Chapter 33 : Operational Amplifiers

Example No. 33_1 Page No. 1072

In [2]:
# Calculate the differential voltage gain, Ad, and the ac output voltage, Vout.

# Given data

Vin = 10*10**-3#    # Input voltage=10 mVolts(p-p)
Rc = 10*10**3#      # Collector resistance=10 kOhms
Ie = 715.*10**-6#    # Emitter current=715 uAmps

re = (25*10**-3)/Ie#

Ad = Rc/(2*re)#
print 'The Differential Voltage Gain =%0.2f'%Ad

Av = Ad

Vo = Av*Vin#
print 'The Ac Output Voltage = %0.2f Volts(p-p)'%Vo

The Differential Voltage Gain =143.00
The Ac Output Voltage = 1.43 Volts(p-p)

Example No. 33_2 Page No. 1073

In [5]:
from math import log10
# calculate the common-mode voltage gain, ACM, and the CMRR (dB).

# Given data

Rc = 10*10**3#     # Collector resistance=10 kOhms
Re = 10.*10**3#     # Emitter resistance=10 kOhms
Ad = 142.86#      # Differential gain=142.86

Acm = Rc/(2*Re)#
print 'The Common-Mode Voltage Gain Acm = %0.2f'%Acm

CMRR = 20*log10(Ad/Acm)#
print 'The Commom-Mode Rejection Ratio = %0.2f dB'%CMRR

The Common-Mode Voltage Gain Acm = 0.50
The Commom-Mode Rejection Ratio = 49.12 dB

Example No. 33_3 Page No. 1074

In [6]:
from math import pi
# Calculate fmax for an op amp that has an Sr of 5 V/us and a peak output voltage of 10 V.

# Given data

Vpk = 10.#       # Peak output voltage=10 Volts
Sr = 5./10**-6#   # Slew rate=5 V/us


fo = Sr/(2*pi*Vpk)#
print 'The Output Frequency = %0.2e Hertz'%fo
print 'i.e 79.6 kHz'

The Output Frequency = 7.96e+04 Hertz
i.e 79.6 kHz

Example No. 33_4 Page No. 1075

In [8]:
# calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.

# Given data

Vin = 1.#        # Input voltage=1 Volts(p-p)
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
Ri = 1.*10**3#    # Input resistance=1 kOhms

Acl = -(Rf/Ri)#
print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl

Vo = -Vin*Acl#
print 'The Output Voltage = %0.2f Volts(p-p)'%Vo
print 'The -ve sign indicates that input and output voltages are 180° out-of-phase'

The Closed-Loop Voltage Gain Acl =-10.00
The Output Voltage = 10.00 Volts(p-p)
The -ve sign indicates that input and output voltages are 180° out-of-phase

Example No. 33_5 Page No. 1076

In [12]:
#If Avol equals 100,000, calculate the value of Vid.

# Given data

Avol = 100000.#  # Open loop voltage gain=100,000
Vo = 10.#        # Output voltage=10 Volts(p-p)

Vid = Vo/Avol#
print 'The Differential Input Voltage = %0.2e Volts(p-p)'%Vid
print 'i.e 100 uVolts(p-p)'

The Differential Input Voltage = 1.00e-04 Volts(p-p)
i.e 100 uVolts(p-p)

Example No. 33_6 Page No. 1078

In [13]:
# calculate Zin and Zout(CL). Assume AVOL is 100,000 and Zout(OL) is 75 Ohms.

# Given data

Avol = 100000.#  # Open loop voltage gain=100,000
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
Ri = 1.*10**3#    # Input resistance=1 kOhms
Zool = 75.#      # Output impedence (open-loop)=75 Ohms

Zi = Ri#
print 'The Input Impedence = %0.2e Ohms'%Zi
print 'i.e 1 kOhms'

Beta = Ri/(Ri+Rf)#

A = Avol*Beta#

Zocl = Zool/(1+A)#
print 'The Closed Loop Output Impedence = %0.2f Ohms'%Zocl

The Input Impedence = 1.00e+03 Ohms
i.e 1 kOhms
The Closed Loop Output Impedence = 0.01 Ohms

Example No. 33_7 Page No. 1083

In [14]:
from math import pi
# Calculate the 5-V power bandwidth.

# Given data

Vo = 10.#            # Output voltage=10 Volts(p-p)
Sr = 0.5/10**-6#     # Slew rate=0.5 V/us

Vpk = Vo/2#

fo = Sr/(2*pi*Vpk)#
print 'The Output Frequency = %0.2e Hertz'%fo
print 'i.e 15.915 kHz'

The Output Frequency = 1.59e+04 Hertz
i.e 15.915 kHz

Example No. 33_8 Page No. 1085

In [15]:
# Calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.

# Given data

Vin = 1#        # Input voltage=1 Volts(p-p)
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
Ri = 1.*10**3#    # Input resistance=1 kOhms

Acl = 1+(Rf/Ri)#
print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl

Vo = Vin*Acl#
print 'The Output Voltage = %0.2f Volts(p-p)'%Vo

The Closed-Loop Voltage Gain Acl =11.00
The Output Voltage = 11.00 Volts(p-p)

Example No. 33_9 Page No. 1089

In [16]:
# Calculate Zin(CL) and Zout(CL). Assume Rin is 2 MOhms, Avol is 100,000, and Zout(OL) is 75 Ohms.

# Given data

Avol = 100000.#  # Open loop voltage gain=100,000
Ri = 2.*10**6#    # Input resistance=2 MOhms
B = 0.0909#     # Beta=0.0909
Zool = 75.#      # Output impedence (open-loop)=75 Ohms

Zicl = Ri*(1+Avol*B)#
print 'The Input Impedence Closed-Loop = %0.2e Ohms'%Zicl
print 'i.e 18 GOhms'

A = Avol*B#

Zocl = Zool/(1+A)#
print 'The Closed-Loop Output Impedence = %0.2f Ohms'%Zocl

The Input Impedence Closed-Loop = 1.82e+10 Ohms
i.e 18 GOhms
The Closed-Loop Output Impedence = 0.01 Ohms

Example No. 33_10 Page No. 1090

In [17]:
# Assume Rin is 2 MOhms, Avol is 100,000, and Zout(OL) is 75 Ohms. Calculate Zin(CL) and Zout(CL)

# Given data

Avol = 100000.#  # Open loop voltage gain=100,000
Ri = 2.0*10**6#    # Input resistance=2 MOhms
B = 1.0#          # Beta=1
Zool = 75.#      # Output impedence (open-loop)=75 Ohms

Zicl = Ri*(1+Avol*B)#
print 'The Input impedence closed-loop = %0.2e Ohms'% Zicl
print 'i.e 200 GOhms'

A = Avol*B#

Zocl = Zool/(1+A)#
print 'The Closed loop Output Impedence = %0.3f Ohms'%Zocl

The Input impedence closed-loop = 2.00e+11 Ohms
i.e 200 GOhms
The Closed loop Output Impedence = 0.001 Ohms

Example No. 33_11 Page No. 1091

In [18]:
# Calculate the closed-loop voltage gain, Acl, and the dc voltage at the op-amp output terminal.

# Given data

V = 15.#         # Voltage at +ve terminal of op-amp=15 Volts
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
Ri = 1.*10**3#    # Input resistance=1 kOhms
R1 = 10.*10**3#   # Resistance1=10 kOhms
R2 = 10.*10**3#   # Rsistance2=10 kOhms

Acl = -(Rf/Ri)#
print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl

Vo = V*(R2/(R1+R2))#
print 'The Output Voltage = %0.2f Volts'%Vo

The Closed-Loop Voltage Gain Acl =-10.00
The Output Voltage = 7.50 Volts

Example No. 33_12 Page No. 1095

In [19]:
# Calculate the output voltage, Vout.

# Given data

V1 = 1#     # Input voltage1=1 Volts
V2 = -5#    # Input voltage2=-5 Volts
V3 = 3#     # Input voltage3=3 Volts

Vo = -(V1+V2+V3)#
print 'The Output Voltage %0.f Volts'%Vo

The Output Voltage 1 Volts

Example No. 33_13 Page No. 1097

In [20]:
# Calculate the output voltage, Vout.

# Given data

V1 = 0.5#       # Input voltage1=0.5 Volts
V2 = -2.0#        # Input voltage2=-2 Volts
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
R1 = 1.*10**3#    # Resistance1=1 kOhms
R2 = 2.5*10**3#  # Rsistance2=2.5 kOhms

A = Rf/R1#
B = Rf/R2#

Vo = -(A*V1+B*V2)#
print 'The Output Voltage = %0.2f Volts'%Vo

The Output Voltage = 3.00 Volts

Example No. 33_14 Page No. 1101

In [21]:
# Calculate the output voltage, Vout, if (a) Vx is 1 Vdc and Vy is -0.25 Vdc, (b) -Vx is 0.5 Vdc and Vy is 0.5 Vdc, (c) Vx is 0.3 V and Vy is 0.3 V.

# Given data

Rf = 10.*10**3#   # Feedback resistance=10 kOhms
R1 = 1.*10**3#    # Resistance1=1 kOhms
Vx1 = 1.#        # Input voltage Vx1 at -ve terminal of op-amp=1 Volts
Vy1 = -0.25#    # Input voltage Vy1 at +ve terminal of op-amp=-0.25 Volts
Vx2 = -0.5#     # Input voltage Vx2 at -ve terminal of op-amp=-0.5 Volts
Vy2 = 0.5#    # Input voltage Vy2 at +ve terminal of op-amp=0.5 Volts
Vx3 = 0.3#        # Input voltage Vx3 at -ve terminal of op-amp=0.3 Volts
Vy3 = 0.3#    # Input voltage Vy3 at +ve terminal of op-amp=0.3 Volts

A = -Rf/R1#

#  Case A

Voa = A*(Vx1-Vy1)#
print 'The Output Voltage of Case A = %0.2f Volts'%Voa

#  Case B

Voa = A*(Vx2-Vy2)#
print 'The Output Voltage of Case B = %0.2f Volts'%Voa

#  Case C

Voa = A*(Vx3-Vy3)#
print 'The Output Voltage of Case C = %0.2f Volts'%Voa

The Output Voltage of Case A = -12.50 Volts
The Output Voltage of Case B = 10.00 Volts
The Output Voltage of Case C = -0.00 Volts

Example No. 33_15 Page No. 1102

In [22]:
# Assume that Rd increases to 7.5 k due to an increase in the ambient temperature. Calculate the output of the differential amplifier. Note: Rb is 5 kOhms.

# Given data

Vi = 5.#         # Voltage input=5 Volts(dc)
Rf = 10.*10**3#   # Feedback resistance=10 kOhms
R1 = 1.*10**3#    # Resistance1=1 kOhms
Ra = 5.*10**3#    # Resistance A at wein bridge=5 kOhms
Rb = 10.*10**3#   # Resistance B at wein bridge=10 kOhms
Rc = 5.*10**3#    # Resistance C at wein bridge=5 kOhms
Rd = 7.5*10**3#  # Resistance D at wein bridge=7.5 kOhms

Vx = Vi*(Ra/Rb)#
Vy = Vi*(Rd/(Rd+Rc))#
A = -Rf/R1

Vo = A*(Vx-Vy)#
print 'The Output of Differential Amplifier = %0.2f Volts'%Vo

The Output of Differential Amplifier = 5.00 Volts

Example No. 33_16 Page No. 1103

In [25]:
from math import pi
# Calculate the cutoff frequency, fc.

# Given data

Rf = 10.*10**3#       # Feedback resistance=10 kOhms
Cf = 0.01*10**-6#    # Feedback capacitance=0.01 uFarad

fc = 1./(2.*pi*Rf*Cf)#
print 'The Cutoff Frequency = %0.2e Hertz'%fc
print 'i.e 1.591 kHz'

The Cutoff Frequency = 1.59e+03 Hertz
i.e 1.591 kHz

Example No. 33_17 Page No. 1104

In [27]:
from math import pi,sqrt
# Calculate the Voltage gain, Acl at (a)0 Hz and (b) 1 MHz

# Given data

f1 = 1.*10**6#        # Frequency=1 MHertz
Rf = 10.*10**3#       # Feedback resistance=10 kOhms
R1 = 1.*10**3#        # Resistance1=1 kOhms
Cf = 0.01*10**-6#    # Feedback capacitance=0.01 uFarad

# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf 

Acl = -Rf/R1#
print 'The Closed-Loop Voltage Gain at 0 Hz =%0.2f'%Acl

# At 1 MHz

Xcf = 1/(2*pi*f1*Cf)#

A = (Rf*Rf)#
B = (Xcf*Xcf)#

Zf = ((Xcf*Rf)/sqrt(A+B))#

Acl1 = -Zf/R1#
print 'The Closed-Loop Voltage Gain at 1 MHz =%0.2f'%Acl1

The Closed-Loop Voltage Gain at 0 Hz =-10.00
The Closed-Loop Voltage Gain at 1 MHz =-0.02

Example No. 33_18 Page No. 1105

In [28]:
from math import log10,pi,sqrt
# Calculate the dB voltage gain, at (a)0 Hz and (b) 1.591 kHz

# Given data

f1 = 1.591*10**3#    # Frequency=1.591 kHertz
Rf = 10.*10**3#       # Feedback resistance=10 kOhms
Ri = 1.*10**3#        # Input resistance=1 kOhms
Cf = 0.01*10**-6#    # Feedback capacitance=0.01 uFarad

# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf 

A = Rf/Ri

Acl = 20*log10(A)#
print 'The Voltage Gain at 0 Hz = %0.2f dB'%Acl

# At 1.591 kHz

Xcf = 1/(2*pi*f1*Cf)#
B = (Rf*Rf)#
C = (Xcf*Xcf)#
Zf = (Xcf*Rf/sqrt(B+C))#
D = Zf/Ri#

Acl1 = 20*log10(D)#
print 'The Voltage Gain at 1.591 kHz = %0.2f dB'%Acl1
print 'approx 17dB'

The Voltage Gain at 0 Hz = 20.00 dB
The Voltage Gain at 1.591 kHz = 16.99 dB
approx 17dB

Example No. 33_19 Page No. 1106

In [30]:
from math import pi
# Calculate the cutoff frequency, fc.

# Given data

Ri = 1.*10**3#       # Input resistance=10 kOhms
Ci = 0.1*10**-6#    # Input capacitance=0.01 uFarad

fc = 1/(2*pi*Ri*Ci)#
print 'The Cutoff Frequency = %0.2f Hertz'%fc
print 'i.e 1.591 kHz'

The Cutoff Frequency = 1591.55 Hertz
i.e 1.591 kHz

Example No. 33_20 Page No. 1118

In [31]:
# Vin is  5 V, R is  1 kOhms , and Rl is  100 Ohms . Calculate the output current, Iout.

# Given data

Vin = 5.#        # Input votage=5 Volts
Ri = 1.*10**3#    # Input resistance=1 kOhms
Rl = 100.#       # Load resistance=100 Ohms

Io = Vin/Ri#
print 'The Output Current = %0.2e Amps'%Io
print 'i.e 5 mAmps'

The Output Current = 5.00e-03 Amps
i.e 5 mAmps

Example No. 33_21 Page No. 1120

In [32]:
# Iin is 1.5 mA, R is 1 kOhms, and Rl is 10 kOhms. Calculate Vout.

# Given data

Iin = 1.5*10**-3#    # Input votage=5 Volts
Ri = 1.*10**3#        # Input resistance=1 kOhms
Rl = 100.#           # Load resistance=100 Ohms

Vo = Iin*Ri#
print 'The Output Voltage = %0.2f Volts'%Vo

The Output Voltage = 1.50 Volts

Example No. 33_22 Page No. 1121

In [34]:
# R1 is 1 kOhms and R2 is 100 kOhms . Calculate UTP, LTP, and VH.

# Given data

R1 = 1.*10**3#        # Resistance1=1 kOhms
R2 = 100.*10**3#      # Resistance2=100 kOhms
Vcc = 15.#           # Applied votage=15 Volts
Vsat = 13.#          # Assume Saturation voltage=13 Volts

Beta = R1/(R1+R2)#

Utp = Beta*Vsat#
print 'The Upper Trigger Point = %0.3f Volts'%Utp
print 'i.e 128.7 mVolts'

Ltp = -Beta*Vsat#
print 'The Lower Trigger Point = %0.3f Volts'%Ltp
print 'i.e -128.7 mVolts'

Vh = Utp-Ltp#
print 'The Hysterisis Voltage = %0.3f Volts'%Vh
print 'i.e 257.4 mVolts'

The Upper Trigger Point = 0.129 Volts
i.e 128.7 mVolts
The Lower Trigger Point = -0.129 Volts
i.e -128.7 mVolts
The Hysterisis Voltage = 0.257 Volts
i.e 257.4 mVolts

Example No. 33_23 Page No. 1124

In [35]:
# Rl is 1 kOhms  and the frequency of the input voltage equals 100 Hz. Calculate the minimum value of C required.

# Given data

f = 100.#        # Applied frequency=100 Hertz
Rl = 1.*10**3#    # Load resistance=1 kOhms

T = 1./f#

C = (10*T)/Rl#
print 'The Minimum value of required Capacitor = %0.2e Farads'%C
print 'i.e 100 uFarad'

The Minimum value of required Capacitor = 1.00e-04 Farads
i.e 100 uFarad