Zulfiqar Ali

Chapter GoalUnderstand behavior and characteristics of ideal differential and op amps.Demonstrate circuit analysis techniques for ideal op amps.Characterize inverting, non-inverting, summing and instrumentation amplifiers, voltage follower and integrator.Learns factors involved in circuit design using op amps.Provide and introduction to active filtersExplore applications of op amps in nonlinear circuits, such as precision rectifiers.Provide examples of multivibrator circuits employing positive feedback.Demonstrate use of ac analysis capability of SPICE.

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Differential Basic Model

Represented by:A= open-circuit voltage gainvid = (v+-v-) = differential input signal voltageRid = amplifier input resistanceRo = amplifier output resistance

Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 1800 out of phase with that applied at - input (inverting) terminal.

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Differential Amplifier Model: With Source and Load

LRoRLR

A += idv*ovSRidR

idR+= svidv

RL = load resistanceRS = Thevenin equivalent resistance of signal sourcevs = Thevenin equivalent voltage of signal source

and

LRoRLR

SRidRidR

vA ++==svov

Op amp circuits are mostly dc-coupled amplifiers. Signals vo and vs may have a dc component representing a dc shift of the input away from Q-point. Op-amp amplifies both dc and ac components.

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Differential Amplifier Model: With Source and Load (Example)

• Problem: Calculate voltage gain• Given Data: A=100, Rid =100kΩ, Ro = 100Ω, RS =10kΩ, RL =1000Ω• Analysis:

• Ideal amplifier’s output depends only on input voltage difference and not on source and load resistances.This can be achieved by using fully mismatched resistance condition (Rid >> RS or infinite Rid and Ro << RL or zero Ro ).

or

A = open-loop gain (maximum voltage gain available from the device)

dB3.386.820100100

1000k100k10

k100100svov

==Ω+Ω

ΩΩ+Ω

Ω=

++==

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

LRoRLR

SRidRidR

vA

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Ideal Operational Amplifier• Ideal op amp is a special case of ideal differential

amplifier with infinite gain, infinite Rid and zero Ro .

and

– If A is infinite, vid is zero for any finite output voltage.– Infinite input resistance Rid forces input currents i+ and i- to be

zero.• Ideal op amp has following assumptions:

– Infinite common-mode rejection, power supply rejection, open-loop bandwidth, output voltage range, output current capability and slew rate

– Zero output resistance, input-bias currents and offset current, input-offset voltage.

0idvlim =∞→AA

ovidv =

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Inverting Amplifier: Configuration

Positive input is grounded.Feedback network, resistors R1 and R2 connected between inverting input and signal source and amplifier output node respectively.

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Inverting Amplifier:Voltage Gain

0ov22i1isv =−−− RRs

1

svsi R=∴

1

2svov

R

RvA −==

But is=i2 and v-=0 (since vid=v+-v-=0)

Negative voltage gain implies 1800 phase shift between dc/sinusoidal input and output signals.Gain greater than 1 if R2 > R1Gain less than 1 if R1 > R2Inverting input of op amp is at ground potential (not connected directly to ground) and is said to be at virtual ground.

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Inverting Amplifier: Input and Output Resistances

1sisv

RinR ==

11i22ixv RR +=

)12(1ixv RR +=∴

0=∴ outR

1sisv

RinR ==

Rout is found by applying a test current (or voltage) source to amplifier output and determining the voltage(or current) and turning off all independent sources. Hence, vs = 0

But i1=i2

Since v- = 0, i1=0 and vx = 0 irrespective of the value of ix .

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Inverting Amplifier: Example• Problem:Design an inverting amplifier• Given Data: Av=20 dB, Rin =20kΩ, • Assumptions: Ideal op amp• Analysis: Input resistance is controlled by R1 and

voltage gain is set by R2 / R1.and Av=-100

A minus sign is added since the amplifier is inverting.Ω== k201 inRR

Ω==→−= MRRR

RvA 211002

1

2

100dB20/dB4010 ==vA

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Non-inverting Amplifier: Configuration

Input signal is applied to the non-inverting input terminal.Portion of the output signal is fed back to the negative input terminal.Analysis is done by relating voltage at v1 to input voltage vs and output voltage vo .

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Non-inverting Amplifier: Voltage Gain, Input Resistance and Output Resistance

211ov1v RR

R+= 1vidvsv =−

1vsv =∴

1

211

21svov

121svov

R

RR

RRvA

RRR

+=+

==∴

+=

∞=+

=isv

inR

Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0

Since i=0 and

But vid = 0

Since i+=0

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Non-inverting Amplifier: Example

• Problem:Determine the characteristics of given non-inverting amplifier

• Given Data: R1= 3kΩ, R2 =43kΩ, vs=+0.1 V • Assumptions: Ideal op amp• Analysis:

A3.33k3k43

V53.1

12ov

oi μ=Ω+Ω

=+= RR

V53.1)V1.0)(3.15(svov

k3k431

1

21

===

ΩΩ+=+=

vA

R

RvA

Since i=0

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Unity-gain Buffer

A special case of non-inverting amplifier, also called voltage follower with infinite R1 and zero R2. Hence Av =1.Provides excellent impedance-level transformation while maintaining signal voltage level.Ideal voltage buffer does not require any input current and can drive any desired load resistance without loss of signal voltage.Unity-gain buffer is used in may sensor and data acquisition systems.

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Summing Amplifier

2v2

31v

1

3ovR

R

R

R−−=

11v

1i R=22v

2i R=

Since i=0 , i3=i1+i2

Since negative amplifier input is at virtual ground,

3ov

3i R−=

Scale factors for the 2 inputs can be independently adjusted by proper choice of R2 and R1.Any number of inputs can be connected to summing junction through extra resistors.This is an example of a simple digital-to-analog converter.

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Difference Amplifier

1v1

2-v1

21)-v1v(1

2-v

21i-v22i-vov

R

R

R

RR

R

R

RR

−+

=−−

−=−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

2v21

2vRR

R

+=+

Also called a differential subtractor, amplifies difference between input signals.Rin2 is series combination of R1and R2 because i+ is zero.For v2=0, Rin1= R1, as the circuit reduces to an inverting amplifier.For general case, i1 is a function of both v1 and v2.

also

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Difference Amplifier: Example

V73.2k100k10

k100V32V21

2-VV =Ω+Ω

Ω=+

==+ RR

R

A227k10

V32.7-V5

1

-V1

V2I1I μ=

Ω=

−==

R

V0.20)k110)(A227(V522I11I1VoV −=Ω−=−−= μRR

A2272IoI μ−=−=

Problem:Determine Vo, V+, V-, Io, I1, I2, I3 .

Given Data: R1= 10kW, R2 =100kW, V1=5 V, V2=3 V Assumptions: Ideal op amp. Hence, V-= V+ and I-= I+= 0.Analysis:Using dc values,

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Integrator

dvo∫ = − 1RC

∫ vsdτ

∴vo(t)=− 1RC

vs(τ )0

t∫ dτ +vo(0)

vo(0)=Vc(0)

Rsv

si =dt

odvCci −=

Feedback resistor R2 in the inverting amplifier is replaced by capacitor C.The circuit uses frequency-dependent feedback.

Voltage at the circuit’s output at time t is given by the initial capacitor voltage integral of the input signal from start of integration interval, here, t=0.Integration of an input step signal results in a ramp at the output.

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Differentiator

Rov

Ri −=dt

sdvCsi =

dtsdv

RCov −=Input resistor R1 in the inverting amplifier is replaced by capacitor C.Derivative operation emphasizes high-frequency components of input signal, hence is less often used than the integrator.

Output is scaled version of derivative of input voltage.

Since iR= is