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A simple question about ClassAB common source amplifier. There are two circuits which I have attached with this question, the one on the left is a standard cross coupled circuit which I have taken from CMOS Analog Circuit design by Allen & Holberg.

Left: Standard Cross-Coupled; Right: Modified version

On the picture to the right I replaced transistors M4 and M3 with the opposite counterpart. Vb1 and Vb2 are bias voltages.

My analysis: For the circuit on the left: As Vin increases, M1 will try to push more current,hence source voltage of M1 will rise (or fall?). I am considering that it will rise since the drain voltage of M6 must rise to accomodate more current. This will in turn increase the |Vgs| of M3 and hence the overdrive. The Vds of M3 should remain constant hence the drain voltage (Vgs) of M6 should rise which will cause M7 to sink current. The same analysis for M5, M4, M8 and M2 when Vin decreases.

For the circuit on the right: As Vin increases, M1 will try to push more current,hence source voltage of M1 will rise. Since Vb2 is constant, and increase in Vds or Vgs of M6 will cause M3 to go into cutoff. Hence it wouldnt work.

Now as for my questions:

  1. For the circuit to the left or right, why should the source voltage of M1 rise when Vin increases? Shouldnt it decrease for M1 to push more current? Since the drain of M1 is constant, the only voltage that has to change is the source voltage of M1 such that the Vgs of M1 increases. If it increases then why does it increase?

  2. Circuit to the right: Is my analysis of the fact that M3 will go into cut-off and that is why it wont work, correct?

Any help will be appreciated, thank you!

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  • \$\begingroup\$ Why can't you simulate this and get your answers? It seems the obvious approach to take. \$\endgroup\$ – Andy aka Sep 15 '20 at 14:39
  • \$\begingroup\$ Well, I would like to understand analytically rather than go for simulations each time. I lack the intuition \$\endgroup\$ – RAN Sep 15 '20 at 15:11
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    \$\begingroup\$ Notice the first circuit (on the left) the M1 is just a source follower (common-drain amplifier, voltage follower) and M3 in this case is a common gate amplifier. Thus if Vin increases the voltage at the M1 and M3 source will increases. Thus Vsg3 will also increase and as a result, the M1 and M3 current will rise and this rise in a current will be reflected by M6, M7 current mirror. On the other hand in the right circuit M1 is still working as a voltage follower but now M3 and M6 are active load (constant current source) for M1. So now any change in Vin will not change the M3,M6 current. \$\endgroup\$ – G36 Sep 15 '20 at 16:22
  • \$\begingroup\$ @G36 thank you for your reply, however, for the circuit on the right you described is a simple source follower circuit, which has a constant current source at the source of the transistor, shouldnt the source voltage of M1 rise, and since the current cannot change through M3 and M6, the drain voltage at M6 also rise? \$\endgroup\$ – RAN Sep 15 '20 at 16:33
  • \$\begingroup\$ But the M6 drain voltage is set to a fixed value by Vb2 value Vd6 = Vb2 - Vgs3 isn't it? \$\endgroup\$ – G36 Sep 15 '20 at 16:38
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  1. Because M1 is in source follower configuration (as is M2). So M1 source follows Vin. LH circuit works as follows : Assuming Vb2 constant, M3 is a common gate amplifier, so its drain voltage (and M6 drain, M6,M7 gate voltages) also increase.

  2. M3 drain is a high impedance input so M1 source voltage can increase all it likes with virtually no change in M3 drain (and source) current. Therefore the small signal gain is approximately 0.

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  • \$\begingroup\$ Right, but wont M3 go into cut-off for the circuit on the right hand side? \$\endgroup\$ – RAN Sep 16 '20 at 7:11
  • \$\begingroup\$ M3 provides constant current to M1. To cut M3 off you would have to get Vgs (for M3) below Vgs(th). How are you going to do that through the drain terminal? \$\endgroup\$ – user_1818839 Sep 16 '20 at 11:18

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