chapter05:Field Effect Transistors (FETs)

Example E1 - Pg 161

In [1]:
#Calculate saturation voltage and saturation current
#given
Vp=-4.#V          #pinch off voltage
Idss=12.*10.**-3.;#A      #drain to source current with gate shorted
Vgs=-2.;#V             #gate to source voltage
Vds=Vgs-Vp;
Id=Idss*(Vds/Vp)**2.;
print '%s %.f %s' %("Saturation Voltage is =",Vds,"V\n");
print '%s %.f %s' %("Saturation current is =",Id*10**3,"mA");

Saturation Voltage is = 2 V

Saturation current is = 3 mA

Example E2 - Pg 162

In [2]:
#Find the value of drain current
#given
Vgso=-5.;#V          #gate to source cut off voltage
Idss=20.*10.**-3.;#A     #drain to source current with gate shorted

#At vgs = -2 V
vgs=-2.;#V input voltage
Id=Idss*(1.-(vgs/Vgso))**2.;          #Schockleys equation
print '%s %.1f %s' %("Drain current is (At vgs = -2 V) =",Id*10**3,"mA\n");

#At vgs = -4 V
vgs=-4.;#V        input voltage
Id=Idss*(1.-(vgs/Vgso))**2.;          #Schockleys equation
print '%s %.1f %s' %("Drain current is (At vgs = -4 V) =",Id*10**3,"mA\n");

#At vgs = -8 V
print '%s' %("Drain current is 0 A (At vgs = -8 V) because gate is biased beyond cut off ");

Drain current is (At vgs = -2 V) = 7.2 mA

Drain current is (At vgs = -4 V) = 0.8 mA

Drain current is 0 A (At vgs = -8 V) because gate is biased beyond cut off

Example E3 - Pg 163

In [3]:
#Calculate Vgs and Vds saturation
#given
import math
Vp=5.#V          #pinch off voltage
Idss=-15.*10.**-3.;#A      #drain to source current with gate shorted
Id=-3.*10.**-3.;#A         #saturation current
Vgs=Vp*(1.-math.sqrt(Id/Idss));
Vds=Vgs-Vp;
print '%s %.3f %s' %("The gate to source voltage (Vgs) is =",Vgs,"V\n");
print '%s %.3f %s' %("The saturation voltage is Vds(sat) =",Vds,"V");

print '\nThe value of Vgs = 2.115V and Vds= -2.885V in book because of the calculation error'

The gate to source voltage (Vgs) is = 2.764 V

The saturation voltage is Vds(sat) = -2.236 V

The value of Vgs = 2.115V and Vds= -2.885V in book because of the calculation error

Example E4 - Pg 167

In [4]:
#Calculate drain current Id for N channel
#given
Vp=5.#V          #pinch off voltage
Idss=18.*10.**-3.;#A     #drain to source current with gate shorted

#For Vgs= - 3 V
Vgs=-3.;#V
Id=Idss*(1.-(Vgs/(-Vp)))**2.;
print '%s %.2f %s' %("The drain current Id(For Vgs= -3V) =",Id*10**3,"mA\n");

#For Vgs= 2.5 V
Vgs=2.5;#V
Id=Idss*(1.-(Vgs/(-Vp)))**2.;
print '%s %.1f %s' %("The drain current Id(For Vgs= 2.5V) =",Id*10**3,"mA");

The drain current Id(For Vgs= -3V) = 2.88 mA

The drain current Id(For Vgs= 2.5V) = 40.5 mA

Example E5 - Pg 167

In [5]:
#Calculate drain current Id for P channel
#given
Vp=-5.#V          #pinch off voltage
Idss=18.*10.**-3.;#A     #drain to source current with gate shorted

#For Vgs= -3V
Vgs=-3.;#V
Id=Idss*(1.-(Vgs/(-Vp)))**2.;
print '%s %.2f %s' %("The drain current Id (For Vgs= -3V) =",Id*10**3,"mA\n");

#For Vgs= 2.5V
Vgs=2.5;#V
Id=Idss*(1.-(Vgs/(-Vp)))**2.;
print '%s %.1f %s' %("The drain current Id (For Vgs= 2.5V) =",Id*10**3,"mA");

The drain current Id (For Vgs= -3V) = 46.08 mA

The drain current Id (For Vgs= 2.5V) = 4.5 mA

Example E6 - Pg 172

In [6]:
#Find the value of drain current
#given
Vt=2.;#V      #threshold voltage
K=0.25*10.**-3.;# A/V**2      #conductivity parameter
Vgs=3.;#V #gate supply
Vds=2.;#V #saturation voltage
Vdsm=Vgs-Vt;        #minimum voltage required to pinch off

# Vds > Vdsm therefore the device is in saturation region

Id=K*(Vgs-Vt)**2.;
print '%s %.2f %s' %("The drain current is =",Id*1000,"mA");

The drain current is = 0.25 mA

Example E7 - Pg 172

In [7]:
#Find the value of Id
#given
Vt=1.5;#V      #threshold voltage
Id=2.*10.**-3.;#A
Vgs=3.;#V #gate supply
Vds=5.;#V #saturation voltage
Vdsm=Vgs-Vt;        #minimum voltage required to pinch off

# Vds > Vdsm therefore the device is in saturation region

# Calculating K
K=Id/((Vgs-Vt)**2.);       # A/V**2      #conductivity parameter

#Calculating Id for Vgs= 5 V and Vds= 6 V
Vgs=5;#V #gate supply
Vds=6;#V #saturation voltage
Id=K*((Vgs-Vt)**2);
print '%s %.2f %s' %("The drain current is =",Id*1000,"mA");

The drain current is = 10.89 mA

Example E8 - Pg 174

In [8]:
#Calculate the dynamic drain resistance
#given
gm=200.*10.**-6.;#S transconductance
u=80.;#amplification factor
rd=u/gm;
print '%s %.f %s' %("The dynamic drain resistance is =",rd/1000,"k ohm");

The dynamic drain resistance is = 400 k ohm