Transcript kul-eldas3-07 - IFI TALKS SOMETHING

Kuliah Minggu 3
Elektronika dasar
Jurusan Teknik Elektro
2007
SUMMING AMPLIFIER
If
Recall inverting
amplifier and
If = I1 + I2 + … + In
VOUT = -Rf (V1/R1 + V2/R2 + … + Vn/Rn)
Vout
 RF
RF
RF 
   V1 
V2 
V3 
R2
R3 
 R1
Summing amplifier is a good example of analog circuits serving as analog
computing amplifiers (analog computers)!
Note: analog circuits can add, subtract, multiply/divide (using logarithmic
components, differentiat and integrate – in real time and continuously.
PENGKONDISI SINYAL
(aplikasi penjumlah)
Vref
R2
Vin
R1
•
•
•
•
R
–
+
+
Vo
–
Masukan Vin
Transduser panas kelvin
00 K  0 Volt
2730 K  2,73 Volt
•
•
•
•
diinginkan
Keluaran V0
00 C  0 Volt
1000 C  -10 Volt
Diketahui :
• Vref = -9Volt
• R, R1, R2 ???
Perhitungan gain
•
•
•
•
Masukan Vin
Transduser panas kelvin
00 K  0 Volt
2730 K  2,73 Volt
•
•
•
•
diinginkan
Keluaran V0
00 C  0 Volt
1000 C  -10 Volt
• Per 0 K
•  0,01 Volt
• diinginkan
• per 0 C
•  0,1 Volt
• Gain  10 kali
Perhitungan R1 dan R
-9 V R2
Vin
R1
R

R
Vo    Vin  Vref 
R2
 R1

R
–
+
• Gain  10 kali
+
Vo
–
R

R
Vo    Vin   9
R2
 R1

R
 10
R1
• Bila R1 =10 KΩ  maka R = 100 KΩ


100000
 9
Vo   10 Vin 
R2


Penentuan R2


100000
 9
Vo   10 Vin 
R2


• Diinginkan Keluaran V0
• 00 C  0 Volt

900000
0   10 2,73 

R2 

Tegangan masukan :
00 C  2730 K = 2,73 Volt
Maka Vin =2,73 Volt
900000
R2 
 32727,2727 
27,3
Hasil Akhir
-9 V R2
Vin
R1
R
–
+
+
Vo
–
• R1=10 KΩ
• R = 100 KΩ
• R2 =32727,27 Ω
• Persoalannya, bagaimana realisasi R2 ?
• Pakai hambatan variabel (potensio), agar aman
Pot + R tetap.
• Misal  pot : 10 K dan R = 27 K
VALIDASI
-9 V R2
Vin
R1
R

R
Vo    Vin  Vref 
R2
 R1

R
–
+
Saat suhu 1000 C
Maka 
masukan = 3730 K
Tegangan Vin = 3,73 V
+
Vo
–
• R1=10 KΩ
• R = 100 KΩ
• R2 =32727,27 Ω
100000
100000
 9
Vo  
 3,73 
32727,27
 10000

Vo  37,3  27,3  10
INTEGRATOR
I2
I1 = (Vi - V)/R1
I2 = C
d V  Vo 
dt
set I1 = I2,
I1
d V  Vo 
(Vi - V)/R1 = C
dt
but V- = V+ = 0
Vi/R1 =  C
Output is the integral of input
signal. CR1 is the time constant
Bila vi konstan maka
Solve for Vo
vi
v0  
t
R1C
dVo
dt

1
vo  
vi dt
R1C
Linier
OUTPUT INTEGRATOR
(dengan tegangan masukan tetap)
t
vi
v0  
t
R1C
v0
-VCC
APLIKASI
• Pembangkitan bentuk
gelombang
• Kemiringan tergantung
besarnya RC
vi
v0  
t
R1C
DIFERENSIATOR
I2
I1
V+in
–
I1  C
R
C
–
+
I2
+
Vo
–
Output is the differential of input
signal. CR is the time constant
d Vin  V 
d Vin 
C
dt
dt

V  V0   V0 


R
R
dVin  V0 
C

dt
R
dVin
V0   RC
dt
Bila input konstan maka  tegangan output = nol
Aplikasi diferensiator
• Kelengkungan tergantung besarnya RC
PEMBANGKIT FUNGSI
• Gelombang gigi gergaji
• Gelombang kotak
• Gelombang segitiga
• Gelombang sinus
Pembangkit gelombang gigi gergaji
Saklar ditutup sebelum opamp jenuh,
kemudian langsung buka lagi
I2
Saat saklar ditutup
I1
t
v0
-VCC
ASTABLE MULTIVIBRATORS
PEMBANGKIT GELOMBANG KOTAK
• A switching oscillator known as Astable Multivibrator can be formed
by adding an RC feedback network to a Schmitt Trigger circuit.
They are useful to generate low frequency square waves.
• The comparator and
feedback resistor
form an inverting
Schmitt Trigger
having threshold
levels of A/2 and
–A/2 assuming A
is the output level
of the comparator.
Graphs from Prentice Hall
Astable multivibrators II
•
The operation of the Astable Multivibrators can be described as follows: at
time 0, the initial voltage on the capacitor is 0, assuming the initial output
voltage is +A (A is the level of the comparator output). Thus, initially the
capacitor is charged through the resistor R toward +A. However, when the
capacitor voltage reaches A/2, the output voltage rapidly switches to –A.
•
Then the capacitor starts to discharge, once the voltage drops below –A/2, the
output again
switches back to A. Thus,
the capacitor voltage
cycles back and forth
between A/2 and –A/2.
•
Voltage across
capacitor resembles
Triangular wave and
comparator output
voltage is symmetrical
square wave.
Astable multivibrators III
• The period and frequency of the output square waveform can be
determined by analyzing the transient response of the RC feedback
network.
• The frequency of oscillation for the Astable Multivibrator shown before
is
1
f 
2 RC ln 3
• In real circuit design, several non-idealities related to the comparator
can affect the frequency, such as the propagation delay of the
comparator and bias current effects.
• To minimize the bias current effects, we usually need to make sure
that the smallest current charging to the capacitor should be much
larger than the bias current, for example, a few hundred times.
Pembangkit gelombang segitiga
Bagaimana memutar knob suatu generator fungsi dapat mengubah frekuensi ?
Rangkaian ini terdiri atas integrator, Schmitt trigger dan transistor.
Vin
PEMBANGKIT
SINUS
Wien Bridge Oscillator
•
•
•
R2
R1
•
V0
Vi
If
ZS
ZP
•
Berbasis pada op amp
Kombinasi R dan C dlm
feedback sehingga factor f
tergantung frekuensi.
Analisis beranggapan opamp
ideal.
– Gain A sangat besar
– Arus masuk sangat kecil.
– Terminal input short.
Analyze like a normal feedback
amplifier.
– Determine input and output
loading.
– Determine feedback factor.
– Determine gain with
feedback.
Shunt-shunt configuration.
Wien Bridge Oscillator
R1
Z S  R  ZC  R 
R2
V0
Vi
1
1 

Z P  R Z C   
R
Z
C 

ZS
If
1 1  sRC

sC
sC

ZP
1
1

  sC 
R

R
1  sCR
Output Loading
Input Loading
ZS
Z1
ZS
V0 = 0
ZP
 1
1 
Z1  Z P Z S  


 ZP ZS 
ZP
1

sC 
R1  sCR 
1  sCR
 R  1  sCR   sCR  (1  sCR ) 2
1
Vi = 0
Z2  Z S  R  ZC 
Z2
1  sRC
sC
1
Wien Bridge Oscillator
I1
I2
Amplifier gain including loading effects
R2
R1
Ar 
V0
Vi
To get
If
IS
Z2
Z1
ZS
Since I   0,
V0
ZP
If
Vo
R1 so
R1  R2
V0 R1  R2
R

 1 2
Vi
R1
R1
If
1
f 


X o Vo
ZS
sC

1  sRC
V0
Vo
, we use I1  I 2 
Vi
R1  R2
Vi  V  V  I1R1 
Feedback factor
Xf
V0 V0 Vi

I S Vi I S
Ar 
Z1 
Vi
 Z1 and
IS
 R 
V0 Vi
 Z1 1  2 
Vi I S
 R1 
R1  sCR 
sCR  (1  sCR ) 2
so
 R  R1  sCR 
Ar  1  2 
2
 R1  sCR  (1  sCR )
and
Wien Bridge Oscillator
Oscillation condition
P hase of  f Ar equal to 180o !
 R 
sCR
Need only  f Ar  1  2 
1
2
R
sCR

(
1

sCR
)
1 

Rewriting
 R 
sCR
 f Ar  1  2 
2
 R1  sCR  (1  sCR )
 R 
sCR
 1  2 
2 2 2
 R1  sCR  1  2sCR  s C R

Loop Gain
sC 

 Ar
1

sCR


 f Ar   
sC  R2 
R 1  sCR 


 
1 
R1  sCR  (1  sCR ) 2
 1  sCR 
 R 
sCR
 1  2 
R1  sCR  (1  sCR ) 2

Gain with feedback is
Arf 
Ar
1   f Ar

 R 
 R 
sCR
1
 1  2 
 1  2 
2 2 2
 R1  1  3sCR  s C R
 R1  3  1  sCR
sCR
 R 
1
 1  2 
 R1  3  j  CR  1 
CR 

T hen phase term 0 at the oscillation frequency
1
RC
Finally,we can get  f Ar  1 by selecting the resistors R1 and R2
o 
appropriately using
 R2  1
R
1    1 or 2  2
R1
 R1  3
Wien Bridge Oscillator - Example
Oscillator specifications: o=1x106 rad/s
Selecting for convenience C  10nF, then from  o 
R
1
1

 100
 oC 10nF (1x106 rad / s )
Choosing R 1  10K then
R2  2(10K )  20K
1
RC
Wien Bridge Oscillator
Final note: No input signal is needed. Noise at the desired oscillation frequency
will likely be present and when picked up by the oscillator, it will start the
oscillator and the output will quickly buildup to an acceptable level.
Wien Bridge Oscillator
• Once oscillations start, a limiting circuit is needed
to prevent them from growing too large in
amplitude
Phase Shift Oscillator
If
VX
IC3 V2 IC2
C
V1
C
R
I C 2  I R1  I C1 
IC1
C
V0
R
IR2
Rf
IR1
• Based on op amp using
inverting input
• Combination of R’s and C’s
in feedback loop so get
additional 180o phase shift.
• Analysis assumes op amp is
ideal.
V  V  0 so I f 
Vo
 I C1
Rf
 V  1  Vo
I R1  1 

R
R  sCR f
V2  V1  I C 2 Z C  


  Vo
 sCRR f

Vo
V 
1  1
 o 1 

sCR f R f  sCR  sC
Vo 
1 
2 

sCR f  sCR 
I R2 
Vo 
 V2
1 

2 

R
sCRR f  sCR 
IC3  I R2  IC 2 
Vo 
1  Vo 
1 
2 

1 

sCRR f  sCR  R f  sCR 

1  1 
1  Vo 
3
1 
1


2


1









2
 sCR  sCR  sCR  R f  sCR ( sCR ) 
Finally

Vo
Rf
V X  V2 
V
V1  V  I C1Z C   o
sCR f
Vo
V
V 
1 
 o  o 1 

sCRR f R f R f  sCR 

IC3
V 
1  Vo
  o 2 

sC
sCR f  sCR  sCR f
Vo 
4
1 

3 

sCR f  sCR ( sCR ) 2 

3
1 

1 
2
 sCR ( sCR ) 
Phase Shift Oscillator
Rearranging
IC3
VX
IC2
V2
V1
C
C
R
IC1
If
C
IR2 R

1 
4

3 
2
 sCR ( sCR ) 
we get for the loop gain
VX  
Rf
IR1
V0
Vo
sCR f
L( )   ( ) A( ) 
Selecting for convenience C  10 nF,
then from  o 
R
1
3 oC

1
3RC
1
 58 
3 10nF (1x106 rad / s)
 2CRR f


1 

1  
4

4  j  3CR  CR 
3  j
2
CR (CR )  



T o get oscillations, we need the imaginary term to go to zero.

Example
Oscillator specifications: ωo=1x106 rad/s
 jCR f
 sCR f
V0

VX 
1 
4

3 
2
 sCR ( sCR ) 
We can achieve this at one frequency  o so
3CR 
1
1
so    0 
CR
3RC
T o get oscillations, we also need L(ωo )  1 so
 0 2CRR f
 1 and substituting for ωo we get
L(ωo ) 
4
R f  12(58 )  0.67K
Rf
 0 2CRR f CRR f
1
 1 so


o
Note: We get 180 phase shift from op amp
4 3R 2C 2 12R
4
since input is to inverting terminal and
R f  12R
T hen
another 180o from the RC ladder.
Summary of Oscillator Design
Osilator Wien Bridge
Osilator Geser Fase
• Telah ditunjukkan komponen reaktif di loop
feedback dapat menimbulkan osilasi.
• Agar dicapai feedback posistip.
– Dengan pemilihan hambatan yang
tepat bisa dipilih sinyal feedback yang
sefase dengan sinyal input.
– Dapat dihasilkan amplitude sinusoidal
yang besar
• Telah dijelaskan dua rangkaian oskilator:
• (Osilator Wien Bridge)
• (Osilator Geser Fase)
• untuk menghasilkan frekuensi tertentu,
nilai resistor dan kapasitor dihitung
berdasarkan persamaan yang ada
• Catatan akhir: Perancangan
osilator semata-mata tergantung
pada rangkaian feedback bukan
pada karakteristik opamp.
FILTER
Passive Low-Pass Filter
H ( j)
Vout
Vin
p
s
C
Vout

R
Vin
RL
 The pass-band is from 0
to some frequency wp.
 Its stop-band extends
form some frequency ws,
to infinity.
 In practical circuit
design, engineers often
choose amplitude gain
of 0.95 for passive RC
filters:
Design of Passive Filters
The amplitude response:
R
C
Vin
Vout
RL
Vout

Vin
1
1   RC 
2
The amplitude gain:
Transfer Function
H  j  
H s  
1
jRC  1
1
RCs  1
ZL
G
ZF  ZL
The 3dB break-point is at:
f 3dB 
1
1

2RC 2
Design of Low Pass Active Filters
The -3 dB cut-off frequency:
C2
fH  1
RF
R1
Vin
A
B
The DC gain:
-
K LP   RF
Vout
+
Transfer Function:
T .F .  K LP
2RF C2 
0
s  0
R1
Example:
Design a low pass filter with
cut-off frequency of 5kHz,
and DC gain of 10:
Two equations, three unknowns
Design of High Pass Active
Filters
The -3 dB cut-off frequency:
fH  1
RF
C1
Vin
R1
A
B
Transfer Function:
T .F .  K HP
The DC gain:
+
s
s  0
2R1C1 
K HP  RF
Vout
R1
Two equations, three unknowns
Select one component based on
other conditions, and
determine the values of the
other two components.