Download v - UD Physics

Topic 3: Op-Amp: Golden Rules of OP Amp
1. iin=0, no current flow into op amp.
2. V+=V•
Typically one end of op amp is connected to ground, therefore,
V+=V-= 0V, virtual ground. Often V+ is connected to ground to
avoid stability problem.
3. Various circuits: amplifier, sum and abstract,
differentiator, integrator, etc.
Example
• What does the following circuit
do? That is, find an expression
for the gain and give the circuit
a suitable name.
Example II
(a) (7pts)The circuit is an integrate circuit.
Derive vout(t). RS=10 kΩ, CF=0.008 µF.
(b) (5pts)If the input signal vin=sin(2000πt)
V, calculate vout(t) and the peak
amplitude.
(c) (3pts)If the input signal vin=10 mV,
calculate vout(t) and vout(t=100ms)
(d) (3pts)Op-amp is saturated when the
output voltage reach the power supply
voltage,
±15V in this case. At what time does the
integration of the DC input cause the
op-amp to
saturated fully?
(e) (7pts) If the input signal is now
vin=0.01+ sin(2000πt) V, describe what
happens to the output waveform till the
op-amp is fully saturated.
a. vout  
1
vin dt  12500  vin dt

RS C F
b. vout  2 cos 2000t (V)
c. vout (t )  125t (V)
vout (100ms)  12.5 (V)
d. t  0.12( Sec)  120 ms
e. vout (t )  125t  2 cos 2000t
Example 4
Design an op-amp circuit to convert the triangular waveform v1 in the following Figure into the
square wave vo shown. Use 0.1 mF capacitor. (Hints: First quantitatively determine the
mathematical expression of vo in terms of v1 ).
v1(t)
v0   RF CS
dv1
dt
 0.5   RF  0.1  106
RF  25k
0.2
1  103
Topic 4: Filters
•
Low-pass filters (LP)
HV  j  
V0  j  
V0  j 
Vi  j 
ZC
Vi  j 
ZC  Z R
1
jC
HV  j  
1
R
jC
1
1
e j0


j tan 1 RC 
2
1
1  jRC
1  RC  e


1
1  RC 
2
1
1  RC 
2
e  j tan
e
1
RC
 j tan 1  / 0
;
0 
1
RC
Active Filter
Vout
ZF

VS
ZS
It is phasor time again
Vout
ZF
 1
VS
ZS
Active Low Pass Filter
A( j ) 
Vout
Z
 F
VS
ZS
Z F  RF ||
1
RF

jCF 1  jRF CF
A( j )  
RF / RS
1  jRF CF
RF/RS=10, 1/RFCF=1
• Amplification: RF/RS
• low pass factor 1/(1+jRFCF)
• Cut off frequency: RFCF=1
Band-pass Filters and Resonant Circuits
HV  j  

  
1  Q 2   0 
 0  
1
LC
0 L
R
  
1  jQ  0 
 0  
1
0 
Q
1

0
B
2
e
 
 arctan Q   0
 0 
Resonant frequency

0
2  1
Quality factor



RLC Resonant
when   0
L 
1
C
jL  
1
jC
Z L   ZC
I  j  

V ( j )
Z R  Z L  ZC
At resonance:
• current is max.
• Zeq =R
• current and voltage
are in phase.
• the higher Q, the
narrower the resonant
peak.
V
1 

R  j  L 


C


V
V
I 

2
R
1 

2
R   L 

C 

Applications:
tuning circuit
Example
a. Write the transfer function for this filter. (Both Vin and Vout are referenced to ground).
H V  
vout
R

vin R  1

jC
1
1 j
RC
b). Design a high-pass RC filter with a characteristic frequency of 80 Hz using a capacitor of 2.0 mF.
c). What is the new characteristic frequency when the filter is loaded with 50 ohm?
50  980
Req 
 47.6
1
1
50  980
RC
1
1
f 
 1.67kHz
6
R

980

2


47
.
6

2

10
2 80  2  106
How do you solve the problem:
Use small R or use active filter
A( j ) 

Vout
Z
 F
VS
ZS
RF
RS 
1
jCS

RF
1
1
RS 1 
jRS CS
Example 2
The circuit shown in an active filter. Determine: The voltage transfer function. The
pass-band gain.
v  v1
v+
RF
Ri
v
v v
 0 
Ri  Z c
RF
v0  v
RF

1
v
Ri 
jC
v0
RF
1
 1
v
Ri 1  1
jRiC
G  1
RF
68
 1
 14.6
Ri
5
Exam 3
The input signal can be written as Vi=10mV(sin10t+sin10,000t). Design a circuit
so that the output signal is V0=-100mVsin10t. In another words, the high
frequency signal has to be much smaller (<1%) than the low frequency part.
A( j ) 
Vout
RF / RS

VS
1  jRF CF
Low frequency
RF 100

 10
RS
10
1RF C F  1,
let' s say : RS  1k; RF  10k
high frequency
RF / RS
1

1  j2 RF CF 10
1
1  2 RF CF 
2

1
100
2 RF CF  100
CF 
Check low freq.
100
 1mF
10  104
4
1RF CF  10  104  106  0.1  1
Topic 5: Electromachine
• Motion  Electricity (generator): Magnetic Induction,
Faraday’s law
• Electricity  Motion (motor): force on a current, torque
• Three phase power: structure, advantages.