Ch-7: Field Effect Transistor

Page No. 180 Example 7.1.

In [1]:
VGS=12.
IG=10**-9
GSR=VGS/IG
GSR1=GSR*10**-6
print "VGS = 12 V, IG = 10**-9 A"
print "Therefore,  gate-to-source resistance  = VGS / IG =%0.2f Mohm"%GSR1

VGS = 12 V, IG = 10**-9 A
Therefore,  gate-to-source resistance  = VGS / IG =12000.00 Mohm

Page No. 181 Example 7.2.

In [2]:
delta_VGS=0.1
delta_ID=0.3*10**-3
print "delta_VGS = 4 - 3.9 = 0.1 V"
print "delta_ID = 1.6 - 1.3 = 0.3 mA"
gm=delta_ID/delta_VGS
gm1=gm*10**3
print "Therefore,  transconductance, gm = delta_ID / delta_VGS =%0.2f m-mho"%gm1

delta_VGS = 4 - 3.9 = 0.1 V
delta_ID = 1.6 - 1.3 = 0.3 mA
Therefore,  transconductance, gm = delta_ID / delta_VGS =3.00 m-mho

Page No. 184 Example 7.3

In [4]:
from math import sqrt
VGSoff=-6
IDSS=8
ID=4
print "ID = IDSS*[1 - (VGS/VGS_off)]**2"
VGS=(1-sqrt(ID/IDSS))*VGSoff
print "Therefore,    VGS =%0.2f V"%VGS
VP=abs(VGSoff)
print "VP = |VGS_off| = %0.2f V"%VP

ID = IDSS*[1 - (VGS/VGS_off)]**2
Therefore,    VGS =-6.00 V
VP = |VGS_off| = 6.00 V

Page No. 184 Example 7.4.

In [1]:
VGS=-2
VP=-5
IDSS=8*10**-3
print "The minimum value of VDS for pinch-off to occur for VGS = -2 V is"
VDSmin=VGS-VP
print "VDSmin = VGS - VP =%0.2f V"%VDSmin
IDS=IDSS*(1-(VGS/VP))**2
IDS1=IDS*10**3
print "IDS = IDSS * [1-(VGS/VP)]**2 =%0.2f mA"%IDS1

The minimum value of VDS for pinch-off to occur for VGS = -2 V is
VDSmin = VGS - VP =3.00 V
IDS = IDSS * [1-(VGS/VP)]**2 =8.00 mA

Page No. 186 Example 7.5.

In [2]:
IDSS=10*10**-3
VGS=-3
ID=4*10**-3
VDD=20.
print "The value of drain current at Q-point,"
IDQ=IDSS/2
IDQ1=IDQ*10**3
print "IDQ = IDSS / 2 =%0.2f mA"%IDQ1
print "and the value of drain-to-source at Q-point,"
VDSQ=VDD/2.
print "VDSQ = VDD / 2 = %0.2f V"%VDSQ
print "Therefore, the operating point is at:"
print "VDS =%0.2f V"%VDSQ
print "ID(mA) =%0.2f mA"%IDQ1
RD=(VDD-VDSQ)/ID
RD1=RD*10**-3
print "Therefore,  RD = %0.2f kohm"%RD1
print "The source voltage or voltage across the source resistor RS is"
VS=-VGS
print "     VS = -VGS = -3 V"
print "Also,VS = ID*RS "
RS=VS/ID
print "Therefore, RS = %0.2f ohm"%RS

The value of drain current at Q-point,
IDQ = IDSS / 2 =5.00 mA
and the value of drain-to-source at Q-point,
VDSQ = VDD / 2 = 10.00 V
Therefore, the operating point is at:
VDS =10.00 V
ID(mA) =5.00 mA
Therefore,  RD = 2.50 kohm
The source voltage or voltage across the source resistor RS is
     VS = -VGS = -3 V
Also,VS = ID*RS 
Therefore, RS = 750.00 ohm

Page No. 186 Example 7.6.

In [3]:
IDSS=40*10**-3
VP=-10
VGSQ=-5
print "We know that,  ID = IDSS * [1 - (VGS/VP)]**2"
print "Substituting the given values, we get"
ID = IDSS*(1-(VGSQ/VP))**2
ID1=ID*10**3
print "  ID =%0.2f mA"%ID1
RS=abs(VGSQ/ID)
print "Therefore,  RS = |VGSQ / ID| =%0.2f ohm"%RS

We know that,  ID = IDSS * [1 - (VGS/VP)]**2
Substituting the given values, we get
  ID =40.00 mA
Therefore,  RS = |VGSQ / ID| =125.00 ohm

Page No. 187 Example 7.7.

In [4]:
from sympy import symbols,solve
VDD=24.
R2=8.57*10**6
R1=12*10**6
VP=-2
IDSS=4*10**-3
RD=910.
RS=3*10**3
print "From fig.7.13.,"
VGG=round(VDD*(R2/(R1+R2)))
print "    VGG = VDD*(R2 / (R1+R2)) = %0.2f V"%VGG
x=symbols('x')
p1=solve(9*x**2-73*x+144,x)
ans1=p1[0]
ans2=p1[1]
print "Therefore,  ID=%0.2f or %0.2f mA"%(ans1,ans2)
print "As ID = 4.72mA > 4mA = IDSS, this value is inappropriate. So, IDQ=3.39 mA is selected."
print "Therefore,"
IDQ=3.39*10**-3
VGSQ=VGG-(IDQ*RS)
print "       VGSQ = VGG - (IDQ*RS) =%0.2f V"%VGSQ
VDSQ=VDD-(IDQ*(RD+RS))
print "and VDSQ = VDD - (IDQ*(RD+RS)) =%0.2f V"%VDSQ
VDGQ = VDSQ - VGSQ
print "Then,  VDGQ = VDSQ - VGSQ = %0.2f V"%VDGQ
print "which is grater than |VP| = 2 V. Hence, the FET is in the pinch-off region."

From fig.7.13.,
    VGG = VDD*(R2 / (R1+R2)) = 10.00 V
Therefore,  ID=3.39 or 4.72 mA
As ID = 4.72mA > 4mA = IDSS, this value is inappropriate. So, IDQ=3.39 mA is selected.
Therefore,
       VGSQ = VGG - (IDQ*RS) =-0.17 V
and VDSQ = VDD - (IDQ*(RD+RS)) =10.75 V
Then,  VDGQ = VDSQ - VGSQ = 10.92 V
which is grater than |VP| = 2 V. Hence, the FET is in the pinch-off region.

Page No. 190 Example 7.8.

In [10]:
from math import sqrt
IDSS=10*10**-3
VP=-3.5
Rth=120*10**3 #R1+R2=120 k-ohm
ID=5*10**-3
VDS=5
RS=0.5*10**3
print "Assume that the JFET is biased in the saturation region. Then the dc drain current is given by"
print "            ID = IDSS*(1-(VGS/VP))**2"
VGS=VP*(1-(sqrt(ID/IDSS)))
print "Therefore,  VGS = %0.2f V"%VGS
# textbook answer is wrong
print "The voltage at the source terminal is"
VS=(ID*RS)-5
print "            VS = (ID*RS) - 5 = %0.2f V"%VS
print "The gate voltage is"
VG=VGS+VS
print "            VG = VGS + VS = %0.2f V"%VG
print "The gate voltage is"
print "               VG = ((R2 / (R1 + R2))*10) - 5"
R2=(Rth*(VG+5))/10
R2_1=R2*10**-3
print "Therefore,         R2 = %0.2f kohm"%R2_1
# textbook answer is wrong
R1=Rth-R2
R1_1=R1*10**-3
print "and            R1(k-ohm) = %0.2f kohm"%R1_1
# textbook answer is wrong
print "The drain-to-source voltage is"
print "VDS = 5 - ID*RD - ID*RS - (-5)"
RD=(10-VDS-(ID*RS))/ID
RD1=RD*10**-3
print "     RD = %0.2f kohm"%RD1
x=VGS-VP
print "VGS - VP = %0.2f "%x  # textbook has taken a different value hence the wrong answer in textbook
print "Here, since VDS > (VGS-VP), the JFET is biased in the saturation region, which satisfies the initial assumption"

Assume that the JFET is biased in the saturation region. Then the dc drain current is given by
            ID = IDSS*(1-(VGS/VP))**2
Therefore,  VGS = -1.03 V
The voltage at the source terminal is
            VS = (ID*RS) - 5 = -2.50 V
The gate voltage is
            VG = VGS + VS = -3.53 V
The gate voltage is
               VG = ((R2 / (R1 + R2))*10) - 5
Therefore,         R2 = 17.70 kohm
and            R1(k-ohm) = 102.30 kohm
The drain-to-source voltage is
VDS = 5 - ID*RD - ID*RS - (-5)
     RD = 0.50 kohm
VGS - VP = 2.47 
Here, since VDS > (VGS-VP), the JFET is biased in the saturation region, which satisfies the initial assumption

Page No. 191 Example 7.9.

In [5]:
from math import sqrt
KN=1*10**-3
lamda=0.01
Ri=100*10**3
IDt=4*10**-3
IDQ=1.5*10**-3
VTN=1.5
VDD=12.
VDSQ=7.
VGSt=sqrt(IDt/KN)+VTN
print "VGSt = %0.2f V"%VGSt
print "Therefore,"
VDSt=VGSt-VTN
print "      VDSt = VGSt - VTN = %0.2f V"%VDSt
RD=(VDD-VDSQ)/IDQ
RD1=RD*10**-3
print "Therefore,  RD(k-ohm) = (VDD - VDSQ) / IDQ = %0.2f kohm"%RD1
print "Then,     IDQ = KN*(VGSQ-VTN)**2"
VGSQ=(sqrt(IDQ/KN))+VTN
print "Therefore,  VGSQ = %0.2f V"%VGSQ
R1=1200/2.73
print "    R1 = %0.2f kohm"%R1
R2=R1/((12/2.73)-1)
print "    R2 = %0.2f kohm"%R2

VGSt = 3.50 V
Therefore,
      VDSt = VGSt - VTN = 2.00 V
Therefore,  RD(k-ohm) = (VDD - VDSQ) / IDQ = 3.33 kohm
Then,     IDQ = KN*(VGSQ-VTN)**2
Therefore,  VGSQ = 2.72 V
    R1 = 439.56 kohm
    R2 = 129.45 kohm

Page No. 192 Example 7.10.

In [6]:
VTN=-2
KN=0.1*10**-3
VDD=5
RS=5*10**3
VGS=0
ID=KN*(-VTN)**2
ID1=ID*10**3
print "ID = %0.2f mA"%ID1
print "The d.c. drain-to-source voltage is"
VDS=VDD-(ID*RS)
print "     VDS = VDD - ID*RS = %0.2f V"%VDS
VDSsat=VGS-VTN
print "Then,  VDSsat = VGS - VTN = %0.2f V"%VDSsat
print "Since VDS > VDSsat, the MOSFET is biased in the saturation region"

ID = 0.40 mA
The d.c. drain-to-source voltage is
     VDS = VDD - ID*RS = 3.00 V
Then,  VDSsat = VGS - VTN = 2.00 V
Since VDS > VDSsat, the MOSFET is biased in the saturation region