feedback type in a circuit

[akbarza]

hi
i attached the pic and i saw it in a pdf in net( i do not have access to it now).
i have two question.
1) first if we disconnect feedback, namely vx is connected to a fix voltage, then is vout in phase with vin? namely does vout increased if vin is increased?
i think in this manner that if vin is increased then Ids of m1 is increased then to supply this increasing current, gate voltage of m3 must reduced. because of connection of gates m3 and m4, current of m4 increased. on the other hand , to Iss be constant, m2 current must decrease and this does by fall of drain voltage of m2. this fall of drain voltage of m2, amplifies current increase in m4. where is this current going?
is my above reasoning right?

2) in this pic is feedback negative?(Prof. Razavi had said it is negative.). how can explained that feedback is negative?
thanks

 

[deep_sea]

Assume at the fixed VX, and VIN (small signal is 0), you have VGS1=VGS2 and I1=I2= ISS/2.
As VIN increases, I1 > I2 and eventually = ISS and I2=0.
As VIN increases and consequently I2 decreases, you have M4 with a fixed |VGS|. The decrease in I2 must be accompanied by a decrease in |VDS4| which means Vout increases towards VDD as VIN increases.
Now regarding the feedback, for a negative feedback, the output must be connected via feedback network to the negative terminal of the input. The reason is that when Vout increases, VIN_Negative increases and consequently Vout decreases. In your case the negative input is M2 as explained previously.
What happens if you connect the output to the positive input? any Vout increase will e increase in the input and again the output until the output reaches the supply rail and the opposite is true when Vout decreases, Vin_plus decreases and again Vout decreases until it reaches the ground rail.

 

[LvW]

2) in this pic is feedback negative?(Prof. Razavi had said it is negative.). how can explained that feedback is negative?
thanks

Short answer: A common-source amplifier (gate input, drain output) is an inverting ampifier with 180deg phase shift (inversion) between gate and drain

 

[akbarza]

Assume at the fixed VX, and VIN (small signal is 0), you have VGS1=VGS2 and I1=I2= ISS/2.
As VIN increases, I1 > I2 and eventually = ISS and I2=0.
As VIN increases and consequently I2 decreases, you have M4 with a fixed |VGS|. The decrease in I2 must be accompanied by a decrease in |VDS4| which means Vout increases towards VDD as VIN increases.
Now regarding the feedback, for a negative feedback, the output must be connected via feedback network to the negative terminal of the input. The reason is that when Vout increases, VIN_Negative increases and consequently Vout decreases. In your case the negative input is M2 as explained previously.
What happens if you connect the output to the positive input? any Vout increase will e increase in the input and again the output until the output reaches the supply rail and the opposite is true when Vout decreases, Vin_plus decreases and again Vout decreases until it reaches the ground rail.

hi
thanks for answering. uou said: As VIN increases and consequently I2 decreases, you have M4 with a fixed |VGS|.
with increase of vin and its current,gate and drain of m3 must reduce( namel ,|vgs| must increase) so m4 does not have a fixed |vgs|.

 

[deep_sea]

Yes you are right. But anyway simply put by LvW, CS means inverting amplifier.