Chapter 12 JFETS

Example 12.1 Page No 303

In [1]:
# given data
R1= 20##  kΩ
R2= 10##  kΩ
R_E= 10##  kΩ
R_D= 8.2##  kΩ
V_G= 10##  V
V_BE= 0.7##  V
V_GS= -2##  V
V_DD= 30##  V
V_B= R2*V_DD/(R1+R2)##  V
I_E= (V_B-V_BE)/R_E##  mA
I_D= I_E##  mA
# The dc voltage from the drain to ground 
V_D= V_DD-I_D*R_D##  V
# The source voltage to ground 
Vs= V_G-V_GS##  V
print "The dc voltage from the drain to ground = %.2f volts"%V_D
print "The source voltage to ground = %.2f volts"%Vs
The dc voltage from the drain to ground = 22.37 volts
The source voltage to ground = 12.00 volts

Example 12.2 Page No 307

In [2]:
# given data
gmo= 3000##  µmhoS
V_GSoff= -4##  V
I_DSS= 10##  mA
print "Part (i) When V_GS= -1"
V_GS= -1##  V
# The value of gm 
gm= gmo*(1-V_GS/V_GSoff)##  µS
print "The value of gm = %.2f µS"%gm
print "Part (ii) When I_D= 2.5 mA"
I_D= 2.5##  mA
# The value of gm 
gm= gmo*2*I_D/I_DSS##  µS
print "The value of gm = %.2f µS"%gm
Part (i) When V_GS= -1
The value of gm = 3000.00 µS
Part (ii) When I_D= 2.5 mA
The value of gm = 1500.00 µS

Example 12.3 Page No 311

In [4]:
# given data
gm= 2000##  µS
gm=gm*10**-6##  S
R_D= 4.7##  kΩ
Vin= 2##  mV
R_L= 10##  kΩ
r_D= R_D*R_L/(R_D+R_L)##  kΩ
r_D= r_D*10**3##  Ω
A= gm*r_D## unit less
# The output voltage 
Vout= A*Vin##  mV
print "The output voltage = %.2f mV"%Vout

# Note: The calculated value of A = %.2f the book is wrong. Correct value of A6.39, So the answer in the book is wrong.
The output voltage = 12.79 mV

Example 12.4 Page No 311

In [5]:
# given data
R_D= 7.5##  kΩ
R_L= 3##  kΩ
r_s= R_D*R_L/(R_D+R_L)##  kΩ
r_s= r_s*10**3##  Ω
gm= 2500*10**-6##  S
# The voltage gain 
A= gm*r_s/(1+gm*r_s)## unit less
print "The voltage gain = %.2f"%A
The voltage gain = 0.84