Download Dynamic Braking Using External Resistors

Dynamic Braking Using
External Resistors
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Overview
Dynamic braking is a method of quickly dissipating the rotational energy stored in a
servomotor/machine system when the power is removed from the servodrive. Without dynamic
braking, the energy is dissipated as frictional losses in the system. In a well-built machine, such
frictional losses may be relatively low, and the time required for the system to come to rest could
be unacceptable from a safety standpoint.
A large number of servodrives have dynamic braking circuitry incorporated in the drive
package. This technical note describes how to implement dynamic braking on a
servodrive/servomotor package that does not incorporate this feature. Please note that this
discussion covers permanent magnet brushless servomotors only.
Dynamic braking of a servomotor with an external contactor and resistor provides an
uncontrolled method for quickly decelerating a motor during emergency stop (E-Stop)
conditions.
After power is removed from the output of a servoamplifier, the motor acts as a generator
and converts the kinetic energy of the rotating motor and load into electrical energy. This
electrical energy is then dissipated in a resistor as heat.
Dynamic braking does not provide a mechanism for locking the motor armature in a fixed
position. A mechanical holding brake must be used for this purpose.
Electrical Design
The figure below shows the general electrical connection of the pertinent components.
The diagram shows a 3-phase brushless motor, but may also be applied to DC brush-type motors
as well.
When the Dynamic Braking
contactor is de-energized (i.e., the
machine enters its E-Stop state),
the servodrive is disconnected
from the servomotor, and the
dynamic braking resistors are
switched across the armature
windings of the motor. This
causes the motor to become a
generator, and the kinetic energy
stored in the machine is dissipated
across the dynamic braking
resistors, and the resistance of the
armature windings.
Note that it is important to include the auxiliary contact on the contactor, so that the
output transistors of the servodrive are disabled when the motor is being dynamically stopped.
The amount of braking torque applied to the shaft of the motor is proportional to the
"back-EMF" (bucking voltage) generated by the motor, which in turn is proportional to the
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Dynamic Braking Using
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angular velocity of the motor shaft. This means that as the motor begins to decelerate, the
braking torque also decreases.
When the contactor is de-energized, the amount of braking torque applied to the shaft of the
motor is calculated such that it does not exceed the machine or motor specifications, and stops
the machine within an acceptable time or distance.
Resistance Calculation
For most applications, the dynamic braking resistor is sized so that the peak deceleration
torques are within the specifications of the machine and motor.
This is accomplished by determining the maximum amount of peak torque that can safely
be applied by the motor shaft, and then reducing this amount of torque by an acceptable safety
margin. For example, if a motor is capable of providing 1000 lb-in. of torque, but the maximum
amount of torque that can be tolerated by the machine is 750 lb-in., and a safety factor of 10% is
desired, then the dynamic braking resistor will be sized such that the maximum amount of torque
provided by the motor shaft is 750 × 0.9 , or 675 lb-in.
The wattage of the resistors may be calculated by using 1/20 to 1/50 of the peak wattage
dissipated in the resistors. The exact amount depends on how often the machine experiences an
E-Stop when at full speed.
The resistance value of the dynamic braking circuit, including motor armature resistance
cable resistance, and resistors R db is calculated by using equation (1):
ke ⋅ kt
(1)
R=
⋅ω
104.7 ⋅ T m
R
Where:
is the resistance in one leg of the dynamic braking circuit. It is equal to the
resistance of two dynamic braking resistors, plus the line-to-line resistance
of the motor cable, plus the line-to-line resistance of the armature winding.
Tm
is the maximum deceleration torque that may be safely provided by the
motor
is the torque constant of the motor armature (lb. in./amp).
kt
is the angular velocity of the motor shaft (rad/sec). The maximum
ω
anticipated motor speed should be used here.
ke
is the voltage constant of the armature winding (volts/KRPM).
Note: R is the value of the total resistance. The armature resistance and cable resistance
must be subtracted to find the resistance of each dynamic braking resistor.
The peak wattage is dissipated in the braking resistor at the instant when the contactor is
closed. It is calculated by:
k ⋅ω 2
(2)
P =  e
⋅ 2 ⋅ R db
104.7 ⋅ R 
Example
For this example, suitable dynamic braking resistors will be sized for the following
system:
Motor: MAC-F700C
Inertia = 0.0744 lb-in-sec2
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k e = 136.2 V/KRPM
k t = 19.9 lb.in./amp
Armature Resistance (line-to-line) = 0.23 ohm
The mechanical specifications are:
Max Motor Speed: 1000 RPM = 104.7 rad/sec.
Max Braking Torque Allowed = 1250 lb-in. Note that this is less than the
maximum torque capability of the motor, and is limited to prevent the motor from
exerting more torque than the mechanics can tolerate. If the maximum torque
allowed is not limited, something will eventually fatigue and break.
Motor Cable Resistance (line-to-line) = 0.25 ohm
Total Inertia at Motor shaft = 0.5044 lb.in.sec2
Substitution into equation (1) yields:
R = 2.17 ohms
Note that this is the total resistance in the circuit, and the armature resistance and cable
resistance must be subtracted. This yields:
R db = 2.17 − 0.23 − 0.25 = 0.845 Ω
2
The peak wattage dissipated in the resistors is:
2
P =  136.2 ⋅ 104.7  ⋅ 2 ⋅ 0.845 = 6649 watts
104.7 ⋅ 2.17
1/20th of the peak power is 332 watts, and 1/50th of the peak power is 133 watts. The
resistors should be sized at either 500 or 125 watts, depending upon the expected frequency that
the dynamic braking circuit is to be used
Mathematical Derivation
The braking torque provided by the motor is given by the equation (3):
−V m ⋅ k t
−k e ⋅ k t
(3)
T m = −I p ⋅ k t =
=
⋅ω
R
104.7 ⋅ R
Where:
Tm
is the deceleration torque provided by the motor
Ip
is the current in the motor armature (amps)
kt
Vm
R
is the torque constant of the motor armature (lb. in./amp)
is the voltage developed across the armature windings (volts)
is the resistance in one leg of the dynamic braking circuit. It is equal to the
resistance of two dynamic braking resistors, plus the line-to-line resistance
of the motor cable, plus the line-to-line resistance of the armature winding.
ω
is the angular velocity of the motor shaft (rad/sec). The maximum
anticipated motor speed should be used here.
ke
is the voltage constant of the armature winding (volts/KRPM).
104.7 is a constant derived from 1000 ⋅ 2 ⋅ π . It converts the units of k e from
60
volts/KRPM to volts per radian/sec.
(4)
Equation (3) can be rewritten to solve for the total resistance, R.
ke ⋅ kt
R=
⋅ω
104.7 ⋅ T m
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Note that this is the value of the total resistance, and the armature resistance and cable
resistance must be subtracted.
The peak wattage is dissipated in the braking resistor at the instant when the contactor is
closed. It is calculated by:
k ⋅ω 2
(5)
P = I m 2 ⋅ 2 ⋅ R db =  e
⋅ 2 ⋅ R db
104.7 ⋅ R 
Decel Time and Distance Calculation
The deceleration time, and distance can be calculated to determine if they are acceptable.
These values can be calculated based upon the electrical properties of the motor, the machine's
velocity and inertia, and the resistance of the braking resistor.
When the motor is in the process of being dynamically braked, the angular velocity of the
motor shaft with respect to time is calculated with equation (6):
ω (t) = ω (0) ⋅ e −K⋅t
(6)
Where:
ω (0)
is the initial velocity of the motor shaft at time t = 0 (in rad/sec.).
ke ⋅ kt
K=
104.7 ⋅ R ⋅ J
J
is the total system inertia (motor inertia plus reflected load inertia).
kt
is the motor torque constant (lb-in/amp).
ke
is the motor voltage constant (volts/KRPM).
104.7 is a constant derived from 1000 ⋅ 2 ⋅ π . It converts the units of k e from
60
volts/KRPM to volts per radian/sec.
is the total dynamic braking resistance (motor armature, cables, resistors).
R
t
is the elapsed time (in seconds).
The distance that the motor will travel (in radians) is given by:
ω (0)
(7)
X=
K
Where:
X is the total distance that the motor moves (in radians) under dynamic
braking.
Note: This distance assumes zero friction in the machine. The actual distance will be
less, depending upon the frictional losses in the machine, and servomotor.
Example
Using the motor and load parameters given in section 3.3, the deceleration time and
distance can be calculated. The first thing to do is to calculate the time constant parameter K .
136.2 ⋅ 19.9
K=
= 23.64
104.7 ⋅ 2.17 ⋅ 0.5044
The time required for the motor to decelerate to 5% of its initial velocity can be
calculated using equation (6) above.
0.05 ⋅ ω (0) = ω (0) ⋅ e −23.64⋅t
Solving for t yields:
t = 0.127 seconds
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The total distance that the motor shaft travels during dynamic braking is
calculated to be:
X = 104.7 = 4.43 rad = 0.7 revs
23.64
Mathematical Derivation
The equation for the deceleration of the motor shaft is given as:
dω −T m
(8)
=
J
dt
dω
Where:
is the angular deceleration of the motor shaft (rad/sec2)
dt
J
is the total inertia at the motor shaft (lb. in. sec2). This is the inertia of the
motor armature, and the inertia of the machine reflected through any
gear/belt reduction.
Tm
is the deceleration torque applied by the motor shaft (lb-in.)
The deceleration torque can be calculated by equation (9):
−V m ⋅ k t −k e ⋅ k t ⋅ ω
(9)
T m = −I p ⋅ k t =
=
104.7 ⋅ R
R
Ip
Where:
is the current in the motor armature (amps)
kt
Vm
R
is the torque constant of the motor armature (lb. in./amp)
is the voltage developed across the armature windings (volts)
is the resistance in one leg of the dynamic braking circuit. It is equal to the
resistance of two dynamic braking resistors, plus the line-to-line resistance
of the motor cable, plus the line-to-line resistance of the armature winding.
ω
is the angular velocity of the motor shaft (rad/sec).
ke
is the voltage constant of the armature winding (volts/KRPM).
104.7 is a constant derived from 1000 ⋅ 2 ⋅ π . It converts the units of k e from
60
volts/KRPM to volts per radian/sec.
Substituting (9) into (8) yields:
(10)
dω
dt
=
−k e ⋅k t ⋅ω
104.7⋅R⋅J
This can be rewritten as:
dω = −k e ⋅ k t dt
(11)
ω 104.7 ⋅ R ⋅ J
Equation (11) above shows how the acceleration of the motor shaft changes with time.
Note that it is also dependent upon the physical properties of the motor, the load inertia, and the
total circuit resistance. It can now be integrated to provide the angular velocity of the motor shaft
with respect to time.
(12)
ω (t) = ω (0) ⋅ e −K⋅t
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Where:
ω (0)
K
is the initial velocity at time=0 (rad/sec). It should be set to the maximum
anticipated motor velocity.
ke ⋅ kt
is the motor time constant (in seconds), which is given by K =
104.7 ⋅ R ⋅ J
.
Note that equation (12) is very similar to the equation that governs the voltage discharge
of a capacitor through a resistor. In this case, the stored energy is in the rotating machinery,
instead of the plates of a capacitor.
Equation (12) can be integrated so that the motor shaft position with respect to time can
be determined. This is performed as follows:
(13)
ω (t) = dx = ω (0) ⋅ e −K⋅t
dt
Where:
x is the position of the motor shaft
(14)
Equation (13) is then rearranged to:
dx = ω (0) ⋅ e −K⋅t dt
This is then integrated to provide:
ω (0) −K⋅t
(15)
x (t) =
⋅e +C
−K
C is a constant of integration, and can be calculated from equation (15) by
Where:
setting x = 0 when t = 0 .
ω (0)
0=
+C
−K
ω (0)
C=
K
(16)
(17)
Equation (15) can now be written as:
ω (0) ω (0) −k⋅t ω (0)
x (t) =
−
⋅e =
⋅ (1 − e −K⋅t )
K
K
K
Equation (16) can be used to find the total deceleration distance by setting t = ∞ :
ω (0)
x tot =
K
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