroup\$ – Bimpelrekkie Jan 23 at 12:47
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How does the current not change through Q2 when the voltage at Q2 increases?

Because the drain current, ID, is directly proportional to the gate-source voltage, VGS: $$I_D = \frac{1}{2} \mu_n \cdot C_{ox} \cdot \frac{W}{L} \cdot (V_{GS} - V_{TH})$$

VTH is the gate-source threshold voltage and the others are geometry- and material-related parameters. Asssuming both MOSFETs are identical (i.e. having identical parameters), their drain currents will be equal since their gate-source voltages are equal (gates are shorted and sources are tied to ground).

Where does the excess voltage go when Vcs is increased?

There's no excess voltage. If the drain voltage of the right-most MOSFET (labeled as VCS in the linked article) increases then the MOSFET will adjust its internal resistance so that the drain current remins the same (because still VGS is constant).

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  • \$\begingroup\$ So, the left hand side MOSFET (connected to resistor) is in saturation mode and the right hand side MOSFET in is active mode(operating in linear region) ? \$\endgroup\$ – Electronic_Maniac Jan 25 at 3:14
  • \$\begingroup\$ Actually, Q2 "should" remain saturated. Otherwise it will not act as a constant current source thus the drain current will change with drain-source voltage. For further information you shoul read about "pinch-off" in MOSFETs. \$\endgroup\$ – Rohat Kılıç Jan 25 at 4:11

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