how to calculate the gain of this circuit?

[wjxcom]

Hi, rfsystem: I have made the equivalent circuit, shown in the file equivalent_circuit.JPG.

If this circuit is correct, I think I must disturb you again. I think the gain has no relation with the M6 and M3, but M1 and M2.

 

[ree]

hello,

i'll say my explantation

gain=Gm* Rout

Gm=gm1

Rout=ro1//r02

and ro1~=ro4 as same current same W/l which mean same bias

the same for ro2

this my guess

 

[wjxcom]

HI, ree: in fact, the gain is -gm6*(ro3||r06) is right, but I do not undersstand how to get this value!!

 

[tekno1]

gyxcom,

mentaly take away the two diode connected transistors (M4 and M5). What you have is now, M6 and resistive load made of M3. This circuit now is a simple gain stage (one can find in any analog book), and its gain is -gM6(ro6//ro3).

Diode connected devices (M4, M5) does not have any gain, think about them as level shifters. And think about the fact that Vout follows sources of M4 and M5 (translinear action Vgs up Vgs down resulting same voltage). Therefore gain here is -gM6(ro^//ro3).

 

[ree]

sorry it a typing mistakke but i meant that
Gm=gm6

 

[wjxcom]

Hi, all: when I calculate the gain of this circuit, I ignored the M4 and M5 (because 1/Gm of M4 and M5 is smaller than r0 of M3 and M6). So the gate of M1 and the gate of M2 are connected each other.

At this situation, the gain that from the gate of M6 to the gate of M1(2) is -Gm6*(r03||r06), at the same, the gain from the gate of M1(2) to OUT is 1.

So the gain of this circuit is -Gm6*(r03||r06). I think this a right way to calculate the gain of this circuit, is it?

 

[allan_guo]

Hi, all: when I calculate the gain of this circuit, I ignored the M4 and M5 (because 1/Gm of M4 and M5 is smaller than r0 of M3 and M6). So the gate of M1 and the gate of M2 are connected each other.

At this situation, the gain that from the gate of M6 to the gate of M1(2) is -Gm6*(r03||r06), at the same, the gain from the gate of M1(2) to OUT is 1.

So the gain of this circuit is -Gm6*(r03||r06). I think this a right way to calculate the gain of this circuit, is it?

I think : "the gain from the gate of M1(2) to OUT is 1 " is under the condition that the size of the mirror pairs is the same,and the RL is much larger than 1/gm. -Gm6*(r03||r06) is just the gain of the first stage.i believe the second stage gain= (gm4+gm5)*(Rl//(1/gm1)//(1/gm2))=(gm4+gm5)/(gm1+gm2)
just my own opinion.

 

[demodb]

i agree with tekno1, if the W/L of M1=M4 and M2=M5 no additional gain is added through those mirrors. So the gain of the complete stage is just -gm6(ro6||ro3).