Transcript Phys 345 Electronics for Scientists

Announcements
• Assignment 0 solutions posted
• Assignment 1 due on Thursday
• DC circuit Lab reports due to Sajan today
and tomorrow
• This week’s lab – AC circuits
Lecture 6 Overview
• AC Circuit Analysis
• Filters
The Story so far…
Generalized Ohm's Law: VS(jω)=ZIS(jω)
1
ZC =
jwC
• Voltage and current not in phase:
• Current leads voltage by 90 degrees
• Impedance of Capacitor decreases with increasing frequency
Z L = jwL
• Voltage and current not in phase:
• Current lags voltage by 90 degrees
• Impedance of Inductor increases with increasing frequency
ZR = R
• Voltage and current in phase
• no frequency dependece
Inductors in AC circuits
Inductive Load
vS  A sin t
vL   L
diL
dt
A sin t  L
(back emf )
diL
dt
from KVL
A
A
iL   sin tdt  
cost
L
L
A
A
iL 
sin(t  90) 
cos(t  180)
L
L
VL ( j )  A(t  90)
A
(t  180)
L
V ( j )
ZL  L
 L90
I L ( j )
cos(90)  j sin(90)  j
Z L  j L
I L ( j ) 
• Voltage and current not in phase:
• Current lags voltage by 90 degrees
• Impedance of Inductor increases with increasing frequency
http://arapaho.nsuok.edu/%7Ebradfiel/p1215/fendt/phe/accircuit.htm
AC circuit analysis
• Effective impedance: example
• Procedure to solve a problem
–
–
–
–
Identify the sinusoid and note the excitation frequency
Convert the source(s) to complex/phasor form
Represent each circuit element by it's AC impedance
Solve the resulting phasor circuit using standard circuit solving
tools (KVL,KCL,Mesh etc.)
– Convert the complex/phasor form answer to its time domain
equivalent
Example
(Z R1  ZC ) I1 ( j )
 ZC I 2 ( j )  VS ( j )
 ZC I1 ( j )  (ZC  Z L  Z R 2 ) I 2 ( j )  0
VS ( j )
 ZC
0
ZC  Z L  Z R 2
( Z C  Z L  Z R 2 )VS ( j )
I1 ( j ) 

2
Z R1  Z C
 ZC
( Z R1  Z C )(Z C  Z L  Z R 2 )  Z C
 ZC
ZC  Z L  Z R 2
1
1
66.7


 66.7 j ()
6
jC j150010
j
Z L  jL  j1500 0.5  750 j ()
ZC 
(75  683j )150
I1 ( j ) 
(100 66.7 j )(75  683j )  4450
(75  683j )150
I1 ( j ) 
(100 66.7 j )(75  683j )  4450
Top:
Bottom:
(75  683j )150  68783.7 150
  tan1
b
683
 tan1
 83.7
a
75
A  a 2  b 2  687
(100 66.7 j )(75  683j )  4450
 7500 45600 5000j  683j  4450 57550 63300j
 8550047.8
68783.7  150
I 1 ( j ) 
8550047.8
 0.1235.9
 0.120.63 radians
i1 (t )  0.12 cos(1500t  0.63) Amps
Transfer Function
H V ( j ) 
VL ( j )
VS ( j )
H I ( j ) 
I L ( j )
IS ( j )
Hv(jω)= Transfer function
Since we are interested in frequency response, use phasors.
VL(jω) is a phase-shifted and amplitude -scaled version of VS(jω)
Hv(jω) describes what the phase shift and amplitude scaling are.
VL ( jw ) = H V ( jw )VS ( jw )
HV = Amplitude of transfer function
ÐHV = Phase shift of transfer function
VL e jf L = HV e jÐHV VS e jf S
VL = HV VS
f L = ÐHV + f S
Low pass filters
Vo ( j )
H V ( j ) 
Vi ( j )
Vo ( j ) 
ZC
Vi ( j )
ZC  Z R
1
H V ( j ) 
• RC low-pass filter: preserves lower
frequencies, attenuates frequencies
above the cutoff frequency ω0=1/RC.
1
0 
RC
j C
1
R
j C
1
1
e j0


j tan1 RC 
2
1
1  jRC
1  (RC) e
1
 j tan1 RC

e
2
1  (RC)

1
1  (RC)
2
e
 j tan1  / 0
Low pass filters
Vo  H V Vi

1
1  (RC) 2
Vi
o   tan1  / 0
Break frequency ω=ω0=1/RC,
HV=1/√2
N.B. decibels:
 X 

X dB  20 log10 
 X0 
 X 

X dB  10 log10 
X
 0
1
20 log10
 3db
2
For voltage
For power
Build other filters by combining
impedance response
Which of the following is a low-pass
filter?
What happens to the output voltage when ω→0 (DC condition)?
Answer: (c)
Which of the following are high-pass or
low-pass filters?
Answers: (b) and (c) are highpass; (a) and (d) are low-pass
RLC Band-pass filters
Measuring voltage output
signal over R, Vr
Low frequencies, C open, L
shorted, Vr minimum
High frequency, C shorted, L
open, Vr minimum
so, at high and low
frequencies, see an open
circuit - Vr minimum
C
L
Band-stop (Notch) filters
Measuring voltage output
signal over L and C
Low frequencies, C open, L
shorted, Vlc maximum
High frequency, C shorted, L
open, Vlcmaximum
so, at high and low
frequencies, see an open
circuit - Vlc maximum
Another Example:
Measuring voltage output
signal over L and C, but this
time in parallel (i.e. at high
and low frequencies, see a
short - V0=0)