EE 174 Fall 2017 Operational Amplifiers - PowerPoint Presentation

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EE 174Fall 2017

Operational Amplifiers

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Contents

Introduction

Brief of History

Fundamentals of Op-Amps

Basic operation

Gain

Offset

Applications

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Introduction

Operational Amplifier (Op-Amp) name comes from the fact that it was originally used to perform mathematical operations.

Op-Amp is an active circuit element which is basic component used to build analog circuits.

Op-Amp is a low cost integrating circuit consisting of transistors, resistors, diodes and capacitors.

Op-Amp amplify an input signal produces an output voltage equal to the difference between the two input terminals multiplied by the gain A.

Op-Amps

are two-port networks in which the output voltage or current is directly proportional to either input voltage or current. Four different kind of amplifiers exits:

Voltage amplifier: A

v

= V

o

/ V

i

Current amplifier: A

i

= I

o

/ I

i

Transconductance amplifier: G

m

= I

o

/ V

i

Transresistance

amplifier: R

m

= V

o

/ I

i

In negative feedback mode, the op-amp always “wants” both inputs (inverting and non-inverting) to be the same value. If they are not, the same value, the op amp output will go positive or negative saturation, depending on which input is higher than the other.

Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters (High, Low, Band Pass and Notch Filters), Schmitt trigger, Comparators, A/D converters.

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Brief History of Op-Amp

Vacuum Tube Op-Amps (1930’s-1940’s)

Dual-supply voltage of +300/-300 V

Output swing +/- 50 voltsOpen-loop voltage gain of 15,000 to 20,000, Slew rate of +/- 12 volts/µsecondMaximum output current of 1 mAGeorge Philbrick

Solid State Discrete Op-Amps (1960’s)Dual-supply voltage of +15/-15 V Output swing +/- 11 voltsOpen-loop voltage gain of 40,000, Slew rate of +/- 1.5 volts/µsecondMaximum output current of 2.2 mA

Monolithic IC Op-Amp

First created in 1963 μA702 by Fairchild Semiconductor μA741 created in 1968, became widely used due to its ease of use 8 pin, dual in-line package (DIP)Further advancements include use of field effects transistors (FET), greater precision, faster response, and smaller packaging

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Internal Op-Amp Circuits

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Op-Amp Ideal, Equivalent Circuit, Characteristics and Features

Ideal Op-Amp

Typical Op-Amp

Op-Amp Characteristics

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The input impedance Ri is infinite - i.e. no current flows into either input.The output impedance Ro is zero - i.e. the op-amp can drive any load impedance to any voltage.The open-loop gain (A) is infinite.The bandwidth is infinite.The output voltage is zero when the input voltage difference is zero.The slew rate is infiniteIn a circuit with negative feedback, the output of the op amp will try to adjust its output so that the voltage difference between the + and − inputs is zero (V+ = V−).

The Ideal Op-Amp Assumptions

vd = v+ – v-vo = Avd = A(v+ – v-)

Op-Amp Golden Rules:

When an op-amp is configured in

any

negative-feedback arrangement, it will obey the following two rules:

i

+

=

i

-

=

0

v

+

= v

-

(broken if V

O

saturation)

Note

: The resistances used in an op-amp circuit must be much larger than

R

o

and much smaller than

R

i

in order for the ideal op-amp equations to be accurate.

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Op-Amp Gain and Bandwidth

The Voltage Gain (A) of the operational amplifier can be found using the following formula:In Decibels or (dB) is given as:

A = 10

5

, BW = 10Hz

Open loop gain:  This form of gain is measured when no feedback is applied to the op amp. Open-loop gain AVOL is very high.Closed loop gain:   This form of gain is measured when the feedback loop is closed and the overall gain of the circuit is much reduced. Loop gain: The difference between the open-loop gain and the closed-loop gain is known as the loop gain. This is useful information because it gives you the amount of negative feedback that can apply to the amplifier system. Gain versus Bandwidth: Applying feedback will reduce the gain but increase the bandwidth.

V

0

+



V

+

V



V

0

+



V

IN

A = 1, BW =1 MHz

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Gain-bandwidth (GBW) product is defined as the op-amp gain A multiplied by the bandwidth BW. The GBW product can be used to calculate the closed-loop gain bandwidth.GBW = A x BW where A is in ratio (not in dB) Closed-loop BW = GBW / AExamples: The GBW @ unity gain (from graph) GBW = 1 x 10 MHz = 10 MHzThe GBW @ closed-loop gain A = 100GBW = 100 x 100 KHz = 10 MHzBW = 1 MHz, closed-loop gain?A = GBW / BW = 10 MHz / 1MHz = 10Closed loop gain 70 dB. Closed-loop BW?Gain: 20 log A = 70 dB  log A = 70/20 = 3.5 or A = 103.5 = 3162  Closed-loop BW = 10 MHz / 3162 = 3163 Hz

Gain Bandwidth Product (GBP)

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Bode Plot Example

Given a transfer function:H(s) = = 3.3 Constant gain = 3.3Gain in dB = 20 log (3.3) 10 dBPole at 30 rad/secPhase = 45o at 30 rad/sec

 

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Op-Amp Two Basic Operations

The Non-inverting Op Amp

The Inverting Op Amp

In negative-feedback configuration, op-amp

always “wants” both inputs (inverting and non-inverting) to be the same value.

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Op-Amp Saturation

The op-amp output voltage is limited by the supply voltages.  In practice the limits are about 1.5 to 2 V below the value of the supply. The op amp has three distinct regions of operation: Linear region: −VEE < Vo < VCC  Vo/Vi = A = constant Positive saturation: Vo > VCC  Vo = VCC Negative saturation: Vo < −VEE  Vo = –VEE Note that the saturation voltage, in general, is not symmetric.For an amplifier with a given gain, A, the above range of Vo translate into a certain range for Vi − VEE < Vo < VCC or − VEE < A Vi < VCC  − VEE / A < Vi < VCC / AAny amplifier will enter its saturation region if Vi is raised above certain limit. The figure shows how the amplifier output clips when amplifier is not in the linear region.Example: For A = 105, –VDD = –12V, VCC = +15V, find range of input Vi to prevent saturation.Solution: –12V / 105 < Vi < 15V / 105 or – 0.12mV < Vi < 0.15mV

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Op-Amp slew rate

Square Wave: The slew rate (SR)of an op amp is the rate of change in the output voltage caused by a step change on the input. It is measured as a voltage change in a given time - typically V / µs. The slew rate may be very important in many applications.For the given graph, it takes 3.16 µs to swing 2.06 V so SR = 0.63V/ µs.

Sine Waves: The maximum rate of change for a sine wave occurs at the zero crossing and for a given vo = VP sin(2π f t), the slew rate SR is:

Example: Slew rate S0 = 1 V/μs; input voltage vin (t) = 1 sin(2π × 105t); closed loop gain AV =10.Sketch the theoretically output and the actual output in the same graph.What is the maximum frequency that will not violating the given slew rate?What is the maximum gain A that will not violating the given slew rate?What is the required SR to prevent distortion output?Solutions:See graphfmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHzVPmax = SR / (f x 2π) = 1 x 106 / (105 x 2π) = 1.59 V  A = 1.59For voltage gain A = 10  vo(t) = 10 sin(2π × 105t)  SR = | = 2π x f x 10 = 2π × 105 x 10 = 6.28 V/μs

 

SR = 0.63V/ µs.

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Another practical concern for op-amp performance is voltage offset.The primary cause of the Op-Amp offset is a slight mismatch of the base-emitter voltages of the differential input stage of the Op-Amp. For an ideal op-amp with both inputs shorted to ground would produce output to zero. If v+ = v- = 0  VO = A(v+ - v-) = 0. However, for a real op-amp  An offset input can drive the output to either negative or positive saturated level even if the op-amp in question has zero common-mode gain (infinite CMRR). This deviation from zero is called offset. The 741 OPAMP have offset voltage null capability by connecting 10kΩ potentiometer between pin 1 and pin 5. By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. For the 741C the offset voltage adjustment range is ± 15 mV.Effect of offset voltage on Op-Amp produces error at the output. VO = – Vin + (1 +) + Desired output Error

 

Op-Amp Offset

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1) Determine the output voltage for the open loop differential amplifier Fig #1.vOS- = 5 µVdc, vOS+ = –7 µVdcvOS- = –10 mVdc, vOS+ = 0VdcSpecifications of the Op-Amp are given below: A = 200,000, Ri = 2 MΩ , RO = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage swing = ± 14V.Solution:a) VO = A(vOS+– vOS-) = 200,000 (– 7 – 5) µV = 2.4 VRemember that vo = 2.4 V dc with the assumption that the dc output voltage is zero when the input signals are zero.b) VO = A(vOS+– vOS-) = 200,000 (0 – (– 10mV)) = 2000 V  Saturation VO = 14VThus the theoretical value of output voltage vo = 2000 V However, the OPAMP saturates at ± 14 V. 2) Given circuit in Fig #2 with Vin = 0.25V, R1 = 1kΩ , R2 = 4 kΩ Vos+ = 5mV. Determine the offset voltage error.Solution:Use superposition to obtain:VO = – Vin + (1 +) + = – 4(0.25V) + 5(0.005V) = (– 1 + 0.025)V = – 0.975V

 

Op-Amp Offset Example

Fig #1

Fig #2

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http://www.ti.com/lit/an/sboa092b/sboa092b.pdf

Op-Amp Applications

I

O

For R

I

=

nRO where n is # of inputs

(c) Voltage follower

(d) Scaling

Summer

(

d1) Voltage Adder

(

d2)

Averager

n

= 3

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(e) Differential Input

For R

1 = R2 = RI and R3 = RO 

Op-Amp

Applications

E

f

Ef

I1 = IO and I2 = I3

(f) Adder-

Subtracter

As an

adder-

subtracter

, unused inputs should be grounded.

R

and

R

I

not necessarily equal

http://

www.ti.com/lit/an/sboa092b/sboa092b.pdf

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Op-Amp Applications

General Form of the Inverting Amplifier

Differentiator

(High pass filter): Bode plot is shown below.

When

X

C = 1/(2πfCI) = RO  Gain = 1 Since highest gain is encountered at high frequencies, this circuit is very susceptible to random noise.

One practical method of reducing noise and preventing instability is shown below. The transition “point” is the frequency at which  XCI = RI

Low frequency cutoff

fO = High frequency cutoff fI = Gain: = –

 

http://

www.ti.com/lit/an/sboa092b/sboa092b.pdf

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Op-Amp Applications

Integrator (Low pass filter): Bode plot is shown below. Gain: = – = – When ZC = 1/(2πfC) = R  Gain = 1

 

Integrator

(Low pass

filter with Gain): Bode plot is shown below. DC Gain: = – AC gain: = – Corner Frequency: f0 =

 

Example:For LPF with gain, find suitable components to achieve a −3-dB frequency of 1 kHz with a dc gain of 20 dB and an input resistance of at least 10 kΩAt what frequency does gain drop to 0 dB?

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i

= 0 A

(b) For

the ideal op amp shown below, what should be the value of resistor

Rf to obtain a gain of 5?

(a) Find

i, if and vo.

Op-Amp Applications

Solution: Want Vo = 5ViFor non-inverting: For Input V+ = and Output VO = 5Vi  Gain = = Gain = 1 +  Rf = 19.5kΩAnother way: i = = Vo = (Rf + )i = (Rf + ) = 5Vi Rf = 19.5kΩ

 

V+

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(a) Given

an op-amp circuit below. If the power supplies for the op-amp are ± VCC = ± 12V. Determine:Overall gain A = VOUT / VIN, VOUT and IOUT If op-amp in saturation, what is voltages at V+ and V-.

Op-Amp Applications

Vx

Solution: (a) Vx = i = Vz = = Vy IOUT = 2 mA + 2 mA = 4 mAVOUT = = 16V  A = VOUT / VIN = 16V / 4V = 4However, VOUT = 16V > Power supply 12V Saturation  VOUT = 12V ClampFor VOUT = 12V  IOUT = 12V / ( + //2i = 3 mA / 2 = 1.5 mA Vz = = Vy = V+ V- = Vx = 2VV+ ≠ V- when op-amp in saturation mode.

 

Vy

 

 

 

V

z

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(a)

Given gain A1 = RF/RS = 100, GBW product = A0 ω0 = K = 4π × 106 . Determine the overall 3-dB bandwidth of the cascade amplifier below:

Solutions: The 3-dB bandwidth for each amplifier is:ω1 = = 4π × 104 rad/s  BW of the cascade = 4π × 104 & cascade gain A = A1A1= 102 × 102 = 104.To achieve the same gain A = 104 with a single-stage amplifier having the same K  BW: ω2 = = 4π × 102 rad/s << 4π × 104 rad/s of cascade

 

Op-Amp Applications

(b) Voltage follower application:

Solutions:

Assume

Rs

=

1

k

Ω

, RL =

100

Ω

 VL = 0.1 Vin

 Huge attenuation of the signal source.

If voltage follower (buffer amplifier) is used.

 VL =

V

out

= V

in

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A comparator is a device that compares two input voltages (inverting and noninverting input) and produces an output indicating the relationship between them. The inputs can be two signals or a signal and a fixed dc reference voltage. The output that usually swings from rail to rail. A typical comparator has low offset, high gain, and high common-mode rejection. Comparators are designed to work as open-loop systems, to drive logic circuits, and to work at high speed, even when overdriven.

Basic Comparators

Non-inverting comparator

Inverting comparator

Zero detection

V

ref

= 0V

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The Comparators with and without Hysteresis

Non-hysteresis uses a voltage divider (Rx and Ry) to set up the reference voltage. The comparator will compare the input signal (Vin) to the reference voltage (Vref). The comparator input signal is applied to the non-inverting input, so the output will have a non-inverted polarity

Hysteresis uses two different threshold voltages to avoid the multiple transitions introduced in the previous circuit. For this non-inverting comparator, the input signal must exceed the upper threshold (VH) to transition high or below the lower threshold (VL) to transition low.

Calculate VH and VL:

1k

Ω // 5.76kΩ = 0.85k ΩVH = = 2.7V VL = = 2.3V

 

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Design of Hysteresis Comparator

Equations (1) and (2) can be used to select the resistors needed to set the hysteresis threshold voltages VH and VL. One value (Rx) must be arbitrarily selected. In this example, Rx was set to 100kΩ to minimize current draw. Rh was calculated to be 575kΩ, so the closest standard value 576kΩ was used

The design requirements are as follows:

Supply Voltage: +5 V

Input: 0V to 5V

VL (Lower Threshold) = 2.3V

VH (Upper Threshold) = 2.7V

VH - VL = 0.4V

Vth ± 0.2V = 2.5V ± 0.2V

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Op-amp Comparator Circuit

Although op amps are not designed to be used as comparators, there are, nevertheless, many applications where the use of an op amp as a comparator is a proper engineering decision.

WHY USE AN OP AMP AS A COMPARATOR? • Convenience • Economy • Low IB • Low VOS There are several reasons to use op amps as comparators. Some are technical, one is purely economic. It is economical to use the spare op amp in a quad as a comparator rather than buying an additional comparator, but this is not a good design practice.

WHY NOT USE AN OP AMP AS A COMPARATOR? • Speed • Inconvenient input structures • Inconvenient logic structures • Stability/hysteresis

There are several reasons not to use op amps as comparators. First and foremost is speed, but there are also output levels, stability (and hysteresis), and a number of input structure considerations.

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References:

http://www.ume.gatech.edu/mechatronics_course/OpAmp_F08.ppthttp://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/operational-amplifier-models/http://www.ti.com/lit/an/slaa068a/slaa068a.pdfhttp://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-slew-rate.phphttp://www.planetanalog.com/author.asp?section_id=483&doc_id=562347http://www-ferp.ucsd.edu/najmabadi/CLASS/ECE60L/02-S/NOTES/opamp.pdfhttp://www-inst.eecs.berkeley.edu/~ee105/fa14/lectures/Lecture06-Non-ideal%20Op%20Amps%20(Offset-Slew%20rate).pdfhttp://nptel.ac.in/courses/117107094//lecturers/lecture_6/lecture6_page1.htmhttp://www.ti.com/ww/en/bobpease/assets/AN-31.pdfhttp://www.cs.tut.fi/kurssit/TLT-8016/Chapter2.pdfhttp://www.electronics-tutorials.ws/opamp/op-amp-comparator.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electronic/schmitt.htmlhttp://lpsa.swarthmore.edu/Bode/BodeExamples.htmlhttp://www.allaboutcircuits.com/worksheets/inverting-and-noninverting-opamp-voltage-amplifier-circuits/Fundamentals of Electrical Engineering, Giorgio Rizzoni, McGraw-Hill Higher Educationhttp://chrisgammell.com/how-does-an-op-amp-work-part-1/http://electronicdesign.com/analog/whats-all-noise-gain-stuff-anyhowhttp://howtomechatronics.com/how-it-works/electrical-engineering/schmitt-trigger/http://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter1.pdfhttp://www.ti.com/lit/an/sloa011/sloa011.pdfhttp://www.analog.com/media/en/technical-documentation/application-notes/AN-849.pdfhttp://www.ti.com/lit/ug/tidu020a/tidu020a.pdf