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I implemented the simple push pull source follower and I understand that it has this distortion problem. That is between -|VTP| ≤ Vin ≤ VTN there is no output because the output terminal floats since both the transistor are not conducting. I hope my understanding is correct.

Simple push pull stage

In order to avoid this distortion problem there is this push pull with quiescent current output stage shown below.

push pull with quiscent current

These are my following questions,

1) How does this new model correct the distortion problem?

2) Should all M3,M4,M5 and M6 be in saturation?

3) Is there any hint on how I could dimension these transisitors?

Also Vss in my case is 0. Any feedback would be much appreciated :)

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1.) The transistors M4 and M5 act like a floating voltage source. M3 is a current source and therefore the voltage across M4 and M5 is constant.

M6 and M3 form a common-source stage that allows to move the floating voltage source up and down. This eliminates the region where neither M1 nor M2 is conducting, therefore distortion is reduced.

2.) All these transistors should be in saturation. M4, M5 will always be in saturation since they are diode-connected.

3.) The dimensioning is usually done for the quiescent condition where the output signal is zero. For this case M4/M1 and M5/M2 act like current mirrors.

The ratio determines the quiescent current.

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  • \$\begingroup\$ Great now I get it ! So if I know the current flowing across the bias transistor M3 I can dimesnision the rest of the transistor depending on that ! But how do I decide how much current M3 would source ? Do i take the maximum current sourced by M1 as the Id for the left side brach ? Also since M4 and M6 have the same current flowing wouldnt their dimension be almost same ? Considering body effect of course M4 and M5 will be a bit different but apart from that wouldnt M4 and M6 be more or less the same and also M3 and M5 would be more or less same . Is my understaning right ? \$\endgroup\$ – Bhuvanesh Narayanan Mar 29 '16 at 12:16
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    \$\begingroup\$ The current in the left branch determines the quiescent current and the bandwidth (speed) of the circuit. The maximum current only depends on the sizing of M1, M2 and their maximum gate-source voltage. \$\endgroup\$ – Mario Mar 29 '16 at 13:17

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