Title: BEE4223 Power Electronics
1BEE4223 Power Electronics Drive Systems
- Chapter 3 AC TO DC CONVERTER (RECTIFIER)
Nor Laili Ismail FKEE,UMP.
2LEARNING 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.
3Overview
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
4Rectifier
- 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
5Rectifier
- 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
6Root-Mean-Squares (RMS)
7Root Mean Squares of f
8Concept of RMS
t
Average of v0
9Ideal 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.
10Ideal 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
11Example 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.
12Example 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.
13Example 1 (Cont)
(iii) average power delivered to the load over
one cycle
- (ii) The rms value of the of Vo and Io
14Example 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.
15Example 2 (Cont)
(ii) The average power absorbed by the load
- Solution
- (i) The average load current
16Example 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.
17Half-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
18Half-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
19Half-wave with R-L load
- The point when the current reaches zero, is when
the diode turns off, given by
20Example 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
21Example 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
22Example 3 (cont)
v) Power absorbed by resistor
iv) rms current
vi) Power factor
23Half-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.
24Half-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
RC.
25Half-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.
26Half-wave with R-C load (Ripple Voltage)
- 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.
27Example 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.
28Example 4 (cont)
- Solution
- For parameter given
(ii) Ripple Voltage
(iii) Peak diode current
(i) Output Voltage
29Example 4 (cont)
(iv) Waveform must be properly labeled according
to data
(v) Capacitor value
30RL 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,
31RL Source Load
32Example 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
33Example 5 (cont)
- Solution
- For parameter given
(i) Current Equation
ii) Power absorbed by resistor
34Example 5 (cont)
- iii) Power absorbed by dc source
v) Power factor
iv) Power supplied
v) Waveform
- Refer notes
35Freewheeling 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.
36Freewheeling 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.
37Freewheeling 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
38Example 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)
39Example 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,
40Example 7 (cont)
Fourier Impedance
- The ac voltage amplitudes,
Note angle note included in calculation
41Example 7 (cont)
Resulting Fourier Terms are as follows
42The 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
43Controlled, Half-wave R load
- A gate signal is applied at ?t ?, where ? is
the delay/firing angle.
44Example 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.
45Example 8 (Cont)
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
46Controlled, Half-wave R-L load
- The analysis of the circuit is very much similar
to that of uncontrolled rectifier.
47Controlled, Half-wave R-L load
48Example 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
-
49Example 9 (cont)
- Solution
- For parameter given
(i) Current Equation
50Example 9 (cont)
iv) Power absorbed by resistor
- ii) Average current and voltage
v) Power factor
iii) rms current