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BEE4223 Power Electronics Drive Systems

- Chapter 3 AC TO DC CONVERTER (RECTIFIER)

Nor Laili Ismail FKEE,UMP.

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.

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

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

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

Root-Mean-Squares (RMS)

Root Mean Squares of f

Concept of RMS

t

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

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

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.

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,

and - (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

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

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

circuit.

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

RC.

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.

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.

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.

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

current

- 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

notes)

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

output.

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

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