Abstract- The following practice will design the control circuit of a single phase full wave rectifier bridge type semi. Which involves taking a sample of the signal is delivered to the rectifier and give proper treatment for comparison with continuous voltage control and thereby produce the firing pulses needed to activate the semiconductor bridge rectifier semi correctly, in this case , SCRs.

Compare key-words cosine Crossing, Shunt, Optocoupler..

I. INTRODUCTION

There are many applications where control of the load voltage is necessary, using thyristors instead of diodes in this circuit it is possible, as will be seen, to regulate the average value of the output voltage by varying the semi-phase difference between the zero crossing of the mains voltage and the firing of the thyristors (angle shot, symbolized by α).

There is, therefore, a control circuit, which, acting on the moment of firing of the thyristors regulates the conversion. In this practice we analyze the single-phase circuit and the control circuit also be for cosine crossing, which is important to consider some aspects to a good response in the load. Let's look at the step by step development of the practice.

II.OBJETIVOS

- Design and build a circuit for controlling the conduction angle of the SCR, a single-phase full-wave rectifier bridge, the ramp-step (linear or cosine) method.

- The load voltage must be controlled with signal Vc varies between 0 and 10 V.

- A resistive-inductive load used. To reduce the risk of electric shock, a step-down transformer is used 120/25V, 60 Hz for power circuit

III. ANÁLISIS PRELIMINAR

1. Principle of operation of single phase controlled rectifier bridge.

In this assembly, the LEDs were uncontrolled rectifier bridge

Report submitted on December 9, 2010.Universidad Francisco de Paula Santander, Cúcuta, Colombia.

are replaced by thyristors SCR, enabling the control phase of a full-wave input signal. The circuit can be seen from Figure 1.

Fig. 1 Controlled rectifier single semi- phase bridge type

Fig. 2. Waveforms fully controlled bridge rectifier with resistive load

The supply current is alternating rectangular, but is zero for 0 <wt <α, and Q2 and D2 conducting and due to the absence of commutation inductance. That means it will trigger the thyristors two at a delayed phase angle α from zero crossing of

Practice 7: Single Phase Controlled Rectifier Full Wave. Control circuit cosine crossing

Vanessa Arenas Gonzalez cod: 1160453, Harold Camacho Guerrero cod: 1160488, Cindy Quiñonez Morales cod: 1160247

1

the input voltage. Figure 2 shows the waveforms of the input current and the output voltage of the rectifier.

The average component of this waveform is determined from:

¿V o≥1π ∫

α

π+α

V m sen ( wt ) d ( wt )=2V m

π(1+cos α )(1)

Therefore, the average output current is

I o=V o

R=

2V m

πR¿

The power delivered to the load is a function of the input voltage, firing angle and load components. To calculate the power in a resistive load will use P=I rms

2 R, where

I rms=√ 1π ∫

α

π+α

(Io )2 d ( wt )

I rms≈ Io(3)

The effective current generator is equal to the effective load current.

With Rl and a discontinuous load current is required to make a different analysis.

For wt = 0 and zero load current, the SCR Q1 and bridge rectifier D2 are directly polarized and Q2 and D1 be polarized in reverse when the generator voltage is made positive. Q2 and D2 were activated when they are applied gate signals for wt = α. When Q1 and D2 are turned on, the charging voltage is equal to the voltage of the generator. For this condition is identical to the circuit controlled half-wave rectifier and the function of the current will be.

io (wt )=V m

Z[sen ( wt−θ )−sen(α−θ)e−(wt−α)/ wτ ](4)

For α ≤ wt ≤ β

Where

Z=√R2+ (wL )2

θ=tan−1( wLR )

τ= LR

The function above is zero current in wt = β. If β <π + α, the

current will be zero until wt = π + α, at which gate signals at T2 and T3, which will be directly polarized and start driving apply. Figure 4 illustrates this mode of operation, called discontinuous stream:

β=α+π→ discontinuous current

Analysis of the full-wave controlled rectifier in the currentdiscontinuous mode is identical to the half-wave controlled rectifier, but the period of the output current is π radians rather than 2π radians.

(5)The power factor is greater than in the bridge type for a highly

inductive load.

2. Principle of operation of the control circuit, the SCR firing:

a. Control circuit for cosine crossing

This circuit is called "cross firing circuit cosine" and its block diagram can be represented by Figure 6

Fig. 3. discontinuous current

Fig. 4. waveform is

2

The operating principle is to monitor the input signal through a step-down transformer, to obtain a sample of the appropriate phase.

This signal shows the input sinusoid:

V i (wt )=V m sin ( wt )(6)

Is derived for a cosine function. We now have, at the output of phase shifter:

V i' ( wt )=V p cos ( wt )(7)

In where Vm is the magnitude of the input signal and Vp is the magnitude of the output signal of the step-down transformer.

If the signal of equation (12) is reversed, then both the "positive" signal and "negative" are conditioned to have a direct offset voltage of the same magnitude of the output signal of the transformer, so the resulting output is above the ground level, the signals are formed:

V 1 (wt )=V pcos (wt )+V p(8)y

V 2 (wt )=−V p cos (wt )+V p(9)

And if the signal Vc control is made to vary only in the range defined by 0 <Vc <Vp, so as to ensure that by a circuit suitable comparison there is always an intersection of these two signals and Vc, then it can define that intersection through:

V p cos (wt )+V p=V c para la ecuación13

and

−V p cos (wt )+V p=V c para la ecuación14

So that if we take as a basis only equation (13), it follows that if wt=,, which is the angle of activation of the optocouplers, then we have the relationship of this with Vc is given by:

α=arccos(Vcd−V m

Vm−1)(10)

If defined optocouplers turn triggering the SCR's, and remember that the equation defining the average value of the output signal of the converter is:

V cd=2V m

π(1+cosα )(11)

Thus we can conclude that a circuit capable of obtaining the required signals and detect the crossing of the cosine has reference to the phase of the input signal, provides a completely linear relationship of the output voltage of a controlled full wave rectifier and voltage control signal.

IV. PLANNING

1. From the proposed block diagram attached, was asked to design a control circuit, using the straight-line method or cosine, using discrete analog components.

V (t): cosine signal source or reduced value synchronized ramp

with the AC power source.

Vc: control DC signal variable between 0 and 10 V, for the driving theoretical angle varies between 0 and 180 degrees.

DESIGN

Then we will see step by step the circuit design.

a) Power.

For a transformer circuit used 120/25/12 5 V.Operational amplifiers (TL 084) are to be fed with ± 12 volt requiring that a voltage divider is made to ensure maximum excursion transformer further comparison voltage will vary between 0 and 10 v.

b) Circuit for 90 ° phase shift

It requires that the cosine signal is therefore the phase shift of 90 ° to the input through an RC filter in follower mode is performed. The following calculations were made:

Vo (wt )=Vi (wt−ϕ )

Fig. 5. Block diagram of the circuit offset control

3

ϕ=2 tan−1 (2 πfoR3C 1 )

The value of C1 = 0.1 uF is assumed. Knowing that fo = 60 Hz and φ = 90 °, then:

R 3=tan(ϕ

2 )2 πfoC 1

=26.53 K Ω≅ 27KΩ

c) Adequacy of signal

It should have a signal excursion of 10 V, and Vc varies between 0 and 10 V, therefore the wavelength is halved and subtracted Vcc / 2 = 5 V, which is made with a voltage divider

As should have two pulses, one 180 ° out of phase from the other, you must generate a positive signal and a negative, that is to say, 180 ° out of phase as well. Therefore requires an investor to the gap.

.

d) Reference voltage

This continuous signal must vary between 0 and 10 V, so a 100KΩ potentiometer used.Comparador

Two Comparators circuits, one for firing at wt = α by comparing the output of the inverting adder with the reference voltage and the other for firing at wt = π + α reference are used. The output voltage is equal to Vcc during the time that V + is higher than V-, so that a step signal is generated.

e) comparison

Two Comparators circuits, one for firing at wt = α by comparing the output of the inverting adder with the reference voltage and the other for firing at wt = π + α reference are used. The output voltage is equal to Vcc during the time that V + is higher than V-, so that a step signal is generated.

f) Shunt

We must consider the design of this circuit for a good pulse to be delivered to the SCR.

The capacitor leads until fully loaded, which will open and will not allow the passage of more current to the resistance. It must:

τ=RC

------ tg = 10 μS

V c=Vin (1−e−tτ )si Vc=Vin

1=1−e−tτ but e− x=1−x

1= tτ

thenτ=t=10 μS

RC=10 μS si C=0.1 μS

R = 100Ω

g) Control Circuit

To insulate the control circuit used to power optocoupler 4N25. Diodes placed between gate and cathode of the SCR to protect it. Also insert a freewheeling diode in anti-parallel with the load to reduce the negative peak and guarantee download coil.

2. Draw a block diagram that includes all stages of the control circuit and power circuit.

Fig. 5. Block diagram of the circuit cross firing cosine

3. Draw the circuit diagram of the power circuit using the transformer 120/12.5/12, 5V, 60hz.

Annex 1.

4. SPICE simulation, the operation of the power circuit for α = 30, 60, 90 and 120 degrees.

Then voltage graphs for different angles without placing a load R freewheeling diode.

a) α = 90

16 Vpico

4

b) α = 60

16 Vpico

c) α = 120

16 Vpico

V. EVALUATION

1. Draw the following graphs:

a. Effective load voltage vs voltage control

b. Angle Shooting vs Control Voltage

c. load voltage for various angles.

a) α = 7

5 Volt/Div

b) α = 30

5 Volt/Div

4 5 6 7 8 9 10 11 127

8

9

10

11

12

13

Voltaje de control (V)

Volta

je e

ficaz

en

la c

arga

(V)

4 5 6 7 8 9 10 11 1215

30

45

60

75

90

105

120

Voltaje de control (V)

Ángu

lo d

e di

spar

o (V

)

5

c) α = 90

5 volt/div

d) α = 120

5 volt/div

Compare the waveform on the oscilloscope obtained with the SPICE circuit simulation.

The waveforms are similar to the simulated. When we apply the angle of 120 ° is significantly negative peak observed in comparison with the simulation.

It was also observed that the need to replace the desired angle in the actual strength is very close to the simulation.

6 Compare Waveforms of the voltages at the output of each of the blocks of the control circuit.En el Anexo 1 encontramos referenciado cada bloque.

To the power circuit come two blocks, one for shooting wt = α, and the other for firing in wt = π + α reference. Let's see step by step wave therapy to get to the power circuit.

a) Input Signal [1]

5 volt/div

b) Signal desfazada (cosine) [2]

5 Volt/div

c) Signal offset voltage [4]

5 volt/div

6

d) Comparator output signal [6]

5 volt/div

e) Pulses [6]

5 volt/div

f) load signal [10]

5 volt/div

VI. CONCLUSIONES

We look through the graph of voltage vs control. Angle shot high linearity in the output response, indicating that the control of single-phase full-wave rectifier for junction cosine behaves with a linear transfer function and the system's response to an increase in the control variable does not depend on firing angle α, which is a desirable feature.

Findings indicate that antiparallel diode load R produces a significant decrease in the negative voltage at the load after zero crossing, and serves to the process of demagnetization coil. Also important is the protective diode connected to the gate of the SCR to protect it.

The moment you connect the firing pulses to the power circuit, you should take special care in assigning that pulse is connected to SCR and diode pair, taking into account the polarity of the AC signal, as this must be positive when the pulses thus applied the voltage anode - cathode of the SCR is positive. Otherwise control will fail.

It is necessary to invert the control voltage since the output signals of adders are displaced negatively, ie a negative offset voltage, to correctly carry out the comparison and properly generate the firing pulses.

VII. REFERENCES

[1] UNIT II. Power Electronics Slides IV. Germán Gallego.

[2] H. MUHAMMAD RASHID. Power Electronics. Edition. Mexico DF Publisher Prentice Hall, 1993. PAG. 118-124

[3] Datasheet C106M Available at: http://www.datasheetcatalog.org/datas heet2/5/0qswy2e8c8us8z1d7h14c pa7psyy.pdf

[4] Converters AC / DC - Rectifiers. Barcelona (2010, Nov 25). [Online]. Available at: http://tec.upc.es/el/TEMA-3% 20EP% 20 (v1) pdf.

[5] Report of Practice and simulations. (2010, November 27). [Online]. Avail-able at: http://www.monografias.com/trabajos12/repract/repract.shtml

7

Anexo 1. Diagram in orcad

8

9

10