Objective of Lecture Discuss analog computing and the application of 1 st order operational...

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Objective of LectureDiscuss analog computing and the

application of 1st order operational amplifier circuits.

Derive the equations that relate the output voltage to the input voltage for a differentiator and integrator.

Explain the source of the phase shift between the output and input voltages.

Mechanical Analog Computers

Designed by Vannevar Bush in 1930 and used to control position of artillery through WWII. Replaced by electrical analog computers towards the end of WWII, which performed the needed calculations much faster. http://www.science.uva.nl/museum/AnalogComputers.html

Why Use an Analog Computer? Calculations performed in real time without the use

of a ‘real’ computer.Can be integrated into the instrumentation circuitry.

Commonly used in control circuits to rapidly monitor and change conditions without the need to communicate back and forth with a digital computer.

Power consumption is not high.Input can be any value between V+ and V-.

Can be designed to handle large (or small) voltages.No digitizing errors.

Analog computers are more robustLess susceptible to electromagnetic radiation damage

(cosmic rays), electrostatic discharge, electromagnetic interference (pick-up of electric noise from the environment), etc.

DisadvantageSlow

Maximum frequency is less than 10 MHz Compare this to the clock speed of your digital

computer.Voltage transfer function is nonlinear over

entire range of input voltages.Timing of inputs needs to be carefully

considered. Any time delays can cause errors in the calculations

performed.

SubsystemsMultipliers

Inverting and non-inverting amplifiers Typically fixed number, which means fixed resistor

values in amplifiers

Adders and SubtractorsSumming and difference amplifiers

DifferentiatorsIntegrators

1st order op amp circuits

Capacitors

)()(1

)(

)(

1

oC

t

t

CC

CC

tvdttiC

tv

dt

dvCti

o

Differentiator

Ideal Op Amp ModelVirtual ground

Op Amp ModelVirtual ground

AnalysisSince current is not

allowed to enter the input terminals of an ideal op amp.

dt

tdvRCtv

dt

dvC

R

vR

vti

dt

dvC

dt

dvCti

tvtv

titi

So

So

oR

SCC

SC

RC

)()(

)(

)(

)()(

)()(

Example #1Suppose vS(t) = 3V u(t-5s)

The voltage source changes from 0V to 3V at t = 5s. Initial condition of VC = 0V when t <5s.

Final condition of VC = 3V when t > 5RC.

Example #1 (con’t)

seVtv

eVVVtv

eVVVtv

Vtv

sstC

sstC

ttCCCC

C

o

initialfinalinitial

5 when t 3)(

t when t 030)(

t when t )(

t when t0)(

8.05

o8.05

o

o

Example #1 (con’t)

ssto

ssto

o

o

o

Co

eVtv

eVFxxs

tv

sVtv

Vtv

Vtvdt

tdvRCtv

8.05

8.0563

o

3)(

)3)(1040)(1020(8.0

1)(

9F))(405(20k5s when t0)(

RC where,5 t when t0)(

5s when t0)(

)()(

Example #2Let R = 2 k, C = 0.1F, and vS(t) = 2V

sin(t) at t = 0s

0s when t0)(

0s when t )500cos(2.0)(

)500cos(50022.0)(

)500sin(2)10)(2000()(

)(

)()( Since

7

Vtv

tVtv

tVmstvdt

tVdFtv

dt

dvRCtv

tvtv

o

o

o

o

So

SC

Cosine to Sine Conversion

, the output voltage lags the input voltage by 90 degrees.

)500sin(2)( As

)90500sin(2.0)(

)18090500sin(2.0)(

)90500sin(2.0)(

)500cos(2.0)(

tVtv

tVtv

tVtv

tVtv

tVtv

S

oo

ooo

oo

o

PSpice Simulation

Shows the 90 degree phase shift as well as the deamplification.

vS(t) vo(t)

IntegratorVirtual ground

Op Amp Model

Integrator

)()(1

)(

0)()(

0)()(

0

)()()(

)()(

12

1

1

2

1

tvdttvRC

tv

RC

tv

dt

tdvdt

tvdC

R

tv

mAii

tvtvvtvdt

dvCi

R

tv

R

vtvi

o

t

t

So

So

oS

CR

ooC

CC

SSR

Example #3Let R = 25 k, C = 5nF, at

t=0s

since vo(t) = -vC(t) and the voltage across a capacitor can’t be discontinuous.

stVts

radkVtV

tVts

radkVtV

dtts

radkV

nFktV

tVdttVRC

tV

oo

o

t

t

o

t

t

o

o

t

t

ino

0 when 85.39024.6sin85.3)(

)(24.6cos85.3)(

24.6sin3)5(25

1)(

)()(1

)(

122

12

2

12

2

1

2

1

2

1

t

s

radkVtvS 24.6sin3)(

PSpice Simulation

Shows that the output voltage leads the input voltage by +90 degree and the voltage offset due to the Vo(t1) term.

Example #4Initial Charge on Capacitor

Example #4 (con’t) If there is an initial charge that produces a

voltage on the capacitor at some time, to:

The voltage on the negative input of the op amp is:v1 = VC + VR1

v1= v2 = 0V

Example #4 (con’t) The current flowing through R1 is the same

current flowing through C.

mAtiVtv

R

tvti

R

tv

R

tvV

R

tvv

R

Vti

dt

tdvCti

CC

oCoR

CCCRR

CC

0)(,0)( , tas

)()(,tat t

)()](0[)]([)(

)()(

1o

111

1

1

11

CR

tt

oCC

C

CR

tt

oCC

CC

CC

RC

o

o

etvRdt

tdvCi

etvtv

CR

tv

dt

tdv

R

tv

dt

tdvC

titi

1

1

1

1

1

1

)(1)(

)()(

0)()(

0)()(

0)()(

R1C is the time constant, .

CR

tt

oCo

ooR

CR

o

etvR

Rtv

R

tv

R

tvVi

ii

1

1

2

222

2

)()(

)()(0

SummaryDifferentiator and integrator circuits are 1st

order op amp circuits.When the C is connected to the input of the op

amp, the circuit is a differentiator. If the input voltage is a sinusoid, the output voltage

lags the input voltage by 90 degrees. The output voltage may be discontinuous.

When the C is connected between the input and output of the op amp, the circuit is an integrator. If the input voltage is a sinusoid, the output voltage

leads the input voltage by 90 degrees. The output voltage must be continuous.