Chapter6-Transistor Characteristics And Applications

Ex1-pg205

In [1]:
##Ex6_1
import math
IB = 40.*10**-6
IC = 3.*10**-3
beta = IC/IB
alpha = beta/(1.+beta)
print'%s %.2e %s'%("IB = ",(IB),"A")##base current 
print'%s %.2e %s'%("IC = ",(IC),"A")##collector current
print'%s %.2f %s'%("beta = IC/IB = ",(beta),"")##current gain in CE configuration
print'%s %.2f %s'%("alpha = beta/(1+beta) = ",(alpha),"")##current gain in CB configuration

IB =  4.00e-05 A
IC =  3.00e-03 A
beta = IC/IB =  75.00 
alpha = beta/(1+beta) =  0.99

Ex2-pg206

In [2]:
##Ex6_2
import math
IE = 1.2*10**-3
beta = 60.
alpha = beta/(1.+beta)
print'%s %.2f %s'%("beta = ",(beta),"")##current gain in CE configuration
print'%s %.2f %s'%("alpha = beta/(1+beta) = ",(alpha),"")##current gain in CB configuraion
print'%s %.2f %s'%("IE = ",(IE),"A")##emitter current
IB = IE/(beta+1.)
IC = beta*IB
print'%s %.2e %s'%("IB = IE/(beta+1) = ",(IB),"A")##base current
print'%s %.2e %s'%("IC = beta*IB = ",(IC),"A")##collector current

beta =  60.00 
alpha = beta/(1+beta) =  0.98 
IE =  0.00 A
IB = IE/(beta+1) =  1.97e-05 A
IC = beta*IB =  1.18e-03 A

Ex3-pg206

In [3]:
##Ex6_3
import math
alpha = 0.98
VBE = 0.7
IE = -2.*10**-3
Re = 100.
RL = 3.3*10**3
print'%s %.2f %s'%("alpha = ",(alpha),"")##current gain
print'%s %.2f %s'%("VBE = ",(VBE),"V")##voltage across base-emitter
print'%s %.2e %s'%("IE = ",(IE),"A")##emitter current
print'%s %.2f %s'%("Re = ",(Re),"ohm")##emitter resistance
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
##now according to circuit given for the question in the textbook
IC = -alpha * IE
print'%s %.2e %s'%("IC = -alpha*IE = ",(IC),"A")##collector current
IB = -IC - IE
print'%s %.2e%s'%("IB = -IC - IE = ",(IB),"A")##base current
VBN = VBE+(abs(IE)*Re)
print'%s %.2f %s'%("VBN = VBE+(IE*Re) = ",(VBN),"V")##voltage across base and ground(N)
##ASSUMING... value for R1 = 30*10^3 ohm
R1 = 30*10**3
print'%s %.2f %s'%("R1 = ",(R1),"ohm")##resistancfe R1 as given in circuit
I = VBN/R1
print'%s %.2e %s'%("I = VBN/R1 = ",(I),"A")##current across resistance R1
##ASSUMING... VCC = 9V
VCC = 9##collector voltage
print'%s %.2f %s'%("VCC = ",(VCC),"V")
VCN = VCC - (RL*(IC+I+IB))
print'%s %.2f %s'%("VCN = VCC - RL*(IC+I+IB)) = ",(VCN),"V")##voltage across collector and ground(N)
## according to the given diagram for the question in the textbook, unknown resistance is,
R = (VCN - VBN)/(I+IB)
print'%s %.2f %s'%("R = (VCN - VBN)/(I+IB) = ",(R),"ohm")##unknown resistance

alpha =  0.98 
VBE =  0.70 V
IE =  -2.00e-03 A
Re =  100.00 ohm
RL =  3300.00 ohm
IC = -alpha*IE =  1.96e-03 A
IB = -IC - IE =  4.00e-05A
VBN = VBE+(IE*Re) =  0.90 V
R1 =  30000.00 ohm
I = VBN/R1 =  3.00e-05 A
VCC =  9.00 V
VCN = VCC - RL*(IC+I+IB)) =  2.30 V
R = (VCN - VBN)/(I+IB) =  20014.29 ohm

Ex4-pg207

In [14]:
##Ex6_4
import math
RC = 2.3*10**3
Re = 1*10**3
VCC = 12.
VCE = 5.
VBE = 0.7
beta = 50.
print'%s %.2f %s'%("RC = ",(RC),"ohm")##collector resistance
print'%s %.2f %s'%("Re = ",(Re),"ohm")##emitter resistance
print'%s %.2f %s'%("VCC = ",(VCC),"V")##supply voltage
print'%s %.2f %s'%("VCE = ",(VCE),"V")##voltage across collector and emitter
print'%s %.2f %s'%("VBE = ",(VBE),"V")##voltage across base and emitter
print'%s %.2f %s'%("beta = ",(beta),"")##current gain
## according to the given circuit, we have
IB = (VCC - VCE)/((beta+1.)*(RC+Re))
print'%s %.2e %s'%("IB = ",(IB),"A")##base current
IC = beta*IB
print'%s %.2e %s'%("IC = ",(IC),"A")##collector current
##from the circuit we have,
Rt = (VCE-VBE)/IB
print'%s %.2e %s'%("Rt = (VCE - VBE)/IB = ",(Rt),"ohm")##resistance Rt as given in circuit

RC =  2300.00 ohm
Re =  1000.00 ohm
VCC =  12.00 V
VCE =  5.00 V
VBE =  0.70 V
beta =  50.00 
IB =  4.16e-05 A
IC =  2.08e-03 A
Rt = (VCE - VBE)/IB =  1.03e+05 ohm

Ex5-pg208

In [15]:
##Ex6_5
import math
VBB = 1.
VCC = 12.
IC = 12.*10**-3
VCE = 4.
beta = 80.
VBE = 0.7
print'%s %.2f %s'%("VBB = ",(VBB),"V")##base supply voltage
print'%s %.2f %s'%("VCC = ",(VCC),"V")##collector supply voltage
print'%s %.2f %s'%("IC = ",(IC),"A")##collector current
print'%s %.2f %s'%("VCE = ",(VCE),"V")##voltage across collector and emitter
print'%s %.2f %s'%("beta = ",(beta),"")##current gain
print'%s %.2f %s'%("VBE = ",(VBE),"V")##voltage across base and emitter
IB = IC/beta
print'%s %.4f %s'%("IB = IC/beta = ",(IB),"A")##base current
RC = (VCC - VCE)/IC
print'%s %.2f %s'%("RC = (VCC - VCE)/IC = ",(int(RC)),"ohm")##collector resistance

VBB =  1.00 V
VCC =  12.00 V
IC =  0.01 A
VCE =  4.00 V
beta =  80.00 
VBE =  0.70 V
IB = IC/beta =  0.0002 A
RC = (VCC - VCE)/IC =  666.00 ohm

Ex6-pg209

In [17]:
##Ex6_6
import math
VCC = 9.
VBB = 3.
IC = 2.*10**-3
beta = 50.
VBE = 0.7
VCE = 4.
print'%s %.2f %s'%("VCC = ",(VCC),"V")##collector supply voltage
print'%s %.2f %s'%("VBB = ",(VBB),"V")##base supply voltage
print'%s %.2e %s'%("IC = ",(IC),"A")##collector current
print'%s %.2f %s'%("beta = ",(beta),"")##current gain
print'%s %.2f %s'%("VBE = ",(VBE),"V")##voltage across base and emitter
print'%s %.4f %s'%("VCE = ",(VCE),"V")##voltage across collector and emitter
IB = IC/beta
print'%s %.2e %s'%("IB = IC/beta = ",(IB),"A")##base current
RB = (VBB - VBE)/IB
print'%s %.2f %s'%("RB = (VBB - VBE)/IB = ",(RB),"ohm")##base resistance according to the given in circuit


## note: misprint in the question, author is asking for IB instead of beta, as beta is already provided.
## note: calculation done in the textbook for the problem is wrong.

VCC =  9.00 V
VBB =  3.00 V
IC =  2.00e-03 A
beta =  50.00 
VBE =  0.70 V
VCE =  4.0000 V
IB = IC/beta =  4.00e-05 A
RB = (VBB - VBE)/IB =  57500.00 ohm

Ex7-pg209

In [20]:
##Ex6_7
import math
VCC = 12.
VBB = 3.
IC = 12.*10**-3
VCE = 5.5
beta = 100.
VBE = 0.7
Re = 50.
print'%s %.2f %s'%("VCC = ",(VCC),"V")##collector supply voltage
print'%s %.2f %s'%("VBB = ",(VBB),"V")##base supply voltage
print'%s %.2e %s'%("IC = ",(IC),"A")##collector current
print'%s %.2f %s'%("VCE = ",(VCE),"V")##voltage across collector and emitter
print'%s %.2f %s'%("beta = ",(beta),"")##current gain
print'%s %.2f %s'%("VBE = ",(VBE),"V")##voltage across base and emitter
print'%s %.2f %s'%("Re = ",(Re),"ohm")##emittter resistance
IB = IC/beta
print'%s %.2e %s'%("IB = IC/beta = ",(IB),"A")##base current
##from base-emitter circuit;
IE = IC+IB
Rb = (VBB - VBE - (IE*Re))/IB
print'%s %.2f %s'%("Rb = (VBB - VBE - IE*Re)/IB = ",(Rb),"ohm")##base resistance
##from collector-emitter circuit, we have
Rc = (VCC - VCE - (IE*Re))/(IC)
print'%s %.2f %s'%("Rc = (VCC - VCE - (IE*Re))/IC = ",(Rc),"ohm")##collector resistance


##NOTE : in textbook the notation used for base and emitter resistance in fig. and in calculation are different



## note : calculation perform in the textbook is wrong for the above problem

VCC =  12.00 V
VBB =  3.00 V
IC =  1.20e-02 A
VCE =  5.50 V
beta =  100.00 
VBE =  0.70 V
Re =  50.00 ohm
IB = IC/beta =  1.20e-04 A
Rb = (VBB - VBE - IE*Re)/IB =  14116.67 ohm
Rc = (VCC - VCE - (IE*Re))/IC =  491.17 ohm

Ex8-pg210

In [21]:
##Ex6_8
import math
VBB = 10.
RB = 500.*10**3
VCC = 15.
RC = 1.2*10**3
beta =100.
print'%s %.2f %s'%("beta = ",(beta),"")##current gain
print'%s %.2f %s'%("VBB = ",(VBB),"V")##base supply voltage
print'%s %.2f %s'%("RB = ",(RB),"ohm")##resistance across base terminal
print'%s %.2f %s'%("VCC = ",(VCC),"V")##collector supply voltage
print'%s %.2f %s'%("RC = ",(RC),"ohm")##resistance across collector terminal
IB = VBB/RB
print'%s %.2e %s'%("IB = VBB/RB = ",(IB),"A")##base current
IC = beta*IB
print'%s %.4f %s'%("IC = beta*IB = ",(IC),"A")##collector current
VCE = VCC - (IC*RC)
print'%s %.2f %s'%("VCE = VCC - IC*RC = ",(VCE),"V")##voltage across collector and emitter


## the answer printed in the textbook for VCE is wrong.

beta =  100.00 
VBB =  10.00 V
RB =  500000.00 ohm
VCC =  15.00 V
RC =  1200.00 ohm
IB = VBB/RB =  2.00e-05 A
IC = beta*IB =  0.0020 A
VCE = VCC - IC*RC =  12.60 V

Ex9-pg211

In [23]:
##Ex6_9
import math
ic = 2.*10**-3
ie = ic## as base current is negligble
VT = 25.*10**-3
re = VT/ie
gm = ie/VT
print'%s %.2e %s'%("ic = ",(ic),"A")##collector current
print'%s %.2e %s'%("ie = ",(ie),"A")##emitter current with negligble base current
print'%s %.2f %s'%("VT = ",(VT),"V")##voltage at room temperature
print'%s %.2f %s'%("re = VT/ie = ",(re),"ohm")##emitter resistance
print'%s %.2f %s'%("gm = ie/VT = ",(gm),"mho")##conductance
rc = 100*10^3##slope of output characteristics
print'%s %.2f %s'%("rc = ",(rc),"ohm")
hoe = 1./rc
print'%s %.2e %s'%("hoe = 1/rc = ",(hoe),"Mho")##output conductance

ic =  2.00e-03 A
ie =  2.00e-03 A
VT =  0.03 V
re = VT/ie =  12.50 ohm
gm = ie/VT =  0.08 mho
rc =  1003.00 ohm
hoe = 1/rc =  9.97e-04 Mho

Ex10-pg211

In [24]:
##Ex6_10
import math
ic = 2.5*10**-3
ib = 50.*10**-6
print'%s %.2e %s'%("ib = ",(ib),"A")##base current
print'%s %.2e %s'%("ic = ",(ic),"A")##collector current
beta = ic/ib
print'%s %.2f %s'%("beta = ic/ib = ",(beta),"")##current gain 

ib =  5.00e-05 A
ic =  2.50e-03 A
beta = ic/ib =  50.00

Ex11-pg211

In [25]:
##Ex6_11
import math
ic = 3.*10**-3
ib = 3.08*10**-3
print'%s %.4f %s'%("ib = ",(ib),"A")##base current
print'%s %.4f %s'%("ic = ",(ic),"A")##collector current
alpha = ic/ib
print'%s %.2f %s'%("alpha = ie/ib = ic/ib = ",(alpha),"")##current gain, assuming ie = ic 

ib =  0.0031 A
ic =  0.0030 A
alpha = ie/ib = ic/ib =  0.97

Ex12-pg211

In [26]:
##Ex6_12
import math
##given, collector voltage swings between 11V to 4V
##thus,
vc = 11.-4.
print'%s %.2f %s'%("vc = 11 - 4 = ",(vc),"V")##PEAK-to-PEAK collector voltage
##given, collector current swings between 5mA to 1.4mA
##thus,
ic = (5. - 1.4)*10**-3
print'%s %.3f %s'%("ic = 5m - 1.4m = ",(ic),"A")##PEAK-to-PEAK collector current

vc = 11 - 4 =  7.00 V
ic = 5m - 1.4m =  0.004 A

Ex13-pg211

In [27]:
##Ex6_13
import math
ic = 4.*10**-3
ib = 80.*10**-6
print'%s %.4f %s'%("ib = ",(ib),"A")##base current
print'%s %.4f %s'%("ic = ",(ic),"A")##collector current
Ai = ic/ib
print'%s %.2f %s'%("Ai = ic/ib = ",(Ai),"")##current gain in CE amplifier

ib =  0.0001 A
ic =  0.0040 A
Ai = ic/ib =  50.00