4 CAPÍTULO III
6.1 ESTUDIO DE LAS MATERIAS PRIMAS E INSUMOS
The action of the series R-L circuit with a single-diode rectifier valve illustrates a lot of the important features of rectifier circuit operation. InFig. 2.9the current is unidirectional and the polarity of the voltage drop eRon the resistor R is always as indicated. The polarity of the instantaneous emf eLof the series inductor L varies cyclically as does the total load voltage eL. Where the application of sinusoi-dal voltage esresults in an instantaneous current iL, this is given by solution of the first-order linear differential equation [Eq. (2.65)].
e⳱ Emsint (2.64)
e i R Ldi
L L dt
= + L (2.65)
Neglecting the voltage drop on the diode, e⳱ iLduring conduction. If no current is initially present conduction begins when the anode voltage of the diode becomes positive, that is, at the beginning of the supply voltage cycle in Fig. 2.10.An expression for the current iL may be thought of, at any instant, as the sum of hypothetical steady-state and transient components iss and it, respectively.
FIG.9 Single-phase, half-wave diode rectifier with series R-L load.
i t i t i t
I t i t
L ss t
m t
ω ω ω
ω ω
( )
=( )
+( )
= sin
(
−)
+( )
(2.66)where⌽ is the phase angle for sinusoidal currents of supply frequency
=tan−1ωL
R (2.67)
At instant t ⳱ 0 in Fig. 2.10 the total current iL(0) ⳱ 0 but the steady-state component iss(0) ⳱ Im sin(0 ⳮ ⌽) ⳱ ⳮIm sin ⌽, so that the instantaneous transient current is
it(0)⳱ Imsin⌽ (2.68)
FIG.10 Current waveform for the single-phase, half-wave circuit of Fig. 2.9:⌽ ⳱ 60.
Fort ⳱ 0 the transient current decays exponentially through the series R-L load
The instantaneous current at any time interval during conduction is therefore iL(t) ⳱ Imsin(t ⳮ ⌽) Ⳮ Imsin⌽εⳮcot⌽䡠t (2.72) It is seen inFig. 2.10that the total current iLconsists of unidirectional, nonsinusoi-dal pulses lying outside the envelope of the steady-state sinusoid. The conduction anglecwhen iL(t) ⳱ 0 in Eq. (2.56) satisfies
sin(cⳮ ⌽) Ⳮ sin⌽ε-cot⌽c⳱ 0 (2.73)
Equation (2.73) is transcendental and must be solved iteratively forc. For values of phase angle⌽ up to about 60 one can estimate cfairly accurately by the relation
c⳱ Ⳮ ⌽ Ⳮ ⌬ (2.74)
where⌬ is of the order a few degrees. With highly inductive loads the conduction angle increases until, at⌽ ⳱ /2, the conduction takes the form of a continuous, unidirectional sinusoidal oscillation of mean value Im⳱ Em/L.
The instantaneous voltage drop eRon the load resistor has the value iLR and must have the same waveform as iL(t) ⳱ 0. Now by Kirchhoff’s law, in
= sin
(
during conduction)
(2.75)At the valuet1, where eL(t) ⳱ eR(t) inFig. 2.11a,eL(t) ⳱ L di/dt ⳱ 0.
The derivative di/dt is zero at a current maximum, and therefore the crossover of the eR(t) curve with the e(t) curve in Fig. 2.11b occurs when eR(t) has its maximum value. Time variations of the circuit component voltages are shown in Fig. 2.11 for a typical cycle. The polarity of the inductor voltage eL(t) is
such as, by Lenz’s law, to oppose the change of inductor current. While the current is increasing, 0 ⬍ t ⬍ t1, the induced emf in the inductor has its positive point nearest the cathode. When the inductor current is decreasing,t1
⬍ t ⬍ t3, the induced emf in the inductor tries to sustain the falling current by presenting its positive pole furthest from the cathode inFig. 2.9.
The average value of the rectified current is the mean value of the iL(t) curve (Fig. 2.10)over 2 radians and is given by
I i t d t
cos cos sin sin sin
cot Eliminating the exponential component between Eq. (2.73) and Eq. (2.76) gives, after some manipulation,
I E
Since the average value of the inductor voltage is zero, Eav =I Rav
The rms value of the load current is found using Eqs. (2.72) and (2.73):
I i t d t
The corresponding ripple factor can be obtained by the substitution of Eqs. (2.78) and (2.79) into Eq. (2.11). Both the average and rms values of the current increase for increased values of phase angle.
2.3.2 Worked Examples
Example 2.6 A single-phase, full-wave diode bridge (Fig. 2.8a) is used to supply power to a resistive load of value R ⳱ 50 ⍀. If the supply voltage
FIG.11 Component voltage waveforms for the single-phase, half-wave circuit ofFig.
2.9:⌽ ⳱ 60.
remains sinusoidal and has a peak value 400 V, calculate the average and rms values of the load current. Calculate the ripple factor for the load current and compare this with half-wave operation.
The average value of the load current (Fig. 2.8b)is given by Eq. (2.54), The corresponding rms value ILis the value of a sinusoidal of the same peak height, as given in Eq. (2.56).
I E
L R
= 1_ m =5 66A.
公2
The ripple factor is
RF I
which compares with a value 1.21 for half-wave operation.
Example 2.7 For the single-phase bridge of Example 2.6 (i.e., Fig. 2.8) evaluate the voltage and current ratings required of the bridge diodes
The diodes in Fig. 2.8a conduct current only in their respective forward directions. When the top terminal of Fig. 2.8a is positive, current flows through D1, R, and D4. The current of diodes D1and D4 is therefore given by
The diode rms current is 1/兹2 or 0.707 of the rms supply current. A diode current waveform consists of only the positive pulses of current and is therefore similar to the waveform of Fig. 2.3.Because a diode current waveform contains only one half the area of the load current its mean current rating is one half the mean load current D3are held in extinction due to reverse anode voltage. The peak value PRV of this reverse voltage is the peak value of the supply voltage.
PRV⳱ Em⳱ 400 V
Example 2.8 A series R-L load with R⳱ 10 ⍀ and XL⳱ 20 ⍀ at f ⳱ 50 Hz and supplied from an ideal single-phase supply e⳱ 300 sin 2ft through
an ideal diode. Determine the peak value of the current pulses, the conduction angle and the average load current
=tan−1X =tan−12=63 4. ° R
L
cot⌽ ⳱ 0.5, cos ⌽ ⳱ 0.447, sin ⌽ ⳱ 0.894
Conduction angle c is determined by iterative solution of the transcendental equation, Eq. (2.73). We start by estimating a value ofcusing Eq. (2.74)
c⳱ 180 Ⳮ 63.4 Ⳮ ⌬ 艑 243.4 Ⳮ ⌬
The algebraic sum of the two parts of Eq. (2.73) is given in the RH column.
From the results obtained for the estimated valuesc⳱ 250 and 249 one can make a linear interpolation that the actual value ofcis 249.25. This is shown in the final row to be almost correct.
c⳱ 249.25
The maximum value of the current pulses occurs when diL/dt ⳱ 0. From Eq.
(2.72) di
dtL =Imcos
(
ωt−)
−Imsincotε−cotωt =0 from which it is seen thatcos(t ⳮ ⌽) ⳱ cot ⌽ εⳮcot ⌽ t
Iterative solution of this, along the lines above, yields ωt di
246° 4.294 0.045 0.117 0.105 0.06
248° 4.33 0.08 0.115 0.103 0.023
250° 4.363 0.115 0.113 0.101 0.014
249° 4.316 0.0976 0.114 0.102 0.0044
249.25° 4.35 0.1019 0.1136 0.1016 0.003
ε−cotΦ⋅θc sinΦε−cotΦ⋅θc
Substitutingt ⳱ 146.2 into Eq. (2.72) gives
ILmax = . Im = . .
+ =
1 2417 1 2417 300
10 20
2 2 16 7A
Now cosc⳱ cos249.25 ⳱ ⳮ0.354. In Eq. (2.77) therefore,
Iav =
×
(
+)
=300
2 10 1 0 354 6 465
π . . A
This compares with the value Im/ or 9.55 A that would be obtained with only the 10-⍀ resistor as load.
If the load consisted of a resistor of the same value as the present load impedance the average current would be 300/( ⳯ 22.36) ⳱ 4.27 A.