Chapter-9 BJT and JFET Frequency Response

Example- 9.1 Page number-539

In [15]:
import math
e=2.718
print "a:",math.log10(10**6)
print "b:",round(math.log(e**3),1)
print "c:",math.log10(10**-2)
print "d:",round(math.log(e**-1),1)
a: 6.0
b: 3.0
c: -2.0
d: -1.0

Example- 9.2 Page number-539

In [7]:
import math
print "a:",round(math.log10(64),3)
print "b:",round(math.log(64),3)
print "c:",round(math.log10(1600),3)
print "d:",round(math.log10(8000),3)
a: 1.806
b: 4.159
c: 3.204
d: 3.903

Example- 9.3 Page number-540

In [14]:
import math
e=2.718
print "a:",round(10**(1.6),2)
print "b:",round(e**0.04,4)
a: 39.81
b: 1.0408

Example- 9.4 Page number-541

In [23]:
import math
x=4000.0/250.0

print "a:",round(math.log10(0.5),1)
print "b:",round(math.log10(x),3)
print "c:",round(math.log10(0.6*30),3)
a: -0.3
b: 1.204
c: 1.255

Example- 9.6 Page number-544

In [25]:
A=100.0                   #Gain in dB

#we know: A=20*log(x),therefore:
x=10**(A/20)

print "Magnitude of gain:",x
Magnitude of gain: 100000.0

Example- 9.7 Page number-544

In [33]:
import math
#from the data given in the question:
Pi=10000                         #input power in watt
Po=500.0                          #output power in watt
Vi=1000                          #input voltage in volts
Zo=20                            #output impedance in ohm

#Calculation:
Gp=10*math.log10(Po/Pi)            #power gain in dB
Vo=math.sqrt(Po*Zo)                 #output voltage in volts
Gv=20*math.log10(Vo/Vi)            #voltage gain in dB


print "a:Power gain:",round(Gp,2),"dB"
print "b:Voltage gain:",Gv,"dB"
a:Power gain: -13.01 dB
b:Voltage gain: -20.0 dB

Example- 9.8 Page number-544

In [34]:
import math
#from the data given in the question:
Po=40.0                          #output power in watt
Zo=10                            #output impedance in ohm
Pi=25                            #input power in dB

#Calculation: 
Piw=Po/(10**2.5)                 #input power in watt
Vo=math.sqrt(Po*Zo)              #output voltage in volts
Vi=Vo/100                        #input voltage in V


print "a:Input power :",round(Piw*1000,1),"dB"
print "b:Input Voltage:",Vi,"V"
a:Input power : 126.5 dB
b:Input Voltage: 0.2 V

Example- 9.10 Page number-553

In [40]:
#from the given figure:
C=0.1*(10**-6)               #capacitance in farad
R=5*(10**3)                  #Resistance in ohm
Avd=-6.0                      #gain in dB

#calculation:

f1=1/(2*3.14*R*C)               #break frequency in Hz
Av=10**(Avd/20)                 #Gain

print "Break frequency:",round(f1,1),"Hz"
print "Gain:",round(Av,3)
Break frequency: 318.5 Hz
Gain: 0.501

Example- 9.11 Page number-558

In [99]:
#from the data given in the question:
Cs=10.0                     #source capacitor in microF
Ce=20.0                     #emitter capacitor in microF
Cc=1.0                      #collector capacitor in microF
Rs=1.0                      #source Resistance in kohm
Re=2.0                      #emitter Resistance in kohm
Rc=4.0                      #collector Resistance in kohm
R1=40.0                     #in kohm
R2=10.0                     #in kohm
Rl=2.2                     #load resistance in kohm
B=100.0
Vcc=20.0                    #supply voltage in volts

#Calculation:

#since,B*Re>>10*R2, we can apply voltage divider configuration:
Vb=(R2*Vcc)/(R2+R1)          #Base voltage in Volts
Ve=Vb-0.7                    #emitter voltage in volts
Ie=Ve/Re                     #emitter current in mA
re=26/Ie                     #in ohm
x=(B*re)/1000                #temporary value
t=(Rc*Rl)/(Rc+Rl)            #effective resistance for Rc||Rl in kohm
Av=-round((t/re)*1000)       #midband gain

Y=(R1*R2)/(R1+R2)            #temporary value
Zi=(Y*x)/(Y+x)               #input impedance in kohm

d=round(Zi/(Zi+Rs),4)        #temporary value
Avs=round(d*Av,2)            #new gain

#calculating effect of capacitors:
Ri=Zi
Fls=1/(2*3.14*(Rs+Ri)*Cs)           #cut-off frequency due to source capacitance in Hz

Flc=1/(2*3.14*(Rc+Rl)*Cc)           #cut-off frequency due to collector capacitance in Hz

Rsnew=(Y*Rs)/(Y+Rs)                  #effective resistance of R1||R2||Rs
g=(Rsnew/B)*1000+re
Re=Re*1000                           #emitter resistance in ohm
Recf=round((g*Re)/(g+Re),2)                   
Fle=1/(2*3.14*Recf*Ce)              #cut-off frequency due to emitter capacitance in Hz

print "Value of re:",round(re,2),"ohm"
print "cut-off frequency due to source capacitance Cs:",round(Fls*1000,2),"Hz"
print "cut-off frequency due to collector capacitance Cc:",round(Flc*1000,2),"Hz"
print "cut-off frequency due to emitter capacitance Ce:",round(Fle*(10**6),1),"Hz"

print "The cut-off frequency of the network:",round(max(Fls*1000,Flc*1000,Fle*(10**6)),1),"Hz"
Value of re: 15.76 ohm
cut-off frequency due to source capacitance Cs: 6.87 Hz
cut-off frequency due to collector capacitance Cc: 25.68 Hz
cut-off frequency due to emitter capacitance Ce: 327.0 Hz
The cut-off frequency of the network: 327.0 Hz

Note: The difference in result obtained is due to different precision of values taken at each step of the solution

Example- 9.12 Page number-566

In [30]:
#from the drawn characterstics graph:
Vgs=-2.0                           #Gate source voltage in volts
Id=2                               #Drain current in mA
Cg=0.01                            #second coupling capacitor in microF
Cs=2.0                             #source capacitor in microF
Cc=0.5                             # second coupling capacitor in microF
Rsig=10.0                          #input Resistance in kohm
Rg=1000.0                          #gate Resistance in Mohm
Rd=4.7                             #drain Resistance in kohm
Rs=1.0                             #source resistancein kohm
Idss=8                             #drain saturation current in mA
Vp=-4.0                            #threshold voltage in volts
Vdd=20                             #supply voltage in volts
Rl=2.2                             #load resistance in kohm

#Calculation:
gmo=(2*Idss)/Vp
gm=gmo*(1-((Vgs/Vp)))
Ro=Rd
#calculating effect of capacitors:
Flg=1/(2*3.14*Cg*(10**-3)*(Rsig+Rg))                              #effect of Coupling capacitor
Flc=1/(2*3.14*Cc*(10**-3)*(Ro+Rl))                                #effect of coupling capacitor

p=-1/gm                                                   #temporary value
Req=((Rs*p)/(Rs+p))*1000                                  
Fls=1/(2*3.14*Req*Cs)                                    #effect of source capacitor

print "cut-off frequency due to source capacitance Cs:",round(Flc,2),"Hz"
print "cut-off frequency due to 1st coupling capacitance Cg:",round(Flg,2),"Hz"
print "cut-off frequency due to 2nd coupling capacitance Cs:",round(Fls*(10**6),1),"Hz"

print "The cut-off frequency of the network:",round(max(Flc,Flg,Fls*(10**6)),1),"Hz"
cut-off frequency due to source capacitance Cs: 46.16 Hz
cut-off frequency due to 1st coupling capacitance Cg: 15.77 Hz
cut-off frequency due to 2nd coupling capacitance Cs: 238.9 Hz
The cut-off frequency of the network: 238.9 Hz

Example- 9.13 Page number-576

In [46]:
#from the data given in the question:
Cs=10.0                      #source capacitor in microF
Ce=20.0                      #emitter capacitor in microF
Cc=1.0                       #collector capacitor in microF
Rs=1.0                       #source Resistance in kohm
Re=2.0                       #emitter Resistance in kohm
Rc=4.0                       #collector Resistance in kohm
R1=40.0                      #in kohm
R2=10.0                      #in kohm
Rl=2.2                       #load resistance in kohm
B=100.0
Vcc=20.0                     #supply voltage in volts

Cbe=36.0                      #base-emitter capacitor in pF
Cbc=4.0                       #base-collector capacitor in pF
Cce=1.0                       #collector-emitter capacitor in pF
Cwi=6                         #in pF
Cwo=8                         #in pF
Ri=1.32                       #in kohm
Avmid=-90.0                   #normal Gain
re=15.76                      #in ohm
#calculation:


y=(R1*R2)/(R1+R2)                #temporary value
z=(Rs*y)/(Rs+y)                  #temporary value
Rthi=(Ri*z)/(Ri+z)               #effective input resistance in kohm
Rtho=(Rc*Rl)/(Rc+Rl)             #effective output resistance in kohm
Ci=Cwi+Cbe+(Cbc*(1-Avmid))       #input capacitance in pF
Co=Cwo+Cce+(1-(1/Avmid))*Cbc     #output capacitance in pF

Fhi=1/(2*Rthi*3.14*Ci*(10**-6))           #cut-off frequency for input network
Fho=1/(2*Rtho*3.14*Co)                    #cut-off frequency for output network

Fb=1/(2*3.14*B*re*(Cbe+Cbc)*(10**-6))
Ft=B*Fb

print "cut-off frequency for input network Fhi:",round(Fhi,2),"kHz"
print "cut-off frequency for output network Fho:",round(Fho*1000,2),"MHz"
print "Beta cut-off frequency Fb:",round(Fb,2),"Hz"
print "Gain Bandwidth Product Ft:",round(Ft,1),"Hz"
cut-off frequency for input network Fhi: 738.36 kHz
cut-off frequency for output network Fho: 8.6 MHz
Beta cut-off frequency Fb: 2.53 Hz
Gain Bandwidth Product Ft: 252.6 Hz

Note: The difference in result obtained is due to different precision of values taken at each step of the solution