ELECTRIC BRAKING

There are three types of electric braking:

 Regenerative braking
 Rheostatic braking
 Plugging or reverse current braking

Regenerative braking implies operating the motor as a generator, while it is still

connected to the supply network. Mechanical energy is converted to electrical
energy, a part of which is fed back to the supply and the rest is lost as heat in the
windings and the bearings.normally it does not involve any switching operation,
unless it as required to change the speed at which it becomes effective. Most
electrical machines pass smoothly from motoring to generating mode, when
overdriven by the load.

Rheostatic braking implies operating the motor as a generator, so that the energy is
dissipated as heat in the rheostat connected across the armature.

Plugging involve reconnecting the power supply so that the motor is driven in the
opposite direction. If left to itself the system will stop and then accelerate in the
opposite direction. To stop the motor it has to be disconnected. This is an
inefficient technique , because in addition to electrical energy converted to
mechanical form, being wasted, the electrical energy is drawn from the supply is
also wasted.

Regenerative Braking of D.C. shunt Motor

When a d.c. shunt motor or a separately excited motor used in hoisting mechanism
is switched on for lowering a load, the developed torque & load torque act in
unison to accelerate the motor. With increase in speed , the induced e.m.f also
increases and becomes equal to the supply voltage, when the speed equals the no
load speed. At this moment armature current and thus Tm becomes zero. The
downward motion is only sustained by the gravitational pull on the load moving
downwards. When the speed become greater than the ideal no-load speed, the
armature current becomes negative . the drive then act as a generator and provides
braking torque. The drive attains a steady state speed when the braking torque
developed is equal to load torque. This type of braking is called regenerative
braking, as the power is fed back to the supply. This type of braking occur, when
the speed become greater than ideal no load speed.

Rheostatic Braking of D.C. Shunt Motor

At speed lower than the no load speed, retarding torque can be produced by
rheostatic braking or dynamic braking. The motor is disconnected from the supply
and shunted across a resistance. The motor develops a generating torque and
intensely brakes itself.

𝐸 = −𝐼 ( 𝑅𝑎 + 𝑅𝑒𝑥 ) where Ra is the armature resistance and Rex is the external

resistance.

𝑘∅𝜔 = − 𝐼 ( 𝑅𝑎 + 𝑅𝑒𝑥 ) 𝑎𝑛𝑑 𝑇 = 𝑘∅𝐼

( 𝑅𝑎 + 𝑅𝑒𝑥 )
𝜔 = −𝑇
𝑘∅2
The speed torque characteristic is a straight line passing through the origin. The
slop of the curve depends on the external resistance.

The basic equations are:

𝑑𝜔
𝑐𝜔 + 𝑖𝑅 = 0 & 𝑐𝑖 = 𝐽 + 𝑇𝐿
𝑑𝑡
𝑐2 𝜔 𝑑𝜔
∴ − = 𝐽 + 𝑇𝐿 , where R = Ra + Rex
𝑅 𝑑𝑡

𝑑𝜔 𝑐 2𝜔
𝐽 + 𝑇𝐿 + =0
𝑑𝑡 𝑅
𝑑𝜔 𝑐 2𝜔
𝐽 = − 𝑇𝐿 −
𝑑𝑡 𝑅
𝐽𝑅 𝑑𝜔 𝑇𝐿 𝑅
= − − 𝜔
𝐶 2 𝑑𝑡 𝑐2
𝐽𝑅 𝑑𝜔 𝑇𝐿 𝑅
= − ∆𝜔 − 𝜔 ; = ∆𝜔 = speed drop determined from the
𝐶 2 𝑑𝑡 𝑐2
dynamic braking speed- torque characteristics at load torque T L

𝑑𝜔 𝜔 ∆𝜔
+ =−
𝑑𝑡 𝑇𝑒𝑚 𝑇𝑒𝑚
𝑡
−𝑇
Solution of this equation is, 𝜔 = − ∆𝜔𝐿 + 𝑐1 𝑒 𝑒𝑚

At t= 0, 𝜔 = 𝜔𝐿 ∴ 𝜔𝐿 = − ∆𝜔𝐿 + 𝑐1 or, 𝑐1 = 𝜔𝐿 + ∆𝜔𝐿

𝑡
−
𝜔 = − ∆𝜔𝐿 + ( 𝜔𝐿 + ∆𝜔𝐿 )𝑒 𝑇𝑒𝑚 curve 1 of Fig. A
𝑡
−𝑇
If TL = 0, ∆𝜔𝐿 = 0 𝜔 = 𝜔𝑜𝑒 𝑒𝑚 for no load curve 2 of Fig A

In case of dynamic braking without load, the speed torque curve

approaches the x-axis as illustrated in curve 2. When dynamic braking
occurs with load the curve approaches the line - ∆𝜔𝐿 in the case the load
torque is active, as for example crane hoist lowering a load. For passive
load the braking ceases as 𝜔 = 0 is reached.
𝑑𝜔 𝐽 𝑑𝜔 𝑇𝐿
𝑐𝑖 = 𝐽 + 𝑇𝐿 𝑎𝑛𝑑 𝑖= + 𝐼𝐿 , where 𝐼𝐿 = = load current
𝑑𝑡 𝑐 𝑑𝑡 𝑐
𝑡 𝑡
−𝑇 𝑑𝜔 𝑐1 −𝑇
Now, 𝜔 = − ∆𝜔𝐿 + 𝑐1 𝑒 𝑒𝑚 ∴ = − 𝑒 𝑒𝑚
𝑑𝑡 𝑇𝑒𝑚

𝐽 𝑐1 − 𝑡
∴ 𝑖= − 𝑒 𝑇𝑒𝑚 + 𝐼𝐿
𝑐 𝑇𝑒𝑚
𝐽 𝑐1 𝐽 𝑐1
At t = 0 i = - Iin ∴ − Iin = − + 𝐼𝐿 ∴ −(Iin + 𝐼𝐿 ) = −
𝑐 𝑇𝑒𝑚 𝑐 𝑇𝑒𝑚

𝑡
𝑐 𝑇𝑒𝑚 (Iin + 𝐼𝐿 ) −
𝐶1 = ∴ 𝑖 = −(Iin + 𝐼𝐿 )𝑒 𝑇𝑒𝑚 + 𝐼𝐿
𝐽

If the load torque is active we get curve 1 as shown is Fig. C . if load torque = 0, IL
𝑡
− cωin
= 0, 𝑖 = −Iin 𝑒 𝑇𝑒𝑚 , where Iin =
R
𝐸 𝜔𝑖𝑛
[ 𝑉 = 𝐸 + 𝐼𝑖𝑛 𝑅; 𝑎𝑠 𝑉 = 0, 𝐸 = −𝐼𝑖𝑛 𝑅 ∴ 𝐼𝑖𝑛 = − 𝑅 = 𝑅
]

If the time taken to brake the motor from 𝜔𝑖𝑛 to any speed 𝜔1 is t, then
𝑡 𝑡
−𝑇 −𝑇
𝜔1 = −∆𝜔𝐿 + ( 𝜔𝑖𝑛 + ∆𝜔𝐿 )𝑒 𝑒𝑚 ∴ 𝜔1 + ∆𝜔𝐿 = ( 𝜔𝑖𝑛 + ∆𝜔𝐿 )𝑒 𝑒𝑚

(𝜔1 + ∆𝜔𝐿) 𝑡
−𝑇
= 𝑒 𝑒𝑚
( 𝜔𝑖𝑛 + ∆𝜔𝐿 )

(𝜔𝑖𝑛 + ∆𝜔𝐿 )
∴ 𝑡 = 𝑇𝑒𝑚 ln
( 𝜔1 + ∆𝜔𝐿 )
Counter Current Braking or Plugging

Plugging of d.c. motor involves reconnecting the motor to the line with reversed
polarity; the motor now produces a torque in opposite direction. The rotor speed
decreases until it becomes zero and then the motor accelerates in the reverse
direction. So, plugging gives a quicker reversal or a rapid stop.

Fig. A shows a schematic diagram. The polarity is reversed by change over switch
Sw. the latter also introduces a resistance R b in the circuit to limit the current. Fig.
B illustrates the transition from motoring to counter current braking operating
condition. If the motor is not disconnected at ‘O1’ the torque developed will greater
than the load torque and the motor accelerates in the opposite direction and speed
𝜔𝑠𝑠 will be reached, this is motor reversal.

During counter current braking

𝑉 𝑅
− 𝑉 = 𝑐𝜔 + 𝑖𝑅 𝑜𝑟 − = 𝜔+𝑖
𝑐 𝑐
𝑑𝜔 𝐽 𝑑𝜔 𝑇𝐿
𝑇 = 𝑐𝑖 = 𝐽 + 𝑇𝐿 𝑜𝑟, 𝑖= +
𝑑𝑡 𝑐 𝑑𝑡 𝑐
 𝑉 𝑅 𝐽 𝑑𝜔 𝑇𝐿 𝑉 𝐽𝑅 𝑑𝜔 𝑇𝐿 𝑅
− = 𝜔+ ( + ) 𝑜𝑟, − = 𝜔+( 2 + 2 )
𝑐 𝑐 𝑐 𝑑𝑡 𝑐 𝑐 𝑐 𝑑𝑡 𝑐
𝑉 𝑇𝐿 𝑅 𝑑𝜔
− − 𝜔− = 𝑇𝑒𝑚
𝑐 𝑐2 𝑑𝑡

𝑑𝜔 𝑇𝐿 𝑅
𝑇𝑒𝑚 = − (𝜔𝑜 + 2 ) − 𝜔
𝑑𝑡 𝑐
𝑑𝜔
𝑇𝑒𝑚 = −(𝜔𝑜 + ∆𝜔𝐿 ) − 𝜔
𝑑𝑡
𝑡
−𝑇
𝜔 = −(𝜔𝑜 + ∆𝜔𝐿 ) + 𝐶1 𝑒 𝑒𝑚

At t =0 𝜔 = 𝜔𝑖𝑛 = 𝜔𝐿 ∴ 𝜔𝐿 = −(𝜔𝑜 + ∆𝜔𝐿 ) + 𝐶1

𝐶1 = 𝜔𝐿 + (𝜔𝑜 + ∆𝜔𝐿 )
𝑡
−𝑇
𝜔 = −(𝜔𝑜 + ∆𝜔) + (𝜔𝐿 + (𝜔𝑜 + ∆𝜔𝐿 ))𝑒 𝑒𝑚
If the motor is running with an active load torque, i.e. the torque TL does
not change sign, the motor speed after reversal will exceed 𝜔𝑜 by an
amount ∆𝜔𝐿 and we get curve 1.
If the motor is reversed without load 𝜔𝑜 = 𝜔𝐿 and ∆𝜔𝐿 = 0, then
𝑡
−
𝜔 = −𝜔𝑜 + 2𝑒 𝑇𝑒𝑚 ; represented by curve 2.
If the motor reversal occurs with passive load , after reaching x-axis it
brakes and approaches as an asymptote to steady state speed −𝜔𝑠𝑠 ;
shown in curve ‘3’
𝑑𝜔 𝐽 𝑑𝜔 𝑇𝐿
𝑐𝑖 = 𝐽 + 𝑇𝐿 𝑎𝑛𝑑 𝑖= + 𝐼𝐿 , where 𝐼𝐿 = = load current
𝑑𝑡 𝑐 𝑑𝑡 𝑐
𝑡 𝑡
−𝑇 𝑑𝜔 𝑐 −
Now, 𝜔 = −(𝜔𝑜 + ∆𝜔𝐿 ) + 𝐶1 𝑒 𝑒𝑚 ∴ 𝑑𝑡
= − 𝑇 1 𝑒 𝑇𝑒𝑚
𝑒𝑚

𝐽 𝑐1 −𝑇 𝑡
∴ 𝑖= − 𝑒 𝑒𝑚 + 𝐼𝐿
𝑐 𝑇𝑒𝑚
1𝐽 𝑐 𝐽 𝑐1
At t = 0 i = - Iin ∴ − Iin = − + 𝐼𝐿 ∴ −(Iin + 𝐼𝐿 ) = −
𝑐 𝑇𝑒𝑚 𝑐 𝑇𝑒𝑚

𝑡
𝑐 𝑇𝑒𝑚 (Iin + 𝐼𝐿 ) −
𝐶1 = ∴ 𝑖 = −(Iin + 𝐼𝐿 )𝑒 𝑇𝑒𝑚 + 𝐼𝐿
𝐽
Curve ‘1’ shows current for active load torque.
𝑡
−
Curve ‘2’ is for TL =0 if load torque = 0, IL = 0, 𝑖 = −Iin 𝑒 𝑇𝑒𝑚 , where 𝐼𝑖𝑛 =
(𝑉+𝑐𝜔𝑜 ) 2𝑉
=− = 𝐼𝑠𝑐
𝑅 𝑅
−(𝑉+𝑐𝜔𝐿 )
[ 𝑉 = 𝐸 + 𝐼𝑖𝑛 𝑅; 𝑎𝑠 𝑉 = −𝑉, 𝐸 = −𝐼𝑖𝑛 𝑅 ∴ 𝐼𝑖𝑛 = 𝑅
]

Curve ‘3’ is for passive load torque.

Energy involved in Transient Process of Shunt Wound DC motor:
Power drawn by the motor from supply consists of:
 Load torque
𝑑𝜔
 Kinetic energy of rotating part 𝑃𝑑𝑦𝑛 = 𝐽𝜔
𝑑𝑡
 Losses of the motor
If we club the fixed losses with load the losses consists of I2R and power
required for field excitation (normally small compared to I2R; is
neglected)
The energy loss with the armature circuit during transient period of
operation is of importance because it influences the power capacity for
which the motor must be selected and the size of the requisite resistance
as well.
Energy loss in the armature circuit during starting:
𝑡𝑠𝑡
∆𝑊𝑠𝑡 = ∫ 𝑖 2 𝑅𝑑𝑡
𝑜

𝑉𝑖 = 𝐸𝑖 + 𝑖 2 𝑅

2
𝑇 𝑇 𝑉 𝐸
𝑖 𝑅 = 𝑉𝑖 − 𝐸𝑖 = 𝑉 − 𝐸 = 𝑇 ( − ) = 𝑇(𝜔𝑜 − 𝜔)
𝑐 𝑐 𝑐 𝑐
 𝑑𝜔 𝑑𝜔
𝑇=𝐽 ; ∴ 𝑑𝑡 = 𝐽
𝑑𝑡 𝑇
𝜔𝑓𝑖𝑛
𝑑𝜔
∴ ∆𝑊𝑠𝑡 = ∫ 𝑇(𝜔𝑜 − 𝜔)𝐽
𝜔𝑖𝑛 𝑇
𝜔𝑓𝑖𝑛
= ∫ 𝐽(𝜔𝑜 − 𝜔)𝑑𝜔
𝜔𝑖𝑛

𝐽𝜔𝑜 2
For starting at no load 𝜔𝑖𝑛 = 0 and 𝜔𝑓𝑖𝑛 = 𝜔𝑜 ∴ ∆𝑊𝑠𝑡 =
2

Braking of DC Series Motor:

If the speed of the D.C. series motor increases, current and flux
decreases. Therefore it is not possible to get an e.m.f greater than the
terminal voltage. Since, there is no ways of making the field current
greater than the armature current , regeneration is not possible. In
electric traction , where regeneration is used, the motors are actually
reconnected as separately excited machine.
Rheosatic braking is possible, but care should be taken to interconnect
the armature winding and the field winding in such a way so as to ensure
that the direction of the current in the field remains the same. Then only,
self excitation will take place. External resistance should be such that the
total resistance is less than the critical resistance. Due to the influence of
the residual flux an e.m.f is induced, under its influence, current, flux
and e.m.f. increases until 𝐸 = 𝐼(𝑅𝑎 + 𝑅𝑓 + 𝑅𝑒𝑥 )

Rheostatic braking is also possible by reconnecting the machine as

separately excited machine.
Counter Current Braking:
When Sw is changed to bottom position, the motor, running with speed
𝜔𝐿 and current 𝐼𝐿 , will change over to counter current braking mode.
The energy losses during starting of a series motor drive for the same
values of inertia and static load torque and the same limits of speed
change, may be greater, equal or less than those of a shunt wound motor
drive.
When the average current during a transient operating conditions, turns
out to be smaller than the motor rated current and the motor torque is
therefore less than the rated torque, the losses in the series wound motor
will be greater than the shunt wound motor due to longer duration of
transient process. If the average armature current is higher than the rated
current, because of shorter duration of transient period, the series wound
motor losses will be lower than the shunt wound motor losses. If the
average current in the transient process is equal, the energy loss will be
same.
For accurate calculation of the losses, it is necessary to plot current as
function of time.
Problems:
1. A 220V DC shunt motor has an armature resistance of 0.062 ohm and
with full field, has an emf of 215 V at a speed of 960 rpm. The motor is
driving an overhauling load withtorque of 172 Nm. Calculate the
minimum speed at which the motor can hold the load by means of
regenerative braking.

At the speed at which the load is balanced, the armature current is given
by
 𝑁
𝑇𝜔 𝑇 2𝜋 60 2𝜋𝑇𝑁 2𝜋 × 172 × 960
𝐼𝑎 = = = = 80.42 𝐴
𝐸𝑏 𝐸𝑏 60𝐸𝑏 60 × 215

Emf induced during regenerative braking

𝐸𝑔 = 𝑉 + 𝐼𝑎 𝑅𝑎 = 220 + 80.42 × .062 = 225𝑉
960×225
Flux remaining constant 𝐸 ∝ 𝜔 ∴ 𝜔 = = 1004.65 𝑟𝑝𝑚
215

2. A DC series motor is subjected to rheostatic braking against a load

torque of 318.3 Nm. Determine the value of the resistance to be
connected in the motor circuit to limit the speed to 480 rpm. The total
resistance of the armature and the field is 0.24 ohm and the
magnetization curve corresponding to 900 rpm is given by:
If 20 40 60 80 100
Eb 261 540 738 882 945
Neglect rotational losses.
2𝜋𝑁𝑇 2𝜋×480×318.3
Input power at 𝑃 = = = 16000𝑊
60 60

Rotational losses are neglected so, P must correspond to Eb.Ia . at 480

rpm we have :
If 20 40 60 80 100
Eb 139.2 288 393.6 870.4 504
EbIa 2784 11520 23616 37632 50400

From Eb.Iavs.Ia curve P=16000W corresponds to Ia=48 A and Eb = 333.3

V
 𝐸𝑏
Total resistance = = 6.94 𝑜ℎ𝑚 ∴ 𝑅𝑒𝑥 = 6.94 − 0.24 = 6.7 𝑜ℎ𝑚
𝐼𝑎

3. A 220V 20 kW Dc shunt motor running at its rated speed of 1200

rpm, is to be braked by reverse current braking. The armature resistance
is 0.1 ohm and the rated efficiency is 88%. Calculate
1. The resistance to be connected to the armature, to limit the initial
braking current to be twice the rated current.
2. The initial braking torque
3. The torque when the speed of the motor falls to 400 rpm.
20000
Rated current = 𝐼𝑎 = = 103.3 𝐴
0.88 ×220
20000×60
Rated torque = 𝑇 = = 159.15 𝑁𝑚
2𝜋×1200

𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅 = 220 − 103.3 × 0.1 = 209.67 𝑉

1. At the moment of braking the voltage across the armature = V + E b
= 220V + 209.67V = 429.67 V
 Initial braking current 2Ia = 2x103.3 A = 206.6 A
429.67
Total resistance in the circuit = 𝑜ℎ𝑚 = 2.08 𝑜ℎ𝑚
206.6

External resistance = = 2.08 𝑜ℎ𝑚 − 0.1 𝑜ℎ𝑚 = 1.98 𝑜ℎ𝑚

2. Flux is constant, so T ∝ 𝐼𝑎
Initial braking torque = 2x159.15 nm = 318.3 Nm
209.67
3. Eb at 900 rpm = × 400 = 69.89 𝑉
1200
220+69.89
Braking current at 400 rpm = = 139.37 𝐴
2.08
159.15
Braking torque = × 139.37 = 214.72 𝑁𝑚
103.3