Disclaimer: These are notes taken before the exam not ment as a readable solution. There is no guarantee for correctness nor completness of the solutions sketched here! Other solutions can also be correct.


Task 1
Points for:
- Different drain voltages/out of saturation
- Different transistor sizes, W/L
extra:
- Mismatch
- Dynamic effects

--

Task 2
function '[Vd]=CMOSdiode(Id,kn,Vtn)
VT=26e-3 ;%V
Is=2*kn*VT^2;
% Id=Is*exp((Vg-Vtn)/VT) % solve for Vg
% log(Id)=log(Is)+(Vg-Vtn)/VT
Vg=VT*(log(Id)-Log(Is))+Vtn;

--

Task 3
gate 1
gate 4
gate 2
gate 3

--

Task 4
Truth table
J K  D
0 0  Q
0 1  0
1 0  1
1 1  notQ

More explicit
J K Q  D
0 0 0  0
0 0 1  1
0 1 X  0
1 0 X  1
1 1 0  1
1 1 1  0

Simplest implementation with 2x3-input AND gates and 1x2-input AND gate in the three cases where D is 1

--

Task 5

function [A] = IntrinsicGain( Id, L, uCox, W, LamL)
%UNTITLED2 Summary of this function goes here
%   Detailed explanation goes here

%Id=100e-6
%uCox=400e-6
% LamL=(1/5)*1e-6
% W=10e-6
% L=0.5e-6


%gm=Vov*kn;
%Id=0.5*kn*Vov^2
gm=sqrt(2*Id*uCox*W/L)
%lambda=lamL/L
ro=L/LamL/Id

%A=sqrt(2*Id*uCox*W/L)*L/lamL/Id
A=sqrt(2*L/Id*uCox*W)/LamL
end


Id=40e-6;
L45=45*2e-9;
L180=180*2e-9;
W45=3*L45;
W180=3*L180;
uCox45=280e-6;
uCox180=270e-6;
LamL45=0.1e-6;
LamL180=0.08e-6;

[A45] = IntrinsicGain( Id, L45, uCox45, W45, LamL45)

[A180] = IntrinsicGain( Id, L180, uCox180, W180, LamL180)


A45 =

    5.8327

A180 =

   28.6378

--

Task 6

A1=20;
A2=8;
e1=3e-3;
e2=6e-3;
etot_a=(A1*e1+e2)/A1

e1=6e-3;
e2=3e-3;
etot_b=(A1*e1+e2)/A1


A1=8;
A2=20;
e1=3e-3;
e2=6e-3;
etot_c=(A1*e1+e2)/A1


e1=6e-3;
e2=3e-3;
etot_d=(A1*e1+e2)/A1



etot_a =

    0.0033


etot_b =

    0.0062


etot_c =

    0.0037


etot_d =

    0.0064

--

Task 7

Id=15e-6
kn=400e-6
Cgs=10e-15;
Cgd=1e-15;

%gm=Vov*kn;
%Id=0.5*kn*Vov^2
gm= sqrt(2*Id*kn)

f_T=gm/(Cgs+Cgd)/2/pi


f_T =

   1.5850e+09

--

Task 8

function [A1,A2]=transferF(s)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
K=10.^(50./20)
A1=K.*1./(1+s./10.^5);
A2=A1.*(1-s./10.^5)./(1+s./10.^5);

end

--

Task 9

%function [Q] = QinSF(Cgs,Cgd,Rsig,RL,CL,gm)
%
%  
%Miller=gm*RL+1;
%b1=(Cgd+Cgs/Miller)*Rsig+(Cgs+CL)/Miller*RL;
%b2=((Cgs+Cgd)*CL+Cgs*Cgd)/Miller*Rsig*RL;
%Q=sqrt(b2)./b1;
%end

Rsig=[200 500 5000 50000 125000]

Cgs=2e-12
Cgd=0.1e-12
CL=1e-12
gm=5e-3
RL=30e3

Miller=gm*RL+1;
Q=sqrt(((Cgs+Cgd)*CL+Cgs*Cgd)/Miller*Rsig*RL) ./ ...
((Cgd+Cgs/Miller)*Rsig+(Cgs+CL)/Miller*RL)

 

Q =

    0.4886    0.7324    1.3005    0.7638    0.5123