Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis: Power Electronic Converter

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Classification of Converters

Classification of Converters (Based on Commutation)
Based on how the power semiconductor devices within the converter are switched, 3 major classes:

Line frequency (naturally commutated) converters
Switching ( including forced commutated) converters
Resonant and quasi-resonant converters

Line frequency converters: the devices are turned off by the utility line voltage at one side of the converter and turned on, phase locked to the line voltage waveform by a triggering circuit. For ac-dc, ac-ac (frequency = input frequency, rms ≤ input rms).

Switching converters: the controllable switches in the converter are turned on and off at frequencies that are high compared to the line frequency. For dc-dc, dc-ac, ac-ac (variable frequency and rms, where the frequency ≤ the input frequency).

Resonant converters: the controllable switches turn on and/or turn off at zero voltage and/or zero current. For dc-dc, dc-ac.

Classification of Converters (Based on Functions)

Converters	Input to Output Conversion
1. AC VOLTAGE CONTROLLER	Fixed to Variable ac (Line Commutation).
2. RECTIFIERS (Uncontrolled).	Fixed ac to Fixed dc (Line Commutation).
3. RECTIFIERS (Controlled).	Fixed ac to Variable dc (Line Commutation).
4. DC-to-DC (Chopper).	Fixed dc to Variable dc (Load or Forced Commutation).
5. INVERTERS (Uncontrolled).	Fixed voltage dc to Fixed ac (Line, Load, Forced).
6. INVERTERS (Controlled).	Fixed voltage dc to Variable ac (Line, Load, Forced).
7. CYCLO CONVERTERS.	Fixed ac voltage ac to Variable ac voltage & Frequency (Line or Forced).

Ac-ac Converter

A. Output frequency is equal to input frequency (output rms voltage can be varied)

TRIAC based Ac Voltage regulator (low voltage and current rating)

Thyristor based AC voltage regulator (high voltage and current rating)

B. Output frequency is less than input frequency (output rms voltage can be varied)
Cycloconverter
${V_{o(rms)}} = \frac{{{V_m}}}{{\sqrt 2 }}{\left[ {\frac{1}{\pi }\left( {\pi - \alpha + \frac{{\sin 2\alpha }}{2}} \right)} \right]^{\frac{1}{2}}}$

Fig. TRIAC BASED Ac-ac Converter and waveforms.

Ac-dc Converter (Rectifier) Two Types

Diode Rectifier (uncontrolled rectifier)
Ac-dc converters (controlled rectifiers)

Diode Rectifiers
${V_{o(average)}} = \frac{{2{V_m}}}{\pi }$

Ac-dc Converter (Controller Rectifier)

${V_{o(average)}} = \frac{{2{V_m}}}{\pi }\left( {1 + \cos \alpha } \right)$

Dc-dc Converter (Chopper)
Three Types:

Buck (step-down)
Boost (step-up)
Buck-boost

${V_{o(average)}} = \delta {V_S}$

Fig. buck converter or chopper

Dc-ac Converter (Inverter)
Output AC may be Single-phase or three-phase.
${V_{o(rms - fundamental)}} = \frac{{4{V_S}}}{{\pi \sqrt 2 }} = 0.90{V_S}$

Fig. single-phase inverter

previous Applications of Power Electronics
next Design Consideration of Power Electronic Equipment

Thyristor Three-Phase, Six-Pulse Converter (PROCEDURE)

Setting up the equipment
(1) Install the Power Supply, the Enclosure / Power Supply, the DC Motor Generator, the Four-pole Squirrel-Cage induction Motor, the Smoothing inductors, the DC Voltmeter/Ammeter, the Three-Phase Wattmeter/Varmeter, the Tandem Rheostats and the Power-Thyristors modules in the Mobile Workstation.
Note: Align the brushes of the DC Motor / Generator in the neutral position by centring the metal tab on the red mark (on the casing).
(2) Install the Thyristor Firing Unit and the Current/Voltage isolators in the Enclosure / Power Supply.
Note: Before installing the Thyristor Firing Unit, make sure that switches SW1 and SW2 (located on the printed circuit board) are in the 0 position.
(3) Make sure that the main power switch of the Power Supply is set to the 0 (OFF) position. Connect the Power Supply to a three-phase wall receptacle.
(4) Plug the Enclosure / Power Supply line cord into a wall receptacle. Set the rocker switch of the Enclosure / Power Supply to the 1 (ON) position.
(5) On the Power Supply, set the 24-V ac power switch to 1 (ON) position.
Rectifier and inverter modes

(6) Set up the circuit of Figure 4.

Figure 4: Three-phase, six-pulse converter circuit.

LINE VOLTAGE (V_ac)	I₁ dc (A)	I₁ (A)	e₁ (V)	E₁ dc (V)	L₁ (H)
120	2.5	10	300	150	0.2 (3 A dc max.)
220	1.5	5	600	300	0.8 (1.5 A dc max.)
240	1.5	5	600	300	0.8 (1.5 A dc max.)

(7) Make the following settings:

On the Power Supply

Voltage Selector . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-5

On the Thyristor Firing Unit

ANGLE CONTROL COMPLEMENT . . . . . . . . . . . . . . . . . 0

ANGLE CONTROL ARC COSINE . . . . . . . . . . . . . . . . . . . 0

FIRING CONTROL MODE . . . . . . . . . . . . . . . . . . . . . . 3~

DC SOURCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MAX.

On the Oscilloscope

Channel-1 Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . 5 V/DIV. (DC coupled)

Channel-2 Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . 1 V/DIV. (DC coupled)

Time Base . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 ms/DIV.

Trigger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LINE

(8) On the Tandem Rheostats, set the control knob to the centre position. On the power Supply, make sure that the voltage control knob is set to the 0 position, then set the main power switch to 1 (ON). The Four-Pole Squirrel-Cage Induction Motor should begin to rotate.

On the Power Supply, adjust the voltage control knob to obtain the line-to-line voltage shown in table 1, as indicated by the voltmeter on the Power Supply.

LINE VOLTAGE	LINE-TO-LINE VOLTAGE
V_ac	V_ac
120	90
220	175
240	175

Table 1: Line-to-line voltage.

Turn the control knob on the Tandem Rheostats to adjust the voltage E of the dc source to the level indicated in Table 2.

LINE VOLTAGE	DC SOURCE VOLTAGE E₁dc
V_ac	V
120	100
220	200
240	200

Table 2: DC source voltage E₁.

Vary the firing angle and observe the effect on the waveforms and on the current delivered to the active load. How does the current vary as the firing angle is reduced to 0°? . . . . . . . . . . . . . . . . . . . . . . . . .

(9) On the Thyristor Firing Unit, adjust the FIRING ANGLE to 0°, adjust the Tandem Rheostats to obtain the current I₁shown in Table 3.

LINE VOLTAGE	CURRENT I₁dc
V_ac	A
120	0.5
220	0.25
240	0.25

Table 3: Current I₁delivered to load.

For each FIRING ANGLE in Table 4 adjust the Thyristor Firing Unit to the given firing angle, then adjust the Tandem Rheostats to obtain the current I₁Shown in Table 3. Observe the waveforms on the oscilloscope. Calculate the theoretical output voltage, record the measured voltage E₁dc as well as reactive power, and calculate the active power delivered to the reversible dc power supply.

Note: Do not increase the firing angle beyond 165° or the current will increase suddenly. If this happens reduce the firing angle to approximately 120° to restore normal operation.

FIRING ANGLE	THEORETICAL VOLTAGE E₀= 1.35 E_s cos	MEASURED VOLTAGE E₁dc	ACTIVE POWER P = E₁× I₁	REACTIVE POWER Q
degree	V	V	W	var
0
15
30
45
60
75
90
105
120
135
150
165

Table 4: Data for three-phase, six-pulse converter.

(10) Turn the knob on the Tandem Rheostats to the centre position, so the field current in the DC Motor / Generator is zero. On the Power Supply, set the voltage control knob to 0 then set the main power switch and the 24-V ac power switch to 0 (OFF).

For what range of firing angle does the converter operate as a rectifier? For what range does it operate as a inverter? Explain. . . . . . . . . . . . . . . . . . . . . .

In Figure 5, plot the output voltage E₁versus the firing angle. Then, in the same figure, plot the theoretical relationship E₀= 1.35 E_scos , where E_sis the line voltage. Compare the two curves.

Thyristor Three-Phase, Six-Pulse Converter

Figure: 5

Describe how the active and reactive power change as the firing angle is varied. . . . . . . . . . . . . . . .

(11) Set the rocker switch on the Enclosure / Power Supply to the 0 position. Remove all leads and cables.

CONCLUSION
In the exercise, you observed that a three-phase, six-pulse converter can operate both as a rectifier and as an inverter. You demonstrated that the static transfer function obtained experimentally is similar to the theoretical curve.
REVIEW QUESTIONS

What are the advantages of the three-phase, six pulse converter over other rectifier/inverter circuits? . . . . . . . . . . . . . . . . . . .
Why is it advantageous to have a high ripple frequency at the output of a rectifier? . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
What is the average output voltage E₀for a three-phase, six pulse converter if the line-to-line source voltage is 240 V and the firing angle is 75°? . . . . . . . . . . . . . . . . . . . . . . . . . . . .
What is the average current in each of the three ac line in a three-phase, six-pulse converter? . . . . . . . . . . . . . . . . . . . . .
What are the advantages and disadvantages of a three-phase bridge made with three thyristors and three diodes? . . . . . . . . .

previous Thyristor Three-Phase, Six-Pulse Converter

Thyristor Three-Phase, Six-Pulse Converter

OBJECTIVE

To study the operation of the three-phase, six-pulse rectifier/inverter.
To plot the static transfer function of the three-phase, six-pulse rectifier/inverter, and compare this with the theoretical curve.

DISCUSSION

The three-phase, six-pulse thyristor converter, or rectifier inverter, shown in Figure 1, is used in power electronics. This type of circuit gives the highest and most regular output voltage with the least amount of ripple. It can function as a rectifier or, when connected to a correctly polarized dc source, as an inverter.

This is the static transfer function of a three-phase, six-pulse converter, and is valid only when the on-time of the thyristors is equal to 120°. That is, when the series inductor is large enough to ensure continuous conduction.

Figure 1: Three-phase, six-pulse converter with resistive load.

Figure 2: Waveforms for the three-phase, six-pulse converter.

With respect to the three-phase restifier/inverter, the six-pulse circuit has the following differences:

the average value of E_D is twice as great
the transfer of active power is two times greater with the same value of current
the ripple frequency of E₀ is 360 Hz instead of 180 Hz
the average value of current I_A , I_B , I_C is zero.

This last difference is important because it prevents saturation of the transformers supplying the converter. According to Kirchhoff’s current law, I_A = I₁ - I₄. We can therefore draw the curve of I_A as shown in Figure 2. I_A changes polarity each half cycle. being equal to I₁ than -I₄ , and then repeating again. I_A flows for 240° or two-third of the cycle.

Three-phase bridge using three thyristors and three diodes

A three-phase bridge can be made using three thyristors and three diodes. Figure 3 shows an example of such circuit.

Figure 3: Three-phase bridge using three common-cathode thyristors.

The free-wheeling diode D₄ is necessary to ensure that the circuit be able to turn off an inductive load. Without this diode, when the gate pulses are stopped, the current may never drop to zero and one thyristor may continue to conduct.The freewheeling diode also relives the thyristors from freewheeling duty, allowing the use of lower power thyristors.

One advantage of the circuit in Figure 3 is that the firing control circuit can be simplified since the cathode of the three thyristors are at a common potential. This circuit is of lower cost than a three-phase six-thyristor bridge of comparable power, and both allow control of power from 0 to 100%. Unlike the six-thyristor bridge, however, this bridge cannot be used to make a line-commutated inverter.

Procedure summary

In the first part of this exercise, you will set up the equipment.

In the second part of this exercise, you will operate the three-phase, six pulse converter in both the rectifier and the inverter modes. You will plot the static transfer function of the converter, and compare it to the theoretical curve.

next Thyristor Three-Phase, Six-Pulse Converter (PROCEDURE)

Thyristor Single-Phase Bridge Rectifier/Inverter (PROCEDURE)

Setting up the equipment

(1) Install the Power Supply, the Enclosure / Power Supply, the DC Motor / Generator, the Four-Pole Squirrel-Cage Induction Motor, the Resistive Load, the Smoothing inductors, the DC Voltmeter/Ammeter, the AC Ammeter, the Three-Phase Wattmeter/Varmeter, the Temdem Rheostates, the Power Thyristors, and the Power Diodes modules in the Mobile Workstation.

Note: Align the brushes of the DC Motor / Generator in the neutral position by centering the metal tab on the red mark (on casing).

(2) Install the Thyristor Firing Unit and and the Current/Voltage Isolators in the Enclosure / Power Supply.

Note: Before installing the Thyristor Firing Unit, make sure that switches SW1 and SW2 (located on the printed circuit board) are in the 0 position.

(3) Make sure that the main power switch of the Power Supply is set to the 0 (OFF) position. Set the voltage control knob to 0. Connect the Power Supply to a three-phase wall receptacle.

(4) Plug the Enclosure / Power Supply line cord into a wall receptacle. Set the rocker switch of the Enclosure / Power Supply to the 1 (ON) position.

(5) On the Power Supply, set the 24-V ac power switch to the 1 (ON) position.

(6) Make sure that the toggle switches on the Power Thyristors and the Resistive Load modules are all set to the 0 (open) position.

Controlled bridge supplying a passive load

(7) Set up the circuit of Figure 12 using the resistive load Z (a). To simplify connecting the thyristors, set both interconnection switches on the Power Thyristors module to the 1 position.

Figure 12: Thyristor bridge circuit.

LINE VOLTAGE (Vac)	I₁ac (A)	I₂dc (A)	I₁ (A)	E₁dc (V)	e₁ (V)	Z₁(a)	Z₁(a)
120	2.5	2.5	10	150	300	R=60Ω	R=60Ω, L=0.2H (3A_dc max)
220	1.5	1.5	5	300	600	R=220Ω	R=220Ω, L=0.8H (1.5A_dc max)
240	1.5	1.5	5	300	600	R=240Ω	R=240Ω, L=0.8H (1.5A_dc max)

(8) Make the following setting:

On the Power Supply
    Voltage Selector . . . . . . . . . . . . . . . . . . . . . . . 4-N
On the Thyristor Firing Unit
    ANGLE CONTROL COMPLEMENT . . . . . . . . . . . . . . . 0
    ANGLE CONTROL ARC COSINE . . . . . . . . . . . . . . . . 0
    FIRING CONTROL MODE . . . . . . . . . . . . . . . . . . . . 1~
    DC SOURCE . . . . . . . . . . . . . . . . . . . . . . . . . . . MIN.
On the Oscilloscope
    Channel-1 Sensitivity . . . . . . . . 5 V/DIV. (DC coupled)
    Channel-2 Sensitivity . . . . . . . . 2 V/DIV. (DC coupled)
    Time Base . . . . . . . . . . . . . . . . . . . . . . . 5 ms/DIV.
    Trigger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LINE

Rectifier: Voltage and current waveforms (α = 45°)

Figure 13: Voltage and current waveforms (α = 45°).

(9) On the power Supply, make sure that the voltage control knob is set to the 0 position then set the main power switch to 1 (ON). Set the voltage control knob so that voltage indicated by the Power Supply voltmeter is equal to 90 % of the nominal line-to-neutral voltage.

On the Thyristor Firing Unit, set the FIRING ANGLE to 45°. Sketch the voltage and current waveforms in Figure 13.

Fill in the first row of Table 1.

LOAD
Z₁
OUTPUT
VOLTAGE

E₁ dc
OUTPUT
CURRENT

I₁ dc
OUTPUT
POWER

P₀ = E₁ × I₁

CONDUCTION

ANGLE

V A W degrees

(a) Resistive

(b) Inductive

Table 1: Measurements for controlled bridge (α = 45°).

On the Power Supply, set the voltage control knob to 0 then set the main power switch to 0 (OFF).

(10) Change the load in the circuit to the inductive load Z₁(b).

On the Power Supply, set the main power switch to 1 (ON). Set the voltage control knob so that the voltage indicated by the Power Supply voltmeter is equal to 90 % of the nominal line-to-neutral voltage. Sketch the voltage and current waveforms in figure 13.

Fill in the second row of Table 1.
On the Power Supply, set the voltage control knob to 0 then set the main power switch to 0 (OFF).
(11) Add a free-wheeling diode to the circuit, as shown in Figure 14. On the Power Supply, set the main power switch to 1 (ON), and set the voltage control knob to 90(%)

Figure 14: Thyristor bridge with free-wheeling diode.

LOAD Z₁	OUTPUT VOLTAGE E₁ dc	OUTPUT CURRENT I₁ dc	OUTPUT POWER P₀ = E₁ × I₁
	V	A	W
(b) Inductive

Table 2: Measurements for controlled bridge with free-wheeling diode.

Single-phase bridge with two thyristors and two diodes

(12) Set up the circuit of Figure 15.

Figure 15: Bridge rectifier wit two thyristors on common ac line.

LINE VOLTAGE (Vac)	I₁ac (A)	I₂dc (A)	I₁ (A)	E₁dc (V)	e₁ (V)	Z₁(a)
120	2.5	2.5	10	150	300	R=60Ω, L=0.2H (3A_dc max)
220	1.5	1.5	5	300	600	R=220Ω, L=0.8H (1.5A_dc max)
240	1.5	1.5	5	300	600	R=240Ω, L=0.8H (1.5A_dc max)

On the Power Supply; set the main power switch to 1 (ON), and set the voltage control knob so that the voltage indicated by the Power Supply voltmeter is equal to 90 % of the nominal line-to-neutral voltage. Vary the firing angle and observe the waveforms. Then set the firing angle to 45° and fill in the first row of Table 3.

CONFIGURATION

OUTPUT
VOLTAGE

E₁ dc

OUTPUT
CURRENT

I₁ dc

OUTPUT
POWER

P₀ = E₁ × I₁

V

A

W

Two thyristors on common ac line

Common-cathode thyristors

Table 3: Measurements for bridge rectifiers with two thyristors and two diodes (α = 45°).

Compare the voltage waveform at the output of this bridge to those obtained with the thyristor bridges of Figure 12 and 14.

On the Power Supply, set the voltage control knob to 0 then set the main power switch to 0 (OFF).

(13) Set up the circuit of Figure 16.

Figure 16: Bridge rectifier with common-cathode thyristors and free-wheeling diode.

On the Power Supply, set the main power switch to 1 (ON), and set voltage control knob so that the voltage indicated by the Power Supply voltmeter is equal to 90 % of the nominal line-to-neutral voltage. Vary firing angle and observe the waveforms. Then set the firing angle to 45° and fill in the second row to Table 3.

Compare the voltage waveform at the output of this bridge to those obtained with the bridge of Figure 15.

(14) On the Power Thyristors, connect the FIRING CONTROL INPUTS DISABLE jack to + 5 V jack on the Enclosure / Power Supply. This disables the gate pulses to all of the thyristors, and shuts off current to the load. What happens? . . . . . .

(15) On the Power Thyristors, disconnect the FIRING CONTROL INPUTS DISABLE jack from the + 5 V jack on the ENCLOSURE / Power Supply.

On the Power Supply, set the voltage control knob to 0 then set the main power switch to 0 (OFF).

Remove the free-wheeling diode from the circuit. Then, on the Power Supply, set the main power switch to 1 (ON), and set the voltage control knob to 90(%).

Could this bridge operate without the free-whiling diode? Explain. . . . . . . . . . . . .

ON the Power Thyristors, connect the FIRING CONTROL INPUTS DISABLE jack to + 5 V jack on the Enclosure / Power Supply. This disables the gate pulses to all of the thyristors. Disconnect then reconnect the plug at the FIRING CONTROL INPUTS DISABLE jack several times. Explain what you observe. . . . . . . . . . . .

On the Power Supply, set the voltage control to 0 then set the main power switch to 0 (OFF).

(16) Set up the circuit of Figure 17.

Figure 17: Controlled bridge rectifier/Inverter circuit.

(17) On the Thyristor Firing Unit, set the DC SOURCE control to MAX.

On the Tandem Rheostats, set the control knob to the centre position. On the Power Supply, make sure that the voltage control knob is set to the 0 position, then set the main power switch to 1 (ON). The Four-Pole Squirrel-Cage Induction Motor should began to rotate.

On the Power Supply, set the voltage control knob so that the voltage indicated by the Power Supply voltmeter id equal to 90 % of the nominal line-to-neutral voltage.

Adjust the Tandem Rheostate to obtain the voltage E₁ shown in Table 4 at the generator terminals of the motor-generator set.

LINE

VOLTAGE

ACTIVE LOAD

VOLTAGE

E₁ dc

V ac

V

120

80

220

160

240

160

Table 4: Active load voltage E₁.

Vary the firing angle and observe the effect on the waveforms and on the current delivered to the active load. How does the current vary as the firing angle is reduced to 0°? . . . . . . . . .

(18) On the Thyristor Firing Unit, adjust the FIRING ANGLE to 0°. Adjust the Tandem Rheostates to obtain the current I₁ shown in Table 5.

LINE

VOLTAGE

CURRENT
I₁ dc

V ac

A

120

1.0

220

0.5

240

0.5

Table 5: Current I₁ delivered to load.

For each FIRING ANGLE in Table 6, adjust the Thyristor Firing Unit to the given firing angle, then adjust the Tandem Rheostates to obtain the current I₁shown in Table 5. Observe the waveforms on the oscilloscope. Calculate the theoretical output voltage, record the measured voltage (E₁dc), and calculate the power delivered to the reversible dc power supply.

FIRING ANGLE

THEORETICAL
VOLTAGE
E₀ = 0.9 E_scos

MEASURED
VOLTAGE
E₁ dc

POWER
P = E₁ × I₁

degrees

V

V

W

0

15

30

45

60

75

90

105

120

135

150

165

Table 6: Data for bridge rectifier/inverter circuit.

(19) Turn the knob on the Tandem Rheostates to the centre position, so the field current of the DC Motor / Generator is zero. On the Power Supply, set the voltage control knob to 0 then set the main power switchand the 24-V ac power switch to 0 (OFF).

For what range of firing angle does the bridge operates as a rectifier?

For what range does it operate as a inverter? Explain. . . . . . . . . . . .

(20) In Figure 18, plot the voltage E₁ versus the firing angle. Then, in the same figure, plot the theoretical relationship E₀ = 0.9 E_s cos , where E_s is the line voltage. Compare the two curves.

Figure 18: Static transfer function for a single-phase bridge.

(21) Set the rocker switch on the Enclosure / Power Supply to the 0 position. Remove all leads and cables.

CONCLUSION

In this exercise, you observed that a single-phase thyristor bridge can operate both as a controlled rectifier and as inverter. You saw that a bridge made with two thyristors and two diodes has same characteristics as a four-thyristor bridge with a free-wheeling diode, but is more economical to build.

REVIEW QUESTIONS

In what direction is active power transferred by a bridge operating in the rectifier mode? . . . . . . . . . .
Under what conditions can a thyristor bridge operate in the inverter mode? . . . . . . . .
In what direction is active power tranferred by a bridge operating in the inverter mode? . . . . . . . . . . . .
Which bridge configuration using two thyristors and two diodes requires the addition of a free-wheeling diode? Explain why. . . . . . . . . . . . . . . .
What is the role of the inductor in the bridge rectifier/inverter circuit? . . . . . . . . . . . . . . . .

CAUTION

High voltages are present in the laboratory exercise! Do not make or modify any banana jack connection with the power on unless otherwise specified!

previous Rectifier and inverter modes

Rectifier and inverter modes

The function of the rectifying circuits seen so far is to convert ac current to dc current. The process is shown in Figure 6. Since the ideal rectifies does not consume power, the active power supplied by the source must be absorbed by the load. Note that the term load is used in its most general sense. A load can even be a voltage source, as in the case of a battery charger circuit.

An inverter is a circuit which performs the opposite function of a rectifier; it converts dc current to ac current. Figure 7 shows a general inverter system.

The active power supplied by the source is consumed by the load, as the ideal inverter does not consume power.
Inverters fall into two categories:

the self-commutated or autonomous inverter in which the frequency of the output ac current is proportional to the triggering rate of the thyristors;
the line-commutated or non-autonomous inverter in which the frequency of the output ac current is imposed by the ac network to which the inverter is connected.

Figure 6: A general rectifier system.

Figure 7: A general inverter system.

The frequency of self-commutated inverters is fixed. You will use the line-commutated inverter in this exercise.

The single-phase thyristor bridge can be used both as a rectifier and as an inverter. The inductor is large enough to ensure continuous conduction over a wide range of firing angles. The power supplied to the battery is P₀ = I × E₀, where E₀ is the average dc voltage at the output of the bridge.

Rectifier: Single-phase bridge with active dc load

Figure 8: Single-phase bridge with active dc load.

The average dc voltage E₀decreases as the firing angle α is increased, as shown in Figure 9 and 10. However, current flows only as long as voltage E₀is more positive than the battery voltage E_B. If α is large, say 75°, voltage E₀is very small and current will only flow if voltage E_B is near zero.

Rectifier: Waveform for single-phase bridge in rectifier mode (α = 0°)

Figure 9: Waveform for single-phase bridge in rectifier mode (α = 0°).

Rectifier: Waveforms for single-phase bridge in rectifier mode (α = 30°

Figure 10: Waveforms for single-phase bridge in rectifier mode (α = 30°)

If the firing angle α is greater than 90°, E_D becomes negative. In this case, current will only flow if the polarity of the battery voltage is reversed, so that E_D is still higher than E_B. This is illustrated in Figure 11. The power supplied to the battery is still P₀ = I × E₀. However, since the voltage E₀ is negative, this power is negative, showing that power is actually transferred from the the battery to the ac source. The bridge is now operating as a line-commutated inverter, converting dc power to ac.

Rectifier: Waveform for single-phase bridge in inverter mode (α = 150°

Figure 11: Waveform for single-phase bridge in inverter mode (α = 150°)

Procedure summary

In the first part of the exercise, you will set up the equipment.

In the second part, you will observe the operation f a controlled bridge supplying a passive load, with both a resistive and an inductive load.

In the third part, you will set up two different bridge configurations having two thyristors and two diodes.

In the fourth part, you will operate a thyristor bridge in both the rectifier and the inverter modes. The reversible dc power supply circuit (see Familiarization with the Reversible DC Power Supply) is used to simulate a battery.

previous Bridge rectifier with two thyristors and two diodes

next Thyristor Single-Phase Bridge Rectifier/Inverter (PROCEDURE)

Bridge rectifier with two thyristors and two diodes

A single-phase bridge can be made using two thyristors and two diodes. Figure 4 shows an example of such a circuit.

This circuit offers the same control as four-thyristor bridge. In addition, the output voltage never becomes negative, even when the load is inductive. This is because the two diodes provide a free-wheeling path: they conduct while the energy stored in the inductor is released. This circuit operates as the circuit of Figure 3, yet has one less component and requires only two thyristors.

Figure 5 shows another bridge with two thyristors and a free-wheeling diode. This circuit could operate without the free-wheeling diode. In this case, the free-wheeling path would change every half-cycle of the ac source voltage. Q₁ and D₁ would free-wheeling during part of one half-cycle, and Q₂ and D₂during part of the second half-cycle.

Rectifier: Bridge rectifier with two thyristors on common ac line

Figure 4: Bridge rectifier with two thyristors on common ac line.

The free-wheeling D₃ is necessary, however, to ensure that the circuit be able to turn off an inductive load. Without this diode, when the gate pulses are stopped, the current may never drop to zero and one thyristor may continue to conduct. The free-wheeling diode also relieves the thyristors from free-wheeling duty, allowing the use of lower power thyristors.

Rectifier: Bridge rectifier using two common-cathode thyristors

Figure 5: Bridge rectifier using two common-cathode thyristors.

One advantage of the circuit in Figure 5 is that the firing control circuit can be simplified since the cathodes of the two thyristors are at a common potential. Both the circuit of Figure 4 and that of Figure 5 are of lower cost than a four-thyristor bridge of comparable power, nad both allow control of power from 0 to 100%. Unlike the four-thyristor bridge, however, these bridge cannot be used in the inverter mode, which is explained in the next section.

previous Thyristor Single-Phase Bridge Rectifier/Inverter

next Rectifier and inverter modes

Thyristor Single-Phase Bridge Rectifier/Inverter

Objectives

To demonstrate the operation of a thyristor single-phase bridge in both rectifier and inverter modes.
To demonstrate the operation of bridge formed with thyristors and two diodes.

Discussion

Thyristor single-phase bridge (Figure 1) operates on the same principle as the diode single-phase bridge rectifier, except that each thyristor begins to conduct only when a current pulse is injected into the gate (providing that the thyristor is forward biased). Once a thyristor begins to conduct, it continues to conduct until the current flowing through it becomes zero. With a resistive load, the current becomes zero the instant the ac source voltage E_spasses through zero volts. Therefore, the output is a full-wave rectified voltage which is always positive (Figure 2(b)).

Rectifier: Thyristor single-phase bridge

Figure 1: Thyristor single-phase bridge.

Since conduction can be initiated at any angle in the waveform between 0° and almost 180°, the average output voltage E₀, and therefore the average current, can be varied between 0 and 100%.

Rectifier: Output waveforms for a thyristor bridge

Figure 2: Output waveforms for a thyristor bridge.

The following equation gives the value of the average output voltage E₀as a function of the firing angle. this equation is only when conduction is continuous, that is, when the on-time of the thyristors corresponds to 180°.

E₀ = 0.9 E_s cos α
where E_sis the voltage of the source [V ac]
α is the firing angle in degrees.

When the load is inductive, the output voltage can be negative for part of the cycle, as shown in figure 2(c). This is because an inductor stores energy in its magnetic field which is later released. Current continuus to flow, and the same thyristors continue to conduct, until all the stored energy is released. Since this occurs some time after the ac source voltage passes through zero, the output voltage becomes negative for part of cycle.

The negative part of the output voltage waveform reduces the average output voltage E₀. As seen the previous exercise, a free-wheeling diode can be placed in the circuit to prevent the output voltage from going negative (Figure: 3). When the output voltage begins to go negative, the free-wheeling diode conducts. This maintains the output voltage at approximately zero while the energy stored in the inductor is released. The output voltage waveform is the same as for a purely resistive load (Figure: 2(b)), and the average output voltage is therefore greater than it would be without the free-wheeling diode. The addition of a free-wheeling diode makes the output current waveform smoother.

Rectifier: Thyristor single-phase bridge with free-wheeling diode

Figure 3: Thyristor single-phase bridge with free-wheeling diode.

next Bridge rectifier with two thyristors and two diodes

Sinusoidal PWM 3-Phase Inverter

Dc-to-ac converters are known as inverters. The function of an inverter is to change the dc input voltage to an ac output voltage of desired magnitude and frequency. The output voltage waveforms of ideal inverters should be sinusoidal. However, the output of practical inverters contains harmonics. For high power applications, low distorted sinusoidal waveforms are required. Harmonic contents could be minimized by the use of high-speed. semiconductor switching techniques

Inverters are widely used in industrial applications-

motor drives, UPS, induction heating, standby power supplies, etc.

input may be a battery, fuel cell, solar cell, or there dc source

Dc-to-ac inverters can make smooth transition into the rectification mode, where the flow of power reverses from the ac side to the dc side. Two types of inverters: single-phase inverters and three-phase inverters.

$n = j{m_f} \pm k$

$j = 1,3,5,...{\rm{ }}for{\rm{ }}k = 2,4,6,...$

$j = 2,4,6,...{\rm{ }}for{\rm{ }}k = 1,5,7,...$

${\hat v_{ab1}} = M\sqrt 3 \frac{{{V_S}}}{2}{\rm{ }}for{\rm{ 0}} \le {\rm{m}} \le {\rm{1}}$

power-electronic-converter-DC-AC CONVERTERS

previous Three-Phase Inverter

Three-Phase Inverter

Used to supply three-phase loads
Three single-phase inverters could be used, however, 12 switches are necessary, as a result, less efficient
Consists of three legs, one for each phase
One of the two switches in a leg is always ON at any instant
Output of each leg depends on Vs and the switching status

Three-Phase PWM Inverter

${v_{ab}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t + \frac{\pi }{6})}$

${v_{bc}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t - \frac{\pi }{2})}$

${v_{ca}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t - \frac{{7\pi }}{6})}$

$\begin{array}{l} {V_{Ln(rms)}} = \frac{{4{V_S}}}{{\sqrt 2 n\pi }}\sin (\frac{{n\pi }}{3})\\ {V_{L1(rms)}} = \frac{{4{V_S}}}{{\sqrt 2 \pi }}\sin (\frac{\pi }{3}) = 0.7797{V_S} \end{array}$

Delta and Y-Connected

Equivalent Circuits

Equivalent Circuits for Y-Load

${v_{ab}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{\sqrt 3 n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t)}$

${v_{bc}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{\sqrt 3 n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t - \frac{{2\pi }}{3})}$

${v_{ca}}(t) = \sum\limits_{n = 1,3,5,..}^\infty {\frac{{4{V_S}}}{{\sqrt 3 n\pi }}\sin (\frac{{n\pi }}{3})\sin n(\omega t - \frac{{4\pi }}{3})}$

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next Sinusoidal PWM 3-Phase Inverter

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Latest Post

Classification of Converters

Thyristor Three-Phase, Six-Pulse Converter (PROCEDURE)

Thyristor Three-Phase, Six-Pulse Converter

Thyristor Single-Phase Bridge Rectifier/Inverter (PROCEDURE)

Rectifier and inverter modes

Bridge rectifier with two thyristors and two diodes

Thyristor Single-Phase Bridge Rectifier/Inverter

Sinusoidal PWM 3-Phase Inverter

Three-Phase Inverter

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CONFIGURATION	OUTPUT VOLTAGE E₁ dc	OUTPUT CURRENT I₁ dc	OUTPUT POWER P₀ = E₁ × I₁
	V	A	W
Two thyristors on common ac line
Common-cathode thyristors

LINE VOLTAGE	ACTIVE LOAD VOLTAGE E₁ dc
V ac	V
120	80
220	160
240	160

FIRING ANGLE	THEORETICAL VOLTAGE E₀ = 0.9 E_scos	MEASURED VOLTAGE E₁ dc	POWER P = E₁ × I₁
degrees	V	V	W
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