MOSFET Constant current - Current mirror circuit

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?

• 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/… – Bimpelrekkie Jan 23 '19 at 12:47

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).

• 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) ? – Freshman Jan 25 '19 at 3:14
• 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. – Rohat Kılıç Jan 25 '19 at 4:11

How does the current not change through Q2 when the voltage at Q2 increases?

The current change through Q2 due to the finite output impedance of the transistors, which is caused by the channel length modulation. Though very basic, high level explanations neglect this effect for the sake of simplicity. This conductance is often negligible compared to the transconductance of the transistor (their contribution to the drain current is important). The depth depletion region of the p-n junction at the drain side of a pinched-off changes with the reverse bias voltage across it and thus the effective length of the transistor channel gets a little shorter. As the depth of the depletion region does not depend on the channel length, but the doping concentration and the applied voltage, this effect is more pronounced in very narrow transistors.

Putting the channel length modulation aside, all MOS transistors can be viewed as a voltage controlled current source. You can think like that of a bipolar transistor as well (though their physical operation depends on current). That's where the mighty power the transistors come from! Their current is independent to the drain voltage in a wide voltage range.

Where does the excess voltage go when Vcs is increased?

Nowhere. It stays there. The output impedance of the current is high, which means that it does not load the circuit where its output is connected to. Or with other words: The voltage VCS is floating, it can take any value (within reasonable limits of course).

There is no such thing as the voltage on a transistor. A voltage always refers to the potential difference between two nodes/terminals. Here I've assumed that you meant voltage at the drain of Q2.

It is arguable, but the abbreviation Qx is usually referred to bipolar tranistors, while Mx --- or Nx, Px depending on the channel type --- are more often used for MOS transistors.