8a electrical braking rev 2 100418(good one)

7/31/2019 8a Electrical Braking Rev 2 100418(GOOD ONE)

1/17

Chapter 8a: Electrical Braking 0903582: Electrical Drives

Copyright held by the author 2010: Dr. Lutfi R. Al-Sharif Page 1 of 17

Chapter 8aElectrical Braking

(Revision 2.0, 18/4/2010)

1 IntroductionIn any motion control system (especially hoisting applications) braking is necessaryto accurately control the position and speed of the load. Braking can be achieved byelectrical or mechanical means. In traditional vehicle applications, braking isachieved by using a hydraulically operated mechanical brake (e.g., as in a car). Inmost modern motion control systems electrical braking is used on the motor, and themechanical brake is only used as a parking brake.

Most modern applications use these two methods as follows:

1. Electrical braking is used to decelerate and bring the load to a standstill. It

has the advantage that it does not lead to any wear in the system. In certaincases the energy can also be recovered and stored or returned to the mainsupply. It cannot be used as a safety brake as it relies on the presence of apower supply that could be lost under certain conditions.

2. Mechanical braking is used as an emergency brake or as a parking brake.

This Chapter discusses electrical braking, while mechanical braking is discussed in8b.

2 Excessive Energy during BrakingIn any system which can overhaul, excess energy will be generated by the motor.

When a load is overhauling (braked lowering) excess energy will be generated. Themotor will act as a generator (e.g., if it is an induction motor it will be running abovethe synchronous speed).

In the case of variable frequency inverter drives, this excess energy will bereturned as charge on the terminals of the capacitor. If not removed, this chargecauses the voltage on the DC link to rise, and could cause damage to the system.Removing this excess charge protects the system, and also achieves the function ofbraking the motor. Two methods exist for removing the charge: dynamic resistorbraking and regenerative braking. These are discussed in this Chapter as well, astwo of the methods of electrical braking.

3 Methods of Electrical BrakingFive methods of electrical braking are discussed in this Chapter:

1. DC injection braking.2. Plugging.3. Eddy current braking.4. Dynamic resistor braking.5. Regenerative braking.

The first method is mainly used in variable voltage AC drives. The last two methodsare mainly used in variable frequency drives.

3.1 Plugging


2/17



Another method of braking is called plugging, which involves applying the reversephase sequence to the winding, as if trying to reverse the motor rotation. A typicalset-up is shown in Figure 1. This method does not require a pole changing motor (asrequired in the dc injection braking method). However, care has to be taken not toswitch on any of the reverse sequence SCRs until the forward sequence SCR has

ceased to conduct; otherwise a short circuit will result, which would damage theSCRs and trip the electrical protection. For these reasons, plugging systems willinvariably has a zero current detector fitted in the path of the motor current to checkthat the current has dropped to zero before reversing the phase sequence. Thismethod also suffers from the disadvantage that it can inject high values of current inthe rotor, and rotor bars have been known to rupture due to the high currents inducedin the rotor bars.

R S T

1 234 561'4' 2' 5'

Figure 1: The use of 5 pairs of back to back SCRs to drive and brake the motor.

3.2 Eddy Current BrakingAnother method of braking which is not widely used is the so-called Eddy currentbraking method. It is based on the principle of inducing a current in a rotating disk,by which the circulating current will induce a back torque opposing the rotation. Thismethod is used to obtain braking torque from the eddy current brake which isattached on one end of the motor shaft. It has the characteristics of the DC dynamicbrake. The braking torque, which is substantially good enough at longer speedranges, becomes zero when the speed becomes zero. (Fukuda, 1979). Thismethod needs a complicated non-standard motor, cannot produce any torque atstandstill and dissipates all the heat in the machine, rather than returning it to thesupply.


3/17



Figure 2 shows an example of an Eddy current brake attached to a lift motorthat is driven by a variable drive. Figure 3 and Figure 4 show a Siemens Eddycurrent brake used for teaching purposes in a laboratory.

Figure 2: Eddy current brake attached to the motor (Eddy current brake can be seen on theright hand side) [97 Cromwell Road].

Figure 3: Eddy current brake used in the lab. Figure 4: Eddy current brake terminals.

3.3 Resistor Dynamic Braking


4/17



This method is used in variable frequency drives. A resistor is used to dissipate theextra charge, whereby the resistor is called a dynamic braking resistor. This isachieved by placing the resistor in parallel with the DC link capacitor, and beingswitched (chopped) by using a transistor. This is shown in Figure 5. The transistor isswitched on and off, where the ratio of the on time to the off time is adjusted in order

to achieve the necessary value of current to remove excess charge from the DC link.

+

-

Power flow

M (G)

InverterRectifier

DC Link

R

S

T

R

Figure 5: Dynamic (resistor) braking system.

The main disadvantage with this method of braking is that the energy is lost as heatin the resistor, and the need to cool the resistor if the amount of braking is excessive.

However, the set-up is simple and cheap. The resistance of the resistor and itspower have to be calculated to suit the amount of braking expected and the size ofthe motor.


5/17


6/17



Relationship between lift system efficiency and speed (geared)

50%

55%

60%

65%

70%

75%

80%

0 0.5 1 1.5 2 2.5 3 3.5 4

Speed (m/s)

Efficiency

Figure 7: System efficiency against speed.

SolutionFrom Figure 7, at a speed of 2.5m/s, the efficiency of the system will be around 67%.The expected power produced by the lift when braking is:

)1(81.905.0Re kQnratingpowersistor

Where:0.05accounts for the fact that braking only takes place for 5% of the total time;

is the efficiency of the lift system;nis the lift speed in m/s;Qis the capacity in kg;k is the counterweight ratio.

Thus, the power rating for the resistor is:

W

ratingpowersistor

440)45.01(81.9)7513(5.2%6705.0

Re

3.4 Regenerative BrakingThe main disadvantage with the resistor braking method, is that the excess energy islost as heat. It will be more efficient to attempt to return the energy back in thesupply. This is called regenerative braking.

This method of braking is achieved by connecting a second inverter in parallel

with the rectifier in variable frequency drive systems. When it is detected that the DClink voltage has exceeded a specified level, the regenerative inverter will go into


7/17



inverting mode and move the energy back onto the 3 phase main supply. This isshown in the block diagram of Figure 8.

+

-

Power flow

M (G)

InverterRectifier

DC Link

Inverter

Power flow

R

S

T

Figure 8: Regenerative braking system.

The size and complexity of this method of braking compared to the more simpleresistor dynamic braking, necessitates that it is only used when a significant amountof regenerated energy (due to overhauling and deceleration) is expected. Thus, thismethod could be used in counterweighted elevator applications, where the systemregenerates for a significant period of time. However, for a luggage conveyor belt inan airport, or a horizontal people mover, it would not be worth installing.

3.5 DC injection braking

The method used to achieve electrical braking is to inject a DC voltage into the lowspeed winding of the motor, as shown in Figure 9.

This method has mainly developed because many of the variable voltage system,when they first appeared where retrofitted on sites which had two speed AC motorson them. It was easy them to use the low speed windings for braking, by injecting aDC current in the low speed winding.

Injecting DC in a motor winding will tend to try to stall the rotor. This is because themagnetic field set up inside the motor is a stationary constant field which is alwayspointing in one direction. This will have a braking effect on the moving rotor, trying to

bring it to the synchronous speed of DC (i.e., which is 0 rpm). The disadvantageswith this method is that the energy is dissipated as heat in the windings, and that


8/17



braking torque approaches zero at the speed approaches zero, making it difficult tobrake the motor at standstill.

R S T

1 234 56

Figure 9: Diagram of a variable voltage variable speed system using a two speed induction

motor and DC injection braking in the low speed winding.

This above point can be seen in practice. If you carefully watch the motor when iffinally stops on a variable voltage system, you will be able to see that the flywheelmight slightly slip after the motor has come to a standstill and before the mechanicalbrake is applied. Thus, the tendency in these systems is to adjust the mechanicalbrake to apply immediately when the motor is at standstill. This is done by theadjuster by trial and error.

The configuration shown in Figure 9 shows the same three pairs of back toback SCRs for controlling the driving voltage to the high speed winding of the motor,and a half wave controlled rectifier for controlling the DC voltage applied to the lowspeed winding of the motor.

Half Controlled Voltage ControllerThe half controlled voltage controller differs from the full wave voltage controller inthat it has three SCRs and three diodes. Thus, firing pulses are only applied to theSCRs, which will be allowed to stay on by virtue of the diodes complementing them.The firing scheme for such a set-up is shown in Figure 10.


9/17



1

2

3

4

5

6

60 120 180 240 300 0 60 1200

Angle (in degrees)

Thyristor

number

R

T'

S

R'

T

S'

Figure 10: Firing timing diagram for a half controlled voltage controller.

Connection Method for DC Injection Braking

Historically, when variable voltage DC injection braking systems were firstintroduced, the braking connection was always made to any two terminals of the lowspeed winding, and the third terminal was not used. However, it has been foundfrom experience that this is not the best connection to use, as it does not make fulluse of the third winding, and it causes overheating of the low speed windings due toexcessive current and heat concentration in the two coils of the winding. Now, mostsystems coming onto the market are invariably connected using all three windings inseries, in order to utilise full potential of the low speed winding. This is explained inthe following discussion.

A criterion of performance for a braking connection is that is should give the highestmagnetic flux for every ampere of current. Thus, a figure of merit, or a performancefactor would be to divide the resultant flux by the flowing braking current. This can bedone relative terms, and not in absolute numbers, due to the difficulty of obtainingvalues of flux and coil resistance.

In order to be able to evaluate and compare different connections and set-upsof low speed windings for brake purposes, we can assume that the applied brakevoltage is the same for all cases, and that the base case (i.e., the case to which allother set-ups will be compared) is the case in which only two coils are connected inseries. We will assume that the resistance of each coil is R (resistance is onlyconsidered here, and not impedance, because the voltage is mainly DC). A resultant

resistance for any connection can be found by combining series and parallel coils.Then, by dividing the voltage by the combined resistance, a resultant current will be


10/17



found which can be compared to the original resultant current. It is important toremember that the three coils of the low speed winding are displaced in space by120 each. Thus, when combining them, they have to be added vectorially (i.e., andnot added as scalars). Then to calculate the figure of merit, the resultant flux vectormagnitude is divided by the value of resultant current. The connection with the

highest value is the connection providing the highest value of magnetic flux per unitof current. In other words, it is the connection which, for the same value of magneticbraking flux (and thus magnetic braking torque), provides the lowest value of current,and thus the least heat into the low speed windings.

Table 1: Different connection methods for the low speed winding in the DC injection brakingconfiguration.

CONNECTION DIAGRAM VOLTAGE RES. CURRENT FLUX RATIO FLUX VECTORS

1

R

RR

V

vw

u

g

I g

100%(Basecase)

2R 100%(BaseCase)

173% 1.73F=173

I=100%

I=100%

2

R

RR

V

vw

ug

I g

100% 1.5R 133% 200% 1.5I=133%

I=67%I=67%

F=200%

3

R

RR

V

vw

ug

I g

100% 3R 67% 133% 2I=67%

I=67%I=67%

F=133%

4

R

R

R V

v

wu

g

I g

100% 0.67R 300% 300% 1I=200

I=100%

I=300%

5

R

R

R V

v

wu

g

I g

100% 0.5R 400% 346% 0.87I=200%I=200%

F=346%

This comparison has been carried out in Table 1, where all the possible connectionmethods are shown (all the useful ones; some other connections might exist which


11/17



will lead to a negligible value of flux have not been show). The first column showsthe connection method. Note that the direction of connection of a winding makes adifference; if the terminals of a winding are reversed, the current will be reversed,leading to a reversal of the direction of the flux vector. This can change the resultantvalue of the flux dramatically.

Note that in order to be able to achieve connection method number 3 in Table1, all six terminals of the windings are needed or the windings should be pre-wired inthat method beforehand. This is because standard motors sometimes do not allowaccess to the star-point of the motor (the low speed winding in this case).

Table 1 shows in the first column the method of connection; in the thirdcolumn the resultant resistance (assuming each coil has a resistance of R); in thefourth column the resultant current (by dividing the voltage by the resistance); in thefifth column the resultant magnitude of the flux, which is based on the vectorialsummation of flux vectors in the last column; and in the sixth column the ratiobetween the flux and the current.

Connection method number 3 gives the highest flux to current ratio, and has

now become the industry de factostandard. All motors are now either re-wired inthat configuration, or all six terminal of the low speed winding provided at the terminalbox.

Example 2By referring to connection number 3, show what would happen if the terminal ofwinding Rw were to be reversed. Does this affect the value of the resultant flux, andthus the flux to current ratio? Is this an acceptable connection? [Remember that theflux has to be summed vectorially.]

R

RR

V

vw

ug

I g

The direction of the fluxes will now be the same as the direction of the arrows shownin the new figure above. The direction of the flux for the Rw coil will be reversed.

Thus, the new flux diagram will become (the magnitude of the current will still be thesame, as the resistance of the windings is still the same):

I=67%

I=67%

I=67%

F=133%

The flux summation is first carried out between the fluxes Rw and Rv. The resultant

vector is horizontal, and is equal to:


12/17



67% 30 67% 30 116% cos( ) cos( )

This, then is summed with the vertical Ru flux, giving a resultant value of:

1 15 0 67 1 33 133%2 2

. . .

Thus, although the orientation of the resultant flux changes, the absolute value of theflux does not change, neither does the flux to current ratio. Thus, this connection isalso an acceptable connection.

Example 3In the same way, show what would happen if the terminal of both coils, Rv and Rwwere reversed. How would this affect the value of the resultant flux, and the flux tocurrent ratio? [Remember that the flux has to be summed vectorially.]

SolutionThe modified connection diagram is an shown in the figure below. Both Rw and Rvwinding connections have been reversed. This modification would not affect thevalue of the current, because the value of the resistance of the coils has notchanged.

R

RR

V

vw

ug

I g

The direction of the vector fluxes will be in the direction of the current arrows shownin the figure above. However, if these vectors are plotted, as shown below, they areall of equal magnitude and pulling in opposing directions. The resultant flux is zero.Thus, no braking torque will be produced with this connection, regardless of themagnitude of current drawn. Thus, this is not an acceptable connection.

I=67%

I=67%I=67%

Braking Voltage Value CalculationsIn order to derive the braking voltage, a voltage source needs to be used. It iscustomary to use the phase to phase voltage rather than the phase to neutralvoltage, for the following reasons:

The phase to phase voltage (so-called line voltage, which usually has a value of380-415V RMS) has a higher value compared to the phase to neutral voltage

(usually called phase voltage, which usually has a value of 220-240V RMS). The


13/17



larger value results in a larger possible braking voltage range, which providesbetter flexibility in operating the drive.

The neutral connection is not always available. In fact, some customers wouldinsist that a neutral not be used, and that any phase voltages are derived using

step down transformers.

Example 4Assuming that the phase voltage from a supply is 230 V RMS, what is the maximumaverage braking DC voltage possible?

SolutionThe full wave rectified DC voltage waveform is shown in Figure 1. If we assume thatthe RMS of the original phase voltage is 230 V, then the RMS value of the linevoltage is:

V V V RMS L 230 3 398

However, the peak value of the line voltage will be:

V V RMS V peak

VV

V mean

P

ave

398 2 563

563

2

358

( )

( )

0

0.2

0.40.6

0.8

1

1.2

1.4

1.6

1.8

030

60

90

120

150

180

210

240

270

300

330

360

Angle (degrees)

Ratio

ofphasevoltage

|R-T|

Figure 1: Waveform of a full wave rectified brake voltage, at 0 degrees firing angle (i.e., fullvoltage).


14/17



By varying the firing angle in each SCR in the full wave bridge rectifier, the mean andRMS value of the resulting brake waveform can be varied, from the full voltage downto zero volts.

The waveforms for various values of firing angles are shown in Figure 2,

Figure 3, Figure 4, Figure 5, Figure.6 and Figure 7. The waveforms assume a pureresistive load, although this in not correct, as the low speed motor windings will havea combined inductance and resistance.


15/17



0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value

Figure 2: 0 degrees.

0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value


0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value


0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value


0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value

Figure.6: 120 degrees.

0

100

200

300

400

500

600

030

60

90

120

150

180

210

240

270

300

330

360

390

420

450

480

Angle (degrees)

Voltage(V)

Average

value


Example 5Calculate the average value of the full wave rectified braking waveform, at a firing

angle of 45.

SolutionIn order to calculate the average value of a waveform, the function should be

integrated throughout the period, and the result divided by 2 radians (i.e., 360).


16/17



N.B.: All calculations involving integration should be applied to radians (and notdegrees).

The function is identical every radians (i.e., 180). Thus, the average can be foundby integrating for 180, and dividing by radians.

As the firing angle is 45 (i.e., /4 radians), the function is zero for the first 45,

and is then equal to sin() from then until 180 (i.e., ).

V dave 1

5631

563

4

4

sin( ) cos( )

V Vave 563

1 0 707 306

.

It has to be borne in mind that the number of poles of the low speed winding of themotor will have an immediate effect on the braking torque available from the motor.

Usually suppliers will not recommend using a low speed winding if it has more than16 poles without checking that the braking torque is sufficient. Most motors employ 4poles for the high speed winding and 16 poles (sometime 24 poles ) for the lowspeed winding.

References & BibliographyBarney, G.C. & Loher, A.G., 1990, Elevator Electric Drives: Concepts and

principles, control and practice, Ellis Horwood.Bird, B.M. & King, K.G., 1983, An introduction to power electronics, John Wiley &

Sons, 1983.Datta, S.K., 1985, Power electronics and control, Reston Publishing Company,

1985.Dewan, S.B. & Straughen, A., 1975, Power Semiconductor Circuits, John Wiley &

Sons.Fukuda, T., 1979b, AC feedback control in Japan: Part V, Elevator World, Feb.

1979.Lander, C.W., 1993, Power Electronics, Third Edition, McGraw Hill, 1993.Shepherd, W., Hulley, L.N. & Liang, D.T.W., 1995, Power electronics and motor

control, Second Edition, Cambridge University Press.

Problems1- Using a method similar to the example in the Chapter, derive a general formulafor the mean voltage of a DC waveform from a full wave controlled rectifier, as a

function of the firing angle, . Assume a pure resistive load. Plot the resultingfunction against the firing angle.

2- In the same way, repeat the last exercise, but derive a general formula for the

RMS value of the waveform as a function of the firing angle, . Assume a pureresistive load. Plot the resulting function against the firing angle.


17/17



3- Discuss the two methods of braking in VF systems, outlining their advantagesand disadvantages. Which method is more suitable for the following:

A lightly loaded counterweighted lift system.

A half loaded counterweighted lift system. A highly loaded down escalator. A highly loaded up escalator.

What are the criteria for using one braking system or the other?

4- By using op-amp applications, design an analogue system which will measurethe amount of voltage on the DC link of an inverter, and switch a choppertransistor in order to discharge the extra charge on the capacitor in the brakingresistor. Ensure that you provide proper isolation between the high voltage DClink, the low voltage op-amp electronics, and the base of the chopper transistor.

5- Calculate the power rating of a braking resistor, for an elevator running at aspeed of 1.6 m/s, carrying 23 passengers, counterweighted at 50%, andrunning at a duty cycle of 7.5%.

8a electrical braking rev 2 100418(good one)

Documents