Chapter 08 : BJT specifications and performance

Example 8.2, Page No 313

In [1]:
import math
#initialisation of variables
P2=25.0*10**-3#when frequency increase to 20KHz
P1=50.0*10**-3#when signal frequency is 5KHz

#Calculations
Po=10*math.log((P2/P1),10)

#Results
print(" output power change in decibels is %.2f dB " %Po)

 output power change in decibels is -3.01 dB

Example 8.3, Page No 314

In [2]:
import math

#initialisation of variables
v1=1# output voltage measured at 5KHz
v2=.707# output voltage measure at 20kHz

#Calculations
Po=20*math.log((v2/v1),10)

#Results
print(" output power change is %.2f dB " %Po)

 output power change is -3.01 dB

Example 8.4 Page No 317

In [3]:
import math

#initialisation of variables
Ic=1.0*10**-3
hfe=50.0
hie=1.3*10**3
fT=250.0*10**6
Cbc=5.0*10**-12
Rc=8.2*10**3
Rl=100.0*10**3

#Calculations
Ie=Ic
Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie
Cbe=(6.1*Ie)/fT
Cin=(Cbe+(1+Av)*Cbc)*10**9

#Results
print(" input capacitance when the circuit operated as CE is %.2fnF "%Cin)

 input capacitance when the circuit operated as CE is 1.49nF

Example 8.5, Page No 319

In [4]:
import math
#initialisation of variables
R1=100*10**3
R2=47.0*10**3
Re=4.7*10**3
Cbc=5.0*10**-12
Cbe=24.4*10**-12
hfe=50
hie=1.3*10**3
hib=24.5
rs=hib
rs=600.0

#Calculations
print(" common emitter circuit")
Rb=(R1*R2)/(R1+R2)
Zi=(Rb*hie)/(Rb+hie)
Cin=1.48*10**-9
f2=1/(2*3.14*Cin*((rs*Zi)/(rs+Zi)))
print("input-capacitance upper cutoff frequency is %dHz " %f2)
print("common base circuit")
Zi=(Re*hib)/(Re+hib)
Cin=(Cbe+Cbc)
f2=(1/(2*3.14*Cin*((rs*Zi)/(rs+Zi))))*10**-6

#Results
print(" input capacitance upper cutoff when operating as CB circuit with base bypassed to ground is %.2f MHz " %f2)

 common emitter circuit
input-capacitance upper cutoff frequency is 265447Hz 
common base circuit
 input capacitance upper cutoff when operating as CB circuit with base bypassed to ground is 231.25 MHz

Example 8.6 Page No 322

In [5]:
import math
#initialisation of variables
fT=50.0*10**6
hfe=50.0
f2o=60.0*10**3
Rc=10.0*10**3

#Calculations
fae=fT/hfe
C4=(1.0/(2*3.14*f2o*Rc))*10**12

#Results
print("capacitance required for C4 to give 60kHz upper cutoff frequency is %.2f pF " %C4)

capacitance required for C4 to give 60kHz upper cutoff frequency is 265.39 pF

Example 8.8 Page No 326

In [6]:
import math

#initialisation of variables
ton=100.0*10**-9
Rs=600.0
Rb=4.7*10**3

#Calculations
C1=(ton/Rs)*10**12
print(" suitable speed up capacitor is %dpF " %C1)
C1=160*10**-12#standard value
PWmin=(5*Rs*C1)
SWmin=5*Rb*C1
fmax=1/(PWmin+SWmin)

#Results
print("maximum signal frequency is %.2f Hz " %(fmax/1000))

 suitable speed up capacitor is 166pF 
maximum signal frequency is 235.85 Hz

Example 8.9, Page No 330

In [7]:
import math
#initialisation of variables
R1=30.0*10**3
R2=30.0*10**3
rs=30.0*10**3
f2=40.0*10**3
f1=100.0
k=1.37*10**-23
R=10.0*10**3
Av=600.0
Ri=3.0*10**3

#Calculations
Rb=(R1*R2)/(R1+R2)
Rg=(rs*Rb)/(rs+Rb)
T=(273+25)
B=f2-f1
en=math.sqrt(4*k*T*B*R)
eni=en*((Ri/(Ri+Rg)))
eno=(Av*eni)*10**6

#Results
print("noise output voltage is %.2f uV " %eno)

noise output voltage is 353.44 uV

Example 8.10 Page No 331

In [8]:
import math 

#initialisation of variables
Ic=30.0*10**-6
Vce=5.0
eno=354.0*10**-6
NF=10.0
F=2.51#F=antilog(NF/10)

#Calculations
Vn=((math.sqrt(F))*eno)*10**6

#Results
print("total noise output volateg for amplifier is %.2f uV " %Vn)

total noise output volateg for amplifier is 560.84 uV

Example 8.11 Page No 333

In [9]:
import math
#initialisation of variables
Pd25=625.0*10**-3
D=5.0*10**-3
Vce=10.0
T2=55.0

#Calculations
Pdt2=Pd25-D*(T2-25)
Pd=Pdt2
Ic=Pd/Vce

#Results
print(" maximum Ic level is %.2fA " %(Ic*1000))

 maximum Ic level is 47.50A

Example 8.13 Page No 335

In [10]:
import math

#initialisation of variables
Pd=80.0
Vce=60.0

#Calculations
Ic=Pd/Vce
print("point 1 Vce=60 and Ic= %.2f A" %Ic)
Vce=40.0
Ic=Pd/Vce
print("point 2 Vce=40 and  Ic= %.2f A" %Ic)
Vce=20.0
Ic=Pd/Vce
print(" point 3 Vce=20 and  Ic= %.2f A" %Ic)
Vce=10.0
Ic=Pd/Vce

#Results
print(" point 4 Vce=10 and  Ic= %.2f A" %Ic)

point 1 Vce=60 and Ic= 1.33 A
point 2 Vce=40 and  Ic= 2.00 A
 point 3 Vce=20 and  Ic= 4.00 A
 point 4 Vce=10 and  Ic= 8.00 A

Example 8.14, Page No 339

In [11]:
import math

#initialisation of variables
Vce=20.0
Ic=1.0
T2=90.0
T1=25.0

#Calculations
Q=Vce*Ic
Qcs=.4
Qjc=1#from table
Qsa=((T2-T1)/Q)-(Qjc+Qcs)

#Results
print("Qsa= %.2f " %Qsa)

Qsa= 1.85