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Resistive Load Current and Voltage Curve.

u d = u 2 u d = iR = 2U 2 sin t


i 2 U 2 sin t

R

- The current will be sinusoidal and in phase with u.

+ In the range from 2 voltage u 2 (-), the voltage of point A (-) compared to point B, diode D is placed in reverse voltage and blocks the current from passing through so i = 0

u d = 0.




1

0



2U 2


U d

=

2U 2 sin td t

=

= 0.45U 2

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- The maximum reverse voltage applied to the diode is:

U 2max = 2U 2

- The average value of rectified voltage in 1 cycle is:

- The average value of the current through the load is:




U d


2U 2


0.45U 2

I d

=

R

=

R

=


R


- The effective value of the secondary current of the transformer is:






U 2

I

=

I 2

=


2R


- Voltage current curve for load R.

Figure 3.2. Current and voltage curve of purely resistive load.

* When the load is R - L.

Because of the inductor, when the current changes, a self-induced electromotive force e = -L di/dt appears in the inductor. According to Kiecchop's second law for the loop, we have:

u 2 + e = Ri

When u d increases, Ri increases slower than u d by an amount e (the crossed-out part) because now di/dt

> 0 so e < 0, so the inductor stores energy



Figure 3.3. Principle of 1-phase half-cycle rectifier circuit RL .

At point a, the current reaches its maximum and then gradually decreases, du/dt < 0, the self-induced electromotive force changes direction. At point b: e = - u 2 , diode D is locked so i = 0. Thus, the current i will be maintained in the range from 2 even though u 2 has changed direction.


Figure 3.4. Current and voltage load curve RL.

2. SINGLE-PHASE TWO HALF-CYCLE RECTIFIER CIRCUIT:

2.1. Draw a schematic diagram of a single-phase, two-half-cycle rectifier circuit:

- Secondary transformer diagram with center tap.

Figure 3.5. 1-phase 2-half-cycle rectifier circuit with center-tapped secondary transformer .

2.2 . Working principle, draw the voltage-current curve for load R, R – L. Secondary voltage:

u 21 = 2U 2 sin t

u 22 = 2U 2 sin t

We have u 21 and u 22 are out of phase with each other. When point A has a (+) voltage, u 21 applies a forward voltage to diode D 1 , then point B has a (-) voltage, u 22 applies a reverse voltage to D 2 . We have

+ In the range (0 ): u 21 (+) D 1 opens u 22 (-) D 2 locks.

The current through D 1 is:

21 D1 d

i= i= i=

u 21

=

2U 2

sin t

R

R

The reverse voltage applied to D 2 is: because D 1 is open, consider the voltage drop on the forward resistance of D 1 to be 0, then the voltage at point A will be applied to the good karma of D 2, so the reverse voltage applied to D 2 is:

u D2 = u 22 – u 21 = -2 2U 2 sin t

The maximum reverse voltage applied to D 2 is: U nm = -2 2U 2

+ In the range from (0 2 ): u 21 (-) and apply reverse voltage to D 1 , u 22 (+) and apply forward voltage to D 2 , D 2 opens and D 1 locks.

22

+ Average value of rectified voltage:


12

1

U

d 2

u dd t

0 0

2 U 2 sin td t U 2


+ Average load current value:



d

I=

U d

=

2 2

U 2

R

R

+ Average value of current through diode:

2

1 2 I

ID ​sin td t

d


Comment:

2 0 R 2

- The average value of rectified voltage and load current is twice as large as in half-wave rectification diagram.

- The maximum source voltage applied to the diode when it is locked is also twice as large as in ½-cycle rectification.

- Bridge diagram:

Figure 3.6. Diagram of a 1-phase, 2-half-cycle rectifier bridge.

Whole diagram operation:

+ In the range from (0 ): u 2 > 0 and has polarity (+) at A, (-) at B, D 1 and D 3 open for current to flow along the path A D 1 R D 3 B, D 2 and D 4 are locked

+ in the range (0 2 ): u 2 < 0 and has polarity (+) at B, (-) at A and D 2, D 4 opens for current to flow along path B D 2 R D 4 A, D 1 and D 3 are locked

The average values ​​of voltage and current across the load are U d and I d as in the case of a centre-tapped secondary transformer.

* Steps and how to do the job :

1. EQUIPMENT, TOOLS, MATERIALS:

(Calculated for a practice session of 20 students)


TT

Equipment Type

Quantity

1

Soldering iron.

01

2

Universal board.

01

3

Clamp.

01

4

Bending pliers.

01

5

Drag

01

6

Damaged material container

01

Multimeter.

01

8

Oscilloscope.

01

9

Tin, rosin, wire.


10

- Accessory:

Semiconductor diode 1A 4

Resistor 30 - 30W1

Single phase transformer S = 15VA 1

small capacity U 2 = 24V

Printed circuit 1

- Oscilloscope 1



-



-


7
















2. IMPLEMENTATION PROCESS:

2.1. General process:

+ How to check: use a multimeter to measure:

- Step 1: Plug the red measuring stick into the (-) socket of the meter (positive battery), plug the red measuring stick into the (+) socket of the meter (negative battery).

- Step 2: Turn the knob to the meter at the x10 (x1) resistance scale, short the two ends of the measuring rod, turn the potentiometer so that the indicator needle is at the 0Ω position.

- Step 3: Place the two ends of the measuring stick on the two diode poles as shown (Figure 1.9a) to read the value R 1

2.2. Specific process:

Step 1: Draw the circuit board and component layout diagram.

- Draw the schematic diagram as a pure resistive load circuit.

- Assembly diagram.

Step 2: Mount components and weld

- Mount components in the correct position and polarity. Select components according to load requirements.

- The weld must be standard and beautiful as required.

+ The weld must be neat, round and have a peak.

+ The connecting wire must be tinned.

Step 3: Measure and draw the voltage waveform on the load using a measuring instrument.

- Use a meter to measure output voltage.

- The output voltage graph will be in the form of a half cycle.

- Use an oscilloscope to measure the output voltage on the load.


3. CHECK:

* Student evaluation form:

TT

Job Description

Score

Point

Evaluate

Note

1

Make a work plan

0.5



2

Recognize symbols and actual shapes of

equipment needed for survey

1



3

Analysis of operating principles

1.5



4

Install and survey according to diagram

4



5

Draw the state diagram of the operation

2



6

Give practical application circuit

1



Total score

10



Classification





LESSON 4: THREE-PHASE UNCONTROLLED POWER RECTIFIER

Article code: MD 23 - 04

Introduce:

Three-phase power rectifiers are used as power sources for controllers in particular and electrical equipment in general where there is a three-phase power source. Understanding the working principle of the power source and assembling rectifier power sources is a necessary task for every electrical engineering student.

Target:

- Understand the schematic diagram of three-phase rectifiers in radial and spherical shapes

- Present the working principle, draw the rectifier output voltage and current graph.

- Demonstrate how to install components according to the schematic diagram

- Identify the type of components in the diagram

- Know how to check components

- Install circuit according to correct process, ensuring technical and aesthetic requirements, on time

- Use measuring tools and equipment properly

- Careful, accurate, strictly follow the process

- Ensure safety.

Main content:

1. THREE-PHASE RECTIFIER CIRCUIT RAY DIAGRAM:

1.1. Draw a schematic diagram of a three-phase radial rectifier circuit:



Figure 4.1. Schematic diagram of a three-phase radial rectifier circuit

The circuit includes: 3-phase isolation transformer, three rectifier diodes with pure resistive load.

1.2. Working principle, draw the voltage-current curve for load R:

The secondary voltage of the transformer is:

u 2a = 2U 2 sin t

u 2b = 2U 2 sin t ( t - 2 /3) u 2c = 2U 2 sin t ( t - 4 /3)

The three diodes D 1, D 2, D 3 have a common cathode, so only the diode with the anode connected to the largest (+) voltage will open.

We have a time graph as shown in figure 4.2.

Considering the time corresponding to 1, we have u A > u B > u C, so D 1 opens to let current flow through. Because D 1 is open, the voltage at point M is: u M = u 2a , so D 2 and D 3 are closed because the cathode voltage is greater than the anode voltage.

Thus from /6 < < 5 /6 D 1 is open, D 2 and D 3 are closed.



Figure 4.2. Voltage-time graph on load

From 5 /6 < < 9 /6 D 2 is open, D 1 and D 3 are locked. From 9 / 6 < < 13 /6 D 3 is open, D 2 and D 1 are locked.

Thus each diode is open for about 1/3 of a cycle.


1 2

5

2 U 2 sin td t 1.17 U

36

3 6 U

U d 2

u d d t 2

0

6

2

2 2

- Average value of voltage across load:

- Maximum reverse voltage applied to the diode:

In the case D 1 is open, the reverse voltage applied to D 2 is

u n = u 2a – u 2b = u ab so the maximum reverse voltage applied to D 2 is:

U nm = 2 3U 2 = 6U 2 = 2.45 U 2

- The average value of the current on the load is:

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