_
+
Kristin Ackerson, Virginia Tech EE
Spring 2002
Table of Contents
The Operational Amplifier______________________________slides 3-4
The Four Amplifier Types______________________________slide 5
VCVS(Voltage Amplifier) Summary:
Noninverting Configuration____________slides 6-9
Inverting Configuration________________slides 10-12
ICIC(Current Amplifier) Summary________________________slide 13
VCIS (Transconductance Amplifier) Summary_____________slides 14-15
ICVS (Transresistance Amplifier) Summary_______________slides 16-18
Power Bandwidth_____________________________________slide 19
Slew Rate____________________________________________slide 20
Slew Rate Output Distortion____________________________ slide 21
Noise Gain___________________________________________slide 22
Gain-Bandwidth Product_______________________________slide 23
Cascaded Amplifiers - Bandwidth________________________slide 24
Common Mode Rejection Ratio__________________________slides 25-26
Power Supply Rejection Ratio___________________________slide 27
Sources_____________________________________________slide 28
Kristin Ackerson, Virginia Tech EE
Spring 2002
The Operational Amplifier
•
Usually Called Op Amps
•
An amplifier is a device that accepts a varying input signal and
produces a similar output signal with a larger amplitude.
•
Usually connected so part of the output is fed back to the input.
(Feedback Loop)
•
Most Op Amps behave like voltage amplifiers. They take an input
voltage and output a scaled version.
•
They are the basic components used to build analog circuits.
•
The name “operational amplifier” comes from the fact that they were
originally used to perform mathematical operations such as
integration and differentiation.
•
Integrated circuit fabrication techniques have made highperformance operational amplifiers very inexpensive in comparison
to older discrete devices.
Kristin Ackerson, Virginia Tech EE
Spring 2002
The Operational Amplifier
+VS
Inverting
_
i(-)
RO
vid
Noninverting
i(+)
Ri
Output
vO = Advid
A
+
-VS
• i(+), i(-) : Currents into the amplifier on the inverting and noninverting lines
respectively
• vid : The input voltage from inverting to non-inverting inputs
• +VS , -VS : DC source voltages, usually +15V and –15V
• Ri : The input resistance, ideally infinity
• A : The gain of the amplifier. Ideally very high, in the 1x1010 range.
• RO: The output resistance, ideally zero
• vO: The output voltage; vO = AOLvid where AOL is the open-loop voltage gain
Kristin Ackerson, Virginia Tech EE
Spring 2002
The Four Amplifier Types
Description
Gain
Symbol
Transfer
Function
Voltage Amplifier
or
Voltage Controlled Voltage Source (VCVS)
Av
vo/vin
Current Amplifier
or
Current Controlled Current Source (ICIS)
Ai
io/iin
Transconductance Amplifier
or
Voltage Controlled Current Source (VCIS)
gm
(siemens)
io/vin
Transresistance Amplifier
or
Current Controlled Voltage Source (ICVS)
rm
(ohms)
vo/iin
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier) Summary
Noninverting Configuration
i(+)
+
vid
+
vin
iO
_
i(-)
_
+
vF
iL
RF RL
_
+
v1
Applying KVL the
following equations
can be found:
v1 = vin
vO = v1 + vF = vin+ iFRF
_
+
iF
R1
i1
+
vO
vL
-
_
vid = vo/AOL
Assuming AOL  
vid =0
Also, with the
assumption that Rin = 
i(+) = i(-) = 0
This means that,
iF = i1
Therefore: iF = vin/R1
Using the equation to the left the output
voltage becomes:
vo = vin + vinRF = vin RF + 1
R1
R1
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier) Summary
Noninverting Configuration Continued
The closed-loop voltage gain is symbolized by Av and is found to be:
Av = vo = RF + 1
vin
R1
The original closed loop gain equation is:
Av = AF = AOL
AF is the amplifier
gain with
1 + AOL
feedback
Ideally AOL   , Therefore Av = 1

Note: The actual value of AOL is given for the specific device and
usually ranges from 50k  500k.
 is the feedback factor and by assuming open-loop gain is infinite:
 =
R1
R1 + R F
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier) Summary
Noninverting Configuration Continued
Input and Output Resistance
Ideally, the input resistance for this configuration is infinity, but the a
closer prediction of the actual input resistance can be found with the
following formula:
RinF = Rin (1 + AOL)
Where Rin is given for the
specified device. Usually Rin is
in the M range.
Ideally, the output resistance is zero, but the formula below gives a
more accurate value:
RoF =
Ro
AOL + 1
Where Ro is given for the
specified device. Usually Ro is in
the 10s of s range.
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier)
Noninverting Configuration Example
i(+)
+
vid
vin
+
_
i(-)
iO
_
+
vF
_
+
v1
_
iL
iF
RF RL
+
vL
_
+
vO
-
Given: vin = 0.6V, RF = 200 k
R1 = 2 k , AOL = 400k
Rin = 8 M  , Ro = 60 
Find: vo , iF , Av ,  , RinF and RoF
R1
i1
Solution:
vo = vin + vinRF = 0.6 + 0.6*2x105 = 60.6 V
R1
2000
Av = RF + 1 = 2x105 + 1 = 101
R1
2000
iF = vin = 0.6 = 0.3 mA
R1 2000
 = 1 = 1 = 9.9x10-3
AOL 101
RinF = Rin (1 + AOL) = 8x106 (1 + 9.9x10-3*4x105) = 3.1688x1010 
RoF =
Ro
=
60
= 0.015 
AOL + 1
9.9x10-3*4x105 + 1
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier) Summary
Inverting Configuration
RF
iF
i1
R1
+
+
+
_
vin
vO
_
RL
-
The same
assumptions used to
find the equations for
the noninverting
configuration are
also used for the
inverting
configuration.
General Equations:
i1 = vin/R1
iF = i1
vo = -iFRF = -vinRF/R1
Av = RF/R1
 = R1/RF
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier) Summary
Inverting Configuration Continued
Input and Output Resistance
Ideally, the input resistance for this configuration is equivalent to R1.
However, the actual value of the input resistance is given by the
following formula:
Rin = R1 +
RF
1 + AOL
Ideally, the output resistance is zero, but the formula below gives a
more accurate value:
RoF =
Note:
=
R1
R1 + RF
Ro
1 + AOL
This is different from the equation used
on the previous slide, which can be confusing.
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCVS (Voltage Amplifier)
Inverting Configuration Example
RF
iF
R1
i1
+
+
_
vin
_
RL
+
vO
-
Given: vin = 0.6 V, RF = 20 k
R1 = 2 k , AOL = 400k
Rin = 8 M  , Ro = 60 
Find: vo , iF , Av ,  , RinF and RoF
Solution:
vo = -iFRF = -vinRF/R1 = -(0.6*20,000)/2000 = 12 V
iF = i1 = vin/R1 = 1 / 2000 = 0.5 mA
Av = RF/R1 = 20,000 / 2000 = 10
Rin = R1 +
RoF =
 = R1/RF = 2000 / 20,000 = 0.1
RF
= 2000 + 20,000
= 2,000.05 
1 + AOL
1 + 400,000
Ro
=
60
= 1.67 m 
1 + AOL 1 + 0.09*400,000 Note:  is 0.09 because using
different formula than above
Kristin Ackerson, Virginia Tech EE
Spring 2002
ICIS (Current Amplifier) Summary
 Not commonly done using operational amplifiers
iL
Load
_
iin
+
1 Possible
ICIS
Operational
Amplifier
Application
iin = iL
Similar to the voltage
follower shown below:
_
+
vin
_
+
vin = vo
+
vO
-
Both these amplifiers have
unity gain:
Av = Ai = 1
Voltage Follower
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCIS (Transconductance Amplifier) Summary
Voltage to Current Converter
iL
i1
R1
+
vin
Load
Load
i1
_
+
OR
R1
_
+
+
vin
_
iL
_
+
vin
_
General Equations:
iL = i1 = v1/R1
v1 = vin
The transconductance, gm = io/vin = 1/R1
Therefore, iL = i1 = vin/R1 = gmvin
The maximum load resistance is determined by:
RL(max) = vo(max)/iL
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCIS (Transconductance Amplifier)
Voltage to Current Converter Example
iL
i1
+
vin
R1
Load
_
Given: vin = 2 V, R1 = 2 k
vo(max) =  10 V
Find: iL , gm and RL(max)
+
_
Solution:
iL = i1 = vin/R1 = 2 / 2000 = 1 mA
Note:
• If RL > RL(max) the op amp
will saturate
• The output current, iL is
independent of the load
resistance.
gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS
RL(max) = vo(max)/iL = 10 V / 1 mA
= 10 k 
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCIS (Transresistance Amplifier) Summary
Current to Voltage Converter
RF
iF
_
+
iin
+
vO
-
General Equations:
iF = iin
vo = -iFRF
rm = vo/iin = RF
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCIS (Transresistance Amplifier) Summary
Current to Voltage Converter
•
Transresistance Amplifiers are used for low-power
applications to produce an output voltage proportional to
the input current.
•
Photodiodes and Phototransistors, which are used in the
production of solar power are commonly modeled as
current sources.
•
Current to Voltage Converters can be used to convert these
current sources to more commonly used voltage sources.
Kristin Ackerson, Virginia Tech EE
Spring 2002
VCIS (Transresistance Amplifier)
Current to Voltage Converter Example
RF
iF
Given: iin = 10 mA
RF = 200 
Find: iF , vo and rm
_
+
iin
+
vO
-
Solution:
iF = iin = 10 mA
vo = -iFRF = 10 mA * 200  = 2 V
rm = vo/iin = RF = 200
Kristin Ackerson, Virginia Tech EE
Spring 2002
Power Bandwidth
The maximum frequency at which a sinusoidal output signal can be
produced without causing distortion in the signal.
The power bandwidth, BWp is determined using the desired
output signal amplitude and the the slew rate (see next slide)
specifications of the op amp.
BWp =
SR
2Vo(max)
SR = 2fVo(max) where SR is the slew rate
Example:
Given: Vo(max) = 12 V and SR = 500 kV/s
Find:
BWp
Solution:
BWp =
500 kV/s = 6.63 kHz
2 * 12 V
Kristin Ackerson, Virginia Tech EE
Spring 2002
Slew Rate
A limitation of the maximum possible rate of change of the
output of an operational amplifier.
As seen on the previous slide,
This is derived from:
SR = 2fVo(max)
SR = vo/tmax
 f is the
frequency in
Hz
Slew Rate is independent of the
closed-loop gain of the op amp.
Example:
Given: SR = 500 kV/s and vo = 12 V (Vo(max) = 12V)
Find: The t and f.
Solution: t = vo / SR = (10 V) / (5x105 V/s) = 2x10-5 s
f = SR / 2Vo(max) = (5x105 V/s) / (2 * 12) = 6,630 Hz
Kristin Ackerson, Virginia Tech EE
Spring 2002
Slew Rate Distortion
v
desired output
waveform
SR = v/t = m (slope)

v
actual output
because of
slew rate
limitation

t
t
The picture above shows exactly what happens when the
slew rate limitations are not met and the output of the
operational amplifier is distorted.
Kristin Ackerson, Virginia Tech EE
Spring 2002
Noise Gain
The noise gain of an amplifier is independent of the amplifiers
configuration (inverting or noninverting)
The noise gain is given by the formula:
AN = R1 + R F
R1
Example 1: Given a noninverting amplifier with the resistance
values, R1 = 2 k and RF = 200 k
Find: The noise gain.
AN = 2 k + 200 k = 101
2 k
 Note: For the
noninverting amplifier AN = AV
Example 2: Given an inverting amplifier with the resistance
values, R1 = 2 k and RF = 20 k
Find: The noise gain.
AN = 2 k + 20 k = 12
2 k
 Note: For the
inverting amplifier AN > AV
Kristin Ackerson, Virginia Tech EE
Spring 2002
Gain-Bandwidth Product
In most operational amplifiers, the open-loop gain begins
dropping off at very low frequencies. Therefore, to make the
op amp useful at higher frequencies, gain is traded for
bandwidth.
The Gain-Bandwidth Product (GBW) is given by:
GBW = ANBW
Example: For a 741 op amp, a noise gain of 10 k corresponds
to a bandwidth of ~200 Hz
Find: The GBW
GBW = 10 k * 200 Hz = 2 MHz
Kristin Ackerson, Virginia Tech EE
Spring 2002
Cascaded Amplifiers - Bandwidth
Quite often, one amplifier does not increase the signal enough
and amplifiers are cascaded so the output of one amplifier is the
input to the next.
The amplifiers are matched so:
BWS = BW1 = BW2 = GBW
AN
where, BWS is the bandwidth of all
the cascaded amplifiers and AN is
the noise gain
The Total Bandwidth of the Cascaded Amplifiers is:
BWT = BWs(21/n – 1)1/2
where n is the number of amplifiers
that are being cascaded
Example: Cascading 3 Amplifiers with GBW = 1 MHz and AN = 15,
Find: The Total Bandwidth, BWT
BWS = 1 MHz / 15 = 66.7 kHz
BWT = 66.7 kHz (21/3 – 1)1/2 = 34 kHz
Kristin Ackerson, Virginia Tech EE
Spring 2002
Common-Mode Rejection Ratio
The common-mode rejection ratio (CMRR) relates to the ability of
the op amp to reject common-mode input voltage. This is very
important because common-mode signals are frequently
encountered in op amp applications.
CMRR = 20 log|AN / Acm|
Acm =
AN
log-1 (CMRR / 20)
We solve for Acm because Op Amp data sheets list the CMRR value.
The common-mode input voltage is an average of the voltages that
are present at the non-inverting and inverting terminals of the
amplifier.
vicm = v(+) + v(-)
2
Kristin Ackerson, Virginia Tech EE
Spring 2002
Common-Mode Rejection Ratio
Example
Given: A 741 op amp with CMRR = 90 dB and a noise gain,
AN = 1 k
Find: The common mode gain, Acm
Acm =
AN
=
log-1 (CMRR / 20)
1000
log-1 (90 / 20)
= 0.0316
It is very desirable for the common-mode gain to be small.
Kristin Ackerson, Virginia Tech EE
Spring 2002
Power Supply Rejection Ratio
One of the reasons op amps are so useful, is that they can
be operated from a wide variety of power supply voltages.
The 741 op amp can be operated from bipolar supplies
ranging from 5V to 18V with out too many changes to
the parameters of the op amp.
The power supply rejection ratio (SVRR) refers to the slight
change in output voltage that occurs when the power
supply of the op amp changes during operation.
SVRR = 20 log (Vs / Vo)
The SVRR value is given for a specified op amp. For the
741 op amp, SVRR = 96 dB over the range 5V to 18V.
Kristin Ackerson, Virginia Tech EE
Spring 2002
Open-Loop Op Amp Characteristics
Table 12.11
Device
LM741C
LF351
OP-07
LH0003
AD549K
Technology
BJT
BiFET
BJT
Hybrid
BJT
BiFET
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
Kristin Ackerson, Virginia Tech EE
Spring 2002
Sources
Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New
Jersey: 2001. (pp 456-509)
1Table
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,
New York: 1998.
Neamen, Donald. Semiconductor Physics & Devices. Basic Principles. McGraw-Hill,
Boston: 1997. (pp 351-357)
Web Sources
www.infoplease.com/ce6/sci/A0803814.html
http://www.infoplease.com/ce6/sci/A0836717.html
http://people.msoe.edu/~saadat/PSpice230Part3.htm
Kristin Ackerson, Virginia Tech EE
Spring 2002