Chapter 10 : Field Effect Transistor

Example 10.1a, Page No 337

In [1]:
import math
#initialisation of variables
VGG = -2.5     #v
Id = 3.0         #mA
Vds = 10.0     #v


#Calculations
Rd=Vds/Id


#Results
print("The value of Rd is  = %.2f kOhm " %Rd)

The value of Rd is  = 3.33 kOhm

Example 10.1b, Page No 337

In [2]:
import math
#initialisation of variables
VGG = -2.5     #v
VDD = 20.0
Id = 2.0         #mA drain current
Vds = 2.5     #v


#Calculations
Rd=(VDD-Vds)/Id


#Results
print("The value of Rd is  = %.2f kOhm " %Rd)

The value of Rd is  = 8.75 kOhm

Example 10.2a, Page No 338

In [3]:
import math
#initialisation of variables

a=3*(10**-4)           #in cm
Nd=10**15              #in electrons/cm^3
q=1.6*(10**-19)       #in C
eo=8.85*(10**-12)     #Permittivity of free space
e=12*eo               #Relative Permittivity

#Calculations
Vp=(q*Nd*a*a*10**6*10**-4)/(2*e)      #in V
#a is in cm so 10^-4 is multiplied and Nd is in electrons/cm^3 so 10^6 is multiplied

#Results
print("Pinch off Voltage = %.2f v " %Vp)

Pinch off Voltage = 6.78 v

Example 10.2b, Page No 338

In [4]:
import math

#initialisation of variables
#Given Values
a=3.0*(10**-4)         #in m
Nd=10.0**15            #in electrons/m**3
q=1.6*(10**-19)        #in C
eo=8.85*(10**-12)      #Permittivity of free space
e=12*eo                #Relative Permittivity

#Calculations
Vp=(q*Nd*a*a*10**6*10**-4)/(2*e)#in V
#a is in cm so 10**-4 is multiplied and Nd is in electrons/cm**3 so 10**6 is multiplied
Vgs= Vp/2
b=a*(1-((Vgs/Vp)**(0.5)))#in cm

#Results
print("Channel Half Width = %.2f cm  x 10^-4" %(b*10**4))

Channel Half Width = 0.88 cm  x 10^-4

Example 10.3a Page No 348

In [5]:
import math

#initialisation of variables
VG=0.5     #v
Rd=1.0      #Kohm
Vdd =3.3   #v

#Calculations
# If VG=0.5vv hen Vgs-Vt = 0.5-1=-.05 which is less than zero and therfore the transistor will be off then
#ID=0
Id=0
Vd=Vdd

#Results
print("The drain current  = %.2f mA" %(Id))
print("The drain Voltage  = %.2f V" %(Vd))

The drain current  = 0.00 mA
The drain Voltage  = 3.30 V

Example 10.3b Page No 348

In [6]:
import math

#initialisation of variables
Vg=2.0        #v
Rd=1.0      #Kohm
Vdd =3.3   #v
Vt=1.0

#Calculations
# If VG=2v hen Vgs-Vt = 2-1=-2 which is more than zero and therfore the transistor will be on then
Id=(1.0/2)*(Vg-Vt**2)
Vd=Vdd-(Rd*Id)

#Results
print("The drain current  = %.2f mA" %(Id))
print("The drain Voltage  = %.2f V" %(Vd))

The drain current  = 0.50 mA
The drain Voltage  = 2.80 V

Example 10.4a Page No 363

In [7]:
import math

#initialisation of variables
Vp=-2#in V
Idss=1.65#in mA
#it is desired to bias the circut at Id=0.8mA
Ids=0.8#in mA
Vdd=24#in V
#Assumption: rd>Rd

#Calculations
Vgs=Vp*(1-(Ids/Idss)**0.5)#in V

#Results
print("Vgs= %.2f v " %Vgs)

Vgs= -0.61 v

Example 10.4b, Page No 363

In [8]:
import math
#initialisation of variables
Vp=-2#in V
Idss=1.65#in mA
#it is desired to bias the circut at Id=0.8mA
Ids=0.8#in mA
Vdd=24#in V
#Assumption: rd>Rd

#Calculations
Vgs=Vp*(1-(Ids/Idss)**0.5)#in V

gmo=-(2*Idss/Vp)

#Results
print("gmo= %.2f mA/V " %gmo)
gm=gmo*(1-(Vgs/Vp))
print("gm= %.2f mA/V " %gm)

gmo= 1.65 mA/V 
gm= 1.15 mA/V

Example 10.4c Page No 363

In [9]:
import math
#initialisation of variables
Vp=-2#in V
Idss=1.65#in mA
#it is desired to bias the circut at Id=0.8mA
Ids=0.8#in mA
Vdd=24#in V
#Assumption: rd>Rd

#Calculations
Vgs=Vp*(1-(Ids/Idss)**0.5)#in V

gmo=-(2*Idss/Vp)
gm=gmo*(1-(Vgs/Vp))

Rs=-(Vgs/Ids)#in ohm

#Results
print("Rs= %.2f K " %Rs)

Rs= 0.76 K

Example 10.4d Page No 363

In [10]:
import math

#initialisation of variables
Vp=-2#in V
Idss=1.65#in mA
#it is desired to bias the circut at Id=0.8mA
Ids=0.8#in mA
Vdd=24#in V
#Assumption: rd>Rd

#Calculations
Vgs=Vp*(1-(Ids/Idss)**0.5)#in V
gmo=-(2*Idss/Vp)
gm=gmo*(1-(Vgs/Vp))
Rs=-(Vgs/Ids)#in ohm
print('20dB corresponds to voltage gain of i0')
Av=10
Rd=Av/gm#in ohm

#Results
print("Rd= %.2f ohm" %Rd)

20dB corresponds to voltage gain of i0
Rd= 8.70 ohm