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 (Vac)	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
Vac	Vac
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
Vac	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 I₁ shown in Table 3.

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

Table 3: Current I₁ delivered 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 I₁ Shown 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 Es 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 E₁ versus the firing angle. Then, in the same figure, plot the theoretical relationship E₀ = 1.35 E_s cos , where E_s is the line voltage. Compare the two curves.

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

1. What are the advantages of the three-phase, six pulse converter over other rectifier/inverter circuits? . . . . . . . . . . . . . . . . . . .
   
2. Why is it advantageous to have a high ripple frequency at the output of a rectifier? . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
   
3. What is the average output voltage E₀ for a three-phase, six pulse converter if the line-to-line source voltage is 240 V and the firing angle is 75°? . . . . . . . . . . . . . . . . . . . . . . . . . . . .
   
4. What is the average current in each of the three ac line in a three-phase, six-pulse converter? . . . . . . . . . . . . . . . . . . . . .
   
5. 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

1. To study the operation of the three-phase, six-pulse rectifier/inverter.
2. 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:

1. the average value of E_D is twice as great
2. the transfer of active power is two times greater with the same value of current
3. the ripple frequency of E₀ is 360 Hz instead of 180 Hz
4. 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 (3Adc max)
220	1.5	1.5	5	300	600	R=220Ω	R=220Ω, L=0.8H (1.5Adc max)
240	1.5	1.5	5	300	600	R=240Ω	R=240Ω, L=0.8H (1.5Adc 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

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 (3Adc max)
220	1.5	1.5	5	300	600	R=220Ω, L=0.8H (1.5Adc max)
240	1.5	1.5	5	300	600	R=240Ω, L=0.8H (1.5Adc 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 I₁ shown 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 Escos	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

1. In what direction is active power transferred by a bridge operating in the rectifier mode? . . . . . . . . . .
2. Under what conditions can a thyristor bridge operate in the inverter mode? . . . . . . . .
3. In what direction is active power tranferred by a bridge operating in the inverter mode? . . . . . . . . . . . .
4. Which bridge configuration using two thyristors and two diodes requires the addition of a free-wheeling diode? Explain why.    . . . . . . . . . . . . . . .
5. 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:

1. the self-commutated or autonomous inverter in which the frequency of the output ac current is proportional to the triggering rate of the thyristors;
2. 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.

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

The average dc voltage E₀ decreases as the firing angle α is increased, as shown in Figure 9 and 10. However, current flows only as long as voltage E₀ is more positive than the battery voltage E_B. If α is large, say 75°, voltage E₀ is very small and current will only flow if voltage E_B is near zero.

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

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. 

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 D₂ during part of the second half-cycle.

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.

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

1. To demonstrate the operation of a thyristor single-phase bridge in both rectifier and inverter modes.
2. 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_s passes through zero volts. Therefore, the output is a full-wave rectified voltage which is always positive (Figure 2(b)).

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%.

Figure 2: Output waveforms for a thyristor bridge.

The following equation gives the value of the average output voltage E₀ as 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_s is 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.

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

next Bridge rectifier with two thyristors and two diodes

Familiarization with the Reversible DC Power Supply

DISCUSSION

The reversible dc power supply

A reversible dc power supply is a device whose voltage and current are reversible. This means that the polarity of the voltage across the dc power supply can be either positive or negative, and that the current can flow through the dc power supply in either direction. F8-1 shows the symbol to represent a reversible dc power supply whose voltage (E) is adjustable. The arrow beside letter E in the symbol points towards the positive terminal of the reversible dc power supply when voltage E is positive.

F8-1: Symbol representing a reversible dc power supply whose voltage is adjustable.

F8-2 shows four simple circuits in which a reversible dc power supply is connected to a fixed-voltage dc power supply.

In F8-2(a), the polarity of voltages E and E₁ is positive, and the voltage E is greater than the voltage E₁. Therefore, the current in the circuit (I) flows in the direction shown in F8-2(a). The polarity of this current is positive because it flows in the direction indicated by the arrow over the letter I in F8-2(a). therefore, the polarity  of the voltage E and current I is positive and power flows from the reversible dc power supply t the fixed-voltage dc power supply.

In F8-2(b), the polarity of the voltages E and E₁ is still positive, but this time the voltage E is lower than the voltage E₁. Therefore, the current I flows in the direction shown in F8-2(b). The polarity of this current is negative because it flows in the direction opposite to the direction indicated by the arrow over the letter I in F8-2(b). Therefore, the polarity of the voltage E is positive, the polarity of the current I is negative, and power flows from the fixed-voltage dc power supply to the reversible dc power supply.

F8-2: Simple circuits using a reversible dc power supply.

In F8-2(c), the voltage E has been adjusted so that its polarity is negative and the connections of the fixed-voltage power supply have been reversed so that the polarity of the voltage E₁ is also negative. Furthermore, the voltage E is lower than the voltage E₁. For example, the voltage E and E₁ could be equal to -30 and -20 V, respectively. Therefore, the current I flows in the direction shown in F8-2(c). The polarity of this current is negative because it flows in the direction opposite to the direction indicated by the arrow over the letter I in F8-2(c). Therefore, the polarity of the voltage E and current I is negative and power flows from the reversible dc power supply to the fixed-voltage dc power supply.

In F8-2(d), the polarity of the voltage E and E₁ is still negative, but this time the voltage E is greater than the voltage E₁. For example, the voltage E and E₁ could be equal to 10 and 20 V, respectively. Therefore, the current I flows in the direction shown in F8-2(d). The polarity of this current is positive because it flows in the direction indicated by the arrow over the letter I in F8-2(d). Therefore, the polarity of the voltage E is negative, the polarity of the current I is positive, and power flows from the fixed-voltage dc power supply to the reversible dc power supply.

F8-3 is a four-quadrant diagram of the current I versus the voltage E which summarizes the operation of the dc power supply. Dot A in this diagram represents the voltage E and current I related to the reversible dc power supply in F8-2(a), dot B represents the voltage E and the current I related to the reversible dc power supply in F8-2(b) and so on. This figure shows that the reversible dc power supply can operate in either one of the four quadrants. For each quadrant, the figure indicates whether the reversible dc power supply sources or sink power.

F8-3: Four-quadrant diagram summarizing the operation of reversible dc power supply.

In brief, current can flow in either direction in the reversible dc power supply regardless the polarity of the voltage across the reversible dc power supply. Furthermore, the reversible dc power supply can either source or sink power.

Implementing a reversible dc power supply

A reversible dc power supply can be implemented using a separately-exited dc motor/generator mechanically coupled to a synchronous or asynchronous motor/generator. F8-4 shows a reversible dc power supply in which the separately-excited dc motor/generator is coupled to a three-phase squirrel-cage induction motor (asynchronous motor).

 F8-4: A reversible dc power supply implemented with a separately-excited dc motor/generator and an asynchronous motor.

The output of the reversible dc power supply is taken across the armature circuit of the separately-excited dc motor generator. The polarity and value of the armature voltage, and therefore, the polarity and value of the voltage E provided by the reversible dc power supply depends on the polarity and value of the current flowing in the exciting coil of the dc motor/generator and the rotation speed of the dc motor/generator. The rotation speed of the dc motor/generator is equal to the rotation speed of the asynchronous motor, which varies little. Therefore, the polarity and value of the voltage E provided by the reversible dc power supply especially depends on the polarity and value of the current flowing in the exciting coil of the dc motor/generator.

The current flowing in the exciting coil of the dc motor /generator depends on the voltage provided by a fixed-voltage dc power supply and tandem rheostats. The tandem rheostats are rheostats which share a single shaft. The tandem rheostats allows the voltage applied to the exciting coil is –E_x. On the other hand, the voltage applied to the exciting coil is E_x when the cursors are set to the lower end of the rheostats.

The direction of the current flowing in the armature circuit, and therefore, the direction of the current I flowing in reversible dc power supply depends on whether the dc motor/generator source or sinks power. When the dc motor/generator sources power, the squirrel-cage induction motor drives the dc motor/generator which operates as a generator. Conversely, when the dc motor/generator sinks power, it operates as a motor and drives the squirrel-cage induction motor which operates as a generator and provides power to the ac power supply.

Familiarization with the IGBT Chopper/Inverter module

DISCUSSION

Description of the module

The IGBT Chopper / Inverter module mainly consists of 7 insulated-gate bipolar transistors (IGBT). An IGBT is a switching transistor that requires voltage on the gate to conduct. The IGBTs of the IGBT Chopper/Inverter module are labeled Q₁ to Q₇. IGBTs Q₁ to Q₆ are grouped in pairs as follows: Q₁ with Q₄, Q₂ with Q₅ and Q₃ with Q₆ and Q₇ is part of the breaking circuit which limits the voltage on the DC bus. The module also contains electronic circuitry that isolates the gate of IGBTs and protects the IGBTs against overheating, overvoltage and over current.

F7-6 shows the front panel of the IGBT Chopper / Inverter Module. It shows many important characteristics about the operation of the module.

The DC voltage is applied through terminals 1 and 2 to supply the DC bus.

The DC bus is linked to IGBT Q₁ and Q₄ to supply loads from terminal 3. The DC bus may also be linked to IGBTs Q₂ and Q₅ through switch S₁ and to IGBTs Q₃ and Q₆ through switch S₂ to supply three- phase loads from terminals 3, 4 and 5.

Capacitor C₁ is used to maintain a smooth DC voltage in spite of the current pulsations produced by the IGBTs. It is connected between terminals 1 and 2.

A power diode is connected between the collector and emitter of each IGBT. The diodes connected in parallel with the IGBTs have a very short recovery time. This important feature allows the IGBT Chopper / Inverter module to be used in high speed power switching circuits.

F7-6: Front panel of the IGBT Chopper / Inverter module.

Protection circuits

The IGBT Chopper / Inverter module contains many electronic circuits that protect the IGBTs against various types of overloads. The operation of each protection circuit is briefly described in the following paragraphs.

The positive and negative branches of the DC bus are individually protected by breakers. When an overload condition occurs on the DC bus, one of the breakers trips. Correct the faulty condition and corresponding pushbutton to reset the breaker.

Over current protection circuit

The over current protection circuit is designed to protect IGBTs Q₁ to Q₆ against instantaneous over current. When an over current condition is detected on either IGBT, the six IGBTs are switched off by setting the gate voltage at 0 V. when this condition occurs, the OVERCURRENT LED turns on. The OVERCURRENT RESET pushbutton must be depressed to reset this protection circuit.

Overvoltage protection circuit

The overvoltage protection circuit is designed to protect IGBTs Q₁ to Q₆. This protection circuit senses the DC bus voltage. When the voltage exceeds a safe value, the six IGBTs are switched off by setting the gate voltage at 0 V and the OVERVOLTAGE LED turns on t indicate the faulty condition. The overvoltage protection circuit is automatically deactivated when the DC bus voltage returns to a safe value.

Overheat protection circuit

IGBTs Q₁ to Q₆ are also protected against overheating. The overheat protection circuit senses the heat sink temperature of the IGBTs. When the temperature exceeds a safe value, the six IGBTs are switched off by setting the gate voltage at 0 V and the OVERHEAT LED turns on to indicate the faulty condition. The overheating protection circuit is automatically deactivated when the temperature returns to safe value.

Breaking circuit

The breaking circuit consists mainly of the diode D₇, resistor R₁ and IGBT Q₇ shown on the module panel. This circuit is designed to dissipate the energy produced by a decelerating motor connected to the IGBT Chopper / Inverter module. When the motor decelerates, it behaves like a generator and some power is returned to the DC bus. When the DC bus voltage exceeds a safe level, the braking circuit transfers some energy from the DC bus to resistor R₁ and the BRAKING LED flashes. Notice that the module continues to operate in this condition. The braking circuit may be disabled by setting the BRAKING switch at 0.

Interconnection with control module

The gate of IGBTs Q₁ to Q₆ is connected to the SWITCHING CONTROL INPUTS connector through a series of internal isolators and amplifiers. This 9-pin connector may be connected to the CONTROL OUTPUTS of the Chopper/Inverter Control Unit module, or to the FIRING CONTROL OUTPUTS of the Thyristro Firing Unit.  

The pin configuration of the SWITCHING CONTROL INPUTS connector is given in F7-6. The inputs (pins 1 to 6) require 0-5 V transition of the control signals. Pin 7 is used to input a synchronization signal coming from the Chopper/Inverter Control Unit or the Thyristor Firing Unit.

The miniature banana jacks identified SWITCHING CONTROL INPUTS 1 to 6 are parallel connected to pins 1 to 6 of the SWITCHING CONTROL INPUTS 9-pin connector, respectively. This allows switching control signals coming from other equipment to be used to control the IGBTs.

The SYNC. OUTPUT miniature banana jack provides the synchronization signal coming from the Chopper/Inverter Control Unit or from the Thyristor Firing Unit. This 0-5 V signal may be used to synchronize an oscilloscope when observing the switching control signals.

The SWITCHING CONTROL INPUTS  DISABLE miniature banana jack allows to switch off IGBTs Q₁ to Q₆ by applying a +5-V voltage to this jack. In this condition, the IGBT gate voltage is 0 V, and the IGBTs cannot be controlled. They can be controlled when the SWITCHING CONTROL INPUTS DISABLE miniature banana jack is simply left open or connected to the common.

A 24-V AC power supply must be connected to either one of the two LOW POWER INPUT jacks to supply the electronic circuits. The POWER ON LED lights when AC voltage is applied.

The chassis terminal (green) on the front panel of the IGBT Chopper / Inverter module is used to prevent harmful electromagnetic emissions from interfering with other components. To do so, the shield of a special connection cable must be connected to this terminal.

Using the IGBT Chopper / Inverter module

The IGBT Chopper / Inverter module is used to build various power electronic circuits such as choppers and inverters. F7-7 and 7-8 show examples of such power electronics circuits.

In the circuit of F7-7, IGBT Q₁ and power diode D₄ are used to build a bulk chopper. IGBT Q₄ (parallel connected to power diode D₄) and power diode D₁ (parallel connected to IGBT Q₁) are not shown in the figure because they are not used in this circuit.

In this circuit, SWITCHING CONTROL INPUT 4 of the IGBT Chopper / Inverter module is connected to common point to prevent IGBT Q₄ from being switched on.

F7-7: Buck chopper built using the IGBT Chopper / Inverter module.

In the circuit of F7-8, the six IGBT Chopper / Inverter module form a three-phase inverter.

In the circuit of F7-7 and 7-8, the Chopper/Inverter Control Unit provides the control signals required to switch the IGBTs on or off.

Notice that in both circuits, a 24-V AC power supply is connected to the LOW POWER INPUT of the IGBT Chopper / Inverter module to supply the internal circuits. Also, notice that the capacitor connected on the DC bus has been intentionally omitted to simplify the diagrams.

The IGBT Chopper / Inverter module can be used with the Thyristor Firing Unit to build the Three-phase inverter shown in F7-8.

F7-8: Three-phase inverter built using the IGBT Chopper / Inverter module.