Chapter 8: FET Amplifiers

Example 8.1, Page Number: 253

In [1]:
# variable declaration
g_m=4.0*10**-3; #gm value
R_d=1.5*10**3;  #resistance

#calculation
A_v=g_m*R_d;     #voltage gain

#result
print "Voltage gain = %.2f" %A_v

Voltage gain = 6.00

Example 8.2, Page Number: 253

In [2]:
# variable declaration
r_ds=10.0*10**3;
R_d=1.5*10**3;    #from previous question
g_m=4.0*10**-3;   #from previous question

#calculation
A_v=g_m*((R_d*r_ds)/(R_d+r_ds));  #voltage gain

#result
print "Voltage gain = %.2f" %A_v

Voltage gain = 5.22

Example 8.3, Page Number:254

In [3]:
# variable declaration
R_s=560;       #resistance in ohm
R_d=1.5*10**3;  #resistance in ohm
g_m=4*10**-3;   #g_m value

#calculation
A_v=(g_m*R_d)/(1+(g_m*R_s))  #voltage gain

#result
print "Voltage gain = %.2f" %A_v

Voltage gain = 1.85

Example 8.4, Page Number: 257

In [4]:
#Variable declaration
vdd=12.0 #volts
Id=1.96*10**-3 #Amp
Rd=3.3*10**3 #ohm
Idss=12.0*10**-3 #Amp
Rs=910           # Ohm
vgsoff= 3         #v
vin=0.1          #V

#calculation
vd=vdd-(Id*Rd)
vgs=-Id*Rs
gm0=2*Idss/(abs(vgsoff))
gm=0.00325         #mS
vout=gm*Rd*vin
vout=vout*2*1.414
#Result
print"Total output ac voltage(peak-to-peak) = %f V \nridig on DC value of %fV "%(vout,vd)

Total output ac voltage(peak-to-peak) = 3.033030 V 
ridig on DC value of 5.532000V

Example 8.5, Page Number: 258

In [5]:
# variable declaration
R_D=3.3*10**3;    #resistance in ohm
R_L=4.7*10**3;    #load resistance in ohm
g_m=3.25*10**-3;   #from previous question
V_in=100.0*10**-3;   #previous question

#calculation
R_d=(R_D*R_L)/(R_D+R_L);    #Equivalent drain resistance
V_out=g_m*R_d*V_in;         #output RMS voltage in volt

#result
print "Output voltage rms value = %.2f Volts" %V_out

Output voltage rms value = 0.63 Volts

Example 8.6, Page Number: 259

In [6]:
# variable declaration
I_GSS=30.0*10**-9;           #current in ampere
V_GS=10.0;                   #ground-source voltage
R_G=10.0*10**6;              #resistance in ohm

#calculation
R_IN_gate=V_GS/I_GSS;                    #gate input resistance
R_in=(R_IN_gate*R_G)/(R_IN_gate+R_G);    #parallel combination

#result
print "Input resistance as seen by signal source = %.2f ohm" %R_in

Input resistance as seen by signal source = 9708737.86 ohm

Example 8.7, Page Number: 260

In [7]:
# variable declaration
I_DSS=200.0*10**-3;
g_m=200.0*10**-3;
V_in=500.0*10**-3;
V_DD=15.0;
R_D=33.0;
R_L=8.2*10**3;

#calculation
I_D=I_DSS;    #Amplifier is zero biased
V_D=V_DD-I_D*R_D;
R_d=(R_D*R_L)/(R_D+R_L);
V_out=g_m*R_d*V_in;

#result
print "DC output voltage = %.2f Volts" %V_D
print "AC output voltage = %.2f volts" %V_out

DC output voltage = 8.40 Volts
AC output voltage = 3.29 volts

Example 8.8, Page Number: 262

In [8]:
# Theoretical example
# result

print "Part A:\nQ point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,"
print "At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is" 
print "the difference of the two drain currents=1.6mA"
print "\nPart B:\nQ point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,"
print "At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is"
print" the difference of the two drain currents=2.8mA"
print "\nPart C:\nQ point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,"
print " At V_GS=7V, I_D=1.7mA. So peak to peak drain current is"
print " the difference of the two drain currents=2.2mA"

Part A:
Q point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,
At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is
the difference of the two drain currents=1.6mA

Part B:
Q point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,
At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is
 the difference of the two drain currents=2.8mA

Part C:
Q point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,
 At V_GS=7V, I_D=1.7mA. So peak to peak drain current is
 the difference of the two drain currents=2.2mA

Example 8.9, Page Number:263

In [9]:
# variable declaration
R_1=47.0*10**3;
R_2=8.2*10**3;
R_D=3.3*10**3;
R_L=33.0*10**3;
I_D_on=200.0*10**-3;
V_GS=4.0;
V_GS_th=2.0;
g_m=23*10**-3;
V_in=25*10**-3;
V_DD=15.0;

#calculation
V_GSnew=(R_2/(R_1+R_2))*V_DD;
K=I_D_on/((V_GS-V_GS_th)**2)
#K=value_of_K(200*10**-3,4,2);
K=K*1000;
I_D=K*((V_GSnew-V_GS_th)**2);
V_DS=V_DD-I_D*R_D/1000;
R_d=(R_D*R_L)/(R_D+R_L);
V_out=g_m*V_in*R_d;

#result
print "Drain to source voltage = %.2f volts" %V_GSnew
print "Drain current = %.2f mA" %I_D
print "Gate to source voltage = %.2f volts" %V_DS
print "AC output voltage = %.2f volts" %V_out
print "Answer in textbook are approximated"

Drain to source voltage = 2.23 volts
Drain current = 2.61 mA
Gate to source voltage = 6.40 volts
AC output voltage = 1.72 volts
Answer in textbook are approximated

Example 8.10, Page Number: 266

In [10]:
# variable declaration
V_DD=-15.0;    #p=channel MOSFET
g_m=2000.0*10**-6;    #minimum value from datasheets
R_D=10.0*10**3;
R_L=10.0*10**3;
R_S=4.7*10**3;

#calculation
R_d=(R_D*R_L)/(R_D+R_L);    #effective drain resistance
A_v=g_m*R_d;
R_in_source=1.0/g_m;
#signal souce sees R_S in parallel with ip rest at source terminal(R_in_source)
R_in=(R_in_source*R_S)/(R_in_source+R_S);   

#result 
print "minimum voltage gain = %.2f" %A_v
print "Input resistance seen from signal source = %.2f ohms" %R_in

minimum voltage gain = 10.00
Input resistance seen from signal source = 451.92 ohms