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Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
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Table of ContentsTable of Contents
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
The Operational Amplifier______________________________slides 3-4The Operational Amplifier______________________________slides 3-4
The Four Amplifier Types______________________________slide 5The Four Amplifier Types______________________________slide 5
VCVS(Voltage Amplifier) Summary:VCVS(Voltage Amplifier) Summary:
Noninverting Configuration____________slides 6-9Noninverting Configuration____________slides 6-9
Inverting Configuration________________slides 10-12Inverting Configuration________________slides 10-12
ICIC(Current Amplifier) Summary________________________slide 13 ICIC(Current Amplifier) Summary________________________slide 13
VCIS (Transconductance Amplifier) Summary_____________slides 14-15VCIS (Transconductance Amplifier) Summary_____________slides 14-15
ICVS (Transresistance Amplifier) Summary_______________slides 16-18ICVS (Transresistance Amplifier) Summary_______________slides 16-18
Power Bandwidth_____________________________________slide 19Power Bandwidth_____________________________________slide 19
Slew Rate____________________________________________slide 20Slew Rate____________________________________________slide 20
Slew Rate Output Distortion____________________________ slide 21Slew Rate Output Distortion____________________________ slide 21
Noise Gain___________________________________________slide 22Noise Gain___________________________________________slide 22
Gain-Bandwidth Product_______________________________slide 23Gain-Bandwidth Product_______________________________slide 23
Cascaded Amplifiers - Bandwidth________________________slide 24Cascaded Amplifiers - Bandwidth________________________slide 24
Common Mode Rejection Ratio__________________________slides 25-26Common Mode Rejection Ratio__________________________slides 25-26
Power Supply Rejection Ratio___________________________slide 27Power Supply Rejection Ratio___________________________slide 27
Sources_____________________________________________slide 28Sources_____________________________________________slide 28
The Operational AmplifierThe Operational Amplifier
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
• Usually Called Op AmpsUsually Called Op Amps
• An amplifier is a device that accepts a varying input signal and An amplifier is a device that accepts a varying input signal and produces a similar output signal with a larger amplitude.produces a similar output signal with a larger amplitude.
• Usually connected so part of the output is fed back to the input. Usually connected so part of the output is fed back to the input. (Feedback Loop)(Feedback Loop)
• Most Op Amps behave like voltage amplifiers. They take an input Most Op Amps behave like voltage amplifiers. They take an input voltage and output a scaled version.voltage and output a scaled version.
• They are the basic components used to build analog circuits.They are the basic components used to build analog circuits.
• The name “operational amplifier” comes from the fact that they The name “operational amplifier” comes from the fact that they were originally used to perform mathematical operations such as were originally used to perform mathematical operations such as integration and differentiation.integration and differentiation.
• Integrated circuit fabrication techniques have made high-Integrated circuit fabrication techniques have made high-performance operational amplifiers very inexpensive in comparison performance operational amplifiers very inexpensive in comparison to older discrete devices.to older discrete devices.
• ii(+)(+), i, i(-)(-) : Currents into the amplifier on the inverting and noninverting lines : Currents into the amplifier on the inverting and noninverting lines
respectively respectively• vvidid : The input voltage from inverting to non-inverting inputs : The input voltage from inverting to non-inverting inputs
• +V+VSS , -V , -VSS : DC source voltages, usually +15V and –15V : DC source voltages, usually +15V and –15V
• RRii : The input resistance, ideally infinity : The input resistance, ideally infinity
• A : The gain of the amplifier. Ideally very high, in the 1x10A : The gain of the amplifier. Ideally very high, in the 1x101010 range. range.• RROO: The output resistance, ideally zero: The output resistance, ideally zero
• vvOO: The output voltage; v: The output voltage; vOO = A = AOLOLvvidid where A where AOLOL is the open-loop voltage gain is the open-loop voltage gain
The Operational AmplifierThe Operational Amplifier
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
+V+VSS
-V-VSS
vvidid
InvertingInverting
NoninvertingNoninverting
OutputOutput
++
__ii(-)(-)
ii(+)(+)
vvOO = A = Addvvidid
RROO
AARRii
The Four Amplifier TypesThe Four Amplifier Types
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
DescriptionGain
SymbolTransfer Function
Voltage Amplifieror
Voltage Controlled Voltage Source (VCVS)Av vo/vin
Current Amplifieror
Current Controlled Current Source (ICIS)Ai io/iin
Transconductance Amplifieror
Voltage Controlled Current Source (VCIS)
gm
(siemens)io/vin
Transresistance Amplifieror
Current Controlled Voltage Source (ICVS)
rm
(ohms)vo/iin
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) Summary
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Noninverting ConfigurationNoninverting Configuration
++
__
vvinin
++++
--vvOO
vvidid
ii(+)(+)
ii(-)(-)
iiOO
iiFF
RRFF RRLL
RR11
ii11
vvidid = v = voo/A/AOLOL
Assuming AAssuming AOLOL
vvidid =0 =0
Also, with the Also, with the assumption that Rassumption that Rinin = =
ii(+)(+) = i = i(-)(-) = 0 = 0
__vvFF
++
__
vv11
++
__
vvLL
++
__
iiLL
Applying KVL the Applying KVL the following equations following equations
can be found:can be found:
vv11 = v = vinin
vvOO = v = v11 + v + vFF = v = vinin+ i+ iFFRRFF
This means that, This means that,
iiFF = i = i11
Therefore: iTherefore: iFF = v = vinin/R/R11
Using the equation to the left the output Using the equation to the left the output voltage becomes:voltage becomes:
vvoo = v = vinin + v + vininRRFF = v = vinin R RFF + 1 + 1
RR11 R R11
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued
The closed-loop voltage gain is symbolized by AThe closed-loop voltage gain is symbolized by Avv and is found to be: and is found to be:
AAvv = v = voo = R = RFF + 1 + 1
vvinin R R11
The original closed loop gain equation is:The original closed loop gain equation is:
AAvv = A = AFF = A = AOLOL
1 + A1 + AOLOL
Ideally AIdeally AOLOL , Therefore A , Therefore Avv = 1 = 1
Note: The actual value of ANote: The actual value of AOLOL is given for the specific device and is given for the specific device and
usually ranges from 50k usually ranges from 50k 500k. 500k. is the feedback factor and by assuming open-loop gain is infinite:is the feedback factor and by assuming open-loop gain is infinite:
= R= R11
RR11 + R + RFF
AAFF is the amplifier is the amplifier
gain with gain with feedbackfeedback
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued
Input and Output ResistanceInput and Output Resistance
Ideally, the input resistance for this configuration is infinity, but the a Ideally, the input resistance for this configuration is infinity, but the a closer prediction of the actual input resistance can be found with the closer prediction of the actual input resistance can be found with the following formula:following formula:
RRinFinF = R = Rinin (1 + (1 + AAOLOL)) Where RWhere Rinin is given for the is given for the
specified device. Usually Rspecified device. Usually Rinin is is
in the Min the M range. range.
Ideally, the output resistance is zero, but the formula below gives a Ideally, the output resistance is zero, but the formula below gives a more accurate value:more accurate value:
RRoFoF = R = Roo Where RWhere Roo is given for the is given for the
AAOLOL + 1+ 1specified device. Usually Rspecified device. Usually Roo is in is in
the 10the 10ss of of ss range. range.
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Noninverting Configuration ExampleNoninverting Configuration Example
++
__
vvinin
++++
--vvOO
vvidid
ii(+)(+)
ii(-)(-)
iiOO
iiFF
RRFF RRLL
RR11
ii11
__vvFF
++
__
vv11
++
__
vvLL
++
__
iiLL Given:Given: vvinin = 0.6V, R = 0.6V, RFF = 200 k = 200 k
RR11 = 2 k = 2 k , A , AOLOL = 400k = 400k
RRinin = 8 M = 8 M , R , Roo = 60 = 60
Find: vFind: vo o , i, iF F , A, Avv , , , R , RinFinF and R and RoFoF
Solution:Solution:
vvoo = v = vinin + v + vininRRFF = 0.6 + 0.6*2x10 = 0.6 + 0.6*2x1055 = = 60.6 V60.6 V i iFF = v = vin in = 0.6 = = 0.6 = 0.3 mA0.3 mA
RR11 20002000 R R11 2000 2000
AAvv = R = RFF + 1 = 2x10 + 1 = 2x1055 + 1 = + 1 = 101101 = 1 = 1 = = 1 = 1 = 9.9x109.9x10-3-3
RR11 20002000 A AOLOL 101 101
RRinFinF = R = Rinin (1 + (1 + AAOLOL) = 8x10) = 8x1066 (1 + 9.9x10 (1 + 9.9x10-3-3*4x10*4x1055) = ) = 3.1688x103.1688x101010
RRoFoF = R = Roo = 60= 60 = = 0.015 0.015
AAOLOL + 1 9.9x10+ 1 9.9x10-3-3*4x10*4x105 5 + 1+ 1
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting ConfigurationInverting Configuration
++
__
RRLL
vvOO
++
--vvinin
++
__
RR11ii11
RRFFiiFFThe same The same
assumptions used to assumptions used to find the equations for find the equations for
the noninverting the noninverting configuration are configuration are also used for the also used for the
inverting inverting configuration.configuration.
General Equations:General Equations:
ii11 = v = vinin/R/R11
iiFF = i = i11
vvoo = -i = -iFFRRFF = -v = -vininRRFF/R/R11
AAvv = R = RFF/R/R11 = R = R11/R/RFF
Input and Output ResistanceInput and Output Resistance
Ideally, the input resistance for this configuration is equivalent to RIdeally, the input resistance for this configuration is equivalent to R11. .
However, the actual value of the input resistance is given by the However, the actual value of the input resistance is given by the following formula:following formula:
RRinin = R = R11 + R + RFF
1 + A1 + AOLOL
Ideally, the output resistance is zero, but the formula below gives a Ideally, the output resistance is zero, but the formula below gives a more accurate value:more accurate value:
RRoFoF = R = Roo
1 + 1 + AAOLOL
Note: Note: = R = R11 This is different from the equation usedThis is different from the equation used
RR11 + R + RFF on the previous slide, which can be confusing.on the previous slide, which can be confusing.Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting Configuration ContinuedInverting Configuration Continued
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Inverting Configuration ExampleInverting Configuration Example
++
__
RRLL
++
--vvinin
++
__
RR11ii11
RRFFiiFF
Given:Given: vvinin = 0.6 V, R = 0.6 V, RFF = 20 k = 20 k
RR11 = 2 k = 2 k , A , AOLOL = 400k = 400k
RRinin = 8 M = 8 M , R , Roo = 60 = 60
Find: vFind: vo o , i, iF F , A, Avv , , , R , RinFinF and R and RoFoF
vvOO
Solution:Solution:
vvoo = -i = -iFFRRFF = -v = -vininRRFF/R/R11 = -(0.6*20,000)/2000 = = -(0.6*20,000)/2000 = 12 V12 V
iiFF = i = i11 = v = vinin/R/R11 = 1 / 2000 = = 1 / 2000 = 0.5 mA0.5 mA
AAvv = R = RFF/R/R11 = 20,000 / 2000 = = 20,000 / 2000 = 10 10 = R = R11/R/RFF = 2000 / 20,000 = = 2000 / 20,000 = 0.10.1
RRinin = R = R11 + R + RF F = 2000 + 20,000 = 2000 + 20,000 = = 2,000.05 2,000.05
1 + A1 + AOLOL 1 + 400,000 1 + 400,000
RRoFoF = R = Roo = 60= 60 = = 1.67 m 1.67 m
1 + 1 + AAOLOL 1 + 0.09*400,0001 + 0.09*400,000
Note: Note: is 0.09 because using is 0.09 because using different formula than abovedifferent formula than above
ICIS (Current Amplifier) SummaryICIS (Current Amplifier) Summary
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Not commonly done using operational amplifiersNot commonly done using operational amplifiers
++
__
LoadLoad
iiinin
iiLL
Similar to the voltage Similar to the voltage follower shown below:follower shown below:
Both these amplifiers have Both these amplifiers have unity gain:unity gain:
AAvv = A = Aii = 1 = 1
++
__
iiinin = i = iLL
vvinin = v = voovvinin
++
__ ++
--vvOO
Voltage FollowerVoltage Follower
1 Possible 1 Possible ICIS ICIS
Operational Operational Amplifier Amplifier
ApplicationApplication
VCIS (Transconductance Amplifier) SummaryVCIS (Transconductance Amplifier) Summary
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Voltage to Current ConverterVoltage to Current Converter
++
__
LoadLoadiiLL
RR11ii11
vvinin
++
__
OROR++
__
LoadLoadiiLL
RR11ii11
vvinin
++
__vvinin
++
__
General Equations:General Equations:
iiLL = i = i11 = v = v11/R/R11
vv11 = v = vinin
The transconductance, gThe transconductance, gmm = i = ioo/v/vinin = 1/R = 1/R11
Therefore, iTherefore, iLL = i = i11 = v = vinin/R/R11 = g = gmmvvinin
The maximum load resistance is determined by:The maximum load resistance is determined by:
RRL(max)L(max) = v = vo(max)o(max)/i/iLL
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCIS (Transconductance Amplifier)VCIS (Transconductance Amplifier)Voltage to Current Converter ExampleVoltage to Current Converter Example
++
__
LoadLoadiiLL
RR11ii11
vvinin
++
__
Given: vGiven: vinin = 2 V, R = 2 V, R11 = 2 k = 2 k
vvo(max)o(max) = = 10 V 10 V
Find: iFind: iL L , g, gmm and R and RL(max)L(max)
Solution:Solution:
iiLL = i = i11 = v = vinin/R/R11 = 2 / 2000 = = 2 / 2000 = 1 mA1 mA
ggmm = i = ioo/v/vinin = 1/R = 1/R11 = 1 / 2000 = = 1 / 2000 = 0.5 mS0.5 mS
RRL(max)L(max) = v = vo(max)o(max)/i/iLL = 10 V / 1 mA = 10 V / 1 mA
= = 10 k 10 k
Note: Note:
• If RIf RLL > R > RL(max)L(max) the op amp the op amp
will saturatewill saturate
• The output current, iThe output current, iLL is is
independent of the load independent of the load resistance.resistance.
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter
General Equations:General Equations:
iiFF = i = iinin
vvoo = -i = -iFFRRFF
rrmm = v = voo/i/iinin = R = RFF
++
__
iiFF
iiinin
RRFF
vvOO
++
--
VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
• Transresistance Amplifiers are used for low-power Transresistance Amplifiers are used for low-power applications to produce an output voltage proportional to applications to produce an output voltage proportional to the input current.the input current.
• Photodiodes and Phototransistors, which are used in the Photodiodes and Phototransistors, which are used in the production of solar power are commonly modeled as production of solar power are commonly modeled as current sources.current sources.
• Current to Voltage Converters can be used to convert these Current to Voltage Converters can be used to convert these current sources to more commonly used voltage sources.current sources to more commonly used voltage sources.
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
VCIS (Transresistance Amplifier)VCIS (Transresistance Amplifier)Current to Voltage Converter ExampleCurrent to Voltage Converter Example
++
__
iiFF
iiinin
RRFF
vvOO
++
--
Given: iGiven: iinin = 10 mA = 10 mA
RRFF = 200 = 200
Find: iFind: iF F , v, voo and r and rmm
Solution:Solution:
iiFF = i = iinin = = 10 mA10 mA
vvoo = -i = -iFFRRFF = 10 mA * 200 = 10 mA * 200 = = 2 V2 V
rrmm = v = voo/i/iinin = R = RFF = = 200200
Power BandwidthPower Bandwidth
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
The maximum frequency at which a sinusoidal output signal can be The maximum frequency at which a sinusoidal output signal can be produced without causing distortion in the signal.produced without causing distortion in the signal.
The power bandwidth, BWThe power bandwidth, BWpp is determined using the desired is determined using the desired
output signal amplitude and the the slew rate (output signal amplitude and the the slew rate (see next slidesee next slide) ) specifications of the op amp.specifications of the op amp.
BWBWpp = SR = SR
22VVo(max)o(max)
SR = 2SR = 2fVfVo(max)o(max) where SR is the slew rate where SR is the slew rate
Example:Example:
Given: VGiven: Vo(max)o(max) = 12 V and SR = 500 kV/s = 12 V and SR = 500 kV/s
Find: BWFind: BWpp
Solution:Solution: BW BWpp = = 500 kV/s500 kV/s = 6.63 kHz = 6.63 kHz
22 * 12 V * 12 V
Slew RateSlew Rate
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
A limitation of the maximum possible rate of change of the A limitation of the maximum possible rate of change of the output of an operational amplifier.output of an operational amplifier.
As seen on the previous slide, As seen on the previous slide, This is derived from:This is derived from:
SR = 2SR = 2fVfVo(max)o(max) SR = SR = vvoo//ttmaxmax
Slew Rate is independent of the Slew Rate is independent of the closed-loop gain of the op amp.closed-loop gain of the op amp.
Example:Example:
Given: SR = 500 kV/s and Given: SR = 500 kV/s and vvo o = 12 V (Vo(max) = 12V)= 12 V (Vo(max) = 12V)
Find: The Find: The t and f.t and f.
Solution: Solution: t = t = vo / SR = (10 V) / (5x10vo / SR = (10 V) / (5x1055 V/s) = 2x10 V/s) = 2x10-5-5 s s
f = SR / 2f = SR / 2VVo(max)o(max) = (5x10 = (5x1055 V/s) / (2 V/s) / (2 * 12) = 6,630 Hz * 12) = 6,630 Hz
f is the f is the frequency in frequency in
HzHz
Slew Rate DistortionSlew Rate Distortion
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
vv
tt
desired output desired output waveformwaveform
actual output actual output because of because of slew rate slew rate limitationlimitation
tt
vv
The picture above shows exactly what happens when the The picture above shows exactly what happens when the slew rate limitations are not met and the output of the slew rate limitations are not met and the output of the
operational amplifier is distorted.operational amplifier is distorted.
SR = SR = v/v/t = m (slope)t = m (slope)
Noise GainNoise Gain
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
The noise gain of an amplifier is independent of the amplifiers The noise gain of an amplifier is independent of the amplifiers configuration (inverting or noninverting)configuration (inverting or noninverting)
The noise gain is given by the formula:The noise gain is given by the formula:
AANN = R = R11 + R + RFF
RR11
Example 1: Given a noninverting amplifier with the resistance Example 1: Given a noninverting amplifier with the resistance values, R values, R11 = 2 k = 2 k and R and RFF = 200 k = 200 k
Find: The noise gain.Find: The noise gain.
AANN = = 2 k2 k + 200 k + 200 k = 101 = 101 Note: For the Note: For the
2 k2 k noninverting amplifier Anoninverting amplifier ANN = A = AVV
Example 2: Given an inverting amplifier with the resistance Example 2: Given an inverting amplifier with the resistance values, R values, R11 = 2 k = 2 k and R and RFF = 20 k = 20 k
Find: The noise gain.Find: The noise gain.
AANN = = 2 k2 k + 20 k + 20 k = 12 = 12 Note: For the Note: For the
2 k2 k inverting amplifier Ainverting amplifier ANN > A > AVV
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Gain-Bandwidth ProductGain-Bandwidth Product
In most operational amplifiers, the open-loop gain begins In most operational amplifiers, the open-loop gain begins dropping off at very low frequencies. Therefore, to make the dropping off at very low frequencies. Therefore, to make the
op amp useful at higher frequencies, gain is traded for op amp useful at higher frequencies, gain is traded for bandwidth.bandwidth.
The Gain-Bandwidth Product (GBW) is given by:The Gain-Bandwidth Product (GBW) is given by:
GBW = AGBW = ANNBW BW
Example: For a 741 op amp, a noise gain of 10 k corresponds Example: For a 741 op amp, a noise gain of 10 k corresponds to a bandwidth of ~200 Hz to a bandwidth of ~200 Hz
Find: The GBW Find: The GBW
GBW = 10 k * 200 Hz = 2 MHzGBW = 10 k * 200 Hz = 2 MHz
Cascaded Amplifiers - BandwidthCascaded Amplifiers - Bandwidth
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Quite often, one amplifier does not increase the signal enough Quite often, one amplifier does not increase the signal enough and amplifiers are cascaded so the output of one amplifier is the and amplifiers are cascaded so the output of one amplifier is the
input to the next.input to the next.
The amplifiers are matched so:The amplifiers are matched so:
BWBWSS = BW = BW11 = BW = BW22 = = GBWGBW where, BWwhere, BWSS is the bandwidth of all is the bandwidth of all
AANN the cascaded amplifiers and Athe cascaded amplifiers and ANN is is
the noise gainthe noise gain
The Total Bandwidth of the Cascaded Amplifiers is:The Total Bandwidth of the Cascaded Amplifiers is:
BWBWTT = BW = BWss(2(21/n1/n – 1) – 1)1/21/2 where n is the number of amplifiers where n is the number of amplifiers
that are being cascadedthat are being cascaded
Example: Cascading 3 Amplifiers with GBW = 1 MHz and AExample: Cascading 3 Amplifiers with GBW = 1 MHz and ANN = 15, = 15,
Find: The Total Bandwidth, BWFind: The Total Bandwidth, BWTT
BWBWSS = 1 MHz / 15 = 66.7 kHz = 1 MHz / 15 = 66.7 kHz
BWBWTT = 66.7 kHz (2 = 66.7 kHz (21/31/3 – 1) – 1)1/21/2 = 34 kHz = 34 kHz
Common-Mode Rejection RatioCommon-Mode Rejection Ratio
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
The common-mode rejection ratio (CMRR) relates to the ability of The common-mode rejection ratio (CMRR) relates to the ability of the op amp to reject common-mode input voltage. This is very the op amp to reject common-mode input voltage. This is very
important because common-mode signals are frequently important because common-mode signals are frequently encountered in op amp applications.encountered in op amp applications.
CMRR = 20 log|ACMRR = 20 log|AN N / A/ Acmcm||
AAcmcm = = AANN
loglog-1-1 (CMRR / 20) (CMRR / 20)
We solve for AWe solve for Acmcm because Op Amp data sheets list the CMRR value. because Op Amp data sheets list the CMRR value.
The common-mode input voltage is an average of the voltages that The common-mode input voltage is an average of the voltages that are present at the non-inverting and inverting terminals of the are present at the non-inverting and inverting terminals of the
amplifier.amplifier.
vvicmicm = v = v(+)(+) + v + v(-)(-)
22
Common-Mode Rejection RatioCommon-Mode Rejection Ratio
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
ExampleExample
Given: A 741 op amp with CMRR = 90 dB and a noise gain, Given: A 741 op amp with CMRR = 90 dB and a noise gain, AANN = 1 k = 1 k
Find: The common mode gain, AFind: The common mode gain, Acmcm
AAcmcm = = A ANN = 1000= 1000
loglog-1-1 (CMRR / 20) (CMRR / 20) log log-1-1 (90 / 20) (90 / 20)
= 0.0316= 0.0316
It is very desirable for the common-mode gain to be small.It is very desirable for the common-mode gain to be small.
Power Supply Rejection Ratio Power Supply Rejection Ratio
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
One of the reasons op amps are so useful, is that they can One of the reasons op amps are so useful, is that they can be operated from a wide variety of power supply voltages.be operated from a wide variety of power supply voltages.
The 741 op amp can be operated from bipolar supplies The 741 op amp can be operated from bipolar supplies ranging from ranging from 5V to 5V to 18V with out too many changes to the 18V with out too many changes to the
parameters of the op amp.parameters of the op amp.
The power supply rejection ratio (SVRR) refers to the slight The power supply rejection ratio (SVRR) refers to the slight change in output voltage that occurs when the power change in output voltage that occurs when the power
supply of the op amp changes during operation.supply of the op amp changes during operation.
SVRR = 20 log (SVRR = 20 log (VVss / / VVoo))
The SVRR value is given for a specified op amp. For the The SVRR value is given for a specified op amp. For the 741 op amp, SVRR = 96 dB over the range 741 op amp, SVRR = 96 dB over the range 5V to 5V to 18V.18V.
Open-Loop Op Amp Characteristics Open-Loop Op Amp Characteristics
Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
Table 12.1Table 12.111
Device LM741C LF351 OP-07 LH0003 AD549K
Technology BJT BiFET BJTHybrid
BJTBiFET
AOL(typ) 200 k 100 k 400 k 40 k 100 k
Rin 2 M 1012 8 M 100 k 1013 || 1 pF
Ro 50 30 60 50 ~100
SR 0.5 V/s 13 V/s 0.3 V/s 70 V/s 3 V/s
CMRR 90 dB 100 dB 110 dB 90 dB 90 dB
SourcesSourcesDailey, Denton. Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated.Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New Prentice Hall, New
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11Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457
Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press, Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press, New York: 1998.New York: 1998.
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Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002
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