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I have a question regarding small-signal analysis of the differential pair with current mirror load.

A differential-input, single-ended-output MOS gain stage

For Small-signal analysis, why is I(d2) = -I(s1) ?

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  • \$\begingroup\$ Where is I(d2)? And why does i(s1) appear in three different places? \$\endgroup\$ – uint128_t Feb 26 '16 at 3:39
  • \$\begingroup\$ I(d2) is the drain current for mosfet Q2 \$\endgroup\$ – kevin Feb 26 '16 at 4:23
  • \$\begingroup\$ This looks like homework. Is it a problem assigned by the professor, or is it just something you want a little help understanding? Or is it not homework at all? \$\endgroup\$ – mkeith Feb 26 '16 at 8:03
  • \$\begingroup\$ Because Ibias is constant (and therefore, in small signal analysis, 0). \$\endgroup\$ – Brian Drummond Feb 26 '16 at 12:22
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It's easier to see when looking at an example. Let's assume that Ibias is 100 (microamps or percent). Just looking at the differential pair and ignoring the current mirror, we have the same current through Q1 and Q1 for \$v_{in}=0\$, which would be 50uA through each transistor.

Now we increase \$v_{in}\$, the gate-source voltage of Q1 increases, Q1 carries more current. Let's assume that we increased the current through M1 to 51uA. Since Ibias is fixed there are only 49uA left for Q2.

When doing small-signal analysis we only look at the changes. So for the equilibrium point (\$v_{in}=0\$) we set everything to zero. We set the current through Q1 and Q2 to zero which means we subtract 50uA from each. An increase to 51uA would now be 1uA and 49uA becomes -1uA. That is I(d2) = -I(s1).

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