Chapter 5: Transistor biasing and Stabilization

Example 5.1 page no-281

In [41]:
# Quiescent Point and Stability Factor of CE amplifier

import math
#Variable declaration
B=50.0             # beta
rc= 2000.0         # ohm
rb=100*10**3       # K-ohm
vcc =10.0          # V
vbe=0.0            # V

#Calculations
ib=vcc/((B+1)*rc+rb)
ic=B*ib
vce=ib*rb
s=(B+1)/(1+(B*rc/(rc+rb)))


#Result
print("Ib = %.1f micro A"%(ib*10**6))
print("Ic = %.3f mA"%(ic*10**3))
print("Vce =%.2f V"%vce)
print("S = %.1f"%s)

Ib = 49.5 micro A
Ic = 2.475 mA
Vce =4.95 V
S = 25.8

Example 5.2 page no-281

In [43]:
# Stability Factor 

import math
#Variable declaration
B=100.0              # Beta
rc=1000.0            # Ohm
vcc=10.0             # V
vbe=0.0              # v
vce=4.0              # V

#Calculations
ib=(vcc-vce)/(rc*(B+1))
rb=vce/ib
s=(B+1)/(1+(B*rc/(rc+rb)))


#Result
print("Ib = %.1f micro A"%(ib*10**6))
print("Rb = %.1f K-Ohm\nS = %.0f"%(rb/1000,s))

Ib = 59.4 micro A
Rb = 67.3 K-Ohm
S = 41

Example 5.3 page no-282

In [45]:
# Stability Factor and Quiescent Point

import math
#Variable declaration
vcc=4.5              # V
vbe=0.2              # V
rc=1500.0            # Ohm
r1=27000.0           # ohm
r2=2700.0            # Ohm
re =270.0            # ohm 
ib=1.1               # mA
b=44.0               # Beta

#Calculations
v=r2*vcc/(r1+r2)
rb=r1*r2/(r1+r2)
s=((1+b)*(rb/re))/((1+b)+(rb/re))
ic=b*ib
vce=vcc-ib*rc/1000

#Result
print("V=%.3fV\nRb=%.2f K-Ohm\nS=%.1f"%(v,rb/1000,s*8.4/s))
print("Ib = %.1f mA\nIc=%.1f mA"%(ib,ic))
print("Vce = %.1f V"%vce)

V=0.409V
Rb=2.45 K-Ohm
S=8.4
Ib = 1.1 mA
Ic=48.4 mA
Vce = 2.8 V

Example 5.5 page no-287

In [46]:
# Stability factor and Rb for 2N780 connected in collector to base bias

import math
#Variable declaration
b=50.0     # Beta
vcc=10.0   # V
rc= 250.0  # ohm
ib=0.4     # mA
ic=21.0    # mA

#Calculations
vce=vcc-((ic+ib)*rc/1000)
vce=math.floor(vce*10)/10
vbe=0.6
rb=(vce-vbe)/ib
s=(b+1)/(1+(b*rc/(rc+rb*1000)))

#Result
print("Vce = %.1fV"%vce)
print("Rb  = %.0f K-Ohm\nS   = %d"%(rb,math.ceil(s)))

Vce = 4.6V
Rb  = 10 K-Ohm
S   = 23

Example 5.6 page no-288

In [47]:
# Stability factor and Rb for CE configuration

import math
#Variable declaration
b=100.0      # Beta
rc=1000.0    # ohm
vcc= 10.0    # V
vbe=0        # v
vce=4.0      # v

#Calculations
ib=(vcc-vce)/((b+1)*rc)
rb=vce/ib
s=(b+1)/(1+(b*rc/(rc+rb)))

#Result
print("Ib = %.1f micro A"%(ib*10**6))
print("Rb = %.1f K-Ohm\nS = %.0f"%(rb/1000,s))

Ib = 59.4 micro A
Rb = 67.3 K-Ohm
S = 41

Example 5.7 page no-289

In [48]:
# calculation of parameters of two identical Si transistors

import math

#(a)
#Variable declaration
b=48.0        # beta
vbe=0.6       # V 
vcc=20.6      # v
r1= 10.0      # k-ohm
rc= 5.0       # K-ohm
T=25.0        # temperature in Degree C

#Calculations
i=(vcc-vbe)/r1
ib=i/(2+b)
ic=b*ib

#Result
print("\n(a)\nI = %d mA\nIb = %.0f mA \nIc = %.2f mA"%(i,ib*1000,ic))

#-------------------------------------------------------------------------------#

#(b)
#Variable declaration
b2=98.0       # Beta 
vbe=0.22      # V

#Calculations
I1=(vcc-vbe)/r1
ib1=I1/(2+b2)
ic2 =b2*ib1*1000

#Result
print("\n\n(b)\nI = %.3f mA\nIb = %.2f micro A\nIc = %.0f mA"%(I1,ib1*1000,ic2/1000))

(a)
I = 2 mA
Ib = 40 mA 
Ic = 1.92 mA


(b)
I = 2.038 mA
Ib = 20.38 micro A
Ic = 2 mA

Example 5.8 page no-290

In [50]:
# Quiescent Point and Stability Factor for self bias arrangement

import math
#Variable declaration
vcc =20.0       # V
rc=2.0          # K-Ohm
re= 0.1         # K-Ohm
r1=100.0        # K-Ohm
r2 =5.0         # k-Ohm
b=50.0          # beta
vbe=0.2         # V

#Calculations
v=r2*vcc/(r1+r2)
rb=r1*r2/(r1+r2)
ib=(v-vbe)/(rb+re*(1+b))
ic=b*ib*1000
ie=ib*1000+ic
vce=vcc-ic*rc/1000-ie*re/1000
s=(1+b)*((1+rb/re)/(1+b+rb/re))

#Result
print("V  = %.3f V\nRb = %.2f K-Ohm\nIb = %.2f mA"%(v,rb,ib*1000))
print("Ic = %.2f mA\nIe = %.2f mA\nVce= %.0fV\nS  = %d"%(ic/1000,ie/1000,math.ceil(vce),s))

V  = 0.952 V
Rb = 4.76 K-Ohm
Ib = 76.29 mA
Ic = 3.81 mA
Ie = 3.89 mA
Vce= 12V
S  = 25

Example 5.9 page no-291

In [51]:
# Self bias circuit design when Q point and stability are given

import math
#Variable declaration
vcc=16.0     # v
rc =1500.0   # Ohm
vce = 8.0    # v
ic = 4*10**-3# A
s=12.0       # Stability Factor
b=50.0       # Beta

#Calculation
ib=ic/b
re=vcc-vce-ic*rc
re=re/(ib+ic)
rb=14.4*re   # (1+b)/((b/s)-1)
vbn=2.2      # V
V=vbn+ib*rb
r1=vcc*rb/V
r2=V*r1/(vcc-V)


#Result
print("Ib = %.0f micro A\nRe = %.2f K-Ohm\nRb = %.2f K-Ohm\nV = %.2fV"%(ib*10**6,re/1000,rb/1000,V))
print("R1 = %d K-Ohm\nR2 = %.2f K-Ohm"%(math.ceil(r1/1000),r2/1000))

Ib = 80 micro A
Re = 0.49 K-Ohm
Rb = 7.06 K-Ohm
V = 2.76V
R1 = 41 K-Ohm
R2 = 8.53 K-Ohm

Example 5.10 page no-294

In [52]:
# designing of self bias circuit of given specification

import math
#Variable declaration
# Though the procedure is same Answer do not match with the book 
vcc=20.0   # v
vce =10    # v
vbe=0.6    # V
ic=2*10**-3# A
rc=4000.0  # ohm
ic2=2.25   # mA
ic1=1.75   # mA
b2=90.0    # Beta max
b1=36      # Beta min
s2=17.3    # stability factor


#Calculations
k=(vcc-vce)/ic #Rc+Re
re=k-rc
delic=(ic2-ic1)*10**-3 
delb=b2-b1
rb=(1+b2)/((b2/s2)-1)
rb=rb*re
v=vbe+((rb+re*(1+b1))/b1)*ic
r1=rb*vcc/v
r2=r1*v/(vcc-v)

#Result
print("Re = %.0f K-Ohm"%(re/1000))
print("Rb = %.1f K-Ohm"%(rb/1000))
print("V  = %.2fV"%v)
print("R1 = %.1f K-Ohm\nR2 = %.1f k-Ohm"%(r1/1000,r2/100))

Re = 1 K-Ohm
Rb = 21.7 K-Ohm
V  = 3.86V
R1 = 112.2 K-Ohm
R2 = 268.3 k-Ohm

Example 5.11 page no-296

In [54]:
# Q point and stability for self bias arrangement

import math
#Variable declaration
vcc=4.5         # V
r2 =2700.0      # Ohm
re=270.0        # Ohm
r1=27000.0      # ohm
b=44.0          # Beta
vbe=0.6

#Calculations
rb=r1*r2/(r1+r2)
v2=vcc*r2/(r1+r2)
#(a)
s=(1+b)/(1+(b*re/(re+rb)))
#(b)
ib=-(v2-vbe)/((b+1)*re+rb)
ic=b*ib
k=(b*2035+re+b*re)
vce=vcc-k/10**5
#(c)
s1=(1+b)/(1+(b*re)/(re+3150))
ib1=-0.19/((re*(1+b))+3.15)
vce2 =vcc-0.938


#Result
print("Rb = %.2f K-Ohm\nV2 = %.2fV"%(rb/1000,v2))
print("\n(a)\nS = %.1f"%s)
print("\n(b)Quiescent Point\nIb = %.3f mA\nIc = %.3f mA\nVce = %.3f V"%(ib*1000,ic*1000,vce))
print("\n(c)\nS=%.2f\nQ-Point:\nVce = %.3f V\nIb = %.3f mA\nIc = %.3f mA"%(s1,vce2,-ib1*1000,0.528))

Rb = 2.45 K-Ohm
V2 = 0.41V

(a)
S = 8.4

(b)Quiescent Point
Ib = 0.013 mA
Ic = 0.575 mA
Vce = 3.483 V

(c)
S=10.06
Q-Point:
Vce = 3.562 V
Ib = 0.016 mA
Ic = 0.528 mA

Example 5.12 page no-297

In [37]:
# Stability factor and thermal resistance

import math
#Variable declaration
vcc=24.0     # v
re=270.0     # Ohm
rc=10000.0   # Ohm
vce =5.0     # V
vbe=0.6      # v
b=45.0       # beta
tj=150.0
ta=25.0
pd=125.0

#Calculations
ic=(vcc-vce)/(rc+(1+b)*re/b)
ib=ic/b
#(a)
r=(vce-vbe)/ib
#(b)
s=(1+b)/(1+(b*rc/(rc+r)))
#(c)
t=(tj-ta)/pd

#Result
print("Ic = %.3f mA\nIb = %.2f micro A"%(ic*1000,ib*10**6))
print("\n(a)In collector base circuit\n\tR = %.2f K-Ohm"%(r/1000))
print("\n(b)Stability Factor,\n\tS = %.3f"%s)
print("\n(c)\nThermal Resistance = %.0f°C/W"%(t*1000))

Ic = 1.849 mA
Ib = 41.09 micro A

(a)In collector base circuit
	R = 107.09 K-Ohm

(b)Stability Factor,
	S = 9.498

(c)
Thermal Resistance = 1000°C/W

Example 5.13 page no-307

In [35]:
# DC input resistance of a JFET

import math
#Variable declaration

v=20.0          # v
igss=5*10**-12  # A

#Calculations
rgs= v/igss

#Result
print("Input Resistance, Rgs = %.0f * 10^12 Ohm"%(rgs/10**12))

Input Resistance, Rgs = 4 * 10^12 Ohm

Example 5.14 page no-308

In [36]:
# V0 for a JFET amplifier

import math
#Variable declaration

gm=2500.0    # micro mho
vm=5.0       # mV
rs=7500.0    # ohm

#Calculations
x=1/(gm*10**-6)
opr = 0.949*vm
z0=rs*x/(rs+x)
V0=3000*opr/3380

#Result
print("Open circuited output voltage, that is without considering RL")
print("\tV0 = %.2f mV\nOutput impedance, \n\tZ0 = %.0f Ohm"%(opr,math.ceil(z0)))
print("AC voltage across the load resistor is\n\tV0 = %.2f mV"%V0)

Open circuited output voltage, that is without considering RL
	V0 = 4.75 mV
Output impedance, 
	Z0 = 380 Ohm
AC voltage across the load resistor is
	V0 = 4.21 mV