op amp tutorial-1
DESCRIPTION
operational amplifierTRANSCRIPT
Kristin Ackerson, Virginia Tech EE Spring 2002 – VTech –Calvin Project
_
+
Operational Amplifiers
Tutorial Series
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
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 high-performance operational amplifiers very inexpensive in comparison to older discrete devices.
• 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
The Operational Amplifier +VS
-VS
vid
Inverting
Noninverting
Output
+
_ i(-)
i(+)
vO = Advid
RO A Ri
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
VCVS (Voltage Amplifier) Summary Noninverting Configuration
+
_
vin
+ +
- vO
vid
i(+)
i(-)
iO
iF
RF RL
R1
i1
vid = vo/AOL
Assuming AOL vid =0
Also, with the assumption that Rin =
i(+) = i(-) = 0
_ vF
+
_
v1
+
_
vL
+
_
iL
Applying KVL the following equations
can be found: v1 = vin
vO = v1 + vF = vin+ iFRF
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
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
1 + AOLβ
Ideally AOL , Therefore Av = 1 β
Note: The actual value of AOL is given for the specific device and usually ranges from 50k 500k.
β is the feedback factor and by assuming open-loop gain is infinite: β = R1
R1 + RF
AF is the amplifier gain with feedback
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 + βAOL) Where Rin is given for the specified device. Usually Rin is in the MΩ range.
Ideally, the output resistance is zero, but the formula below gives a more accurate value:
RoF = Ro Where Ro is given for the βAOL + 1 specified device. Usually Ro is in the 10s of Ωs range.
VCVS (Voltage Amplifier) Noninverting Configuration Example
+
_
vin +
+
- vO
vid
i(+)
i(-)
iO
iF
RF RL
R1
i1
_ vF
+
_
v1
+
_
vL
+
_
iL Given: vin = 0.6V, RF = 200 kΩ R1 = 2 kΩ , AOL = 400k Rin = 8 M Ω , Ro = 60 Ω
Find: vo , iF , Av , β , RinF and RoF
Solution: vo = vin + vinRF = 0.6 + 0.6*2x105 = 60.6 V iF = vin = 0.6 = 0.3 mA R1 2000 R1 2000
Av = RF + 1 = 2x105 + 1 = 101 β = 1 = 1 = 9.9x10-3
R1 2000 AOL 101
RinF = Rin (1 + βAOL) = 8x106 (1 + 9.9x10-3*4x105) = 3.1688x1010 Ω
RoF = Ro = 60 = 0.015 Ω βAOL + 1 9.9x10-3*4x105 + 1
VCVS (Voltage Amplifier) Summary Inverting Configuration
+
_
RL
vO
+
- vin
+
_
R1 i1
RF iF 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 β = R1/RF
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 = Ro
1 + βAOL Note: β = R1 This is different from the equation used
R1 + RF on the previous slide, which can be confusing.
VCVS (Voltage Amplifier) Summary Inverting Configuration Continued
VCVS (Voltage Amplifier) Inverting Configuration Example
+
_
RL
+
- vin
+
_
R1 i1
RF iF Given: vin = 0.6 V, RF = 20 kΩ R1 = 2 kΩ , AOL = 400k Rin = 8 M Ω , Ro = 60 Ω
Find: vo , iF , Av , β , RinF and RoF vO
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 β = R1/RF = 2000 / 20,000 = 0.1
Rin = R1 + RF = 2000 + 20,000 = 2,000.05 Ω 1 + AOL 1 + 400,000
RoF = Ro = 60 = 1.67 m Ω
1 + βAOL 1 + 0.09*400,000
Note: β is 0.09 because using different formula than above
ICIS (Current Amplifier) Summary Not commonly done using operational amplifiers
+
_ Load
iin
iL
Similar to the voltage follower shown below:
Both these amplifiers have unity gain:
Av = Ai = 1
+
_
iin = iL
vin = vo vin
+
_ +
- vO
Voltage Follower
1 Possible ICIS
Operational Amplifier
Application
VCIS (Transconductance Amplifier) Summary Voltage to Current Converter
+
_
Load iL
R1 i1
vin
+
_
OR +
_
Load iL
R1 i1
vin
+
_ 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
VCIS (Transconductance Amplifier) Voltage to Current Converter Example
+
_
Load iL
R1 i1
vin
+
_
Given: vin = 2 V, R1 = 2 kΩ vo(max) = 10 V
Find: iL , gm and RL(max)
Solution:
iL = i1 = vin/R1 = 2 / 2000 = 1 mA
gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS
RL(max) = vo(max)/iL = 10 V / 1 mA
= 10 k Ω
Note:
• If RL > RL(max) the op amp will saturate
• The output current, iL is independent of the load resistance.
VCIS (Transresistance Amplifier) Summary Current to Voltage Converter
General Equations:
iF = iin
vo = -iFRF
rm = vo/iin = RF
+
_
iF
iin
RF
vO
+
-
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.
VCIS (Transresistance Amplifier) Current to Voltage Converter Example
+
_
iF
iin
RF
vO
+
-
Given: iin = 10 mA
RF = 200 Ω Find: iF , vo and rm
Solution:
iF = iin = 10 mA
vo = -iFRF = 10 mA * 200 Ω = 2 V
rm = vo/iin = RF = 200
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 2πVo(max)
SR = 2πfVo(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 2π * 12 V
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 = 2πfVo(max) SR = vo/tmax
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 / 2πVo(max) = (5x105 V/s) / (2π * 12) = 6,630 Hz
f is the frequency in
Hz
Slew Rate Distortion v
t
desired output waveform
actual output because of slew rate limitation
t
v
The picture above shows exactly what happens when the slew rate limitations are not met and the output of the
operational amplifier is distorted.
SR = v/t = m (slope)
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 + RF
R1 Example 1: Given a noninverting amplifier with the resistance values, R1 = 2 kΩ and RF = 200 kΩ Find: The noise gain. AN = 2 kΩ + 200 kΩ = 101 Note: For the 2 kΩ noninverting amplifier AN = AV Example 2: Given an inverting amplifier with the resistance values, R1 = 2 kΩ and RF = 20 kΩ Find: The noise gain. AN = 2 kΩ + 20 kΩ = 12 Note: For the 2 kΩ inverting amplifier AN > AV
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
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 where, BWS is the bandwidth of all AN 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
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
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 = 1000
log-1 (CMRR / 20) log-1 (90 / 20) = 0.0316 It is very desirable for the common-mode gain to be small.
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.
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 MΩ 1012 Ω 8 MΩ 100 kΩ 1013 Ω || 1 pF
Ro 50 Ω 30 Ω 60 Ω 50 Ω ~100 Ω
SR 0.5 V/µs 13 V/µs 0.3 V/µs 70 V/µs 3 V/µs
CMRR 90 dB 100 dB 110 dB 90 dB 90 dB
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