Op-AmpsOp-Amps

Microprocessor Interface Microprocessor Interface

Vd

+

Vo

Rin~inf Rout~0

Input 1

Input 2

Output

+Vcc

-Vcc

Operational Amplifier Operational Amplifier (Op-Amp)(Op-Amp)

Very high differential Very high differential gaingain

High input impedanceHigh input impedance Low output impedanceLow output impedance Provide voltage changes Provide voltage changes

(amplitude and polarity)(amplitude and polarity) Used in oscillator, filter Used in oscillator, filter

and instrumentationand instrumentation Accumulate a very high Accumulate a very high

gain by multiple stagesgain by multiple stages

ddo VGV

510say large,y ver

normally gain aldifferenti : dG

IC ProductIC Product

+

1

2

3

4

8

7

6

5

OFFSET NULL

-IN

+IN

V

N.C.

V+

OUTPUT

OFFSET NULL

+

1

2

3

4

8

7

6

5

OUTPUT A

-IN A

+IN A

V

V+

OUTPUT B

-IN B

+IN B+

DIP-741 Dual op-amp 1458 device

DistortionDistortion

Vd

+

Vo

+Vcc=+5V

Vcc= 5V

0

+5V

5V

The output voltage never excess the DC voltage supply of the Op-Amp

Op-Amp PropertiesOp-Amp Properties(1) Infinite Open Loop gain

- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically, Gd = 20,000 to 200,000

(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-

amp

(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically, Rout ~ 20-100

+

V1

V2

Vo

+

Vo

i1~0

i2~0

+

Rout

Vo'Rload

outload

loadoload RR

RVV

Frequency-Gain Frequency-Gain RelationRelation

• Ideally, signals are amplified from DC to the highest AC frequency

• Practically, bandwidth is limited

• 741 family op-amp have an limit bandwidth of few KHz.

• Unity Gain frequency f1: the gain at unity

• Cutoff frequency fc: the gain drop by 3dB from dc gain Gd

(Voltage Gain)

(frequency)f1

Gd

0.707Gd

fc0

1

GB Product : f1 = Gd fc

20log(0.707)=3dB

GainBandwidth ProductGainBandwidth ProductExample: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV

Sol:

Since f1 = 10 MHz

By using GB production equation

f1 = Gd fc

fc = f1 / Gd = 10 MHz / 20 V/mV

= 10 106 / 20 103

= 500 Hz

(Voltage Gain)

(frequency)f1

Gd

0.707Gd

fc0

1

10MHz

? Hz

Ideal Op-Amp Ideal Op-Amp ApplicationsApplicationsAnalysis Method :Analysis Method :Two ideal Op-Amp Properties:Two ideal Op-Amp Properties:

(1)(1) The voltage between The voltage between VV++ and and VV is zero is zero VV++ = = VV

(2)(2) The current into both The current into both VV++ and and VV termainals is zero termainals is zero

For ideal Op-Amp circuit:For ideal Op-Amp circuit:(1)(1) Write the kirchhoff node equation at the Write the kirchhoff node equation at the

noninverting terminal noninverting terminal VV++

(2)(2) Write the kirchhoff node eqaution at the inverting Write the kirchhoff node eqaution at the inverting terminal terminal VV

(3)(3) Set Set VV++ = = VV and solve for the desired closed-loop and solve for the desired closed-loop gaingain

Noninverting Amplifier(1) Kirchhoff node equation at V+

yields,

(2) Kirchhoff node equation at V yields,

(3) Setting V+ = V– yields

or

+Vin

Vo

RaRf

iVV

00

f

o

a R

VV

R

V

0

f

oi

a

i

R

VV

R

V

a

f

i

o

R

R

V

V1

+

vi

Ra

vov-

v+

Rf

+

vi

Ra

vov-

v+

Rf

R2R1

+

vi vov-

v+

Rf

+

vi

vov-

v+

Rf

R2R1

Noninverting amplifier Noninverting input with voltage divider

ia

fo v

RR

R

R

Rv ))(1(

21

2

Voltage follower

io vv

ia

fo v

R

Rv )1(

Less than unity gain

io vRR

Rv

21

2

Inverting AmplifierInverting Amplifier(1)(1) Kirchhoff node equation at Kirchhoff node equation at VV++

yields,yields,

(2)(2) Kirchhoff node equation at Kirchhoff node equation at VV yields, yields,

(3)(3) Setting Setting VV++ = = VV–– yields yields

0V

0_

f

o

a

in

R

VV

R

VV

a

f

in

o

R

R

V

V

Notice: The closed-loop gain Vo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage.

+

~

Rf

Ra

Vin

Vo

Multiple InputsMultiple Inputs(1)(1) Kirchhoff node equation at Kirchhoff node equation at VV++

yields, yields,

(2)(2) Kirchhoff node equation at Kirchhoff node equation at VV yields, yields,

(3)(3) Setting Setting VV++ = = VV–– yields yields

0V

0_

c

c

b

b

a

a

f

o

R

VV

R

VV

R

VV

R

VV

c

aj j

jf

c

c

b

b

a

afo R

VR

R

V

R

V

R

VRV

+

Rf

Va

VoRb

Ra

RcVb

Vc

Inverting IntegratorInverting IntegratorNow replace resistors Ra and Rf by complex

components Za and Zf, respectively, therefore

Supposing• The feedback component is a capacitor C,

i.e.,

• The input component is a resistor R, Za = RTherefore, the closed-loop gain (Vo/Vin) become:

where

What happens if Za = 1/jC whereas, Zf = R Inverting differentiator

ina

fo V

Z

ZV

CjZ f

1

dttvRC

tv io )(1

)(

tjii eVtv )(

+

~

Zf

Za

Vin

Vo

+

~

R

Vin

Vo

C

Op-Amp IntegratorOp-Amp IntegratorExample:

(a) Determine the rate of changeof the output voltage.

(b) Draw the output waveform.

Solution:

+

R

VoVi

0+5V

10 k

0.01FC

Vo(max)=10 V

100s

(a) Rate of change of the output voltage

smV/

F k

V

50

)01.0)(10(

5

RC

V

t

V io

(b) In 100 s, the voltage decrease

VμssmV/ 5)100)(50( oV

0+5V

0

-5V

-10V

Vi

Vo

Op-Amp DifferentiatorOp-Amp Differentiator

RCdt

dVv i

o

+

RC

VoVi

0to t1 t2

0

to t1 t2

Slew RateSlew Rate

FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage divided by the change in timedivided by the change in time

The maximum possible rate at which an amplifier’s output voltage can change, in volts per second, is called its slew rate.