BEE4223 Power Electronics - PowerPoint PPT Presentation


PPT – BEE4223 Power Electronics PowerPoint presentation | free to view - id: 1b945-MjgyO


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation

BEE4223 Power Electronics


To supply a dc source from an ac source. The diode will remain off as long as the voltage of ac source is less than dc voltage. ... – PowerPoint PPT presentation

Number of Views:854
Avg rating:3.0/5.0
Slides: 51
Provided by: Owne8


Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: BEE4223 Power Electronics

BEE4223 Power Electronics Drive Systems

Nor Laili Ismail FKEE,UMP.
  • Upon completion of the chapter the student
    should be able to
  • State the operation and characteristics of diode
  • Discuss the performance parameters and use
    different technique for analyzing and design of
    diode rectifier circuits.
  • Simulate different arrangement of diode
    rectifiers by using PSpice.

Single-phase, full wave
rectifier R load R-L load, Controlled
R, R-L Load continuous and
discontinuous current mode Three-phase
rectifier uncontrolled controlled
  • Single-phase, half wave rectifier
  • Uncontrolled
  • R load
  • R-L load
  • R-C load
  • Controlled
  • Free wheeling diode

  • DEFINITION Converting AC (from mains or other AC
    source) to DC power by using power diodes or by
    controlling the firing angles of
    thyristors/controllable switches.
  • Basic block diagram

  • Input can be single or multi-phase (e.g.
  • Output can be made fixed or variable
  • Applications
  • DC welder, DC motor drive, Battery charger, DC
    power supply, HVDC

Root-Mean-Squares (RMS)
Root Mean Squares of f
Concept of RMS
Average of v0
Ideal RectifierSingle-Phase, Half-Wave R-Load
  • Considering the diode is ideal, the voltage at
    R-load during forward biased is the positive
    cycle of voltage source, while for negative
    biased, the voltage is zero.

Ideal RectifierSingle-Phase, Half-Wave R-Load
  • We observe that
  • DC voltage is fixed at 0.318 or 31.8 of the peak
  • RMS voltage is reduced from 0.707 (normal
    sinusoidal RMS) to 0.5 or 50 of peak value.
  • Half wave is not practical because of high
    distortion supply current. The supply current
    contains DC component that may saturate the input

Example 1
  • Consider the half-wave rectifier circuit with a
    resistive load of 25? and a 60 Hz ac source of
  • Calculate the average values of Vo and Io.
    Justify the significant value of Vo and Io.
  • Calculate the rms values of Vo and Io.
  • Calculate the average power delivered to the load.

Example 1 (Cont)
  • Solution
  • (i) The average values of Vo and Io are given by

In this case, for the particular circuit,
possible dc output voltage obtained from the
circuit is 49.52V and dc output current is 1.98A.
That means, for any dc application within this
value, this circuit can be used.
Example 1 (Cont)
(iii) average power delivered to the load over
one cycle
  • (ii) The rms value of the of Vo and Io

Example 2
  • For the half-wave rectifier, the source is a
    sinusoid of 120Vrms at a frequency of 60Hz. The
    load resistor is 5?. Determine
  • (i) the average load current,
  • (ii) the average power absorbed by the load,
  • (iii) the power factor of the circuit.

Example 2 (Cont)
(ii) The average power absorbed by the load
  • Solution
  • (i) The average load current

Example 2 (Cont)
(iii) Power Factor
Note The power factor at the input of the
rectifier circuit is poor even for resistive
load and decreases as triggering angle for
controlled rectifier is delayed.
Half-wave with R-L load
  • Industrial load typically contain inductance as
    well as resistance.
  • By adding an inductor in series with the load
    resistance causes an increase in the conduction
    period of the load current, hence resulting the
    half-wave rectifier circuit working under an
    inductive load.
  • That means, the load current flows not only
    during Vs 0, but also for a portion of Vs This is due to

Half-wave with R-L load
  • Until certain time (VR (hence VL Vs-VR
    is positive), the current builds up and inductor
    stored energy increases.
  • At maximum of VR, VsVR hence, VL 0V.
  • Beyond this point, VL becomes negative (means
    releasing stored energy), and current begins to
  • After T?, the input, Vs becomes negative but
    current still positive and diode is still
    conducts due to inductor stored energy. The load
    current is present at certain period, but never
    for the entire period, regardless of the inductor
  • This will results on reducing the average output
    voltage due to the negative segment. The larger
    the Inductance, the larger negative segment

Half-wave with R-L load
  • The point when the current reaches zero, is when
    the diode turns off, given by

Example 3
  • For half-wave rectifier with R-L load, R100?,
    L0.1H, ?377rad/s, and Vs100V. Determine
  • An expression for the current in this circuit
  • The point where diode turns off
  • The average current
  • The rms current
  • The power absorbed by the R-L load, and
  • The power factor

Example 3 (cont)
  • Solution
  • For parameter given

(ii) ? (diode stop)
Using numerical root finding, ? is found to be
3.50 rads or 201o
(i) Current Equation
Example 3 (cont)
v) Power absorbed by resistor
  • iii) Average current

iv) rms current
vi) Power factor
Half-wave with R-C load
  • In some applications in which a constant output
    is desirable, a series inductor is replaced by a
    parallel capacitor.
  • The purpose of capacitor is to reduce the
    variation in the output voltage, making it more
    like dc.
  • The resistance may represent an external load,
    while the capacitor is a filter of rectifier

Half-wave with R-C load
  • Assume the capacitor is
  • uncharged,and as source
  • positively increased,
  • diode is forward biased
  • As diode is on, the output voltage is the same as
    source voltage, and capacitor charges.
  • Capacitor is charged to Vm as input voltage
    reaches its positive peak at ?t ?/2.
  • As source decreases after ?t ?/2, the capacitor
    discharges into load resistor. As diode is
    reversed biased, the load is isolated from
    source, and the output voltage (capacitive
    charge) decaying exponentially with time constant

Half-wave with R-C load
  • The angle ?t ? is the point when diode turns
  • The diode will stay off until the capacitor and
    input voltages become equal again.
  • The effectiveness of capacitor filter is
    determined by the variation in output voltage, or
    expressed as maximum and minimum output voltage,
    which is peak-to-peak ripple voltage.

Half-wave with R-C load (Ripple Voltage)
  • The ripple
  • if V??Vm and ???/2, then ripple can be
    approximated as
  • The output voltage ripple is reduced by
    increasing the filter capacitor, C. Anyhow, this
    results in a larger peak diode current.

Example 4
  • The half-wave rectifier has 120Vrms source at
    60Hz, R500?, C100?F and delay when diode turns
    on is given 48. Determine
  • The expression of output voltage
  • Ripple voltage
  • Peak diode current
  • Sketch and label the output waveform
  • Value of C as ripple voltage is 1 of Vm, and
    hence find new ? under this condition.

Example 4 (cont)
  • Solution
  • For parameter given

(ii) Ripple Voltage
(iii) Peak diode current
(i) Output Voltage
Example 4 (cont)
(iv) Waveform must be properly labeled according
to data
(v) Capacitor value
RL Source Load
  • To supply a dc source from an ac source
  • The diode will remain off as long as the voltage
    of ac source is less than dc voltage.
  • Diode starts to conduct at ?t?. Given by,

RL Source Load
Example 5
  • The RL half-wave rectifier has 120Vrms source at
    60Hz, R2?, L20mH, Vdc 100V with extinction
    angle given by 193o. Determine
  • The expression of current in the circuit
  • Power absorbed by resistor
  • Power absorbed by dc source
  • Power supplied by ac source
  • Power factor
  • Draw the waveform

Example 5 (cont)
  • Solution
  • For parameter given

(i) Current Equation
ii) Power absorbed by resistor
Example 5 (cont)
  • iii) Power absorbed by dc source

v) Power factor
iv) Power supplied
v) Waveform
- Refer notes
Freewheeling Diode (FWD)
  • Note that, previously discussed uncontrolled
    half-wave RL load rectifier allows load current
    to present at certain period (current decreasing
    by time since opposing negative cycle of input),
    hence reducing the average output voltage due to
    the negative segment.
  • In other word, for single-phase, half wave
    rectifier with R-L load, the load (output)
    current is NOT CONTINUOUS.
  • A FWD (sometimes known as commutation diode) can
    be placed in parallel to RL load to make the load
    (output) current continuous.

Freewheeling Diode (FWD)
  • Note that both D1and D2 cannot be turned on at
    the same time.
  • For a positive cycle voltage source,
  • D1 is on, D2 is off
  • The voltage across the R-L load is the same as
    the source voltage.
  • For a negative cycle voltage source,
  • D1 is off, D2 is on
  • The voltage across the R-L load is zero.
  • However, the inductor contains energy from
    positive cycle. The load current still circulates
    through the R-L path.

Freewheeling Diode (FWD)
  • negative cycle voltage source (cont),
  • But in contrast with the normal half wave
    rectifier, negative cycle of FWD does not consist
    of supply voltage in its loop.
  • Hence the negative part of Vo as shown in the
    normal half-wave disappear.
  • Irms is determined from Fourier component of

- same as uncontrolled RLoad Rectifier
Example 6
  • Uncontrolled R-L load rectifier, has a problem of
    discontinuous load current. Suggest a solution to
    the problem by justifying your answer through its
    principles of operation and waveform.

Solution Operation of FWD and its waveform (refer
Example 7
  • Determine the average load voltage and current,
    and determine the power absorbed by the resistor
    in the FWD circuit, where R2? and L25mH,
    Vm100V 60Hz.
  • Solution
  • The average load voltage and current,

Example 7 (cont)
Fourier Impedance
  • The ac voltage amplitudes,

Note angle note included in calculation
Example 7 (cont)
Resulting Fourier Terms are as follows
  • Power Absorbed
  • rms current

The Controlled Half-wave Rectifier
  • Previously discussed are classified as
    uncontrolled rectifiers.
  • Once the source and load parameters are
    established, the dc level of the output and power
    transferred to the load are fixed quantities.
  • A way to control the output is to use SCR instead
    of diode. Two condition must be met before SCR
    can conduct
  • The SCR must be forward biased (VSCR0)
  • Current must be applied to the gate of SCR

Controlled, Half-wave R load
  • A gate signal is applied at ?t ?, where ? is
    the delay/firing angle.

Example 8
  • Design a circuit to produce an average voltage of
    40V across 100? load resistor from a 120Vrms 60
    Hz ac source. Determine the power absorbed by the
    resistor and the power factor.
  • Briefly describe what happen if the circuit is
    replaced by diode to produce the same average

Example 8 (Cont)
  • Solution

In such that to achieved 40V average voltage,
the delay angle must be
  • If an uncontrolled diode is used, the average
    voltage would be
  • That means, some reducing average resistor to the
    design must be made. A series resistor or
    inductor could be added to an uncontrolled
    rectifier, while controlled rectifier has
    advantage of not altering the load or introducing
    the losses

Controlled, Half-wave R-L load
  • The analysis of the circuit is very much similar
    to that of uncontrolled rectifier.

Controlled, Half-wave R-L load
Example 9
  • For controlled RL rectifier, the source is
    120Vrms at 60Hz, R20?, L0.04H, delay angle is
    45o and extinction angle is 217o. Determine
  • An expression for i(?t)
  • Average current and voltage
  • Power absorbed by load
  • Power factor

Example 9 (cont)
  • Solution
  • For parameter given

(i) Current Equation
Example 9 (cont)
iv) Power absorbed by resistor
  • ii) Average current and voltage

v) Power factor
iii) rms current