Date post: | 22-Dec-2015 |
Category: |
Documents |
Upload: | avis-sharyl-charles |
View: | 217 times |
Download: | 0 times |
2
Instantaneous power p(t) flowing into the box
)()()( titvtp Circuit in a box, two wires
)(ti
)(tv+
−
)(ti
)()()()()( titvtitvtp bbaa )(tvaCircuit in a box,
three wires
)(tia
+
−
)(tib
+
−
)(tvb
)()( titi ba Any wire can be the voltage reference
Works for any circuit, as long as all N wires are accounted for. There must be (N – 1) voltage measurements, and (N – 1) current measurements.
4
Two-wire sinusoidal case
)sin()sin()()()( tItVtitvtp oo
)cos(22
)cos(2
)(1
IVVIdttp
TP
Tot
otavg
),sin()( tVtv o )sin()( tIti o
2
)2cos()cos()(
tVItp o
)cos( rmsrmsavg IVP Displacement power factor
Average power
zero average
5
Root-mean squared value of a periodic waveform with period T
Tot
otavg dttp
TP )(
1
R
VP rmsavg
2
Tot
otrms dttv
TV )(
1 22
Apply v(t) to a resistor
Tot
ot
Tot
ot
Tot
otavg dttv
RTdt
R
tv
Tdttp
TP )(
1)(1)(
1 22
Compare to the average power expression
rms is based on a power concept, describing the equivalent voltage that will produce a given average power to a resistor
The average value of the squared voltage
compare
6
Root-mean squared value of a periodic waveform with period T
Tot
otorms dttV
TV )(sin
1 222
Tot
oto
oTot
otorms
tt
T
Vdtt
T
VV
2
)(2sin
2)(2cos1
2
222
,2
22 V
Vrms
Tot
otrms dttv
TV )(
1 22
For the sinusoidal case
2
VVrms
),sin()( tVtv o
7
RMS of some common periodic waveforms
22
0
2
0
22 1)(
1DVDT
T
VdtV
Tdttv
TV
DTT
rms
DVVrms
Duty cycle controller
DT
T
V
0
0 < D < 1
By inspection, this is the average value of
the squared waveform
8
RMS of common periodic waveforms, cont.
TTT
rms tT
Vdtt
T
Vdtt
T
V
TV
0
33
2
0
23
2
0
22
3
1
T
V
0
3
VVrms
Sawtooth
9
RMS of common periodic waveforms, cont.
Using the power concept, it is easy to reason that the following waveforms would all produce the same average power to a resistor, and thus their rms values are identical and equal to the previous example
V
0
V
0
V
0
0
-V
V
0
3
VVrms
V
0
V
0
10
RMS of common periodic waveforms, cont.
Now, consider a useful example, based upon a waveform that is often seen in DC-DC converter currents. Decompose the waveform into its ripple, plus its minimum value.
minmax II
0
)(tithe ripple
+
0
minI
the minimum value
)(ti
maxI
minI=
2
minmax IIIavg
avgI
11
RMS of common periodic waveforms, cont.
2min
2 )( ItiAvgIrms
2minmin
22 )(2)( IItitiAvgIrms
2minmin
22 )( 2)( ItiAvgItiAvgIrms
2min
minmaxmin
2minmax2
22
3I
III
IIIrms
2minmin
22
3III
II PP
PPrms
minmax IIIPP Define
12
RMS of common periodic waveforms, cont.
2minPP
avgI
II
222
223
PP
avgPPPP
avgPP
rmsI
III
II
I
423
22
222 PP
PPavgavgPP
PPavgPP
rmsI
IIII
III
I
222
2
43 avgPPPP
rms III
I
Recognize that
12
222 PPavgrms
III
avgI
)(ti
minmax IIIPP
2
minmax IIIavg
13
RMS of segmented waveformsConsider a modification of the previous example. A constant value exists during D of the cycle, and a sawtooth exists during (1-D) of the cycle.
DTot
ot
Tot
DTot
Tot
otrms dttidtti
Tdtti
TI )()(
1)(
1 2222
avgI PPI
)(ti
DT (1-D)T
oI
DTot
ot
Tot
DTotrms dtti
TDTDdtti
DTDT
TI )(
)1(
1)1()(
11 222
avgIIn this example, is defined as the average value of the sawtooth portion
14
RMS of segmented waveforms, cont.
ToverDToverrms tiAvgTDtiAvgDTT
I D)-1( 2
22 )( )1()(
1
ToverDToverrms tiAvgDtiAvgDI D)-1( 2
22 )( )1()(
12)1(
2222 PPavgorms
IIDIDI a weighted average
DTot
ot
Tot
DTotrms dtti
TDTDdtti
DTDT
TI )(
)1(
1)1()(
11 222
So, the squared rms value of a segmented waveform can be computed by finding the squared rms values of each segment, weighting each by its fraction of T, and adding
15
150V
T T2
T 0
0V
Practice Problem
The periodic waveform shown is applied to a 100Ω resistor. What value of α yields 50W average power to the resistor?
16
Fourier series for any physically realizable periodic waveform with period T
11)90cos()sin()(
k
okokavg
kkokavg tkIItkIIti
ooo ffT
1
2
22
Tt
tavgo
odtti
TI )(
1
Tok dttkti
Ta
0cos)(
2
Tok dttkti
Tb
0sin)(
2
22)sin(
kk
kk
ba
a
22)cos(
kk
kk
ba
b
k
k
k
kk b
a
)cos(
)sin()tan(
22kkk baI
When using arctan, be careful to get the correct quadrant
17
Two interesting properties
Half-wave symmetry,
)()2
( tiT
ti
then no even harmonics
(remove the average value from i(t) before making the above test)
1)sin()(
kkok TtkITti
Time shift,
1sin
kokok ktkI
Thus, harmonic k is shifted by k times the fundamental angle shift
where the fundamental angle shift is .Too
20
Half-wave rectified cosine wave
I
T/2
T
T/2
tkk
It
IIti o
k
ko
cos
1
11
2cos
2)(
,6,4,22
12/
ttt
It
IIoooo
6cos
35
14cos
15
12cos
3
12cos
2
21
Triac light dimmer waveshapes(bulb voltage and current waveforms are identical)
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cu
rre
nt
α = 30º
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cu
rre
nt
α = 90º
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cu
rre
nt
α = 150º
22
Fourier coefficients for light dimmer waveform
,sin21
pV
a
2sin2
111 pVb
,...7,5,3,)1cos()1cos(1
1)1cos()1cos(
1
1
kkk
kkk
k
Va
pk
,...7,5,3,)1sin()1sin(1
1)1sin()1sin(
1
1
kkk
kkk
k
Vb
pk
Vp is the peak value of the underlying AC waveform
23
RMS in terms of Fourier Coefficients
1
222
2k
kavgrms
VVV
avgrms VV which means that
and that
2k
rmsV
V for any k
24
Bounds on RMS
From the power concept, it is obvious that the rms voltage or current can never be greater than the maximum absolute value of the corresponding v(t) or i(t)
From the Fourier concept, it is obvious that the rms voltage or current can never be less than the absolute value of the average of the corresponding v(t) or i(t)
26
Some measured current waveforms
-4
-2
0
2
4A
mp
eres
RefrigeratorTHDi = 6.3%
240V residential air conditionerTHDi = 10.5%
277V fluorescent light (magnetic ballast)
THDi = 18.5%
277V fluorescent light (electronic ballast)THDi = 11.6%
27
Some measured current waveforms, cont.
Microwave ovenTHDi = 31.9%
PCTHDi = 134%
Vacuum cleanerTHDi = 25.9%
28
Resulting voltage waveform at the service panel for a room filled with PCs
THDV = 5.1%(2.2% of 3rd, 3.9% of 5th, 1.4% of 7th)
-200
-150
-100
-50
0
50
100
150
200
Vo
lts
THDV = 5% considered to be the upper limit before problems are noticed
THDV = 10% considered to be terrible
29
Some measured current waveforms, cont.
5000HP, three-phase, motor drive
(locomotive-size)
Bad enough to cause many power electronic loads to malfunction
30
Now, back to instantaneous power p(t)
)()()( titvtp Circuit in a box, two wires
)(ti
)(tv+
−
)(ti
)()()()()( titvtitvtp bbaa )(tvaCircuit in a box,
three wires
)(tia
+
−
)(tib
+
−
)(tvb
)()( titi ba Any wire can be the voltage reference
31
Average power in terms of Fourier coefficients
1)sin()(
kkokavg tkVVtv
1)sin()(
kkokavg tkIIti
11)sin()sin()(
kkokavg
kkokavg tkIItkVVtp Messy!
Tt
tavgo
odttp
TP )(
1
32
Average power in terms of Fourier coefficients, cont.
Tt
tavgo
odttp
TP )(
1
1)cos(
22kkk
kkavgavgavg
IVIVP
rmskV ,
Cross products disappear because the product of unlike harmonics are themselves harmonics whose averages are zero over T!
rmskI ,
321 PPPPP dcavg
Due to the DC
Due to the 1st
harmonic
Due to the 2nd
harmonic
Due to the 3rd
harmonic
Harmonic power – usually
small wrt. P1
Not wanted in an AC system
33
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
• Determine the order of the harmonic
• Estimate the magnitude of the harmonic
• From the above, estimate the RMS value of the waveform,
• and the THD of the waveform
Fund. freq
Harmonic+
Consider a special case where one single harmonic is superimposed on a fundamental frequency sine wave
Using the combined waveform,
Combined
34
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
• Count the number of cycles of the harmonic, or the number of peaks of the harmonic
T
17
Single harmonic case, cont.Determine the order of the harmonic
35
• Estimate the peak-to-peak value of the harmonic where the fundamental is approximately constant
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
Single harmonic case, cont.Estimate the magnitude of the harmonic
Imagining the underlying fundamental, the peak value of the fundamental appears to be about 100
Viewed near the peak of the underlying fundamental (where the fundamental is reasonably constant), the peak-to-peak value of the harmonic appears to be about 30
Thus, the peak value of the harmonic is about 15
36
Single harmonic case, cont.Estimate the RMS value of the waveform
22
02
217
21
1
22
22 VVVVV
k
kavgrms
VVrms 5.71
222
51132
10225
2
15
2
100V
Note – without the harmonic, the rms value would have been 70.7V (almost as large!)
37
Single harmonic case, cont.Estimate the THD of the waveform
21
217
21
217
21
2
2
2
2
2
2
2
V
V
V
V
V
V
THD k
k
15.0100
15
1
17 V
VTHD
38-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
Voltage
Current
Given single-phase v(t) and i(t) waveforms for a load
• Determine their magnitudes and phase angles
• Determine the average power
• Determine the impedance of the load
• Using a series RL or RC equivalent, determine the R and L or C
39-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
Voltage
Current
Determine voltage and current magnitudes and phase angles
Voltage sinewave has peak = 100V, phase angle = 0º
Current sinewave has peak = 50A, phase angle = -45º
, 0100~
VV AI 4550~
Using a sine reference,
40
The average power is
1)cos(
22kkk
kkavgavgavg
IVIVP
)cos(22
00 1111 IV
Pavg
)45(0cos2
50
2
100avgP
WPavg 1767
41
The equivalent series impedance is inductive because the current lags the voltage
eqeqeq LjRI
VZ
4524550
0100~
~
414.1)45cos(2eqR
414.1)45sin(2eqL
where ω is the radian frequency (2πf)
If the current leads the voltage, then the impedance angle is negative, and there is an equivalent capacitance
42
C’s and L’s operating in periodic steady-state
Examine the current passing through a capacitor that is operating in periodic steady state. The governing equation is
dt
tdvCti
)()( which leads to
tot
oto dtti
Ctvtv )(
1)()(
Since the capacitor is in periodic steady state, then the voltage at time t o is the same as the voltage one period T later, so
),()( oo tvTtv
The conclusion is that
Tot
otoo dtti
CtvTtv )(
10)()(or
0)( Tot
ot
dtti
the average current through a capacitor operating in periodic steady state is zero
which means that
43
Now, an inductor
Examine the voltage across an inductor that is operating in periodic steady state. The governing equation is
dt
tdiLtv
)()( which leads to
tot
oto dttv
Ltiti )(
1)()(
Since the inductor is in periodic steady state, then the voltage at time t o is the same as the voltage one period T later, so
),()( oo tiTti
The conclusion is that
Tot
otoo dttv
LtiTti )(
10)()(or
0)( Tot
ot
dttv
the average voltage across an inductor operating in periodic steady state is zero
which means that
44
KVL and KCL in periodic steady-state
,0)(
loopAroundtv
,0)(
nodeofOutti
0)()()()( 321 tvtvtvtv N
Since KVL and KCL apply at any instance, then they must also be valid in averages. Consider KVL,
0)()()()( 321 titititi N
0)0(1
)(1
)(1
)(1
)(1
321
dtT
dttvT
dttvT
dttvT
dttvT
Tot
ot
Tot
otN
Tot
ot
Tot
ot
Tot
ot
0321 Navgavgavgavg VVVV
The same reasoning applies to KCL
0321 Navgavgavgavg IIII
KVL applies in the average sense
KCL applies in the average sense
45
KVL and KCL in the average sense
Consider the circuit shown that has a constant duty cycle switch
R1
LV
+
VLavg = 0
−
+ VRavg −+ VSavg −
Iavg
A DC multimeter (i.e., averaging) would show
R2
0 A
Iavg
and would show V = VSavg + VRavg
46
KVL and KCL in the average sense, cont.
Consider the circuit shown that has a constant duty cycle switch
R1
CV
+
VCavg
−
+ VRavg −+ VSavg −
Iavg
A DC multimeter (i.e., averaging) would show
R2
0
and would show V = VSavg + VRavg + VCavg
Iavg
47
+ Vin –
–
Vout +
iL L C iC
Iout
id iin
4a. Assuming continuous conduction in L, and ripple free Vout and Iout , draw the “switch
closed” and switch open” circuits and use them to develop the in
out
V
V equation.
4b. Consider the case where the converter is operating at 50kHz, Vin = 40V, Vout = 120V, P =
240W. Components L = 100µH, C = 1500µF. Carefully sketch the inductor and capacitor currents on the graph provided.
4c. Use the graphs to determine the inductor’s rms current, and the capacitor’s peak-to-peak
current. 4d. Use the graphs to determine the capacitor’s peak-to-peak ripple voltage.
Practice Problem