Ece 232 Lab1 supplementary notes: LM741ece232.cankaya.edu.tr/uploads/files/ece232_lab1_2014...6...
Transcript of Ece 232 Lab1 supplementary notes: LM741ece232.cankaya.edu.tr/uploads/files/ece232_lab1_2014...6...
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