Chapter 4 :Transistor Characteristics

Example 4.1 page no-203

In [70]:
# minimum base current to work transistor in saturation region

import math
#Variable declaration
vcc=12.0     # V
rl=4.0       # Ohm
alfa=0.98

#Calculations
ic=vcc/rl
B=alfa/(1-alfa)
ibmin=ic/B

#Result
print("Ic(saturation)= %d mA\nBeta = %.0f \nIb(min) = %.1f micro A"%(ic,B,ibmin*1000))

Ic(saturation)= 3 mA
Beta = 49 
Ib(min) = 61.2 micro A

Example 4.5 page no-206

In [71]:
# maximum allowable value of RB for transistor in cut off

import math
#Variable declaration
t1=75.0
t2=25.0
icbo=2.0         # at T1=25
vbe=0.1
vbb=5

#Calculations
icbo2=icbo*2**((t1-t2)/10)
Rb=(vbb-vbe)/icbo2

#Result
print("Icbo at 75°C = %.0f micro A\nRb = %.1f K-Ohm"%(icbo2,Rb*1000))

Icbo at 75°C = 64 micro A
Rb = 76.6 K-Ohm

Example 4.6 page no-207

In [72]:
# temperature increase before transistor comes of cut off

import math
#Variable declaration
vbb=-1.0      # V
Rb=50.0       # K-Ohm
vbe=-0.1

#Calculations
Icbo=(vbe-vbb)/Rb
t=math.log(Icbo*1000/2)*10/(math.log(2))

#Result
print("Icbo = %.0f micro A"%(Icbo*1000))
print("Delta_T = %d°C \nHence, T = %d°C"%(math.ceil(t),math.ceil(t)+25))

Icbo = 18 micro A
Delta_T = 32°C 
Hence, T = 57°C

Example 4.7 page no-207

In [73]:
# calculation of ib ic and vbc for transistor AF 114

import math
#Variable declaration

vce = - 0.07   # V 
vbe = - 0.21   # V
vcc=-9.0
rc=1.0         # K-Ohm
rb=30.0        # K-Ohm

#Calculations
ic=(vcc-vce)/rc
ib=(vcc-vbe)/rb
vbc=vbe-vce

#Result
print("Ic = %.2f mA\nIB = %.3f mA\nVbc = %.2f V"%(ic,ib,vbc))

Ic = -8.93 mA
IB = -0.293 mA
Vbc = -0.14 V

Example 4.8 page no-208

In [74]:
# calculation of resistance in CE configuration

import math
#Variable declaration

alfa=0.98
Ie=-2.0      # in mA IE is negative because it is NPN transistor
Ic=-alfa*Ie
Ib=(1-alfa)*(-Ie)
vbe=0.6      # V
vcc=12.0     # V
re=100.0     # ohm
r2= 20000.0  # ohm
r1=3.3       # k-Ohm

#Calculations
vbn=vbe-(Ie*re*10**-3)
Ir2=vbn*10**3/r2
Ir1=Ir2+Ib
vr1=vcc-((Ir1+Ic)*r1)-vbn
R1=vr1/Ir1

#Result
print("Ic = %.2f mA\nIb = %.0f micro A\nV_BN =%.1f V"%(Ic,Ib*1000,vbn))
print("IR1 = %.0f micro A\nIR2 = %.0f micro A\nIrc = %.2f mA"%(Ir1*1000,Ir2*1000,Ir1+Ic))
print("R1 = %d K-Ohm"%(math.ceil(R1)))

Ic = 1.96 mA
Ib = 40 micro A
V_BN =0.8 V
IR1 = 80 micro A
IR2 = 40 micro A
Irc = 2.04 mA
R1 = 56 K-Ohm

Example 4.9 page no-208

In [75]:
# Barrier Potential

import math
#Variable declaration
eps=12/(36*math.pi*10**11)   # F/cm
mup=500.0                    # cm^2/V-Sec

#Calculations
Vb=(2.54/1000)**2/(2*eps*mup)

#Result
print("VB = %.1f*10^3*W^2/rho_B"%(Vb/1000))

VB = 6.1*10^3*W^2/rho_B

Example 4.10 page no-210

In [76]:
# Av Ai and Ap of transistor in CB configuration

import math
#Variable declaration
alfa=0.96
Rl=5000.0
x=80.0

#Calculations
Av=alfa*Rl/x
pg=Av*alfa

#Result
print("Power Gain = %.1f"%pg)

Power Gain = 57.6

Example 4.11 page no-211

In [77]:
# Av Ai and Ap of Transistor in CE configuration

import math
#Variable declaration
alfa = 0.96
B=alfa/(1-alfa)
x=80.0
Rl=75000.0   # ohm

#Calculations
Av=B*Rl/x
Ap=Av*B

#Result
print("power gain = %.0f"%Ap)

power gain = 540000

Example 4.12 page no-211

In [78]:
# Junction voltages for open collector transistor

import math
#Variable declaration
ico=2.0          # micro A
ieo=1.6          # micro A
alfa = 0.98
ie=2.0           # micro A
T=301.6

#Calculations
vt=T/11600.0
ve=vt*math.log(1+(ie/ieo))
vc=vt*math.log(1+(alfa*ie/ico))

#Result
print("Ve = %.4f V"%ve)
print("Vc = %.4f V\nV_CE = %.4f V"%(vc,vc-ve))

Ve = 0.0211 V
Vc = 0.0178 V
V_CE = -0.0033 V

Example 4.13 page no-212

In [79]:
# variation in Vi corresponding to variation in Vo

import math
#Variable declaration

rs=200.0       # Ohm
vz=100.0       # V
rz=20.0        # Ohm
il=50.0        # mA
iz=0.01        # mA
ilmax=100.0    # mA

#calculations
izmin=0.1*ilmax
vl=vz+iz*rz
v1=vl+((il/1000)+iz)*rs
vldash=vl+1
izdash=(vldash-100)/rz
it=(il/1000)+izdash
vt=vldash+(rs*it)

#Result
print("V_L = %.1f V"%vl)
print("V1 = %.1fV"%v1)
print("Increase in Iz = %.2f mA"%izdash)
print("Total Current = %.1f A\nTotal Voltage = %.1f V\nchange in V1 =%.0fV"%(it,vt,vt-v1))
print("\nA change of 11 V in V, on the input side produces a change of")
print("1V on the output side due to zener diode action")

V_L = 100.2 V
V1 = 112.2V
Increase in Iz = 0.06 mA
Total Current = 0.1 A
Total Voltage = 123.2 V
change in V1 =11V

A change of 11 V in V, on the input side produces a change of
1V on the output side due to zener diode action

Example 4.14 page no-226

In [80]:
# Design of bias circuit for zero drain current drift

import math
#Variable declaration
vp=-3.0       # V
idss=1.75     # mA
rd=5.0        # K-Ohm
gmo=1.8       # mA/V

#Calculations
#(a)
id=idss*(1-(vgs/vp))**2
#(b)
vgs=vp-0.63   # V
#(c)
rs=-vgs/0.08
#(d)
gm=gmo*(vgs-vp)/vp
Av=gm*rd

#Rezult
print("(a)Id for zero drift current\nId = %.2f mA"%id)
print("\n(b)\nVgs = %.2f V\n\n(c)\nRs = %d K-Ohm\n\n(d)\ngm = %.3f mA/V\nAv = %.2f"%(vgs,rs,gm,Av))

(a)Id for zero drift current
Id = 0.01 mA

(b)
Vgs = -3.63 V

(c)
Rs = 45 K-Ohm

(d)
gm = 0.378 mA/V
Av = 1.89

Example 4.15 page no-228

In [81]:
# pinch off voltage

import math
#Variable declaration
a=2*10**-4  # cm
rho = 10.0  # Ohm-cm

#Calculations
eps=12.0/(36*math.pi*10**11) 
mup = 500.0 #cm^2/V-sec
ena=1/(rho*mup)
vp= (ena*a**2)/(2*eps)


#Result
print("Vp = %.2f V"%vp)

Vp = 3.77 V

Example 4.16 page no-231

In [82]:
print("Same as problem 4.15 in the same chapter")

Same as problem 4.15 in the same chapter

Example 4.17 page no-231

In [83]:
# pinch off voltage and channel half width

import math
#Variable declaration

a=3*10**-4         # cm
nd=10**15          # electrons/cm^3
e=1.6*10**-19      # C

#Calculations
eps=12.0/(36*math.pi*10**11)
vp=e*nd*a**2/(2*eps)
b=a*(1-(1.0/2)**(1.0/2))

#Result
print("(a)\nVp = %.1f V"%vp)
print("\n(b)Vgs=Vp/2\nb = %.2f * 10^-4 cm"%(b*10**4))

(a)
Vp = 6.8 V

(b)Vgs=Vp/2
b = 0.88 * 10^-4 cm

Example 4.20 page no-241

In [84]:
# design of self bias circuit

import math
#Variable declaration

vdd=30.0   # v
rl=4.7     # k-ohm
vd=20.0    # v

#Calculations
id=(vdd-vd)/rl
del_id=1/rl
delv=vdd-vd
deli=2.5 
rs=delv/(deli)

#Result
print("Id = %.1f mA"%id)
print("\nfor Vd to be constant, it should be within ±1V.")
print("Delta_Id = ± %.1f mA\nId(min) = %.3f mA\nId(max) = %.3f mA"%(del_id,id-del_id,id+del_id))
print("Rs = %d K-Ohm"%rs)

Id = 2.1 mA

for Vd to be constant, it should be within ±1V.
Delta_Id = ± 0.2 mA
Id(min) = 1.915 mA
Id(max) = 2.340 mA
Rs = 4 K-Ohm

Example 4.21 page no-243

In [85]:
# Voltage gain and output impedance of common source amplifier

import math
#Variable declaration
rd=100*10**3    # Ohm
gm=3000*10**-6
rl=10000.0      # Ohm
f=10**6         # Hz
c=3*10**-12     # F
r0= 9.09        # K-Ohm
#Calculations
Av=(-gm*rd*rl)/(rd+rl)
xc=1/(2*math.pi*f*c)
z0 = (r0*xc)/math.sqrt(r0**2 + (xc/1000)**2)

#Result
print("(a)\nAv = %.1f"%Av)
print("\n(b)\nXc = %d K-Ohm"%(xc/1000))
print("Z0 = %.2f K-Ohm"%(z0/1000))

(a)
Av = -27.3

(b)
Xc = 53 K-Ohm
Z0 = 8.96 K-Ohm

Example 4.22 page no-245

In [86]:
# calculation of Vgs Id and Vds

import math
#Variable declaration

idss=5*10**-3        # mA
vp = -5.0            # V
rs =5000.0           # Ohm
rl=2.0               # k-ohm
vdd=10.0

#Calculations
#Vgs^2+11Vgs+25=0 fro equation of Id and Vgs
vgs=(-11+math.sqrt(121-100))/2
id=idss*(1-(vgs/vp))**2
x=id*rl*1000
y=id*rs
vds =vdd-x-y

#Result
print("Vgs = %.2fV\nId = %.2f mA\nVds = %.1f V\nThe FET must be conducting."%(vgs,id*1000,vds))
print("\nIf VGS = -7.8V, the FET in cut off. Therefore Vp = -5V.")
print("Therefore VGS is chosen as -3.2V")

Vgs = -3.21V
Id = 0.64 mA
Vds = 5.5 V
The FET must be conducting.

If VGS = -7.8V, the FET in cut off. Therefore Vp = -5V.
Therefore VGS is chosen as -3.2V