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I am reading about common source amplifier with a current-mirror active load from the book Analog Integrated Circuit Design by Tony Chan Carusone, David Johns, Kenneth Martin. I am confused by the yellow sentence below:

For a fixed bias drain current,ID , the effective overdrive voltage is reduced by increasing the device width W.

From the formula that relates between Veff and W/L it is clear that Veff is reduced if W is increased. However, from the circuit on the picture, the basing network for Q1 is not shown. How to make the basing network for Q1 so that Veff of Q1 decreases when W increases?

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The bias current is set by the current source Ibias, it is mirrored to the active load Q2 and the current through Q1 has to be about equal to the current delivered by Q2.

This is not done by some sort of bias network but through application of feedback. Imagine this stage as the second stage of an OpAmp as shown below. enter image description here

The CS Stage consists of M3 and M8, biased by M1.

In order to use this Opamp feedback is required. For example OUT could be fed back to IN-. The first stage of the OpAmp will take care of the biasing. The voltage at the gate of M8 will be such, that OUT is almost equal to IN+.

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    \$\begingroup\$ Yes, this is correct. \$\endgroup\$ – Mario Sep 23 '16 at 13:32
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    \$\begingroup\$ W of M8 goes up, Vout decreases, is fed back to IN-, current trough M4 increaes, current through M5 decreases, the voltage at the gate of M8 goes down. \$\endgroup\$ – Mario Sep 23 '16 at 14:50
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    \$\begingroup\$ A wider transistor can sink more current, Vds has to go down for a constant Id. \$\endgroup\$ – Mario Sep 23 '16 at 15:09
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    \$\begingroup\$ Yes, you are right. \$\endgroup\$ – Mario Sep 23 '16 at 16:11
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    \$\begingroup\$ The circuit is usually designed to operate in or close to a operating point where all transistors are in saturation and high gain is possible even under open loop conditions. However, due to device mismatch and thermal effects this operating point is not stable and some feedback is required. In any case the input of the OpAmp has to be at a defined potential, i.e. the input terminals must not be left floating. \$\endgroup\$ – Mario Sep 24 '16 at 5:47

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