8/10/2019 Example Question - 1

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R1180

R2

180

C1

10F

C2

10FQ1

100um 100um

V_DD

v_in

v_out

FIRST: We would be given a transfer functionthat gives us:I_D = k(V_GS - V_T)^2 (1)Because we have the decoupling capacitors,we can ignore the AC aspect.

Using DC Analysis we get:

V_DD - I_D*R1 = V_GS (2)

Then using load line analysis,

we intersect the two equations and solve foreither V_GS or I_D and then get the other one

R3180

V_DD

Q2

100um 100um

THIS IS THE DC ANALYSIS CIRCUIT

ORIGINAL CIRCUIT

Initial information is values for all components,V_T and transfer function

Assumption is that the capacitors are largeenough that you can sue them as decouplingcapacitors

R_2 is R_Gand R_1 is R_D

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The next step is to use small signal analysis:This is for the load-line analysis where we take smalldeviations in V_DD in order to do AC Analysis

R4180

Q3

100um 100um

R5

180

v_in

v_out

For the AC analysis circuit, we short the capactiorswhich removes them from the circuitThe reason why we keep R_5 is because currentcan go from v_in into R_5

i_i

V_GS

We know that the voltage across G and Sfor the transistor is V_GSand that there is a current i_igoing from v_in into R_5

i_X

We know that there is a current going from R_Dinto the transistor which equals to i_Xwe also have a current i_D going fromD into S in the

we get: v_gs = v_i

i_DThis then gives us the expression for v_out:

v_out = - i_X * R_D (3)

the reason why there is a negative is becausecurrent flows in opposite direction

We also know that the currents add as:

i_i + i_X = i_D (4)

The expression for i_i is:

i_i = (v_i - v_o)/R_5

this is because the current is split betweenoutput and input.

AC Analysis Circuit

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This transconductance is also called g_m, so we have:

g_m = 2k*(V_GS - V_T) (5)

we then use (3) and (5) to get:

i_X = v_gs * g_m - i_i

v_o = [v_gs * gm - (v_i - v_o)/R_G ]*R_D

= -v_i *g_m*R_D + v_i*R_D/R_G - v_o*R_D/R_G

v_o * (1+R_D/R_G) = v_i*(R_D/R_G - g_m*R_D)

v_o*((R_G+R_D)/R_G) = v_i * ((R_D-g_m*R_D*R_G)/R_G)

therefore v_o/v_i = (R_D-g_m*R_D*R_G)/(R_G+R_D)

therefore A_v = (R_D-g_m*R_D*R_G)/(R_G+R_D)

We have the gain, we now need to calculat the outputand input impedence, which is as follows:

The output impedence is defined as:open output voltage / short current

We know that for open output voltage is just

v_o = A_V*v_i.

The short current is i_o, which is:

i_o = v_i/R_G

the reason why is that the voltage from R_D to transistoris zero (since ground at top) and the voltage fromground to transistor is also zero. These mean no current comesfrom there, and only current going to outputis i_i and it equals v_i/R_G

Therefore the output impedence is Z_o:

Z_o = A*v_i/(v_i/R_G)

therefore Z_o = A*R_G

8/10/2019 Example Question - 1

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