I am trying to get into understanding of current mirrors, amplifiers and other basic electronic blocks, and I have come across this a lot, but I couldn't find an explanation how that works anywhere (maybe I was Googling the wrong words). For example, concerning this basic differential input circuit: enter image description here

How can there be any current flowing through the MOSFETs, when the Gate is grounded? I was under an impression that MOSFETs (the standard ones) conduct Drain to Source (or vice versa) current only when there is a Gate to Source voltage applied (and an inversion layer forms the channel and so on).

Futhermore I came across that you can bias a MOSFET in pretty much the same way, by placing a current source to the Source terminal. How does that work? I thought you can bias only by setting a Gate-Source voltage. Can a MOSFET be opened by forcing the current between the Drain and Source?

Man, I am confused.

  • 4
    \$\begingroup\$ You're thinking the source is grounded too - it's not - your source pins are connected through the current source to -V, not ground. \$\endgroup\$
    – brhans
    Commented Jun 5, 2017 at 15:17
  • \$\begingroup\$ The current will flow because the Ibais is connected the negative voltage -V. Hence the voltage at the MOSFET source is -Vgs. \$\endgroup\$
    – G36
    Commented Jun 5, 2017 at 15:17

1 Answer 1


There will still be a positive gate-to-source voltage if the source is negative with respect to ground. Note that the current source isn't grounded; it's connected to a negative voltage. So there's room for some GS voltage.

EDIT: Yes, the magnitude of the current through the source can affect other parts of the circuit. It's common to think of \$V{gs}\$ as the input to a MOSFET, and the drain current as the output, but really they're just two variables with a defined relationship to each other. You can put it in a circuit that forces a \$Vgs\$ and allows drain current to adjust to where it wants, or you can put it in a circuit that forces the drain current and allows the gate voltage to adjust. Provided you're operating within the allowable range of parameters of the device, the two variables should always obey the equations. But which is "input" and which is "output" is just a point of view.

  • \$\begingroup\$ Yeah, you're right, I completely overlooked the voltage at the bottom. Anyways, when the GS voltage is set, the MOSFET is opened (able to conduct DS current) - does the current forced from the current source somehow influence the Q-point, like change the GS voltage? I know the relation between Id and Ugs, but does it work both ways? \$\endgroup\$ Commented Jun 5, 2017 at 15:49
  • \$\begingroup\$ In this case, I believe that changing the current changes the source voltage (Vs). Since MOSFET's Gate voltage is fixed, varying the current will vary the source voltage (one step down Vt). \$\endgroup\$
    – sundar
    Commented Oct 1, 2017 at 15:22

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