Kristin Ackerson, Virginia Tech EE Spring 2002 – VTech –Calvin Project

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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