# Solved Examination Paper¶

## Exa 1.b - page 476¶

In [2]:
# Given data
R1= 10 # in kΩ
R2= 10 # in kΩ
Rf= 50 # in kΩ
V= 2 # in V
V1= V*R1/(R1+R2) # in V
V01= -Rf/R1*V1 # in V
print "The value of V1 = %0.f Volts" %(V1)
print "The value of V01 = %0.f Volts" %(V01)

The value of V1 = 1 Volts
The value of V01 = -5 Volts


## Exa 2.a - page 479¶

In [4]:
# Given data
V_P= -4 # in V
I_DSS= 10 # in mA
V_GS= 0 # in V
R_D= 1.8 # in kΩ
V_DD= 20 # in V
I_D= I_DSS*(1-V_GS/V_P)**2 # in mA
# Applying KVL to the circuit, we get V_DD= I_D*R_D+V_D
V_D= V_DD-I_D*R_D # in V
print "The value of I_D = %0.f mA" %(I_D)
print "The value of V_D = %0.f Volts" %V_D

The value of I_D = 10 mA
The value of V_D = 2 Volts


## Exa 2.c - page 480¶

In [5]:
# Given data
V_GS= 3 # in V
Vth= 1 # in V
unCox= 25 # in mA/V**2
unCox= unCox*10**-3 # in A/V**2
W=3 # in µm
L=1 # in µm
r_DS= 1/(unCox*W/L*(V_GS-Vth)) # in Ω
print "The value of r_DS = %0.2f Ω " %r_DS

The value of r_DS = 6.67 Ω


## Exa 3.b - page 481¶

In [7]:
# Given data
I_CQ= 10 # in mA
I_CQ= I_CQ*10**-3 # in A
V_CQ= 5 # in V
V_CC= 10 # in V
R_C= 0.4 # in kΩ
R_C= R_C*10**3 # in Ω
V_BE= 0.075 # in V
V_BB= 0.175 # in V
beta=100
beta_max=120
beta_min= 40
# Applying KVL we get, V_CQ= V_CC-I_C*(R_C+R_E)
R_E= (V_CC-V_CQ)/I_CQ-R_C # in Ω
print "The value of R_E = %0.f Ω" %( R_E)
I_B= I_CQ/beta # in A
R_B= (V_BB-V_BE)/I_B # in Ω
print "The value of R_B = %0.f kΩ" %(R_B*10**-3)
I_Cmax= beta_max*I_B # in A
I_Cmin= beta_min*I_B # in A
delta_I_CQ= I_Cmax-I_Cmin # in A
print "The value of delta_I_C = %0.f mA" %(delta_I_CQ*10**3)

The value of R_E = 100 Ω
The value of R_B = 1 kΩ
The value of delta_I_C = 8 mA