Page 22 - Rashid, Power Electronics Handbook

1 Introduction 5

FIGURE 1.5 Input and output waveforms for Example 1.4.

not zero when the voltage ®rst becomes negative. If the
switch attempts to turn off, it must drop the inductor
current to zero instantly. The derivative of current in the
inductor di=dt would become negative in®nite. The
inductor voltage Lðdi=dtÞ similarly becomes negative
in®nite Ð and the devices are destroyed. What really
happens is that the falling current allows the inductor
to maintain forward bias on the diode. The diode will
turn off only when the current reaches zero. A diode has
de®nite properties that determine circuit action, and
both voltage and current are relevant. Figure 1.7 shows
FIGURE 1.6 Half-wave recti®er with L-R load for Example 1.5.
the input and output waveforms for a time constant t
equal to 1=3 of the ac waveform period.
the ac voltage be V cosðotÞ. From Kirchhoff's voltage
0
law (KVL),
1.4.2 The Method of Energy Balance
di
V cosðotÞ¼ L þ Ri ð1:2Þ Any circuit must satisfy conservation of energy. In a lossless
0
dt
power electronic circuit, energy is delivered from source to
Let us assume that the diode is initially off (this load, which is possible through an intermediate storage step.
assumption is arbitrary, and we will check it out as the The energy ¯ow must balance over time such that the energy
example is solved). If the diode is off, the diode current drawn from the source matches that delivered to the load. The
i ¼ 0, and the voltage across the diode will be v . The converter in Fig. 1.8 serves as an example of how the method
ac
diode will become forward-biased when v becomes of energy balance can be used to analyze circuit operation.
ac
positive. The diode will turn on when the input voltage
EXAMPLE 1.5. The switches in the circuit of Fig. 1.8 are
makes a zero-crossing in the positive direction. This
controlled cyclically to operate in alternation: when the
allows us to establish initial conditions for the circuit:
left switch is on, the right one is off, and so on. What
iðt Þ¼ 0, t ¼ÿp=ð2oÞ. The differential equation can
0 0 does the circuit do if each switch operates half the time?
be solved in a conventional way to give
The inductor and capacitor have large values.
When the left switch is on, the source voltage V
in
oL ÿt p
iðtÞ¼ V 0 exp ÿ appears across the inductor. When the right switch is on,
R þ o L t 2ot
2
2 2
the output voltage V out appears across the inductor. If

R oL this circuit is to be a useful converter, we want the
þ cosðotÞþ sinðotÞ ð1:3Þ
2 2
2
2 2
2
R þ o L R þ o L inductor to receive energy from the source, then deliver
it to the load without loss. Over time, this means that
where t is the time constant L=R. What about diode energy does not build up in the inductor (instead it
turn-off? One ®rst guess might be that the diode turns ¯ows through on average). The power into the inductor
off when the voltage becomes negative, but this is not therefore must equal the power out, at least over a cycle.
correct. We notice from the solution that the current is Therefore, the average power in should equal the average