6

Vin=Sinusoidal

FREQ = 500 HzVAMPL = 1V

3

2

CH 1

R2

12k

-

+

741

OPAMP

+

-

OUT

CH 2- 12

R1

1.2k

+ 12

47

Ece 232

Lab1 supplementary notes:

LM741

1)

Vin ! AR1

=A!VoutR2

, A = 0Volt

Vout = !10Vin

if Vin (t) = Sin(2! ft) then Vout (t) = !10Sin(2! ft)

   

CH 1

741

OPAMP

+

-

OUT3

2

67

+

R2

12k

- 12CH 2

C1

47n

-

4

+ 12Vin=Square wave

FREQ = 500 HzVAMPL = 1 V

2)

C1d(Vin ! A)

dt=A!VoutR2

Vout = !C1R2dVindt

C1R2 = 564!10*6

a) If Vin (t) = Sqr( ft) (square wave with frequency f and period T)

Vin (t) =1Volt, 0 ! t ! T

2

"1Volt, T2! t ! T

#

$%%

&%%

'

(%%

)%%

(square wave)

Vout (t) =!564"10!6!(t ! n T

2), n even

!564"10!6!(t ! n T2), n odd

#

$%%

&%%

'

(%%

)%%

(impulse train)

   

b) IfVin (t) = Tri( ft) (triangular wave with frequency f and period T)

Vin (t) =

4Tt, 0 ! t ! T

4

"4Tt + 2, T

4! t ! 3T

44Tt " 4, 3T

4! t ! T

#

$

%%%

&

%%%

'

(

%%%

)

%%% (triangular wave)

Vout (t) =

!564"10!6 4T, 0 # t # T

4

564"10!6 4T, T

4# t # 3T

4

!564"10!6 4T, 3T

4# t # T

$

%

&&&

'

&&&

(

)

&&&

*

&&& (square wave)

   

741

OPAMP

+

-

OUT +

Vin=Square wave

FREQ = 500 HzVAMPL = 1 V

-

CH 1

3

2

6

R2

1M

C

10n

74

+ 12

R1

100k

- 12CH 2

Example: If f=500Hz and T=2 msec

Vout (t) =

!1.128Volt, 0 " t " T4

1.128Volt, T4" t " 3T

4

!1.128Volt, 3T4" t " T

#

$

%%%

&

%%%

'

(

%%%

)

%%%

3.)

Vin ! AR1

=A!VoutR2

+C d(A!Vout )dt

Since R2 is very high the above equation can be written as,

Vin ! AR1

"C d(A!Vout )dt

Putting A=0 Volt we will have

dVoutdt

= !1000Vin

If the applied input is a square wave

Vin (t) =1Volt, 0 ! t ! T

2

"1Volt, T2! t ! T

#

$%%

&%%

'

(%%

)%%

where Vin(t) is periodic with period T

   

Vin =1when 0 ! t !T2hence

dVoutdt

= !1000

Vout (t) = !1000t + X

Assuming Vout(0)=V Volt and f=500 Hz

Vout (t) = !1000t +V

Vout (T2) = !1000T

2+V = !1000 1

500"2+V =V !1

Vin = !1whenT2" t " T hence

dVoutdt

=1000

Vout (t) =1000t + X

Knowing that Vout(0.5T)=V-1 Volt and f=500 Hz

Vout (t) =1000t +V ! 2

As the results

Vout (t) =!1000t +V Volt, 0 " t " T

2

1000t +V ! 2Volt, T2" t " T

#

$%%

&%%

'

(%%

)%%

where Vin(t) is periodic with period T where

T = 1f=1500

= 2msec

   

Graphs

1) Inverting amplifier

0 1 2 3 4 5 6 7 8

x 10-3

-10

-8

-6

-4

-2

0

2

4

6

8

10

Time (sec)

Mag

nitu

de

Inverting amplifier

inputoutput

 

 

 

 

 

   

2a)  Differentitor  (input  is  square  wave)  

0 1 2 3 4 5 6 7 8

x 10-3

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (sec)

Mag

nitu

deDifferetiator

inputoutput

 

   

2b) Differetiator (iput is triangular wave)

0 1 2 3 4 5 6 7 8

x 10-3

-1.5

-1

-0.5

0

0.5

1

1.5

Time (sec)

Mag

nitu

deDifferentiator

inputoutput

   

3) Integrator

0 1 2 3 4 5 6 7 8

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time (sec)

Mag

nitu

deIntegrator

outputinput