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For part (c) of this problem, I am not sure how the input capacitance is defined for a differential pair since we have two inputs. Would I have to find two different input capacitances (one for differential mode and the other for common mode)? So for a basic differential pair such as the one shown below,

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Would the input capacitance for differential mode be the same as that of a common source amplifier?

Also I would appreciate if someone could give me hints on how to answer part (d) and (e)...

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To measure (i.e. analyse) differential input capacitance, consider that you apply a small +deltaV/2 to the in+ and a -deltaV/2 to the in- and measure the charge required. Note that the common source node won't change in this small-signal analysis -- it's not a lot different from a common-source circuit.

For part e), consider the deltaV across the new blue capacitors -- the effect may surprise you !

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  • \$\begingroup\$ Would the capacitors have no effect since it would induce equal and opposite change on the vout+ and vout- nodes, and therefore vout = (vout+) - (vout-) is not affected? \$\endgroup\$
    – user207787
    Commented Apr 16, 2022 at 0:37
  • \$\begingroup\$ It's a differential amplifier and so there will (nearly) always be equal and opposite effects on VOUT+ and VOUT- -- but you care about VOUT+ - VOUT-, so there is no cancellation there. \$\endgroup\$
    – jp314
    Commented Apr 16, 2022 at 1:43
  • \$\begingroup\$ But then what happens? I can't wrap my head around this haha. \$\endgroup\$
    – user207787
    Commented Apr 16, 2022 at 3:42
  • \$\begingroup\$ If VIN+ rises up, then VOUT- falls. Similarly, VOUT+ rises. Thus if the blue caps match the parasitic CDG of the diff. pair, then the coupling to VIN- cancels... \$\endgroup\$
    – jp314
    Commented Apr 16, 2022 at 5:36

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