How Do You Calculate the Overall Voltage Gain of Two Cascaded CS Amplifiers?

[asdf12312]

Homework Statement

two identical CS amplifiers connected in cascade. 1st stage fed with v_sigand R_sig=100k ohm. load R_L= 10k ohm connected to drain of 2nd stage. each MOSFET biased at I_D = 0.25mA and V_OV=0.25V. assume V_A is very large. Each stage has R_D=10k ohm.

a) sketch equivalent circuit of two-stage amplifier
b) find overall voltage gain G_v

Homework Equations

I_D = (1/2)*k_n*(V_OV)² <-- sat.
g_m = k_nV_OV
r_o = |V_A|/I_D
G_V=A_V=-gm*R_D || R_L = -gm*R_O
V_o = -(g_mv_gs)(R_D||r_o)

The Attempt at a Solution

i sketched the small signal circuit on paper but I'll probably show it later. solving for k_n i get 8mA/V². and g_m=2mA/V. If V_A is infinite then r_o is infinite, and R_O=r_o||R_D=R_D. I get for 1st stage A_Vo1=G_Vo1=-20V.

To find the relationship for 2nd stage I did V_o1=V_i2=-(g_mV_i1)(R_O) and V_o2=-(g_mV_i2)(R_O)=(g_m)²(R_D||r_o)²V_i2 then G_Vo2=V_o2/V_i2=400V. but is that right? I get a positive gain so I think i did it wrong.

 

[rude man]

Need circuit diagram.

 

[asdf12312]

I wasn't thinking, and i did it wrong xD
the gain for 2nd cascade stage, would actually be -gm* RD|| RL = -2 * 10||10 = -10 V/V. what I don't get though, how they get an answer of total gain of -20 V/V (1st stage) * -10 V/V = 200 V/V. are you allowed to multiply them like that?

sorry, here's the image for 2 stage cascade CS amplifiers
http://postimg.org/image/cnqur1y6h/

 

[rude man]

I guess from what you wrote that g_m = 2 mA/V.
If I have two amplifier in cascade as you have, why should you not be able to multiply their gains? The input to the 2nd stage is the output of the 1st stage.
The answer they give is correct.