Chapter 5 Transistor Amplifier

Exa 5.1 page No 289

In [6]:
from math import pi,sqrt
from __future__ import division
# Given data
R_C = 10# k ohm
R_C = R_C * 10**3# ohm
f = 2# kHz
f = f * 10**3# Hz
R = 8#  k ohm
R = R * 10**3# ohm
#Formula,  f = 1/(2*pi*R*C*sqrt(6+((4*R_C)/R)))
C = 1/(2*pi*R*f*sqrt(6+((4*R_C)/R)))# F
C= C*10**6# µF
print "The value of capacitor = %.4f µF"%C
h_fe= 23+29*R/R_C+4*R_C/R
print "The value of h_fe is : %.2f"%h_fe
The value of capacitor = 0.0030 µF
The value of h_fe is : 51.20

Exa 5.2 page No 293

In [8]:
from math import pi,sqrt
from __future__ import division
# Given data
f = 10# kHz
f = f * 10**3# Hz
R = 100# k ohm
R = R * 10**3# ohm
#Formula used, f = 1/(2*pi*R*C)
C = 1/(2*pi*R*f)# F
C = round(C * 10**12)# pF
print "The value of capacitor = %.2f pF"%C
The value of capacitor = 159.00 pF

Exa 5.3 page No 299

In [9]:
from math import pi,sqrt
from __future__ import division
# Given data
C = 0.06# pF
C = C * 10**-12# F
L = 0.5# H
R = 10# k ohm
R = R * 10**3# ohm
Cm = 1# pF
Cm = Cm * 10**-12# F
# The parallel resonant frequency 
f_p = (1/(2*pi))*( sqrt( (C+Cm)/(L*C*Cm) ) )# Hz
f_p= f_p*10**-3# kHz
print "The parallel resonant frequency = %.2f kHz"%f_p
# The series resonant frequency 
f_s = 1/(2*pi*(sqrt(L*C)))# Hz
f_s= f_s*10**-3# kHz
print "The series resonant frequency = %.2f kHz"%f_s
omega_s = 2*pi*f_s*10**3# rad/sec
# The Q factor of the crystal 
Q = (omega_s*L)/R
print "The Q factor of the crystal = %.2f"%Q
The parallel resonant frequency = 946.05 kHz
The series resonant frequency = 918.88 kHz
The Q factor of the crystal = 288.68

Exa 5.4 page No 300

In [1]:
from __future__ import division
# Given data
R1 = 6# ohm
R2 = 4# ohm
R3 = 4# ohm
h11 = R1+( (R2*R3)/(R2+R3) )# ohm
print "The value of h11 = %.2f ohm"%h11
# i2 = -i*1/2 and h21 = i2/i1 = (-i1/2)/2, So
h21 = -1/2
print "The value of h21 = %.2f"%h21
# V1 = V2/2 and h12 = V1/V2 = (V2/2)/2, SO
h12 = 1/2
print "The value of h12 = %.2f"%h12
Zo = R2+R3## output resistance  = %.2f ohm
h22 = 1/Zo# mho
print "The value of h22 = %.2f mho"%h22
The value of h11 = 8.00 ohm
The value of h21 = -0.50
The value of h12 = 0.50
The value of h22 = 0.12 mho

Exa 5.5 page No 303

In [3]:
from __future__ import division
# Given data
R1 = 3# ohm
R2 = 6# ohm
R3 = 2# ohm
# V1 = (R1*I1)+  (R2*(I1+I2)) = ((R1+R2)*I1)+(R2*I2)   (i)
# V2 = (R3*I2) + (R2*(I1+I2)) = (R2*I1) + ((R1+R3)*I2) (ii)
# Standard h-parameter equation 
#V1= h11*I1 +h12*I2 and V2= h21*I1 + h22*I2
# Comparing eq (i) and (ii) with standard equaation, we get
h11= R1+R2
print "The value of h11 = %.2f"%h11
h12= R2
print "The value of h12 = %.2f"%h12
h21= R2
print "The value of h21 = %.2f"%h21
h22= R2+R3
print "The value of h22 = %.2f"%h22
The value of h11 = 9.00
The value of h12 = 6.00
The value of h21 = 6.00
The value of h22 = 8.00

Exa 5.7 page No 304

In [5]:
from __future__ import division
# Given data
h_ic = 1.1#  k ohm
h_ic = h_ic * 10**3# ohm
h_rc = 1
h_fc = -51
h_oc = 25# µA/V
h_oc = h_oc * 10**-6# A/V
R_L = 10# k ohm
R_L = R_L * 10**3# ohm
R_S = R_L# ohm
# The current gain,
Ai = -h_fc/(1+(h_oc*R_L))
print "The value of Ai = %.2f"%Ai
# The value of Ri 
Ri = h_ic+(h_rc*Ai*R_L)# ohm
Ri= Ri*10**-3# k ohm
print "The value of Ri = %.2f k ohm"%Ri
Ri= Ri*10**3# ohm
# The value of A_V 
A_V = (Ai*R_L)/Ri
print "The value of A_V = %.2f"%A_V
Gamma_o = h_oc - ( (h_fc*h_rc)/(h_ic+R_S) )# mho
# The value of Ro 
Ro = round(1/Gamma_o)# ohm
print "The value of Ro = %.2f ohm"%Ro
A_VS = (A_V*Ri)/(Ri+R_S)
print "The value of A_VS = %.2f"%A_VS
The value of Ai = 40.80
The value of Ri = 409.10 k ohm
The value of A_V = 1.00
The value of Ro = 216.00 ohm
The value of A_VS = 0.97

Exa 5.10 page No 306

In [6]:
from __future__ import division
# Given data
h_ie = 1.5# ohm
h_fe = 100
h_re = 3 * 10**-4
h_oe = 25 * 10**-6#  mho
V_CC = 20# V
R1 = 50 * 10**3# ohm
R2 = 5 * 10**3# ohm
R_C = 5 * 10**3# ohm
R_L = 10 * 10**3# ohm
R_S = 1# k ohm
Ri= 1.4*10**3# ohm
R_B = (R1*R2)/(R1+R2)# ohm
R_Ldesh= R_L*R_C/(R_L+R_C)# ohm
# Current gain for transisor,
Ai= -h_fe/(1+h_oe*R_Ldesh)
# Overall current gain, A_I= Ai+Ib/I= Ai+R_B/(R_B+Ri) 
A_I= Ai*R_B/(R_B+Ri)
print "The overall current gain is : %.2f"%A_I
#Part (ii)
R_i= R_B*Ri/(R_B*Ri)# k ohm
print "The value of Ri = %.2f k ohm"%R_i
# Part (iii)
Av= A_I*R_Ldesh/Ri## voltage gain for transistor
# overall voltage gain,
A_VS=  Av*R_i/(R_S+R_i)
print "The value of A_VS is : %.2f"%A_VS
# Part (iv)
R_S=R_S*10**3# ohm
R_Sdesh= R_S*R_B/(R_S+R_B)# ohm
gamma_o= h_re*h_fe/(h_ie+R_Sdesh)-h_oe# mho
gamma_o= round(gamma_o*10**6)# µ mho
print "The value of gamma_o = %.2f µ mho"%gamma_o

# Note: In the book, all these calculated value is not accurate.
The overall current gain is : -70.57
The value of Ri = 1.00 k ohm
The value of A_VS is : -84.01
The value of gamma_o = 12.00 µ mho

Exa 5.11 page No 307

In [9]:
from __future__ import division
# Given data
R_S = 800# ohm
R_L = 2# k ohm
R_L = R_L * 10**3# ohm
h_ie = 1.1# k ohm
h_ie = h_ie * 10**3# ohm
h_fe = 50
h_oe = 25# µohm
h_oe = h_oe * 10**-6# ohm
h_re = 2.5*10**-4

# Part (a) : Exact analysis
print "Part (a) : Exact analysis"
# The current gain
Ai_exact = -h_fe/(1+(h_oe*R_L))
print "The current gain = %.2f"%Ai_exact
# The input resistance 
Ri = h_ie + (h_re*Ai_exact*R_L)# ohm
Ri= Ri*10**-3# k ohm
print "The input resistance = %.2f k ohm"%Ri
Ri= Ri*10**3# ohm
A_V = (Ai_exact*R_L)/Ri
# The voltage gain,
A_VS = (A_V*Ri)/(Ri+R_S)
print "The voltage gain = %.2f"%A_VS
Gamma_o = h_oe - ( (h_re*h_fe)/(h_ie+R_S) )# mho
# The output resistance 
Ro = 1/Gamma_o# ohm
Ro= Ro*10**-3## k ohm
print "The output resistance = %.2f k ohm"%Ro
Ro= Ro*10**3## ohm
# The power gain,
Ap = Ai_exact*A_V
print "The power gain = %.2f"%Ap
# The output terminal resistance 
Rot = (Ro*R_L)/(Ro+R_L)# ohm
Rot= Rot*10**-3# k ohm
print "The output terminal resistance = %.2f k ohm"%Rot
Rot= Rot*10**3# ohm

# Part (b) : Approximate analysis
print "Part (b) : Approximate analysis"
h_re =0
h_oe = 0
Ai_app = -h_fe/(1+(h_oe*R_L))
print "The value of Ai = %.2f"%Ai_app
Ri = h_ie+(h_re*Ai_app*R_L)
print "The value of Ri = %.2f ohm"%Ri
A_V= Ai_app*R_L/Ri
A_VS = (-A_V*h_ie)/(R_S+h_ie)
print "The value of A_VS = %.2f"%A_VS
Gamma_o = 18.42# = %.2f µohms
Gamma_o = Gamma_o * 10**-6# mho
# # The output resistance 
Ro = 1/Gamma_o# ohm
Ro= Ro*10**-3## k ohm
print "The output resistance = %.2f k ohm"%Ro
Ro= Ro*10**3## ohm
Ap = Ai_app*A_V
print "The value of Ap = %.2f"%Ap
#The output terminal resistance, Rot = (Ro*R_L)/(Ro+R_L)= 20# k ohm
Rot= 20# k ohm
print "The output terminal resistance = %.2f k ohm"%Rot
Rot= Rot*10**3# ohm
# Percentage error
Per_error= abs((Ai_exact-Ai_app)/Ai_exact*100)# %
print "The percentage error = %.2f %%"%Per_error
Part (a) : Exact analysis
The current gain = -47.62
The input resistance = 1.08 k ohm
The voltage gain = -50.76
The output resistance = 54.29 k ohm
The power gain = 4214.08
The output terminal resistance = 1.93 k ohm
Part (b) : Approximate analysis
The value of Ai = -50.00
The value of Ri = 1100.00 ohm
The value of A_VS = 52.63
The output resistance = 54.29 k ohm
The value of Ap = 4545.45
The output terminal resistance = 20.00 k ohm
The percentage error = 5.00 %

Exa 5.13 page No 310

In [10]:
from __future__ import division
# Given data
R_S = 600# ohm
r_i = 400# ohm
Vs = 1# mV
Vs = Vs * 10**-3# V
#Input voltage
Vi = Vs*(r_i/(R_S+r_i))# V
# Input current
Ii =Vs/(R_S+r_i)# A
Vo = 100# mV
Vo = Vo * 10**-3# V
R_L = 5#  k ohm
R_L = R_L * 10**3# ohm
# Output current 
Io = Vo/R_L# A
# voltage gain
A_V =Vo/Vi # 
print "The voltage gain",A_V
# current gain
A_I = Io/Ii
print "The current gain",A_I
# Power gain
P =A_V*A_I
print "The power gain",P
The voltage gain 250.0
The current gain 20.0
The power gain 5000.0

Exa 5.14 page No 312

In [12]:
from __future__ import division
# Given data
A = 125
Beta = 1/10
# Gain of negative feedback
Af = A/(1+(A*Beta))
print "The gain of negative feedback : %.3f"%Af
The gain of negative feedback : 9.259

Exa 5.15 page No 312

In [13]:
from __future__ import division
# Given data
Af = 100
Vi = 0.6# V
Vo =Af*Vi# V
Vi = 50# mV
Vi = Vi * 10**-3# V
# Internal gain of amplifier,
A = Vo/Vi
print "The value of A is : %.2f"%A
B= ((A/Af)-1)/A
print "The value of B : %.2f"%B
The value of A is : 1200.00
The value of B : 0.01

Exa 5.16 page No 313

In [15]:
from __future__ import division
# Given data
A = 60# dB
A =10**(A/20)
Beta = 0.005
dAbyA = -12/100##gain reduction = %.2f %
dAf_byAf = (1/(1+(A*Beta)))*dAbyA## change = %.2f overall gain
dAf_byAf = dAf_byAf * 100# %
print "The change in overall gain = %.2f %%"%dAf_byAf
print "Thus, the overall gain will be reduced by 2%"
The change in overall gain = -2.00 %
Thus, the overall gain will be reduced by 2%

Exa 5.17 page No 313

In [16]:
from __future__ import division
# Given data
Zo =  12.6# k ohm
Zo = Zo * 10**3# ohm
A = 60# dB
A = 10**(A/20)
Zof = 500# ohm
# Zof = Zo/(1+(A*Beta))
Beta = ((Zo/Zof)-1)/A
print "The value of feed back factor = %.2f"%Beta
# Part (ii)
dAbyA= 20/100## change = %.2f gain of basic amplifier
dAf_byAf =dAbyA*1/(1+A*Beta)*100##change = %.2f overall gain  = %.2f  %
print "The change in overall gain for 20 %% change in gain of the basic amplifier = %.2f %%"%dAf_byAf

# Note: In the book, there is calculation error to find the value of dAf/Af
The value of feed back factor = 0.02
The change in overall gain for 20 % change in gain of the basic amplifier = 0.79 %

Exa 5.18 page No 314

In [18]:
from __future__ import division
# Given data
Vo = 36# V
Vi = 0.028# V
A = Vo/Vi
Vf = 1.2
Vo = 100
Beta = Vf/Vo
# Gain with feedback
Af = A/(1+(A*Beta))
Vs = Vi# V
# Output voltage
Vo =Af*Vs# V
print "The output voltage = %.2f V"%Vo
#Df =D/(1+(A*Beta))
DbyDf = 7/1
ABeta = (DbyDf)-1
Af =A/(1+(ABeta))
Vo = 36# V
# The input voltage,
Vs =Vo/Af
print "The input voltage = %.2f"%Vs
The output voltage = 2.19 V
The input voltage = 0.20

Exa 5.19 page No 315

In [19]:
from __future__ import division
# Given data
dAf_byAf = 10/100
dAbyA = 10
A = 1000
# dAf_byAf = dAbyA*(1/(1+(A*Beta)))
# The required feed back 
Beta = ((dAbyA/dAf_byAf)-1)/A
print "The required feed back = %.2f"%Beta
Af = A/(1+(A*Beta))## closed loop voltage gain 
print "The closed loop voltage gain = %.2f"%Af
The required feed back = 0.10
The closed loop voltage gain = 10.00

Exa 5.20 page No 315

In [21]:
from __future__ import division
# Given data
Vi= 10*10**-3## input voltage = %.2f V
A=1000## open loop voltage gain
Do= 10##second harmonic distortion = %.2f %
feedback= 40##feedback = %.2f dB
feedback= 10**(feedback/20)
# feedback= 1+A*bita or
bita= (feedback-1)/A
Af= A/(1+A*bita)
# New value of second harmonic distortion,
Df= Do/(1+A*bita)# %
print "The new value of second harmonic distortion = %.2f %%"%Df
# New value of input required,
Vs= Vi*(1+A*bita)# V
print "The new value of input required = %.2f volts"%Vs
The new value of second harmonic distortion = 0.10 %
The new value of input required = 1.00 volts

Exa 5.21 page No 316

In [24]:
from __future__ import division
# Given data
D = 10/100
Df = 1/100
A = 200
Vs = 10# mV
Vs = Vs * 10**-3# V
# Df = D/(1+A*Beta)
Beta = (((D/Df)-1)/A)
# gain with feedback 
Af = A/(1+(A*Beta))
print "The gain with feedback is : %.2f"%Af
# The output voltage 
Vo = Af*Vs# V
print "The output voltage = %.2f V "%Vo
# The input voltage 
Vin = Df+(-Beta*Vo)# V
print "The input voltage = %.4f V"%Vin
The gain with feedback is : 20.00
The output voltage = 0.20 V 
The input voltage = 0.0010 V

Exa 5.22 page No 316

In [25]:
from __future__ import division
# Given data
A = 10000
Beta = 0.02
Zi = 1# k ohm
Zi = Zi * 10**3# ohm
Zo = 10# k ohm
Zo = Zo * 10**3# ohm
# The input impedance 
Zif = Zi*(1+(A*Beta))# ohm
Zif= Zif*10**-3# k ohm
print "The input impedance = %.2f k ohm"%Zif
# The output impedance 
Zof = Zo/(1+(A*Beta))# ohm
Zof= Zof*10**-3# k ohm
print "The output impedance = %.2f k ohm"%Zof

# Note:In the book, there is calculation error to find the value of Zof because 10/201 will be 0.04975 not 0.4975
The input impedance = 201.00 k ohm
The output impedance = 0.05 k ohm

Exa 5.23 page No 317

In [27]:
from __future__ import division
# Given data
A = 400
Beta = 0.01
# The gain with feedback
Af =A/(1+(A*Beta))
print "The gain with feedback = %.2f"%Af
f_L = 200# Hz
# The Lower cut-off frequency with feedback 
f_LF = f_L/(1+(A*Beta))# Hz
print "The Lower cut-off frequency with feedback = %.2f Hz"%f_LF
f_H = 40# kHz
f_H = f_H * 10**3# Hz
# The Upper cut-off frequency with feedback 
f_HF = f_H*(1+(A*Beta))# Hz
f_HF=f_HF*10**-3# k Hz
print "The Upper cut-off frequency with feedback = %.2f kHz"%f_HF

# Note:In the book, there is calculation error to find the value of gain with feedback i.e. Af, so the answer = %.2f the book is wrong.
The gain with feedback = 80.00
The Lower cut-off frequency with feedback = 40.00 Hz
The Upper cut-off frequency with feedback = 200.00 kHz

Exa 5.24 page No 317

In [28]:
from __future__ import division
# Given data
A = 4000
R1 = 1# k ohm
R1 = R1 * 10**3# ohm
R2 = 9# k ohm
R2 = R2 * 10**3# ohm
Beta = R1/(R1+R2)## feedback fraction 
print "The feedback fraction = %.2f"%Beta
# The overall voltage gain with feedback 
Af = A/(1+(A*Beta))
print "The overall voltage gain with feedback = %.2f"%Af
Vs = 2# mV
#Af = Vo/Vs
# The output voltage 
Vo = Af*Vs# mV
print "The output voltage = %.2f mV"%Vo
The feedback fraction = 0.10
The overall voltage gain with feedback = 9.98
The output voltage = 19.95 mV

Exa 5.25 page No 318

In [31]:
from math import log
from __future__ import division
# Given data
A = 54.8
A = 20 * log(A)
Beta = 1/50## feedback factor
# gain with feedback
Af = A/(1+(A*Beta))
#Distortion with feedback, Df = D/(1+(A*Beta))
Df = 1
D = 12
Pd = (Df/D)*100## percenatge change = %.2f distortion = %.2f %
print "The percentage reduction in harmonic distortion = %.2f  %%"%Pd
The percentage reduction in harmonic distortion = 8.33  %

Exa 5.26 page No 318

In [3]:
from __future__ import division
# Given data
# Part(iii)
Vf_byVo= 1/3
bita= Vf_byVo
# A*bita>=1 or
# The minimum gain required for oscillation 
Amin= 1/bita
print "The minimum gain required for oscillation is : %.2f"%Amin
The minimum gain required for oscillation is : 3.00

Exa 5.27 page No 320

In [6]:
from math import sqrt,tan,pi
from __future__ import division
# Magnitude of loop gain, Mag_Tf= bita*100/(sqrt(1+(f/10**5)**2))**3         (i)
# Phase of loop gain, P_Tf= -3*atand(f/10**5)                                               (ii)
f_180= tan(pi/180*180/3)*10**5## from eq(ii), frequency at which phase becomes -180°
bita= 0.20
T_f_180= bita*100/(sqrt(1+(f_180/10**5)**2))**3## from eq (i), the magnitude of loop gain at f_180 for bita= 0.20
print "The magnitude of loop gain for beta = 0.20 is : %.2f"%T_f_180
bita= 0.02
T_f_180= bita*100/(sqrt(1+(f_180/10**5)**2))**3## from eq (i), the magnitude of loop gain at f_180 for bita= 0.20
print "The magnitude of loop gain for beta = 0.20 is : %.2f"%T_f_180
print "Hence system is unstable for beta=0.20 and stable for beta= 0.02 because at beta= 0.20 gain is greater than 1."
The magnitude of loop gain for beta = 0.20 is : 2.50
The magnitude of loop gain for beta = 0.20 is : 0.25
Hence system is unstable for beta=0.20 and stable for beta= 0.02 because at beta= 0.20 gain is greater than 1.

Exa 5.29 page No 322

In [7]:
from math import sqrt,pi
from __future__ import division
# Given data
L = 0.01# H
C = 10# pF
C = C * 10**-12# F
# The frequency of oscillation 
f = 1/(2*pi*sqrt(L*C))# Hz
f = f * 10**-3# kHz
print "The frequency of oscillation = %.2f kHz"%f

# Note:In the book, the calculation is wrong.
The frequency of oscillation = 503.29 kHz

Exa 5.30 page No 322

In [13]:
from math import sqrt,pi
from __future__ import division
# Given data
L = 150# µH
L = L * 10**-6# H
# f = 1/(2*pi*sqrt(L*C)) or C = 1/(4*pi**2*f**2*L)
#For f= 500 kHz
f1 = 500# kHz
f1 = f1 * 10**3# Hz
C1 = 1/(4*pi**2*f1**2*L)# F
#For f= 1500 kHz
f2 = 1500# kHz
f2 = f2 * 10**3# Hz
C2 = 1/(4*pi**2*f2**2*L)# F
C1= C1*10**12# pF
C2= C2*10**12# pF
print "The required capacitor range is : %.2f"%(C2),"pF to %.2f"%(C1),"pF"

# Note: The answer in the book is wrong.
The required capacitor range is : 75.05 pF to 675.47 pF

Exa 5.31 page No 323

In [12]:
from math import sqrt,pi
from __future__ import division
# Given data
L = 100# µH
L = L * 10**-6# H
A=10
C1 = 0.001# µF
C1 = C1 * 10**-6# F
C2 = 0.01# µF
C2 = C2 * 10**-6# F
C = (C1*C2)/(C1+C2)# F
f = 1/(2*pi*sqrt( L*C ))# Hz
f = round(f * 10**-3)# kHz
print "The operating frequency = %.2f kHz"%f
Beta = C1/C2## feedback fraction
print "The feed back fraction = %0.2f"%Beta
#Minimum gain to sustain oscillations, Amin*Beta = 1
Amin = 1/Beta
print "The minimum gain to sustain oscillation = %.2f"%Amin
# A = R_C/R_E
R_C = 2.5# ohm
R_E = R_C/A# ohm
print "The emitter resistance = %.2f ohm"%R_E
The operating frequency = 528.00 kHz
The feed back fraction = 0.10
The minimum gain to sustain oscillation = 10.00
The emitter resistance = 0.25 ohm

Exa 5.32 page No 324

In [11]:
from math import pi
from __future__ import division
# Given data
L1 = 0.1# mH
L1 = L1 * 10**-3# H
L2 = 10# mH
L2 = L2 * 10**-3# H
L = 150# µH
L = L * 10**-6# H
f = 4110# kHz
f = f * 10**3# Hz
#Frequency of oscillations, f = 1/(2*pi*sqrt(L*C)), So
C = 1/(4*(pi**2)*(f**2)*L )# F
C = round(C * 10**12)# pF
print "The value of capacitor = %.2f pF"%C
The value of capacitor = 10.00 pF

Exa 5.33 page No 324

In [14]:
from math import sqrt,pi,tan
from __future__ import division
# Given data
f = 1# kHz
f = f * 10**3# Hz
R = 10# k ohm
R = R * 10**3# ohm
omega = 2*pi*f##in rad/sec
phi = 60# degree
# tand(phi) = Alpha_C/R = 1/(omega*C*R)
C = 1/(omega*R*tan(pi/180*phi))# F
C = C * 10**6# µF
print "The value of C = %.2f µF"%C
The value of C = 0.01 µF

Exa 5.34 page No 324

In [15]:
from __future__ import division
from math import sqrt,pi
# Given data
R2 = 200# k ohm
R2 = R2 * 10**3# ohm
R1 = R2# ohm
C2 = 250# pF
C2 = C2 * 10**-12# F
C1 = C2# F
R = R2# ohm
C = C2# F
# Frequency of oscillations,
f = 1/(2*pi*R*C)# Hz
print "The frequency = %.2f Hz"%f

# Note: The answer in the book is not accurate.
The frequency = 3183.10 Hz

Exa 5.36 page No 326

In [16]:
from __future__ import division
from math import sqrt,pi

# Given data
C1 = 300# pF
C2 = 100# pF
Ceq = (C1*C2)/(C1+C2)# pF
Ceq = Ceq * 10**-12# F
L = 50# µH
L = L * 10**-6# H
# The frequency of oscillation 
f = 1/(2*pi*sqrt(L*Ceq))# Hz
f = f * 10**-6# MHz
print "The frequency of oscillation = %.2f MHz"%f
# For maintaining oscillation, A_loop >=1 and Aopenloop*Beta = 1
# Beta = C2/C1
Aopenloop = C1/C2
print "The minimum gain for maintaining oscillation = %.2f"%Aopenloop
The frequency of oscillation = 2.60 MHz
The minimum gain for maintaining oscillation = 3.00