Half-Wave Rectifiers
3.8 HALF-WAVE RECTIFIER WITH A CAPACITOR FILTER
88 C H A P T E R3 Half-Wave Rectifiers
3.8 HALF-WAVE RECTIFIER WITH A CAPACITOR
3.8 Half-Wave Rectifier With a Capacitor Filter 89
downward rate of change of the source exceeds that permitted by the time con- stant of the RCload. The angle t is the point when the diode turns off in Fig. 3-11b. The output voltage is described by
(3-37)
where (3-38)
The slopes of these functions are
(3-39) and
(3-40) At t , the slopes of the voltage functions are equal:
Solving for and expressing so it is in the proper quadrant, we have
(3-41) In practical circuits where the time constant is large,
(3-42) When the source voltage comes back up to the value of the output voltage in the next period, the diode becomes forward-biased, and the output again is the same as the source voltage. The angle at which the diode turns on in the second period, t2 , is the point when the sinusoidal source reaches the same value as the decaying exponential output:
Vmsin(2 )(Vmsin)e (2 )>RC L
2
and
Vmsin L Vm
tan 1( RC) tan 1(RC) Vmcos aVmsin
RCbe ( )>RCVmsin
RC Vmcos
Vmsin 1 RC 1
tan 1
RC d
d(t)
A
Vmsine (t )>RCB
Vmsina 1RCbe (t )>RC d
d(t)[Vmsin(t)]Vmcos(t) VVmsin
vo(t)cVmsint
diode on
Ve (t )>RC diode off
90 C H A P T E R3 Half-Wave Rectifiers
or
(3-43) Equation (3-43) must be solved numerically for .
The current in the resistor is calculated from iR voR. The current in the capacitor is calculated from
which can also be expressed, using tas the variable, as
Using vofrom Eq. (3-37),
(3-44)
The source current, which is the same as the diode current, is
(3-45) The average capacitor current is zero, so the average diode current is the same as the average load current. Since the diode is on for a short time in each cycle, the peak diode current is generally much larger than the average diode current.
Peak capacitor current occurs when the diode turns on at t2 . From Eq. (3-44),
(3-46) Resistor current at t2+ is obtained from Eq. (3-37).
(3-47) Peak diode current is
(3-48) The effectiveness of the capacitor filter is determined by the variation in out- put voltage. This may be expressed as the difference between the maximum and minimum output voltage, which is the peak-to-peak ripple voltage. For the half- wave rectifier of Fig. 3-11a, the maximum output voltage is Vm. The minimum
ID, peak CVmcos Vmsin
R VmaCcos sin R b iR(2t )Vmsin(2t )
R Vmsin
R IC, peak CVmcos(2 ) CVmcos
iSiDiRiC iC(t)d
aVmsin
R be (t )>RC for
t2 (diode off) CVmcos(t) for 2
t2 (diode on)
iC(t) C dvo(t) d(t) iC(t)C dvo(t)
dt
sin (sin)e (2 )>RC0
har80679_ch03_065-110.qxd 12/17/09 2:09 PM Page 90
3.8 Half-Wave Rectifier With a Capacitor Filter 91
output voltage occurs at t2+ , which can be computed from Vmsin . The peak-to-peak ripple for the circuit of Fig. 3-11ais expressed as
(3-49) In circuits where the capacitor is selected to provide for a nearly constant dc output voltage, the RCtime constant is large compared to the period of the sine wave, and Eq. (3-42) applies. Moreover, the diode turns on close to the peak of the sine wave when L/2. The change in output voltage when the diode is off is described in Eq. (3-37). In Eq. (3-37), if VLVmand L/2, then Eq. (3-37) evaluated at /2 is
The ripple voltage can then be approximated as
(3-50) Furthermore, the exponential in the above equation can be approximated by the series expansion:
Substituting for the exponential in Eq. (3-50), the peak-to-peak ripple is approx- imately
(3-51) The output voltage ripple is reduced by increasing the filter capacitor C. As Cincreases, the conduction interval for the diode decreases. Therefore, increas- ing the capacitance to reduce the output voltage ripple results in a larger peak diode current.
Vo L Vma 2
RCb Vm
fRC e 2>RC L 1 2
RC
Vo L Vm Vme 2>RCVm
A
1 e 2>RCB
vo(2 )Vme (2>2 >2)RCVme 2>RC VoVm Vmsin Vm(1 sin)
EXAMPLE 3-9
Half-Wave Rectifier with RCLoad
The half-wave rectifier of Fig. 3-11ahas a 120-V rms source at 60 Hz, R500 , and C100 F. Determine (a) an expression for output voltage, (b) the peak-to-peak volt- age variation on the output, (c) an expression for capacitor current, (d) the peak diode cur- rent, and (e) the value of Csuch that Vois 1 percent of Vm.
■ Solution
From the parameters given,
Vm12022169.7 V
RC(260)(500)(10) 618.85 rad
92 C H A P T E R3 Half-Wave Rectifiers
The angle is determined from Eq. (3-41).
The angle is determined from the numerical solution of Eq. (3-43).
yielding
(a) Output voltage is expressed from Eq. (3-37).
(b) Peak-to-peak output voltage is described by Eq. (3-49).
(c) The capacitor current is determined from Eq. (3-44).
(d) Peak diode current is determined from Eq. (3-48).
(e) For Vo0.01Vm, Eq. (3-51) can be used.
Note that peak diode current can be determined from Eq. (3-48) using an estimate of from Eq. (3-49).
From Eq. (3-48), peak diode current is 30.4 A.
■ PSpice Solution
A PSpice circuit is created for Fig. 3-11ausing VSIN, Dbreak, R, and C. The diode Dbreak used in this analysis causes the results to differ slightly from the analytic solution based on the ideal diode. The diode drop causes the maximum output voltage to be slightly less than that of the source.
L sin 1a1 Vo
Vm b sin 1a1 1
fRCb81.9°
C L Vm
fR(Vo) Vm
(60)(500)(0.01Vm) 1
300F = 3333F ID, peak12(120)c377(10) 4cos0.843 sin8.43
500 d
4.260.344.50 A
iC(t)c
0.339e (t 1.62)>18.85
A
t2
6.4cos(t)
A
2 t 2
VoVm(1 sin)169.7(1 sin0.843)43 V vo(t)c
169.7 sin (t)
2 t2
169.5e (t 1.62)>18.85
t2
0.843 rad48°
sin sin(1.62)e (2 1.62>18.85)0 tan 1(18.85) 1.62rad93°
Vmsin 169.5 V har80679_ch03_065-110.qxd 12/17/09 2:09 PM Page 92
3.8 Half-Wave Rectifier With a Capacitor Filter 93
The Probe output is shown in Fig. 3-12. Angles and are determined directly by first modifying the x-variable to indicate degrees (x-variable time*60*360) and then using the cursor option. The restrict data option is used to compute quantities based on steady-state values (16.67 to 50 ms). Steady state is characterized by waveforms begin- ning and ending a period at the same values. Note that the peak diode current is largest in the first period because the capacitor is initially uncharged.
■ Results from the Probe Cursor
Quantity Result
360 408( 48)
98.5
Vomax 168.9 V
Vomin 126 V
Vo 42.9 V
ID,peak 4.42 A steady state; 6.36 A first period
IC,peak 4.12 A steady state; 6.39 A first period
■ Results after Restricting the Data to Steady State
Quantity Probe Expression Result
ID, avg AVG(I(D1)) 0.295 A
IC, rms RMS(I(C1)) 0.905 A
IR, avg AVG(W(R1)) 43.8 W
Ps AVG(W(Vs)) 44.1 W
PD AVG(W(D1)) 345 mW
In this example, the ripple, or variation in output voltage, is very large, and the capacitor is not an effective filter. In many applications, it is desirable to produce an output that is closer to dc. This requires the time constant RCto be large compared to
Figure 3-12 Probe output for Example 3-9.
INPUT AND OUTPUT VOLTAGES
DIODE CURRENT
Time 200 V
-200 V
-0.0 A
0 s 10 ms
4.0 A 8.0 A 0 V
20 ms 30 ms 40 ms 50 ms
1(D) V(1) V(2)
94 C H A P T E R3 Half-Wave Rectifiers
the period of the input voltage, resulting in little decay of the output voltage. For an effective filter capacitor, the output voltage is essentially the same as the peak voltage of the input.