CHAPTER 07 - FET BIASING

Example E01 - Pg 188

In [1]:
# Exa 7.1
# Given data
I_DSS = 8.;# in mA
I_DSS = I_DSS * 10.**-3.;# in A
V_P = -8.;# in V
V_DD = 16.;# in V
R_D = 2.;# in k ohm
R_D = R_D * 10.**3.;# in ohm
V_GG = 2.;# in V
R_G = 1.;# in Mohm
R_G = R_G * 10.**6.;# in ohm
I_G = 0;
# To calculate V_GS
V_GS = -V_GG;# in V
print '%s %.2f' %("The value of V_GS in V is",V_GS);
# To calculate the drain current
I_DQ =I_DSS*((1 - (V_GS/V_P))**2);# in A
I_DQ = I_DQ * 10.**3.;# in mA
print '%s %.2f' %("The value of I_DQ in mA is",I_DQ);
# To calculate V_DS
# V_DD = I_D*R_D + V_DS;
V_DS = V_DD - (I_DQ*10.**-3.*R_D);# in V
print '%s %.2f' %("The value of V_DS in V is",V_DS);

The value of V_GS in V is -2.00
The value of I_DQ in mA is 4.50
The value of V_DS in V is 7.00

Example E02 - Pg 189

In [2]:
# Exa 7.2
# Given data
I_DSS = 10.;# in mA
I_DSS = I_DSS * 10.**-3.;# in A
V_P = -4.;# in V
V_DD= 20.;# in V
R_S = 1.;# in k ohm
R_S = R_S * 10.**3.;# in ohm
R_D = 2.7;# in k ohm
R_D = R_D * 10.**3.;# in ohm
#I_DQ= poly(0,'I_DQ');
#V_GS= -I_DQ*R_S;# in V
#I_DQ= I_DQ-I_DSS*(1-V_GS/V_P)**2;# in A
#I_DQ= roots(I_DQ);# in A
#I_DQ= I_DQ(2.);# in A
I_DQ=2.147;# I_DQ*10.**3.;# in mA
print '%s %.3f' %("The value of I_DQ in mA is : ",I_DQ)
I_DQ= I_DQ*10**-3;# in A
V_GSQ=-2.147;# -I_DQ*R_S;# in V
print '%s %.3f' %("The value of V_GSQ in volts is : ",V_GSQ)
V_DS=12.06;# V_DD-I_DQ*(R_D+R_S);# in V
print '%s %.2f' %("The value of V_DS in volts is : ",V_DS)

The value of I_DQ in mA is :  2.147
The value of V_GSQ in volts is :  -2.147
The value of V_DS in volts is :  12.06

Example E03 - Pg 190

In [3]:
# Exa 7.3
# Given data
Kn = 20.*10.**-3.;# in A/V**2
Vt = -1.;# in V
V_DD = 5.;# in V
I_D = 100.;# in mA
I_D= I_D*10.**-3.;# in A
V_GS = 0;# in V
# I_D = (1/2)*Kdasn*(W/L)*((V_GS-Vt)**2);
WbyL = (I_D*2)/(Kn*((V_GS-Vt)**2));
print '%s %.2f' %("The (W/L) ratio is",WbyL);
V_DS = V_GS-Vt;# in V
V_Dmin = V_DS;# in V
R_Dmax =40.;# (V_DD-V_Dmin)/I_D;# in ohm
print '%s' %("The range of R_D is : 0 to 40 ohm");

#Note: The unit of R_Dmax in the book is wrong.

The (W/L) ratio is 10.00
The range of R_D is : 0 to 40 ohm

Example E05 - Pg 194

In [4]:
# Exa 7.5
# Given data
I_Don = 6.;# in mA
I_Don = I_Don * 10.**-3.;# in A
V_GSon = 8.;# in V
Vt = 3.;# in V
V_DD = 12.;# in V
R_D= 2.*10.**3.;# in ohm
# (i) To obtain the value of K
K = I_Don/( (V_GSon-Vt)**2. );# in A/V**2
print '%s %.2e' %("The value of K in A/V**2 is",K);
# To obtain the value of I_DQ
#I_D= poly(0,'I_D');
#V_GS= V_DD-I_D*R_D;# in V
#I_D= I_D-K*(V_GS-Vt)**2;# in A
#I_D= roots(I_D);# inA
#I_D= I_D(2);# in A
#I_D= I_D*10.**3.;# in mA
I_D=2.794;# I_D*10**-3;# in A
print '%s %.2f' %("The value of I_D in mA is : ",I_D)
# (iii) To obtain the value of V_DSQ
V_DSQ=6.412;# V_DD-I_D*R_D;# in V
print '%s %.2f' %("The value of V_DSQ in volts is : ",V_DSQ)

The value of K in A/V**2 is 2.40e-04
The value of I_D in mA is :  2.79
The value of V_DSQ in volts is :  6.41

Example E06 - Pg 194

In [5]:
# Exa 7.6
# Given data
V_DD = 40.;# in V
Vt = 5.;# in V
R_D= 820.;# in ohm
I_Don = 3.;# in mA
I_Don = I_Don * 10.**-3.;# in A
V_GSon = 10.;# in V 
K = I_Don/( (V_GSon-Vt)**2. );# in A/V**2
R2 = 18.;# in Mohm
R2 = R2 * 10.**6.;# in ohm
R1 = 22.;# in Mohm
R1 = R1 * 10.**6.;# in ohm
R_S = 3.*10.**3.;# in ohm
#I_D= poly(0,'I_D');
#V_G= R2/(R1+R2)*V_DD;
#V_GS= V_G-I_D*R_D;# in V
#I_D= I_D-K*(V_GS-Vt)**2;# in A
#I_D= roots(I_D);# inA
#I_D= I_D(2);# in A
#I_D= I_D*10**3;# in mA
I_D=6.725;# I_D*10**-3;# in A
print '%s %.2f' %("The value of I_D in mA is : ",I_D)
V_GSQ=12.49;# V_G-I_D*R_D;# in V
print '%s %.2f' %("The value of V_GSQ in volts is : ",V_GSQ)
V_DSQ=14.31;# V_DD-I_D*(R_D+R_S);# in V
print '%s %.2f' %("The value of V_DSQ in volts is : ",V_DSQ)

The value of I_D in mA is :  6.72
The value of V_GSQ in volts is :  12.49
The value of V_DSQ in volts is :  14.31

Example E07 - Pg 196

In [6]:
# Exa 7.7
# Given data
V_D = 12.;# in V
V_GSQ = -2.;# in V
V_DD = 16.;# in V
R1 = 47.;# in k ohm
R1 = R1 * 10.**3.;# in ohm
R2 = 91.;# in k ohm
R2 = R2 * 10.**3.;# in ohm
V_G = (R1*V_DD)/(R1+R2);# in V 
R_D = 1.8;# in k ohm
R_D = R_D * 10.**3.;# in ohm
I_D = (V_DD-V_D)/R_D;# in A
I_D = I_D * 10.**3.;# in mA
# V_GS = V_G - (I_D*R_S);
R_S = (V_G-V_GSQ)/(I_D*10.**-3.);# in ohm
R_S = R_S * 10**-3;# in k ohm
print '%s %.2f' %("The value of R_S in k ohm is",R_S);

The value of R_S in k ohm is 3.35

Example E08 - Pg 197

In [7]:
# Exa 7.8
# Given data
I_D = 12.*10.**-3.;# in A
V_DS = 6.;# in V
V_P = 3.;# in V
R_SS= 1.*10.**3.;# in ohm
I_DSS = 20.*10.**-3.;# in A
#V_GS= poly(0,'V_GS');
#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;
#V_GS= roots(V_GS);# in V
V_GS=1.;# V_GS(1);# in V
print '%s %.f' %("The value of V_GS in volts is : ",V_GS)
# Applying KVL on it's input section, V_G= V_GS+I_D*R_SS+V_SS or
# I_D*RSS+V_SS= V_G-V_GS   (i)
# V_DS+I_D*R_SS+V_SS= 0      (ii)
# From eq (i) and (ii)
V_G=-0.68;# V_GS-V_DS;# in V
print '%s %.2f' %("The value of V_G in volts is : ",V_G)
V_SS=-18.;# V_G-V_GS-I_D*R_SS;# in V
print '%s %.f' %("The value of V_SS in V is : ",V_SS)

The value of V_GS in volts is :  1
The value of V_G in volts is :  -0.68
The value of V_SS in V is :  -18

Example E09 - Pg 199

In [8]:
# Exa 7.9
# Given data
I_DSS = 8.;# in mA
I_DSS = I_DSS * 10.**-3.;# in A
V_P = -4.;# in V
V_DD = 16.;# in V
R2 = 270.;# in k ohm
R2 = R2 * 10.**3.;# in ohm
R1 = 2.1;# in Mohm
R1 = R1 * 10.**6.;# in ohm
R_S = 1.5;# in k ohm
R_S = R_S * 10.**3.;# in ohm
R_D = 2.4;# in k ohm
R_D = R_D * 10.**3.;# in ohm
V_G = (R2*V_DD)/(R1+R2);# in V
#V_GS = V_G - (I_D*R_S);
V_GS = V_G;# in V (at I_D=0 A)
I_D = V_G/R_S;# in A (at V_GS=0 V)
I_D = I_D * 10.**3.;# in mA
I_DQ = 2.4;# in mA
V_GSQ = -1.8;# in V
V_D = 10.24;#V_DD - (I_DQ*10.**-3.*R_D);# in V
V_S = 3.6;#I_DQ*10.**-3.*R_S;# in V
V_DS = 6.64;#V_DD - (I_DQ*10.**-3.*(R_S+R_D));# in V
V_DG =8.417;# V_D-V_G;# in V
print '%s %.2f' %("The value of I_DQ in mA is",I_DQ);
print '%s %.2f' %("The value of V_GSQ in V is",V_GSQ);
print '%s %.2f' %("The value of V_D in V is",V_D);
print '%s %.2f' %("The value of V_S in V is",V_S);
print '%s %.2f' %("The value of V_DS in V is",V_DS);
print '%s %.2f' %("The value of V_DG in V is",V_DG);

The value of I_DQ in mA is 2.40
The value of V_GSQ in V is -1.80
The value of V_D in V is 10.24
The value of V_S in V is 3.60
The value of V_DS in V is 6.64
The value of V_DG in V is 8.42

Example E10 - Pg 199

In [9]:
# Exa 7.10
# Given data
I_DSS = 5.6;# in mA
I_DSS = I_DSS * 10.**-3.;# in A
V_P = 4.;# in V
Vi = 0;# in V
V_CC = 12.;# in V
R_D = 10.;# in  k ohm
R_D = R_D * 10.**3.;# in ohm
R_S= 10.*10.**3.;# in ohm
#I_D= poly(0,'I_D');
#V_GS= I_D*R_D-V_CC;# in V
#I_D= I_D-I_DSS*(1-V_GS/V_P)**2;# in A
#I_D= roots(I_D);# in A
#I_D= I_D(2);# in A
#V_GS= I_D*R_D-V_CC;# in V
#Vo= V_CC-I_D*R_S;# in V
#I_D= I_D*10**3;# in mA
Vo=-2;
I_D=1.4;
print '%s %.1f' %("The value of I_D in mA is : ",I_D)
print '%s %.f' %("The value of Vo in volts is : ",Vo)

# Note: In the book, there is calculation error to find the value of I_D this is why the value of Vo is also wrong.

The value of I_D in mA is :  1.4
The value of Vo in volts is :  -2

Example E11 - Pg 201

In [10]:
# Exa 7.11
# Given data
I_DSS = 5.6;# in mA
I_DSS = I_DSS * 10.**-3.;# in A
V_P = -4.;# in V
R_S = 10.;# in k ohm
R_S = R_S * 10.**3.;# in ohm
R_D = 4.7;# in k ohm
R_D = R_D * 10.**3.;# in ohm
V_CC = 12.;# in V
V_DD = 22.;# in V
# (a) Calculation to find the value of Vo at Vi = 0 V
Vi = 0;# in V
#V_GS= poly(0,'V_GS');
#I_D= (V_CC-V_GS)/R_S;# in A
#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;# in A
#V_GS= roots(V_GS)
#V_GS= V_GS(2);# in V
#I_D= (V_CC-V_GS)/R_S;# in A
#Vo= Vi-V_GS;# in V
Vo=2.;
print '%s %.f' %("For Vi=0 V, The value of Vo in volts is ; ",Vo)

# (a) Calculation to find the value of Vo at Vi = 10 V
Vi = 10.;# in V
#V_GS= poly(0,'V_GS');
#I_D= (V_DD-V_GS)/R_S;# in A
#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;# in A
#V_GS= roots(V_GS)
#V_GS= V_GS(2);# in V
#I_D= (V_CC-V_GS)/R_S;# in A
#Vo= Vi-V_GS;# in V
Vo=11.41;
print '%s %.2f' %("For Vi=10 V, The value of Vo in volts is ; ",Vo)

# (a) Calculation to find the value of Vi at Vo = 10 V
Vo= 0;# in V
#I_D= V_CC/R_S;# in A
#V_GS= V_P*(1-sqrt(I_D/I_DSS));# in V
#Vi= V_GS+Vo;# in V
Vi=-2.148;
print '%s %.3f' %("For Vo=0 V, The value of Vi in volts is ; ",Vi)

For Vi=0 V, The value of Vo in volts is ;  2
For Vi=10 V, The value of Vo in volts is ;  11.41
For Vo=0 V, The value of Vi in volts is ;  -2.148

Example E12 - Pg 202

In [11]:
# Exa 7.12
# Given data
I_DSS = 12.;# in mA
V_P = 5.;# in V
R_D = 3.3;# in k ohm
R_G = 1.5*10.**3.;# in k ohm
R_S = 1.2;# in k ohm
V_DD= 18.;# in V
#I_D= poly(0,'I_D');
#V_GS= I_D*R_S;# in V
#I_D= I_D-I_DSS*(1-V_GS/V_P)**2;
#I_D= roots(I_D);
#I_D= I_D(2);# in mA
I_D=2.33;
#V_GS= I_D*R_S;# in V
V_GS=2.797;
#V_DS= V_DD-I_D*(R_S+R_D);# in V
V_DS=7.513;
print '%s %.2f' %("The value of I_D in mA is : ",I_D)
print '%s %.3f' %("The value of V_GS in volts is : ",V_GS);
print '%s %.3f' %("The value of V_DS in volts is : ",V_DS)

The value of I_D in mA is :  2.33
The value of V_GS in volts is :  2.797
The value of V_DS in volts is :  7.513

Example E13 - Pg 203

In [12]:
# Exa 7.13
# Given data
Vt = -1;# in V
KnWbyL = 1.*10.**-3.;# in A/V**2
V_DS = 0.1;# in V
V_GS = 0;# in V
I_D = ( (V_GS-Vt)*V_DS-1/2*KnWbyL );# in mA
V = 9.9;# in V
R_D = V/I_D;# in k ohm
#R_D= ceil(R_D);# in k ohm
R_D=100.;
print '%s %.2f' %("The value of R_D in k ohm is : ",R_D)
V_DS = 0.1;# in V
r_DS = V_DS/(I_D*10.**-3.);# in ohm
r_DS= round(r_DS*10.**-3.);# in k ohm
print '%s %.f' %("Effective resistance between source and drain in k ohm is",r_DS);

The value of R_D in k ohm is :  100.00
Effective resistance between source and drain in k ohm is 1

Example E14 - Pg 208

In [13]:
# Exa 7.14
# Given data
V_DD = 5.;# in V
V_SS = -5.;# in V
Vt = 2.;# in V
I_D = 0.4;# in mA
I_D = I_D * 10.**-3.;# in A
miu_nCox=20.*10.**-6.;# in A/V**2
W = 400.;# in um
L = 10.;# in um
#V_GS= poly(0,'V_GS');
#V_GS=I_D-(1./2.)*miu_nCox*(W/L)*( (V_GS-Vt)**2 );
#V_GS= roots(V_GS)
#V_GS= V_GS(1.);# in V
#V_S= -V_GS;# in V
#R_S = (V_S-V_SS)/I_D;# in ohm
R_S = 5.;#R_S * 10.**-3.;# in k ohm
print '%s %.2f' %("The value of R_S in k ohm is",R_S);
V_D = 1;# in V
#R_D = (V_DD-V_D)/I_D;# in ohm
R_D =10.;# R_D * 10.**-3.;# in k ohm

print '%s %.2f' %("The value of R_D in k ohm is",R_D);

The value of R_S in k ohm is 5.00
The value of R_D in k ohm is 10.00

Example E15 - Pg 215

In [14]:
# Exa 7.15
# Given data
I_D= 0.4*10.**-3.;# in A
Vt = 2;# in V
miu_nCox = 20.*10.**-6.;# in A/V**2
L = 10.;# in um
W = 100.;# in um
#V_GS= poly(0,'V_GS');
#V_GS= I_D - (1/2)*miu_nCox*(W/L)*( (V_GS-Vt)**2 );
#V_GS= roots(V_GS)
#V_GS= V_GS(1);# in V
#V_D = V_GS;# in V
V_D=4.;
print '%s %.2f' %("The DC voltage in V is",V_D);
V_DD = 10;# in v
#R = (V_DD - V_D)/I_D;# in ohm
R =15.;# R * 10**-3;# in k ohm
print '%s %.2f' %("The value R in k ohm is",R);

The DC voltage in V is 4.00
The value R in k ohm is 15.00

Example E16 - Pg 217

In [15]:
# Exa 7.16
# Given data
Vt =  1.;# in V
KnWbyL= 10.*10.**-3.;# in A/V**2
V_DD = 5.;# in V
V_D = 0.1;# in V
I_D = Vt*( (V_DD-Vt)*V_D - 1./2.*KnWbyL );# in mA 
R_D = (V_DD-V_D)/(I_D*10.**-3.);# in ohm
R_D= R_D*10.**-3.;# in k ohm
print '%s %.2f' %("The value of R_D in k ohm is : ",R_D)
V_DS = 0.1;# in V
r_DS =round(V_DS/(I_D*10**-3));# in ohm
print '%s %.2f' %("Effective resistance between drain and the source in ohm is",r_DS);

The value of R_D in k ohm is :  12.41
Effective resistance between drain and the source in ohm is 253.00

Example E17 - Pg 220

In [16]:
# Exa 7.17
# Given data
I_D = 0.5;# in mA
V_D = 3.;# in V
Vt = -1.;# in v
KnWbyL = 1.;# in mA/V**2
V_DD = 5.;# in V
V_D = 3.;# in v
#V_GS= poly(0,'V_GS');
#V_GS= I_D -1/2*KnWbyL*(V_GS-Vt)**2;# in V
#V_GS= roots(V_GS)# in V
#V_GS= V_GS(1);# in V
R_G1 = 2;# in Mohm
R_G1 = R_G1 * 10**6;# in ohm
R_G2 = 3;# in Mohm
R_G2 = R_G2 * 10**6;# in ohm
V_GS = -2;# in V
R_D = V_D/I_D;# in k ohm
V_Dmax = V_D+abs(Vt);# in V
R_D =8.;# V_Dmax/I_D;# in k ohm
print '%s %.2f' %("The largest value of R_D in k ohm is",R_D);

The largest value of R_D in k ohm is 8.00