BEE4223 Power Electronics

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BEE4223 Power Electronics Drive Systems

• Chapter 3 AC TO DC CONVERTER (RECTIFIER)

Nor Laili Ismail FKEEUMP.
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LEARNING OBJECTIVES

• Upon completion of the chapter the student
  should be able to
• State the operation and characteristics of diode
  rectifier.
• 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.

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Overview
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

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Rectifier

• 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

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Rectifier

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

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Root-Mean-Squares (RMS)
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Root Mean Squares of f
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Concept of RMS
t
Average of v0
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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.

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Ideal RectifierSingle-Phase Half-Wave R-Load

• We observe that
• DC voltage is fixed at 0.318 or 31.8 of the peak
  value
• 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
  transformer

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Example 1

• Consider the half-wave rectifier circuit with a
  resistive load of 25 and a 60 Hz ac source of
  110Vrms.
• 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.

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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.
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Example 1 (Cont)
(iii) average power delivered to the load over
one cycle

• (ii) The rms value of the of Vo and Io

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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
  and
• (iii) the power factor of the circuit.

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Example 2 (Cont)
(ii) The average power absorbed by the load

• Solution
• (i) The average load current

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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.
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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

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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
  decrease.
• 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
  size.
• This will results on reducing the average output
  voltage due to the negative segment. The larger
  the Inductance the larger negative segment

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Half-wave with R-L load

• The point when the current reaches zero is when
  the diode turns off given by

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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

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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
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Example 3 (cont)
v) Power absorbed by resistor

• iii) Average current

iv) rms current
vi) Power factor
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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
  circuit.

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Half-wave with R-C load

• Assume the capacitor is
• unchargedand 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
  RC.

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Half-wave with R-C load

• The angle t is the point when diode turns
  off.
• 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.

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Half-wave with R-C load (Ripple Voltage)

• The ripple

• if VVm 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.

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Example 4

• The half-wave rectifier has 120Vrms source at
  60Hz R500 C100F 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.

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Example 4 (cont)

• Solution
• For parameter given

(ii) Ripple Voltage
(iii) Peak diode current
(i) Output Voltage
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Example 4 (cont)
(iv) Waveform must be properly labeled according
to data
(v) Capacitor value
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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

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RL Source Load
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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

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Example 5 (cont)

• Solution
• For parameter given

(i) Current Equation
ii) Power absorbed by resistor
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Example 5 (cont)

• iii) Power absorbed by dc source

v) Power factor
iv) Power supplied
v) Waveform
- Refer notes
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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.

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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.

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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
  current

- same as uncontrolled RLoad Rectifier
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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
notes)
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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

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Example 7 (cont)
Fourier Impedance

• The ac voltage amplitudes

Note angle note included in calculation
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Example 7 (cont)
Resulting Fourier Terms are as follows

• Power Absorbed

• rms current

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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

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Controlled Half-wave R load

• A gate signal is applied at t where is
  the delay/firing angle.

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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
  output.

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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

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Controlled Half-wave R-L load

• The analysis of the circuit is very much similar
  to that of uncontrolled rectifier.

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Controlled Half-wave R-L load
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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

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Example 9 (cont)

• Solution
• For parameter given

(i) Current Equation
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Example 9 (cont)
iv) Power absorbed by resistor

• ii) Average current and voltage

v) Power factor
iii) rms current