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ECE2262 Electric Circuits

Chapter 4: Operational Amplifier (OP-AMP) Circuits

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4.1 Operational Amplifiers

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4. Voltages and currents in electrical circuits may represent signals and circuits can perform mathematical operations on these signals such as:

scaling (amplification) !

signal follower (buffer) !

inverting ! addition and subtraction !

differentiation and integration ! Op - Amps provide solutions to all these problems

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An operational amplifier ("op-amp") is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. In this configuration, an op-amp produces an output potential (relative to circuit ground) that is typically hundreds of thousands of times larger than the potential difference between its input terminals.

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4.2 Op-Amp Terminals

!i+

i"

=VCC

= "VCC

Terminals of Op Amp

• inverting input (! )• noninverting input (+ )• output • positive power supply (V + =VCC )• negative power supply (V ! = !VCC )

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v+v!

i+

i!v0

Terminal Voltage Variables Terminal Current Variables

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i+i!

v!v+

v0v+v!

The op amp with power supply terminals removed When the amplifier is operating within its linear region, the dc voltages ±VCC do not enter into the circuit equations

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4.3 Op-Amp Characteristic and Circuit Models

The voltage transfer characteristic of an op amp

v+ ! v!

v0

v0 =

!VCC if A v+ ! v!( ) < !VCC

A v+ ! v!( ) if !VCC!A v+ ! v!( ) "VCCVCC if VCC < A v+ ! v!( )

#

$%%

&%%

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v+ ! v!

v0

• typical values for VCC ! 20 V

• the gain A is no less than 10, 000 ! !106

• in the linear region: v+ ! v! " 20 /104 = 2mV

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v+ ! v!

v0

• in the linear region: v+ ! v! " 20 /104 = 2mV

• node voltages in the circuits we study are much larger than 2mV

• thus, if an op amp is constrained to its linear operation region we can assume that

v+ = v! ! there is a “virtual short” between the input terminals !

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• An analysis of the op amp integrated circuit reveals that the equivalent resistance (Ri ) seen by the input terminals of the op amp is at least 1 M!

This implies that

i+ = i! = 0

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The model of ideal op amp

i+ = 0

i! = 0

v!v+ = v!v+

v0

v+ = v! i+ = i! = 0

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!10 < 6 <10

Example (a) Find v0 if va = 1 V, vb = 2 V. Assume ideal op amp

v+v!

i+i!

v0

•

• v+ = vb = 2V, since v+ = v! ! v! = vb = 2V

• i25 =va ! v!25k

= 1! 225k

= ! 125mA • i100 =

v0! v!100k

= v0! 2100

mA

• KCL at • : i25 + i100 ! i! = 0 ! i100 = !i25 ! v0! 2100

= 125

! v0 = 6V

!10 < 6 <10

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(b) For va=1.5 V specify the range of vb that avoids op amp saturation

v+

v!

i+i!

v0

•

Since i100 = !i25 !

v0! vb100k

= ! va ! vb25k

va=1.5

vb =15v0+ 6( )

The op amp is in the linear region if !10V " v0 "10V ! !0.8V " vb " 3.2V

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The linear (more realistic) circuit model of the op amp

i+

i!

v+

v!

A v+! v!( )

v0

v0 = A v+ ! v!( )+ Roi0

In the linear region of operation this model represents a more realistic approximation of an op amp. It includes: 1: A finite input resistance Ri . 2: A finite open-loop gain A . 3: A nonzero output resistance R0

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• Ri - very large • R0 - very small • A - very large

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4.4 Negative Feedback

How does a circuit maintain a virtual short (v+ = v! ) at the input of an op amp, to ensure operation in the linear region ?

Negative feedback

The portion of the output voltage is applied to the inverting input, causing v+ ! v! to decrease and hence Vo to decrease, until the op amp operates in its linear region.

Positive feedback: the output quickly becomes saturated, i.e., Vo is stuck at one extreme or the other ( ±VCC ) and does not respond to the input.

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v0 =VS

4.5 Voltage Follower and Noninverting Amplifier

A. Voltage Follower : Use the ideal op amp approximation

v+v!

i!

i+

!VS + i+RS + v+ = 0 ! v+ =VS ! v! =Vs ! negative feedback

! v0 = v! !

v0 =VS

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• Ordinary circuits connection

RLV0 =

RLRS + RL

VS

Such a connection changes the behavior of the circuits, e.g., the voltage

V0 =RL

RS + RL

VS is reduced compared to VS

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• Connection of circuits through a voltage follower

RLV0 !VS

RL >> R0 /A

The voltage follower prevents “loading down” the source voltage, i.e., the entire source voltage VS appears across the load resistance RL . Hence, RL does not draw current from the source network (the current is supplied by the op-amp).Generally, the connection of one circuit to another through a voltage follower allows both circuits to continue to operate as designed.

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Example

Vs = 5V

50! i = 20mA

200!

Function Generator

Load

+ "1V

Vo = 4V 200!

A function generator has an output resistance of 50! and so its output will experience a significant internal voltage drop when the attached load draws a “large” current, resulting in a drop in the output terminal voltage.

Therefore, one needs to “buffer” the function generator with an amplifier which presents a high input resistance to the source and which also provides a low output resistance to the load.

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An ideal buffer amplifier with a gain 1, when placed in between the function generator and the load, delivers the full source voltage to the load.

i = 0

Vi = 5V

i = 25mA

Vo = 5V

+

!

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B. Noninverting Amplifier: Use the ideal op amp approximation

i!i+

v+v!

v0Rg

• Since i+ = 0 ! v+ = vg ! v! = vg

• KCL at v! : v!Rs

+ v! ! v0Rf

= 0 ! v! =Rs

Rs + Rf

v0 ! v0 = 1+Rf

Rs

!"#

$%&vg

Note: Rf ! 0 ! voltage follower

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v0 = 1+Rf

Rs

!"#

$%&vg

Operation in the linear region requires that

v0 !VCC ! 1+Rf

Rs! VCC

vg

A noninverting amplifier multiplies the input voltage vg by a gain 1+Rf

Rs

that is independent of the source resistance Rg . Hence, the gain remains unchanged when the circuits are terminated by an external load.

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Example Find the output voltage when Rx = 60k!

v!

i!i+

v+v!

•

• KCL at • : v+Rx

+ i+ +v+ !VS15k

= 0 ! v+115k

+ 1Rx

!"#

$%&= VS15k

! v+ = 320mV

• KCL at v! : v!4.5k

+ v! ! v063k

= 0 ! v! =4.5k

63k + 4.5kv0 ( v! = v+ )

! v0 =67.5k4.5k

v+ = 4.8V ! v0 ! 5V

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v0 = !Rf

Rsvs

4.6 Inverting Amplifier: assume an ideal op amp

i!

v! v+

•

• KCL at • : is + i f ! i! = 0

• v+ = 0 ! v! = v+ = 0 ! is =vsRs

and i f =v0 ! v!Rf

= v0Rf

• Since i! = 0 ! i f = !is ! v0 = !

Rf

Rsvs

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v0 = !Rf

Rsvs

Inverting-Amplifier Equation

The upper limit on the gain Rf

Rs is determined by the power supply voltage VCC

and the value of the signal voltage vs :

v0 !VCC ! Rf

Rsvs !VCC !

Rf

Rs! VCC

vs

v0 = !Rf

Rsvs

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If Rf is removed, the negative feedback path is opened and the amplifier is operating in open loop.

v!

Opening the feedback path drastically changes the behavior of the circuit. It can be shown (use the realistic model of op-amp) that the output voltage is v0 = !Av!

Hence, we can call A the open-loop gain of the op amp.

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Realistic Op Amp Model

i+

i!

v+

v!

A v+! v!( )

v0

v0 = A v+ ! v!( )+ Roi0

v0 = !A + Ro

Rf

RsRf

1+ A + RoRi

+ RoRL

"#$

%&'+ 1+ Ro

RL

"#$

%&'1+ Rs

Ri

"#$

%&'+ RoRf

vs

! v0 = !Rf

Rsvs if Ri !" , A!" , Ro ! 0

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4.7 Summing and Difference-Amplifier Circuits

A. Summing-Amplifier Circuit (SAC)

v!

i!•

• KCL at the inverting terminal: v! ! vaRa

+ v! ! vbRb

+ v! ! vcRc

+ v! ! v0Rf

+ i! = 0

• Since v+ = 0 ! v! = 0 and i! = 0 ! solving the above w.r.t. v0 we obtain

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v0 = !

Rf

Rava +

Rf

Rbvb +

Rf

Rcvc

"#$

%&'

Inverting-Summing Amplifier Equation

v!

i!•

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• If Ra = Rb = Rc = Rs ! v0 = !Rf

Rsva + vb + vc( )

• The scaling factors in summing-amplifier circuits are entirely determined by the external resistors: Rf ,Ra , Rb ,……

v!

i!•

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Example (a) Find v0 if va = 0.1 V, vb = 0.25 V.

v+

v!•

• KCL at v! : v! ! va5k

+ v! ! vb25k

+ v! ! v0250k

= 0 ( i! = 0, v! = 0)

• v0 = ! 50va +10vb{ } = -7.5 V ! v0 !10V

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(b) If vb = 0.25 V, how large can va be before the op-amp saturates ?

v+

v!•

• From the previous solution in (a), we have v0 = ! 50va +10vb{ }

• v0 = ! 50va + 2.5{ } ! !10 " v0 "15 is satisfied if

!0.35V " va " 0.15V

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SAC

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B. Difference -Amplifier Circuit

v!

v+

i!•

• KCL at v! : v! ! vaRa

+ v! ! v0Rb

+ i! = 0 ! v! ! vaRa

+ v! ! v0Rb

= 0

• KCL at v+ : v+! vbRc

+ v+Rd

+ i+ = 0 ! v+! vbRc

+ v+Rd

= 0 ! v! ! vbRc

+ v!Rd

= 0

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v! ! vaRa

+ v! ! v0Rb

= 0

v! ! vbRc

+ v!Rd

= 0

"

#$$

%$$

v0 =

RbRa

1+RaRb

!"#

$%&

RdRc + Rd

vb 'RbRava

v!

v+

i!•

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If Rd = Rb and Rc = Ra !

v0 =RbRa

1+ RaRb

!"#

$%&

RdRc + Rd

vb 'RbRava = Rb

Ra1+ Ra

Rb

!"#

$%&

RbRa + Rb

vb 'RbRava

v!

v+

i!•

Rb

Ra v0 =RbRa

vb ! va( )

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v!

v+

i!•

Rb

Ra v0 =RbRa

vb ! va( )

The scaling is controlled by the external resistors. The relationship between the output voltage and the input voltage is not affected by connecting a nonzero load resistance across the output of the amplifier.

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Difference -Amplifier - Another Perspective

1. differential mode input: vdm = vb ! va

2. common mode input: vcm = va + vb2

! va = vcm !vdm2

and vb = vcm +vdm2

! v0 = Acmvcm +Admvdm

Acm = RaRd ! RbRcRa Rc + Rd( ) and Adm =

Rd Ra + Rb( )+ Rb Rc + Rd( )2Ra Rc + Rd( )

common mode gain differential mode gain

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If RaRb

= RcRd

! Acm = 0 ! v0 = Admvdm ! v0 = Adm vb ! va( )

v!

v+

i!•

An ideal difference amplifier ( perfect matching: RaRb

= RcRd

) amplifies only the

differential mode portion of the input signal voltage, and eliminates the common mode portion ! provides immunity to noise common to both inputs

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Example

200 ! R

Since v! = v+ and i! = i+ = 0 then VD yields

v1 =R

6k + Rv0 ! Gain = v0

v1= 6k + R

R = 31 if R = 200 !

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Example

v! = v+ =Vin , VD: Vin =R1

R1 + R2V0 ! V0 =

R1 + R2R1

Vin

! Gain = V0Vin

= 1+ R2R1

= 1+ 20k3.3k

= 7.06

V0 =Gain!Vin = 14.12 V ! I0 =V023.3k

= 14.1223.3k

= 606µA i! = 0

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Example

VV

• KCL at v+ : V ! v2R2

+ VRI

+ i+ = 0 ! v2 !VR2

= VRI

! V =RI

RI + R2v2

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• KCL at v! : V ! v1R1

+ V ! v0RF

+ i! = 0 ! v1 !VR1

= V ! v0RF

!

! v0 = ! RFR1v1 + 1+ RF

R1

"#$

%&'V V =

RI

RI + R2v2

v0 = !RFR1v1 + 1+

RFR1

"#$

%&'

RI

RI + R2v2

Note: If RFR1

= RI

R2=! ! v0 =! v2 " v1( ) ! ideal difference amplifier

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Summary: Signal Processing Circuits

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50

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If R1Rf

= R2Rg

then Vout =Rf

R1V2 !V1( )

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Op-Amp Integrator

vs

Op-Amp Differentiator

vs

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v0 = A v+ ! v!( )

v+ = v! i+ = i! = 0

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4.8 Problems ! Set 4