UPS
Half-Wave Rectifiers
INTRODUCTION
Arectifier converts ac to dc. The purpose of a rectifier may be to produce an output
that is purely dc, or the purpose may be to produce a voltage or current waveform
that has a specified dc component.
In practice, the half-wave rectifier is used most often in low-power applications
because the average current in the supply will not be zero, and nonzero average
current may cause problems in transformer performance. While practical
applications of this circuit are limited, it is very worthwhile to analyze the halfwave
rectifier in detail. Athorough understanding of the half-wave rectifier circuit
will enable the student to advance to the analysis of more complicated circuits
with a minimum of effort.
The objectives of this chapter are to introduce general analysis techniques
for power electronics circuits, to apply the power computation concepts of the
previous chapter, and to illustrate PSpice solutions.
RESISTIVE LOAD
• Creating a DC Component Using an Electronic Switch
• A basic half-wave rectifier with a resistive load is shown in Fig. 3-1a.
The source
• is ac, and the objective is to create a load voltage that has a nonzero
dc component.
• The diode is a basic electronic switch that allows current in one
direction only. For
• the positive half-cycle of the source in this circuit, the diode is on
(forward-biased).
• Considering the diode to be ideal, the voltage across a forward-
biased diode is zero and the current is positive.
• For the negative half-cycle of the source, the diode is reverse-
biased, making the current zero.
• The voltage across the reverse-biased diode is the source voltage,
which has a negative value.
• The voltage waveforms across the source, load, and diode are
shown in Fig. 3-1b. Note that the units on the horizontal axis are in
terms of angle (t).
• This representation is useful because the values are independent of
frequency.
• The dc component Vo of the output voltage is the average value of a
half-wave
• rectified sinusoid
In the preceding discussion, the diode was assumed to be ideal. For a real
diode, the diode voltage drop will cause the load voltage and current to be reduced,
but not appreciably if Vm is large. For circuits that have voltages much
larger than the typical diode drop, the improved diode model may have only
second-order effects on the load voltage and current computations.
RESISTIVE-INDUCTIVE LOAD
• Industrial loads typically contain inductance as well as
resistance.
• As the source voltage goes through zero, becoming
positive in the circuit of Fig. 3-2a, the
diode becomes forward-biased.
The Kirchhoff voltage law equation that describes the
current in the circuit for the forward-biased ideal diode
is:
The forced response for this circuit is the current that exists after the
natural response has decayed to zero.
In this case, the forced response is the steady-state sinusoidal current
that would exist in the circuit if the diode were not present.
This steady-state current can be found from phasor analysis, resulting
in:
THE FREEWHEELING DIODE
Creating a DC Current
A freewheeling diode, D2 in Fig. 3-7a, can be connected across an RL load
as shown.
The behavior of this circuit is somewhat different from that of the half wave
rectifier of Fig. 3-2.
The key to the analysis of this circuit is to determine when each diode
conducts.
First, it is observed that both diodes cannot be forward-biased at the same
time.
Kirchhoff’s voltage law around the path containing the source and the two
diodes shows that one diode must be reverse biased.
Diode D1 will be on when the source is positive, and diode D2 will be on
when the source is negative.
