Chapter 8 Metal Oxide Semiconductor Field Effect Transistor

Exa 8.1 page No 441

In [1]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show
from __future__ import division
# Given data
I_DSS= 10# mA
V_P= -4# V
V_GS= arange(0,-0.1+V_P,-0.1) # V
# The value of I_D,
I_D= I_DSS*(1-V_GS/V_P)**2# mA
plot(V_GS,I_D)
xlabel("V_GS in volts")
ylabel("I_D in mA")
title("Transfer characteristics for an n-channel depletion-type MOSFET")
show()
print "Transfer characteristics for an n-channel depletion-type MOSFET shown in figure."
Transfer characteristics for an n-channel depletion-type MOSFET shown in figure.

Exa 8.2 page No 442

In [2]:
from __future__ import division
# Given data
V_DD= 20# V
V_GS = 0# V
I_DSS = 10# mA
I_P = I_DSS# mA
I_D = I_DSS# mA
R_D = 1.5# k ohm
# The drain to source voltage,
V_DS = V_DD - (I_D*R_D)# V
print "The value of V_DS = %.2f V"%V_DS
The value of V_DS = 5.00 V

Exa 8.3 page No 442

In [3]:
from __future__ import division
# Given data
I_Don = 5# mA
V_GS = 8# V
V_GST = 4# V
K = I_Don/((V_GS-V_GST)**2)# mA/V**2
# Drain current when V_GS= 6 V
V_GS= 6# V
I_D = K*((V_GS-V_GST)**2)# mA
print "The drain current = %.2f mA"%I_D
The drain current = 1.25 mA

Exa 8.4 page No 442

In [5]:
from __future__ import division
# Given data
I_Don = 10# mA
V_GS = -12# V
V_Ton = -3# V
if V_GS<0:
    print "Since the value of V_GS is negative, hence the device is P-channel"

K = I_Don/((V_GS-V_Ton)**2)# mA/V
V_GS = -6# V
# The drain current,
I_D = K*( (V_GS-V_Ton)**2 )# mA
print "The value of I_D = %.2f mA"%I_D

# Note: The answer in the book is not accurate.
Since the value of V_GS is negative, hence the device is P-channel
The value of I_D = 1.11 mA

Exa 8.5 page No 443

In [7]:
from __future__ import division
# Given data
I_Don = 5# mA
V_GSon = 7# V
V_T = 4# V
V_DD = 9# V
R_D = 1.2# k ohm
R_S = 0.5# k ohm
K = (I_Don)/((V_GSon-V_T)**2)# mA/V**2
# The value of drain current,
I_D = K*((V_GSon-V_T)**2)# mA
print "The value of I_D = %.2f mA"%I_D
#The drain to source voltage, V_DS = V_DD - (I_D*R_D) - (I_D*R_S)
V_DS = V_DD - (I_D*(R_D+R_S))# V
print "The value of V_DS = %.2f V"%V_DS

# Note: The answer in the book is not accurate.
The value of I_D = 5.00 mA
The value of V_DS = 0.50 V