For the first red box , why is it - to + for vgs2 and not + to - ?
For the second red box , why is there a negative sign? vgs2 = - (....)(ro1)


Same here , why is the voltage gain in negative?

 

Why is the voltage gain negative?

Vout decreases as Vin increases. Hence Vout and Vin are 180° out of phase.

 

Hi
But from the circuit how do I know that it is negative?

 

For the first red box , why is it - to + for vgs2 and not + to - ?

To open the N channel MOSFET Vgs voltage must be positive. But in this circuit the M2 gate is at AC GND and this is why we have + (gate) at GND.

For the second red box , why is there a negative sign? vgs2 = - (....)(ro1)

Notice that the voltage between the M1 drain and gnd is equal to Vsg2 ---> voltage between M2 source and the gate.
And this voltage is equal to current through ro1 times the ro1 (Vds1 = Vsg2 = I_ro1*ro1)
And I_ro1 = Id2 - Id1 = Id2 - gm1*Vgs1 therefore Vds1 = Vsg2 = (Id2 - Id1)*ro1 = ( Id2 - gm1*Vgs1 )*ro1
So it should be obvious to you that Vgs2 = -Vsg2 and this is why we have this minus sign.

But from the circuit how do I know that it is negative?

Every CS (command source) amplifier well have a negative gain because CE amplifier can only "sink" the current.

 

Hi
Thanks for the clear explanation but I am still not sure about this statement
Notice that the voltage between the M1 drain and gnd is equal to Vsg2 ---> voltage between M2 source and the gate.

Why is Vgs2 = Vsg2 ? and why does the M1 drain and gnd equal to Vsg2?

 

Why is Vgs2 = Vsg2 ?

No, wrong Vgs = - Vsg you do not see the difference ??

and why does the M1 drain and gnd equal to Vsg2?

Because M2 source is connected to M1 drain and the M2 gate is at "AC ground". So the Vsg2--> the voltage difference between M2 source and M2 gate is also equal to Vds1 (M1 source is also at "AC ground" exactly the same as M2 gate).

 

Because we can redraw the circuit like this

 

Because we can redraw the circuit like this
View attachment 112728

Hmm. This may sound stupid. But why is it vds1 = Vsg2 ? and not vds1= vgs2? How does the AC ground at the M2 gate affect it?

Why is it At AC ground it is + and not - ?

 

The one in red circule


Sorry for asking so many stupid questions

 

Why is it At AC ground it is + and not - ?

Because for N channel MOS the voltage between gate and source we assumed that the Vgs is positive (voltage at gate must be higher with respect to source).
But as you can see this is not the case here, and that's why Vds1 = -Vgs2 . The voltage at M2 source is higher with respect to gate, the M2 gate voltage is lower then the voltage at source.

 

Because for N channel MOS the voltage between gate and source we assumed that the Vgs is positive (voltage at gate must be higher with respect to source).
But as you can see this is not the case here, and that's why Vds1 = -Vgs2 . The voltage at M2 source is higher with respect to gate, the M2 gate voltage is lower then the voltage at source.

I see. Thank you so much for your explanation. I would like ask one more question.
What is Vds2 = to ? Vd2 is grounded so I think Vds2 = Vds1 = -Vgs2 ?

 

What is Vds2 = to ? Vd2 is grounded so I think Vds2 = Vds1 = -Vgs2 ?

You are wrong. Vds1 is not equal to Vds2.
Do you see the difference between Vds and Vsd ??

Vds - is a voltage difference between drain and the source

And

Vsd - is a voltage difference between source and the drain

If for example voltage at drain with respect to reference point is 5V and the source voltage with respect to reference point is 2V we have as follow :

Vds = Vd - Vs = 5V - 2V = 3V

Vsd = Vs - Vd = 2V - 5V = -3V

I hope that know you see that Vds = -Vsd

Vds = -Vsd = -(-3V) = 3V