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E. ELECTRICITY AND MAGNETISM
18. Alternating currents
1
Outline 18.1 Alternating current through a resistor 18.2 Alternating currents through an inductor 18.3 Alternating currents through a capacitor 18.4 R-C and R-L circuits in series
2
Objectives (a) explain the concept of the r.m.s. value of an alternating
current, and calculate its value for the sinusoidal case only (b) derive an expression for the current from V = V0 sin t (c) explain the phase difference between the current and voltage
for a pure resistor (d) derive and use the formula for the power in an alternating
current circuit which consists only of a pure resistor (e) derive an expression for the current from V = V0 sin t (f) explain the phase difference between the current and voltage
for a pure inductor (g) define the reactance of a pure inductor (h) use the formula XL = L (i) derive and use the formula for the power in an alternating
current circuit which consists only of a pure inductor 3
Objectives j) derive an expression for the current from V = V0 sin t k) explain the phase difference between the current and voltage for
a pure capacitor l) define the reactance of a pure capacitor m) use the formula XC = 1/ C n) derive and use the formula for the power in an alternating
current circuit which consists only of a pure capacitor o) define impedance p) use the formula q) sketch the phasor diagrams of R-C and R-L circuits
4
Introduction Alternating currents and voltages vary with time and periodically change their direction
5
Sine waves by far the most important form of alternating quantity
important properties are shown below
6
Instantaneous value shape of the sine wave is defined by the sine function
y = A sin in a voltage waveform
v = Vp sin
7
Angular frequency frequency f (in hertz) is a measure of the number of cycles per second each cycle consists of 2 radians therefore there will be 2 f radians per second this is the angular frequency (units are rad/s)
= 2 f
8
the angular frequency can be thought of as the rate at which the angle of the sine wave changes at any time
= t therefore
v = Vp sin t or v = Vp sin 2 ft
similarly
i = Ip sin t or i = Ip sin 2 ft
Equation of a sine wave
9
Example Determine the equation of the following voltage
signal. From diagram:
Period is 50 ms = 0.05 s Thus f = 1/T =1/0.05 = 20 Hz Peak voltage is 10 V Therefore
tt
ftpVv
126sin10202sin10
2sin
10
18.1 Alternating currents through a resistor
11
If alternating voltage is applied across a resistor, the voltage and current across the resistor obeys
V = IR
12
Phase angles the expressions given above assume the angle of the sine wave is zero at t = 0 if this is not the case the expression is modified by adding the angle at t = 0
13
Phase difference two waveforms of the same frequency may have a constant phase difference we say that one is phase-shifted with respect to the other
14
Average value of a sine wave average value over one (or more) cycles is clearly zero however, it is often useful to know the average magnitude of the waveform independent of its polarity
we can think of this as the average value over
of the rectified signal p
p
p
pav
VV
V
VV
637.02
cos
dsin1
0
0
15
Average value of a sine wave
16
r.m.s. value of a sine wave
the instantaneous power (p) in a resistor is given by
therefore the average power is given by
where is the mean-square voltage
R
vp
2
2v
R
v
R
vavP
2] of mean) (or average[ 2
17
r.m.s. value of a sine wave While the mean-square voltage is useful, more often we use the square root of this quantity, namely the root-mean-square voltage Vrms
where Vrms =
we can also define Irms =
it is relatively easy to show that
2v
2i
pprms VVV 707.02
1pprms III 707.0
21
18
r.m.s. value of a sine wave r.m.s. values are useful because their relationship to average power is similar to the corresponding DC values
rmsrmsavIVP
RIPrmsav
2
RV
P rmsav
2
19
Form factor for any waveform the form factor is defined as
for a sine wave this gives
value averagevalue r.m.s.factor Form
11.1 0.637 0.707
factor FormpVpV
20
Peak factor for any waveform the peak factor is defined as
for a sine wave this gives
value r.m.s.value peakfactor Peak
414.1 0.707factor PeakpV
pV
21
Square Waves Frequency, period, peak value and peak-to-peak value have the same meaning for all repetitive waveforms
22
Phase angle we can divide the period into 360 or 2 radians useful in defining phase relationship between signals in the waveforms shown here, B lags A by 90 we could alternatively give the time delay of one with respect to the other
23
Average and r.m.s. values the average value of a symmetrical waveform is its average value over the positive half-cycle
thus the average value of a symmetrical square wave is equal to its peak value similarly, since the instantaneous value of a square wave is either its peak positive or peak negative value, the square of this is the peak value squared, and
pVavV
pVrmsV24
Form factor and peak factor from the earlier definitions, for a square wave
0.1value averagevalue r.m.s.factor Form
pVpV
0.1value r.m.s.value peakfactor Peak
pVpV
25
Power The instantaneous power (for some frequency, w) delivered at time t is given by:
The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.
To evaluate the average on the right, we first expand the sin( t- ) term.
Power in Alternating Current Circuits
26
Power (Product of even and odd function = 0)
sin tcos t
t 0
0
+1
-1
sin2 t
t 0
0
+1
-1 tttItP pp cossinsinsincos)( 2
cos21)( ppItP
0 1/2
Expanding,
Taking the averages,
)sin(sin tt
0cossin ttGenerally: 2
0
22
21sin
21sin xdxx
Putting it all back together again,
Power in Alternating Current Circuits
)sincoscos(sinsin ttt
27
Power
prms2
1prms II
21
cos)( rmsrmsItP
Phase depends on the values of L, C, R, and w Therefore...
This result is often rewritten in terms of rms values:
cos)( rmsrms ItP
Power in Alternating Current Circuits
28
Power
cosR
I mp
22 2
rms( ) cosrmsP t I RR
We can write this in the following manner (which we
2222
22
)1()(
xQxx
RtP rms
Power, as well as current, peaks at w = w0. The sharpness of the resonance depends on the values of the components. Recall:
Power in Alternating Current Circuits
29
Average and r.m.s. values Let us first consider the general case of a current the waveform of which cannot be represented by a simple mathematical expression. For instance, the waveform shown in Fig. below.
i1 i2 i3 i4
i7 i8 i9 in
t
Fig.4.4 30
i1 i2 i3 i4
i7 i8 i9 in
t
Fig.4.4
Average value of current
= Iav = nni3i2i1i
For a pure sinusoidal wave, Iav = 0.637 Im Vav = 0.637 Vm
Average and r.m.s. values
31
RI smr2
...
2n
22
21 )(
nRiRiRi
Average and r.m.s. values In a.c. work, however, the average value is of comparatively little importance. This is due to the fact that it is the power produced by the current that usually matters. Thus, if the current represented in Fig.4.4 is passed through a resistor having resistance R ohms, the heating effect of i1 is (i1)2R, that of i2 is (i2)2R etc., as shown in Fig. 4.5. Therefore, the average heating effect in half-cycle is :
32
So, Ir.m.s. = root-mean-square of the current
niiii 2
n23
22
21
For a pure sinusoidal wave,
Ir.m.s. = 0.707 Im
Vr.m.s. = 0.707 Vm
Average and r.m.s. values
33
t
Heating effect
(i3)2 (i8)2
r.m.s. value
(i7)2
(i9)2 (i2)2 (i4)2
Average and r.m.s. values
34
Average v value of a sine wave average value over one (or more) cycles is clearly zero
however, it is often useful to know the average magnitude of the waveform independent of its polarity we can think of this as the average value over
of the rectified signal p
p
p
pav
VV
V
VV
637.02
cos
dsin1
0
0
35
Average v value of a sine wave
36
r.m.s. voltage value of a sine wave the instantaneous power (p) in a resistor is given by
therefore the average power is given by
where is the mean-square voltage
R
vp
2
2v
R
v
R
vavP
2] of mean) (or average[ 2
37
Average v value of a sine wave While the mean-square voltage is useful, more often we use the square root of this quantity, namely the root-mean-square voltage Vrms
where Vrms =
we can also define Irms =
it is relatively easy to show that (see text for
analysis)
2v2i
pprms VVV 707.02
1
pprms III 707.02
1
38
Average v value of a sine wave r.m.s. values are useful because their relationship to average power is similar to the corresponding DC values
rmsrmsavIVP
RIPrmsav
2
RV
P rmsav
2
39
Average v value of a sine wave Form factor
for any waveform the form factor is defined as for a sine wave this gives
value averagevalue r.m.s.factor Form
11.1 0.637 0.707
factor FormpVpV
40
Average v value of a sine wave Peak factor
for any waveform the peak factor is defined as for a sine wave this gives
value r.m.s.value peakfactor Peak
414.1 0.707factor PeakpV
pV
41
Square Waves Frequency, period, peak value and peak-to-peak value have the same meaning for all repetitive waveforms
42
Phase angle we can divide the period into 360 or 2 radians
useful in defining phase relationship between signals in the waveforms shown here, B lags A by 90 we could alternatively give the time delay of one with respect to the other
43
Average and r.m.s. values the average value of a symmetrical waveform is its average value over the positive half-cycle
thus the average value of a symmetrical square wave is equal to its peak value similarly, since the instantaneous value of a square wave is either its peak positive or peak negative value, the square of this is the peak value squared, and
pVavV
pVrmsV
44
Form factor and peak factor from the earlier definitions, for a square wave
0.1value averagevalue r.m.s.factor Form
pVpV
0.1value r.m.s.value peakfactor Peak
pVpV
45
Pure Resistor in the AC Circuit
46
I V
Phasor diagram
VR I
47
The current flows in the resistor is II 0 sin
The voltage across the resistor VR at any instant is
IRVR
RIV 0R sin 00 VRI
VVV 0R sin voltageSupply :V
The phase difference between V and I is
0 In pure resistor, the voltage V is in phase with the
current I and constant with time.(the current and the voltage reach their maximum values at the same time). 48
The resistance in a pure resistor is The instantaneous power, o
o
rms
rms
IV
IVR
tVIP
tVtIPR
VRIIVP
oo
oo
2
22
sin
sinsin
The average power,
o
oo
o
rmsave
P
IV
RI
RIP
212121 2
2
A resistor in ac circuit dissipates energy in the form of heat
t0
0P
TT21 T2T
23
)(PPower
18.2 Alternating currents through an inductor
49
Inductor Circuits An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor.
L| = L dI/dt
50
Pure Inductor in the AC Circuit
51
IV
Phasor diagram
II VL
rad2
Pure inductor means that no resistance and capacitance effect in the a.c. circuit.
52
When a sinusoidal voltage is applied across a inductor, the voltage reaches its maximum value one quarter of a cycle before the current reaches its maximum value,( ) rad
2 The current flows in the ac circuit is II 0 sin When the current flows in the inductor, the back emf
caused by the self induction is produced and given by
dtdILB
IdtdL 0B sin
LI 0B cos
53
At each instant the supply voltage V must be equal to the back e.m.f B (voltage across the inductor) but the back e.m.f always oppose the supply voltage V.
Hence, the magnitude of V and B , LIV 0B cos or
2sin
2sin0
VV
LIV
o
where oo LIV
B
B
VIRV0
54
The phase difference between V and I is
2
2
In pure inductor, the voltage V leads the current I by /2 radians or the current I lags behind the voltage V by /2 radians.
The inductive reactance in a pure inductor is
2
rms o oL
rms o o
L
V V LIXI I I
X L fL
55
The inductive reactance is defined as 2LX L fL
The instantaneous power, The average power, 0aveP
tPP
tVIP
tVtIPR
VRIIVP
o
oo
oo
2sin21
2sin21
cossin
22
t0
2P0
2P0
TT21 T2T
23
)(PPower
For the first half of the cycle where the power is positive, the inductor is saving the power. For the second half cycle where the power is negative, the power is returned to the circuit.
Example 1 A coil having an inductance of 0.5 H is connected to a 120 V, 60 Hz power source. If the resistance of the coil is neglected, what is the effective current through the coil.
56
A 0.642 fLV
XVI rms
L
rmsrms
Example 2 A 240 V supply with a frequency of 50 Hz causes a current of 3.0 A to flow through an pure inductor. Calculate the inductance of the inductor.
57
H 0.25 2
80
LfLXIVX
L
L
18.3 Alternating currents through a capacitor
58
Pure capacitor means that no resistance and self-inductance effect in the a.c. circuit.
Pure Capacitor in the AC Circuit
59
I
VPhasor diagram
I I
rad2
VR
60
CVVV )2
sin(0
When an alternating voltage is applied across a capacitor, the voltage reaches its maximum value one quarter of a cycle after the current reaches its maximum value,( ) rad
2 The voltage across the capacitor VC at any instant
is equal to the supply voltage V and is given by
The charge accumulates on the plates of the capacitor is
CCVQ)
2sin(0CVQ
dtdQI The current flows in the ac circuit is
61
2sin0CV
dtdI
2sin0 dt
dCVI
)2
cos(0CVI 00 IC Vand
II sin0
or
The phase difference between V and I is
2
2
62
In pure capacitor, the voltage V lags behind the current I by /2 radians or the current I leads the voltage V by /2 radians.
The capacitive reactance in a pure capacitor is
fCCX
CVV
IV
IVX
C
o
o
o
o
rms
rmsC
211
The capacitive reactance is defined as
fCCX C 2
11
63
The instantaneous power,
tPP
tVIP
tVtIPR
VRIIVP
o
oo
oo
2sin21
2sin21
cossin
22
t0
2P0
TT21 T
23
)(PPower
2P0
T2
The average power,
0aveP
For the first half of the cycle where the power is negative, the power is returned to the circuit. For the second half cycle where the power is positive, the capacitor is saving the power.
Example An 8.00 F capacitor is connected to the terminals of an AC generator with an rms voltage of 150 V and a frequency of 60.0 Hz. Find the capacitive reactance rms current and the peak current in the circuit.
64
Capacitive reactance, 332fCC
X c 211
Rms current,
A0.452C
rmsrms X
VI
Peak current
AIpI rms
320.0452.0707.02
18.4 R-C and R-L circuits in series
65
RC in series circuit
In the circuit diagram: VR and VC represent the instantaneous voltage across the resistor and the capacitor. In the phasor diagram: VR and VC represent the peak voltage across the resistor and the capacitor. 66
R
IRV
V
CV
C
CVI
RV
V
Phasor diagram
voltagesupply angle phase:
V
67
Note CV
IRV
V
Phasor diagram
2
222
22
22
22
22
222
222
1C
RIV
XRIV
XRIV
XRIV
XIRIV
XIRIV
VVV
C
Crmsrms
Coo
Looo
Looo
CoRoo
2by side both ...divide
68
R
IRV
V
CV
C
CVI
RV
V
Phasor diagram
voltagesupply angle phase:
V
The total p.d (supply voltage), V across R and C is equal to the vector sum of VR and VC as shown in the phasor diagram.
2C
2R
2 VVV2
C22 IXIRV
2C
222 XRIV X C1
222
C1RIV
where
IRVR
CC IXV
69
CVI
RV
V
Phasor diagram
CX
R
Z
The impedance in RC circuit,
222
222
1
1
CRZ
IC
RI
IVZ
rms
rmsR
C
VVtan
IRIX Ctan
RX Ctan
1tan
From the phasor diagrams,
or
I leads V by
Impedance diagram
70
fCX C 2
1
Graph of Z against f
Z
f 0
R
Example An alternating current of angular frequency of 1.0 x 104 rad s-1 flows through a 10 k resistor and a 0.10 F capacitor which are connected in series. Calculate the rms voltage across the capacitor if the rms voltage across the resistor is 20 V.
71
R
C
VVtan
RX CtanFrom and
0.1 tanCRR
X C 1
V2.00.120
0.1
C
R
C
VVV
RL in series circuit
72
R
IRV
V
L
LVLV
IRV
V
Phasor diagram
tagesupply vol angle phase:
V
The voltage across the resistor VR and the inductor VL are
IRVR
LL IXV
73
R
IRV
V
L
LV LV
IRV
V
Phasor diagram
voltagesupply angle phase:
V
The total p.d (supply voltage), V across R and L is equal to the vector sum of VR and VL as shown in the phasor diagram.
2L
2R
2 VVV2
L22 IXIRV
2L
222 XRIV X L
222 LRIVwhere
74
LV
IRV
V
Phasor diagram
LXZ
RImpedance diagram
The impedance in RC circuit,
From the phasor diagrams,
222
222
LRZ
ILRI
IVZ
rms
rmsR
L
VVtan
IRIX Ltan
RX Ltan
Rtan
or
V leads I by
75
fLX L 2
Graph of Z against f
Z
f 0
R
RLC in series circuit
76
I
V
R
RV
L
LVCV
C
77
I
V
R
RV
L
LV CV
CLV
IRV
V
Phasor diagram CV
CL VV
78
I
V
R
RV
L
LV CV
C LV
IRV
V
Phasor diagram CV
CL VV
The voltage across the inductor VL , resistor VR and capacitor VC are
LL IXV IRVR CC IXV
79
I
V
R
RV
L
LV CV
CLV
IRV
V
Phasor diagram CV
CL VV
The total p.d (supply voltage), V across L, R and C is equal to the vector sum of VL ,VR and VC as shown in the phasor diagram.
2CL
2R
2 VVVV2
CL22 IXIXIRV
2222CL XXRIV
2CL
2 XXRIV
80
CL XX
LX
R
Z
CX
LV
IRV
V
Phasor diagram CV
CL VV
Impedance diagram
The impedance in RLC circuit,
From the phasor diagrams,
22
22
CL
CL
rms
rms
XXRZ
IXXRI
IVZ
V leads I by
R
CL
VVVtan
IRXXI CLtan
RXX CLtan
R
1
tan
VS
R L C
tot L CX X X
RLC in series circuit RLC in series circuit When a circuit contains an inductor and capacitor in series, the reactance of each tend to cancel. The total reactance is given by The total impedance is given by The phase angle is given by 1tan totX
R
2 2tot totZ R X
81
Rea
ctan
ce
f
XC XL
Series resonance
XC=XL
XC>XL XL>XC
Variation of XL and XC In a series RLC circuit, the circuit can be capacitive or inductive, depending on the frequency. At the frequency where XC=XL, the circuit is at series resonance. Below the resonant frequency, the circuit is predominantly capacitive. Above the resonant frequency, the circuit is predominantly inductive. 82
VS
R L C
1.0 kW XC = 5.0 kW
XL = 2.0 kW
The total reactance is 2.0 k 5.0 k 3.0 ktot L CX X X
The total impedance is 2 2 2 21.0 k +3.0 ktot totZ R X 3.16 k
The circuit is capacitive, so I leads V by 71.6o.
The phase angle is 1 1 3.0 ktan tan1.0 k
totXR
71.6o
Example 1 What is the total impedance and phase angle of the series RLC circuit if R= 1.0 k , XL = 2.0 k , and XC = 5.0 k ?
83
R L C
VS
330 H
f = 100 kHz
470 2000 pF
2 2 100 kHz 330 H 207 LX fL
1 1 796 2 2 100 kHz 2000 pFCX
fC
207 796 589 tot L CX X X
2 2 = 470 589 Z 753
Example 2 What is the magnitude of the impedance for the circuit?
84
XL
f
XC
X
XL
XC
Impedance of Series RLC Depending on the frequency, the circuit can appear to be capacitive or inductive. The circuit in the Example-2 was capacitive because XC > XL
85
R L C
VS
330 mH f = 400 kHz
470 W 2000 pF
The circuit is now inductive.
2 2 400 kHz 330 H 829 LX fL1 1 199
2 2 400 kHz 2000 pFCXfC
829 199 630 tot L CX X X
2 2 = 470 630 Z 786
Example 3 What is the total impedance for the circuit when the frequency is increased to 400 Hz?
86
XL
f
XC
X
By changing the frequency, the circuit in Example-3 is now inductive because XL>XC
XL
XC
Impedance of Series RLC
87
Voltages in a series RLC circuits
Notice that VC is out of phase with VL. When they are algebraically added,
0
VC
VL
This example is inductive.
Voltages in Series RLC The voltages across the RLC components must add to the source voltage in accordance with KVL. Because of the opposite phase shift due to L and C, VL and VC effectively subtract.
88
89
Exercise A series RLC circuit has a resistance of 25.0 , a capacitance of 50.0 F, and an inductance of 0.300 H. If the circuit is driven by a 120 V, 60 Hz source, calculate
a)The total impedance of the circuit b)The rms current in the circuit c)The phase angle between the voltage and the
current.
90
64.9 , 1.85 A, 67.3o
Summary 18.1 Alternating currents through a resistor Power, Prms = Vrms Irms R.m.s. value, Irms = 0.707I0, Vrms = 0.707V0 18.2 Alternating currents through an inductor V leads I by /2 radians I = I0sin t, V = V0cos t, P=1/2 I0Vosin(2 t) 18.3 Alternating currents through a capacitor I leads V by /2 radians I = I0cos t, V = V0sin t, P=1/2 I0Vosin(2 t) 18.4 R-C and R-L circuits in series Impedance, 91
MORE ABOUT
92
Half-wave Rectifiers
Forward biased
93
Half-wave Rectifiers
Reverse biased
Output result
94
Half-wave Rectifier Note that the frequency stays the same Strength of the signal is reduced Vavg = Vp(out)/ = 0.318 x Vp(out) [31.8 % of Vp] Vp(out) = Vp(in) VBar For Silicon VBar = 0.7 V
Half-wave Rectifier
Vp(out) Vp(in)
Vavg
2 95
Half-wave Rectifier - Example
Output:
Peak Inverse Voltage (PIV) The peak voltage at which the diode is reverse biased In this example PIV = Vp(in)-
Hence, the diode must be rated for PIV = 100 V
Draw the output signal Vp(out) = Vp(in) 0.7 Vavg = 99.3/What happens to the
frequency?
96
Transformers (Review) Transformer: Two inductors coupled together separated by a dielectric
When the input magnetic field is changing voltage is induced on the second inductor The dot represents the + (voltage direction)
Applications: Step-up/down Isolate sources
Turns ratio (n) n = Sec. turns / Pri. turns = Nsec/Npri
Vsec = n. Vpri depending on value of n : step-up or step-down
Center-tapped transformer Voltage on each side is Vsec/2
97
Half-wave Rectifier - Example Example:
Assume that the input is a sinusoidal signal with Vp=156 V & T = 2 msec; assume Nsec:Npri = 1:2 Draw the signal Find turns ratio; Find Vsec; Find Vout.
n = ½ = 0.5 Vsec = n.Vpri = 78 V Vout = Vsec 0.7 = 77.3 V
78-0.7
98
Full-wave Rectifier Note that the frequency is doubled Vavg = 2Vp(out)/ = 0.637 x Vp(out)
99
Full-wave Rectifier Circuit
Center-tapped full-wave rectifier Each half has a voltage = Vsec/2
Only one diode is forward biased at a time The voltages at different halves are opposite of each other
100
Full-wave Rectifier Circuit Center-tapped full-wave rectifier
Each half has a voltage = Vsec/2
Only one diode is forward biased at a time The voltages at different halves are opposite of each other
101
Full-wave Rectifier Circuit Vout = Vsec /2 0.7 Peak Inverse Voltage (PIV)
PIV = (Vsec/2 0.7)- (-Vsec/2) = Vsec 0.7 Vout = Vsec/2 0.7
Assuming D2 is reverse-biased No current through D2
102
Full-wave Rectifier - Example Assuming a center-tapped transformer
Find the turns ratio Find Vsec Find Vout Find PIV Draw the Vsec and Vout What is the output freq?
Vsec
n=1:2=0.5 Vsec=n*Vpri=25 Vout = Vsec/2 0.7 PIV = Vsec-0.7=24.3 V
103
Full-wave Rectifier - Multisim XFrmr can be virtual or real Use View Grapher to see the details of your results The wire-color can determine the waveform color Make sure the ground is connected to the scope.
104
Bridge Full-wave Rectifier Uses an untapped transformer larger Vsec Four diodes connected creating a bridge
When positive voltage D1 and D2 are forward biased When negative voltage D3 and D4 are forward biased
Two diodes are always in series with the load
Vp(out) = Vp(sec) 1.4V The negative voltage is inverted
The Peak Inverse Voltage (PIV) PIV=Vp(out)+0.7 105
Bridge Full-wave Rectifier - Example Assume 12 Vrms secondary voltage for the standard 120 Vrms across the primary
Find the turns ratio Find Vp(sec) Show the signal
direction when Vin is positive Find PIV rating
n=Vsec/Vpri = 0.1 10:1 Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V Vp(out) = V(sec) (0.7 + 0.7) = 15.6 V through D1&D2 PIV = Vp(out) + 0.7 = 16.3 V
120Vrms
Note: Vp-Vbr ; hence, always convert from rms to Vp 106
Bridge Full-wave Rectifier - Comparison
120Vrm
s
Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc load current
107
Smoothing by capacitors Filters and Regulators
108
Filters and Regulators
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Filters and Regulators
110
-Ripple voltage depends on voltage variation across the capacitor - Large ripple means less effective filter
Filters and Regulators
peak-to-peak ripple voltage
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Too much ripple is bad! Ripple factor = Vr (pp) / VDC Vr(pp) = (1/ fRLC) x Vp(unfiltered) VDC = (1 1/ fRLC) x Vp(unfiltered)
Filters and Regulators
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Summary
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