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Aplikasi Op amp
Pertemuan ke 10Kuliah Elektronika Dasar
Prodi Teknik ElektroJTETI-UGMMei 2010
Arah arus ?
INV and SUMMING
V to I
Buktikan !
V to I
INTEGRATORI1 = (Vi - V)/R1
I2 =
set I1 = I2,
(Vi - V)/R1 =
but V- = V+ = 0
Vi/R1 =
Solve for Vo
Output is the integral of input signal. CR1 is the time constant
I1
I2
dt
VVdC o
dt
VVdC o
dt
dVC o
dtvCR
v io1
1
Bila vi konstan maka tCR
vv i
10 Linier
OUTPUT INTEGRATOR(dengan tegangan masukan tetap)
tCR
vv i
10
-VCC
t
v0
Pembangkit gelombang gigi gergaji
I1
I2
Saklar ditutup sebelum opamp jenuh,kemudian langsung buka lagi
-VCC
t
v0
Saat saklar ditutup
APLIKASI
• Pembangkitan bentuk gelombang
• Kemiringan tergantung besarnya RC
tCR
vv i
10
DIFERENSIATOR
Output is the differential of input signal. CR is the time constant
dt
VdC
dt
VVdCI inin
1
Vo
I1
I2
–
+Vin+
–
C
R
+–
R
V
dt
dVC in 0
R
V
R
VVI 00
2
dt
dVRCV in0
Bila input konstan maka tegangan output = nol
Aplikasi diferensiator
• Kelengkungan tergantung besarnya RC
Integrator
+
~
R
Vin
Vo
C
Tegangan Input
Tegangan Output
Ref:080114HKN Operational Amplifier 16
Tegangan Output 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
Differentiator
18
Astable Multivibrator or Relaxation Oscillator
Circuit Output waveform
19
Equations for Astable Multivibrator
21
2
21
2 ;RR
RVV
RR
RVV sat
LTsat
UT
1
2121
2ln2
R
RRttT
Assuming|+Vsat| = |-Vsat|
If R2 is chosen to be 0.86R1, then T = 2RfC and
where = RfC
CRf
f2
1
Function Generator - Build your own!Ever wonder how turning a knob on a function generator (which is a variable resistor) can change frequencies? This circuit employs the last two - the integrator and the Schmitt trigger. There is also another old friend - the transistor switch. Lets look at how this circuit works.
First look at the Schmitt trigger part. When does it change? Look at the resistors in the +feedback - this is not the same as our example.
Vin
Function Generator - Build your own! (2)Now look at the first part. When the output is “HI”, meaning close to the power supply voltage, the transistor is OFF, so if effectively is not there. Then if the cap has more that Vin/2 across it, it will discharge through the 100k resistor. If the output is low, the inverting input is effectively grounded through a resistor and it will charge the cap. Changing Vin will change the charge/discharge rate.
Vin