Chapter 7 - Amplifiers¶

Example E1 - Pg 369¶

In [1]:
#Ex7_1 Pg-369
#calculate Lowest frequency
import math
print '%s' %("Refer to the figure 7.19")
print '%s' %("For good coupling")
print '%s' %("Xc <= 0.1*R")
R=10.*10.**(3.) #resistor R in ohm
f=1./(2.*math.pi*C*0.1*R)
print '%s' %("Lowest frequency,f")
print '%s %.2f' %("=Hz",f)
print '%s' %("or lowest frequency is approximately 3 Hz")

Refer to the figure 7.19
For good coupling
Xc <= 0.1*R
Lowest frequency,f
=Hz 2.34
or lowest frequency is approximately 3 Hz


Example E2 - Pg 369¶

In [2]:
#Ex7_2 Pg-369
#calculate
import math
print '%s' %("Refer to the figure 7.20")
print '%s' %("For good coupling")
print '%s' %("Xc <= 0.1*R")
R=10.*10.**(3.) #resistor R in ohm
f=1./(2.*math.pi*C*0.1*R)
print '%s' %("Lowest frequency,f")
print '%s %.2f' %("=Hz",f)
print '%s' %("or lowest frequency is approximately 1 Hz")

Refer to the figure 7.20
For good coupling
Xc <= 0.1*R
Lowest frequency,f
=Hz 0.72
or lowest frequency is approximately 1 Hz


Example E3 - Pg 369¶

In [3]:
#Ex7_3 Pg-369
#calculate Base voltage,Emitter voltage,Emitter current,AC emitter diode resistance
Beta=250. #transistor gain
Vcc=15. #supply voltage
R1=2000. #resistor R1 in ohm
R2=470. #resistor R2 in ohm
Vce=0.7 #voltage drop in V
Re=220. #emitter resistor in ohm
Vb=(Vcc*R2)/(R1+R2) #base voltage in V
print '%s' %("(1) Base voltage Vb")
print '%s %.2f' %("=V",Vb)
Ve=Vb-Vce #emitter voltage in V
print '%s' %("Emitter voltage Vb")
print '%s %.2f' %("=V",Ve)
Ie=Ve/Re #emitter current
print '%s' %("Emitter current ")
print '%s %.3f' %("Ie=*10**-2 A",Ie*10**2)
print '%s' %("For small signal operation, ie <= 0.1*Ie")
ie=0.1*Ie
print '%s %.3f' %("=mA",ie*10**3)
print '%s' %("(2) AC emitter diode resistance =25mV/ie")
Re_ac=25.*10.**(-3.)/ie #AC emitter diode resistance
print '%s %.0f' %("=ohm",Re_ac)
print '%s' %("(3) Z''vm = Beta*r''e")
Re_ac=26. #AC emitter diode resistance assumed 26 ohm not 25.53 ohm
Zvm=Beta*Re_ac
print '%s %.0f' %("=ohm",Zvm)

(1) Base voltage Vb
=V 2.85
Emitter voltage Vb
=V 2.15
Emitter current
Ie=*10**-2 A 0.979
For small signal operation, ie <= 0.1*Ie
=mA 0.979
(2) AC emitter diode resistance =25mV/ie
=ohm 26
(3) Z''vm = Beta*r''e
=ohm 6500


Example E4 - Pg 370¶

In [4]:
#Ex7_4 Pg-370
#calculate Base voltage,Emitter voltage,Emitter current,AC emitter diode resistance,Voltage gai,Zin stage,Input voltage,Output voltage
import math
Beta=100. #transistor gain
Vcc=10.#supply voltage
R1=10.*10.**(3.) #resistor R1 in ohm
R2=2200. #resistor R2 in ohm
Vce=0.7 #voltage drop in V
Re=2000. #emitter resistor in ohm
Rs=600. #source resistor in ohm
Vb=(Vcc*R2)/(R1+R2) #base voltage in V
print '%s' %("Base voltage Vb")
print '%s %.1f' %("=V",Vb)
Ve=Vb-Vce #emitter voltage in V
print '%s' %("Emitter voltage Vb")
print '%s %.1f' %("=V",Ve)
Ie=Ve/Re #emitter current
print '%s' %("Emitter current")
print '%s %.2f' %("=mA",Ie*10**3)
print '%s' %("AC emitter diode resistance =25mV/ie")
re=25.*10.**(-3.)/Ie #AC emitter diode resistance
print '%s %.0f' %("=ohm",re)
rc=((3.6*10.)/(3.6+10.))*10.**(3.) #Collector dioed resistance
A=rc/re #voltage gain(value in text book is wrong)
print '%s' %("Voltage gain A")
print '%s %.0f' %("=",A)
zin_1=((10.*2.2)/(10.+2.2))
zin=((zin_1*Beta*A)/(zin_1+(Beta*A)))*1000
print '%s' %("Zin stage")
print '%s %.3f' %("=kohm",zin*10**-3)
Vin=(zin/(Rs+zin))*10.**(-3.) #input voltage (value in text book is wrong)
print '%s' %("Input voltage")
print '%s %.2f' %("=mV",Vin*10**3)
Vout=A*Vin #output voltage (value in textbook is wrong)
print '%s' %("Output voltage")
print '%s %.2f' %("=mV",Vout*10**3)

Base voltage Vb
=V 1.8
Emitter voltage Vb
=V 1.1
Emitter current
=mA 0.55
AC emitter diode resistance =25mV/ie
=ohm 45
Voltage gain A
= 58
Zin stage
=kohm 1.803
Input voltage
=mV 0.75
Output voltage
=mV 43.82


Example E5 - Pg 371¶

In [5]:
#Ex7_5 Pg-371
#calculate
hfe=50. #current gain
Rs=500. #source resistor in ohm
hie=10.**(3.) #input resitance in ohm
A=hfe*Rl/(Rs+hie) #volatge gain
print '%s %.1f' %("Voltage gain=",A)
Vs=0.02 #source voltage in V
Vout=A*Vs #output voltage
print '%s %.2f' %("Output voltage=V",Vout)

Voltage gain= 333.3
Output voltage=V 6.67


Example E6 - Pg 372¶

In [6]:
#Ex7_6 Pg-372
#calculate Voltage gain,Power gain
import math
Vout=5. #output voltage
Vin=0.5 #input voltage
Ri=10.*10.**(3.) #input resistance in ohm
Ro=10. #output resistance
A=Vout/Vin #voltage gain
print '%s %.0f' %("Voltage gain =%.0f \n",A)
Pi=Vin**2./Ri #input power
Po=Vout**2./Ro #output power
Pow_gain=10.*(math.log10(Po)-math.log10(Pi)) #power gain
print '%s %.0f' %(" Power gain(in decibel) = %.0f dB \n\n",Pow_gain)
print '%s' %("When Ri=Ro")
Ri=Ro
A=Vout/Vin #voltage gain
Pi=Vin**2./Ri #input power
Po=Vout**2./Ro #output power
Pow_gain=10.*(math.log10(Po)-math.log10(Pi)) #power gain
print '%s %.0f' %(" Power gain(in decibel) = %.0f dB",Pow_gain)

Voltage gain =%.0f
10
Power gain(in decibel) = %.0f dB

50
When Ri=Ro
Power gain(in decibel) = %.0f dB 20


Example E7 - Pg 372¶

In [7]:
#Ex7_7 Pg-372
#calculate
import math
Ri=500. #input resistance in ohm
print '%s %.0f' %("Lowest cut-off frequency=Hz",f)

Lowest cut-off frequency=Hz 32


Example E8 - Pg 372¶

In [8]:
#Ex7_8 Pg-372
#calculate Coupling Capacitor
import math
Rl=20.*10.**(3.) #loaH resistance in ohm
Ri=5000. #input resistance in ohm
f=33. #lower cut-off frequency in Hz
f2=33.*10.**(3.) #higher cut-off frequency
C=(1./(2.*math.pi*f*(Rl+Ri))) #coupling capacitance (value in textbook is wrong)
print '%s %.1f' %("Coupling Capacitor=uF",C*10**6)
Req=(Rl*Ri)/(Rl+Ri) #equivalent resistance
Cs=1/(2*math.pi*f2*Req) #shunting capacitance (value in textbook is wrong)
print '%s %.0f' %("Coupling Capacitor=uF",Cs*10**12)

Coupling Capacitor=uF 0.2
Coupling Capacitor=uF 1206


Example E9 - Pg 373¶

In [9]:
#Ex7_9 Pg-373
#calculate Voltage Amplification,Output voltage
Rd=3000. #source resistance in ohm

Voltage Amplification= -8.44