EXAMPLE 1
A half-wave rectifier has R = 100 , L = 0.1 H, f = 60
Hz, and Vm = 100 V.
Find
(a) an expression for the current in circuit,
(b) the average current,
(c) the rms current,
(d) the power absorbed by R-L load, and
(e) the power factor.
EXAMPLE 2
A
204
EXAMPLE 1
A half-wave rectifier has R = 100 , L = 0.1 H, f = 60
Hz, and Vm = 100 V.
Find
(a) an expression for the current in circuit,
(b) the average current,
(c) the rms current,
(d) the power absorbed by R-L load, and
(e) the power factor.
EXAMPLE
1
ECE 4200/5200 POWER ELECTRONICS TOPICS
AC-DC UNCONTROLLED RECTIFIERS
Single-Phase Rectifiers
Single-Phase Half-Wave
Single-Phase Full-Wave
R Load
R - L Load
L Source (E) Load
R - L Source (E) Load
Three-Phase Rectifiers With
R Load
R L Load
R L Source(E
DC-DC CONVERTERS
SUMMARY OF DC-DC CONVERTERS
BUCK CONVERTER:
Va = kVs
(5.56)
Is = kIa
(5.57)
Inductor Ripple Current
I =
Va ( Vs Va ) Vs k ( 1 k )
=
fLVs
fL
(5.59), (5.60)
Inductor Current
IL =
Va
R
Maximum and Minimum Inductor Current
I L max = I L +
I V
4
CHAPTER 3
DIODE RECTIFIERS
SUMMARY OF DIODE RECTIFIER LOADS
AVERAGE VALUES:
Voltage
Vav (Vo)
Current
Iav (Io)
RMS VALUES:
Voltage
Vrsm (V)
Current
Irms (I)
POWER:
Real Power, Apparent Power, Power Factor
Use of Fourier Series to also Compute Vav, Iav, V
106
CHAPTER 6
PULSE-WIDTH-MODULATION INVERTERS
SUMMARY OF CONTROLLED RECTIFIERS LOADS
RMS VALUES:
POWER:
Voltage
Vrsm (V)
Current
Irms (I)
Real Power, Apparent Power, Power Factor
Use of Fourier Series to also Compute Vrms, Irms.
LOADING CONDITIONS
A.
SIN