APPLICATION

VSD advantages, disadvantages,

selection criteria and installation tips
by R G van der Merwe, Motiontronix and C Hoogendoorn, Circuit Breaker Industries

The purpose of this document is to assist and guide users when they acquire a variable speed drive (VSD) and to help them through the many
precarious pitfalls to successfully select, install and operate their VSD.

This document seeks to help users to

understand and simplify the process of
selection, installation and the use of a VSD.
Users should see it as a long-term relationship
and investment when acquiring a VSD.

Introduction

Electric motors are the workhorses of industry

and commerce. Each day large numbers of
industrial electric motors are sold throughout
the world. These motors are reliable and
inexpensive.

However they are fixed speed devices, and

most applications require a variable speed
output from the motor. Frequently this is
obtained by coupling some mechanical
device (clutch, gear, belt and pulley, etc.)
to the motor. These methods are inefficient
in terms of mechanical wear and energy
consumption.

VSD advantages

VSDs have many advantages and

manufacturers of VSDs claims different
advantages for their units. This paper will cover
a couple of basic functions which we use with
Fig. 1: Acceleration and deceleration curves.
most installations.

Speed Control at the design speeds of 2, 4, or 6 poles. At motor is approximately 400 radians/second.
high frequency command speeds, care In a closed loop system with a standard
A fundamental function of a VSD is to adjust
should be taken as torque loss may be squirrel cage induction motor, feedback is
the speed of an electric motor. The basic
experienced. approximately 600 radians/second. A full
command frequency for VSDs is normally
servo system is approximately 1000 radians/
from 0 Hz to 50 Hz, but mostly with the Torque control
second. 1 radian/second = 9.55 rpm or 2π
capability to be adjusted up to 400 Hz. If
Basic torque control is possible in an open radians (rad.) in 360° or 1 radian = 57.3 °.
the base frequency of a motor is 50 Hz then
the final speed will be 8 times the base loop system; however, the actual system
Smooth controllable starting and stopping
frequency of the motor with the command response required must be considered. In
an open loop system the VSD monitors the A Simple adjustment of the time required
frequency set at 400 Hz.
motor current and adjusts the voltage to to accelerate the motor from rest to full
Due to their design, it is not practically normal speed (Starting), normally 50 Hz, and from full
perform torque control, depending on the
for standard induction motors to operate speed to rest (Stopping), ensures a smooth
installation. If the current of the motor does
at these high frequencies. In practice a controllable start and stop sequence. This
not vary sufficiently, very inaccurate results reduces mechanical wear on the machine.
command frequency set point of between
will be obtained. Various types of starting and stopping
25 Hz and 75 Hz is acceptable without
compromising performance or introducing curves are available by setting the correct
Position control
any mechanical damage to the motor. At parameters in the VSD as illustrated in figure
low frequency set points, care must be taken With the aid of an optional interface card 1a to c.
that there is enough cooling for the motor most VSDs have the ability to be used as
Energy saving
produced by the mechanical fan. a low cost position controller. Items to be
taken into consideration are the dynamic We know that a Direct On Line (DOL) starter will
At high frequency set points mechanical response of the motor and control system. supply full voltage to the motor at the supply
failure may occur due to the mechanical As a rule of thumb an open loop system frequency with the current uncontrollable. The
design of the motor bearings normally rated with a standard squirrel cage induction motor will use as much current as the load

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 APPLICATION
requires - normally between 600 to 700% of etc. This allows the user
the full load current of the motor. to customise the inverter
for his application. Most
Before the days of Soft Starters and VSDs our
VSDs have advanced
alternative to control the starting current was
units that could copy
with Star-Delta starters, which reduced the
parameters from one unit
starting current to approximately 200%. Our
next best device today to limit the starting to another. Apart from
current is to fit a Soft Starter. With Soft Starters this basic function, most
we use the Phase Angle principle to control units available today
the voltage and therefore reduce the starting are supplied with serial Fig. 2: Load torque vs. speed curves
current while at the same time producing a communication ports to
smooth controllable start. The limitation is that interface with personal
these units are very basic and have limited computers that allow users
adjustable time settings, normally from 0 to to analyse the behavior of
60 seconds. their system.

The current limiting features on VSDs ensure Ability to interface with

that when you accelerate a motor from other intelligent control
rest, you will not exceed more than 100% of systems
the full load current of the motor. Replacing
Our demanding society
DOL starters with VSDs will reduce the Current
forces managements to
Demand when starting motors. VSDs will
know what is happening in
deliver maximum torque at the motor shaft
their plant and processes.
while limiting the current to the full load
Information from VSDs is Fig. 3: Lift load torque vs. speed curves.
current setting of the motor in the VSD.
not only for the engineer’s
It is the responsibility of every individual to use benefit, but also allows
forms of energy effectively and efficiently. managements to see if
It is already well known in the Heating and they could increase their
Ventilation industr y that volumes, flows production safely without
and pressures of centrifugal fans, pumps overloading the process or
and compressors can be controlled by plant. This is normally done
mechanical means to match the demands via the serial interface. It is
of the system. Many of them do not consider also possible to integrate
the immense amount of energy (i.e. watts, i.e. units into a complex network
money) that can be saved by using modern system.
reliable electronic technology.
Mechanical wear and
If the efficiency of the system can be improved tear
the power demand drops proportionally with
It is to the advantage Fig. 4: Motor connections.
the increased efficiency. Almost all fans,
of the users that where
pumps and compressors are over designed
mechanical wear is part advise users on how to install and operate
(just in case) and seldom do they work at their
of the process, users could speed up or their units.
maximum designed efficiency point.
slow down their application to deliver the
Audible noise
Fitting an inverter to a fan, pump or compressor necessary production.
motor varies the motor speed which then Various stages of the switching frequency
In a case study it was found that due to
varies the characteristics of the fan, pump produce audible noise from the motor.
or compressor to operate at a different excessive wear, the life of a pump was
Although this is not harmful to the motor, in
efficiency, producing energy savings. approximately six months. A VSD was fitted
most instances the noise is not acceptable.
to regulate the flow as the flow reduced due
The Affinity Laws say that flow is proportional The noise is unpleasant and irritating in
to impeller wear. This enabled the user to
to speed and kilowatts vary as the cube of quiet offices, hospitals and other such
increase the speed of the pump to produce
the speed. environments. To overcome this problem
the required flow, increasing the life of the
most VSDs’ switching frequency could be
Flexibility pump from six months to twelve months. The
increased to a higher value, which will
other benefit was that when the pump was still
The flexibility to set up and configure a eliminate the noise problem, but this will
new the speed could be reduced according
VSD for various applications, e.g. Constant introduce harmonics. Therefore proper design
to the necessary flow required, saving energy
torque, Variable torque, Hoisting and of an installation should be done before
and decreasing the wear on the impeller.
many others, allow users to customise using VSDs.
units to suit their needs. See illustrations in VSD disadvantages
RFI
Figs. 2 and 3.
VSDs Due do not have many disadvantages
Radio Frequency Interference generated by
User friendly with respect to industrial standards and
regulations. Most high quality VSDs comply VSDs can be very problematic, introducing
Most VSD are supplied with basic LCD or LED with standards and regulations. Manufacturers faults on other equipment in close proximity
display keypads, with which the user can improve their products with ongoing research to the installed unit. Most drives can be
adjust parameters such as acceleration and development programmes. To eliminate expected to meet the immunity requirements
time, deceleration time, full load current, any disadvantages, manufacturers will also of the CENELEC generic standard EN50082-2.

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Harmonics and let something that I do not understand W = Watts
and is very expensive, control my process?
VSDs, like most other electronic equipment, do π = Pi (Mathematical constant = 3.142)
Even worse, now I need to employ a highly
not draw their current as a smooth sinusoidal
qualified technical engineer to maintain my M = Torque (Nm)
supply. The supply current waveform is
plant”. These remarks, and ignorance of the
generally referred to in terms of the harmonics Example:
instruction manual, create a bad image of a
of the supply frequency, which it contains. The
product that improves life for all of us. A mechanical engineer designs a machine
harmonic current causes harmonic voltage
that requires 405 Nm and a speed
to be experienced by other equipment Selection criteria
range from 100 to 175 rpm. By fitting a
connected to the same supply. Because
Summarised is some of the basic criteria to 10/1 ratio gearbox to the machine he
harmonic voltage can cause disturbance
successfully install a VSD reduces the input torque required to
or stress to other electrical equipment
40,5 Nm, while the minimum and maximum
connected to the same supply. Because Supply Voltage
input speeds increase to 700 and 1750rpm
harmonic voltage can cause disturbance
Always ensure that the correct voltage is respectively. A four-pole 7,5 kW motor
or stress to other electrical equipment
available. In many cases user’s interpretation (1500 rpm @ 50 Hz) produces 47,8 Nm. We
connected to the same supply system,
of a VSD is that you could supply the unit need to calculate if he will produce enough
there are regulations in place to control it. If
with single-phase 220 V AC and control a torque at the maximum speed.
installations contain a large number of VSDs
and/or other power electronic equipment three-phase motor rated for 380 V AC. Most
To reduce the speed to 1000 rpm is not a
such as UPSs, then they may have to be standard induction motors could operate with
problem, as long as he keeps the motor
shown to satisfy the supply authorities’ three phase 380 V AC with all six leads from
speed above 50% of the base speed to
harmonic guidelines before permission the windings available and connected in a
produce enough cooling.
to connect is granted. As well as obeying Star configuration. The same motor could
regulations, users of drives need to ensure operate with 3 Phase 220 V AC if the leads A VSD will produce the full load torque of the
that the harmonic levels within their own plant from the windings are connected in a Delta motor up to the base frequency by changing
are not excessive. configuration. However, consult the motor the voltage to produce the necessary torque.
manufacturer if it is not indicated on the Once the motor reaches its base speed and
Some of the practical problems, which may
motor nameplate. supply voltage the VSD can only change the
arise from excessive harmonic levels, are:
frequency supplied to the motor to increase
Kilowatt size
 Poor power factor, i.e. high current for a the speed - the VSD cannot supply a higher
given power It is not totally correct to select an inverter voltage than the supply voltage.
 Interference to equipment, which is according to motor capacity in “kW”. It is
better to select an inverter based on the To calculate the torque produced by the
sensitive to voltage waveform 7,5 kW at 1750 rpm we have to manipulate
rated current of a motor. If the inverter and
 Excessive heating of neutral conductors the motor have the same capacity (kW), the above formula.
(single-phase loads only) an increase in the number of motor poles
 Excessive heating of induction motors reduces the efficiency and power factor
of the motor increasing the rated current
 High acoustic noise from transformers, value.
bus bars, switchgear etc.
Torque requirements
 Abnormal heating of transformers and
associated equipment If we look at the following calculations we will
 Damage to power factor correction understand why torque loss happens when
running a motor above base speed. This will
capacitors
also explain some of the basic requirements
An important property of harmonics is that why torque is an important factor when Therefore the 7,5 kW motor with a VSD fitted
they tend to be cumulative on a power selecting a VSD. can produce the necessary torque at the
system, i.e. the contributions of the various correct speed. From this we can see that it
harmonic sources add up, to some degree. Motor speed:
is always necessary to check if the speed/
This is different from other high-frequency torque range is within the capability of the
electromagnetic compatibility (EMC) effects, (1) Inverter and motor.
which are generally localised and not
DC injection braking
significantly cumulative. It is important to
differentiate harmonics from high-frequency where: In most hoisting applications, the motor must
EMC effects, which tend to cause interference be kept at zero speed and in position for a
to sensitive data and measuring circuits by n = Motor speed (rpm) short period of time allowing the mechanical
stray coupling paths. With few exceptions, if brake to open or close. To keep the motor in
60 = Seconds (s)
harmonics cause disturbance it is through this position the inverter injects DC into the
direct electrical connection and not through ƒ = Supply frequency (Hz) motor that causes it to produce torque at
stray paths. Screening is rarely a remedial standstill (zero speed). This type of braking is
P = Pairs of motor poles (A four pole motor
measure for harmonic problems. sometimes misunderstood as DC Bus braking,
will have 2 pairs) which is explained in the next section. When
Ignorance
Motor torque: selecting a VSD, and the applications requires
The biggest disincentive to use VSDs and enjoy this function, ensure that it is the function
their ability to improve our lives is ignorance required.
(2)
and unwillingness to change. “My motors DC Bus braking / resistive braking
worked for the last 20 years with Direct on
Line starters - why should I complicate my life where: DC Bus braking is able to control the

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deceleration of induction motors without Installation tips and tricks
activating the over voltage protection
Environmental requirements
function on VSDs. When applications require
a fast deceleration function or the load is Most VSDs are supplied with
very unstable it could be controlled with this a protection rating of IP20
function. There are various methods to solve (finger proof). This normally
the problem depending on the application. requires that the inverter needs
It could be done with a regenerative system, to be mounted in a floor-
feeding energy back to the mains or with a standing enclosure or wall
brake unit and brake resistors dissipating the mounted panel, to increase
energy through external resistors. the degree of protection.
Refer to Table 1 for standard
The main features of an AC regenerative
rating of protection.
system are:
Cooling
 Energy saving
 Expensive to install The reliable trouble-free
operation of all industrial
 The input current waveform is sinusoid
equipment is dependent upon
 The input current has a near unity power Fig. 5: Inverter’sinternal protection. operation in an environment
factor for which that product was
 The output voltage for the motor can without any effect on the DC Bus voltage designed. The single most
be higher than the available AC mains and hence on the operation of the motor significant reason for the premature failure
voltage drives of a VSD controller is operation in excessive
ambient temperature.
 The regenerative unit will synchronise  The regenerative and motor drives are
to any frequency between 30 and identical The design of the enclosure in which the drive
100 Hz, provided that the supply voltage is housed is therefore of critical importance.
When using either the internal braking
is between 380 V – 10 % and 480 V +
10 % system of the VSD with resistors or an external The following guidance covers the basic
brake unit with resistors we waste energy calculations necessary to ensure that heat
 Under conditions of AC mains instability,
unnecessarily. This, however, is the cheapest generated by a drive can be satisfactorily
a Mitsubishi drive regenerative system
solution and unfortunately selected by most transferred to the air surrounding the
can continue to function down to
approximately 270 V AC supply voltage customers. cubicle.

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DEGREE OF PROTECTION (IP) IEC529 If the enclosure is too large for the available
space it can be made smaller by:
1st Numeral Digit 2nd Numeral Digit
Protection against contact + penetration of Protection against penetration of liquids  Reducing the power dissipation in the
solid bodies enclosure
0 No protection 0 No protection  Reducing the ambient temperature
1 Protection against solid objects 1 Protection against vertical drops of water outside the enclosure
more than 50 mm
 Increasing the permissible ambient
2 Protection against solid objects 2 Protection against drops of water falling
temperature inside the cubicle if possible
more than 12 mm up to 150
by derating equipment in line with
3 Protection against solid objects 3 Protection against spraying water manufacturer’s recommendations
more than 2,5 mm
 Increasing the number of unobstructed
4 Protection against solid objects 4 Protection against splashing water
more than 1 mm surfaces of the cubicle
5 Protection against dust (limited 5 Protection against water jets  Circulating the air flow in a ventilated
ingress) enclosure
6 Total protection against dust 6 Protection against heavy seas In this case the dimensions of the enclosure
(dust-tight) are determined only by the requirements to
7 Protection against the effects of accommodate the equipment making sure
immersion to provide any recommended clearances.
Table 1: Degrees of protection. The equipment is cooled by forced airflow.
This being the case it is important is such
When making the calculation, remember to arrangement to ensure that the air flows over
accommodate the following:
take account of all power dissipated inside the heat-generating components to avoid
the cubicle, and not simply that generated  Two Mitsubishi 3,7 kW drives localised hot spots.
by the drive. Further, in the internal layout of  EMC filter for each drive The minimum required volume of ventilating
the cubicle, where possible, avoid placing  Braking resistors mounted outside the air is given by:
electronic components at the top (hot air enclosure
rises!), and where possible provide fans to
 Maximum ambient temperature inside
circulate internal air. Remember, as a rule of (4)
the enclosure 40 ºC
thumb, an electronic product’s lifetime halves
for every 7 °C temperature rise.  Maximum ambient temperature outside where:
the enclosure 30 ºC
The enclosure itself transfers the internally V = Cooling air flow (m3hr -1),
 Maximum dissipation of each drive
generated heat into the surrounding air by P = Power dissipated by all heat sources
= 190 W
natural convection, or external forced airflow. in the enclosure (W),
The greater the surface area of the enclosure  Maximum dissipation of each EMC filter
= 25 W Tamb = Maximum expected ambient
walls, the better is the dissipation capability.
Remember also, that only walls which are Total dissipation = 2 x (190 + 25) = 430 W temperature outside the enclosure
not obstructed (not in contact with walls, The enclosure is to be made from painted 2 (ºC),
floor or another hot enclosure) can dissipate mm sheet steel having a heat transmission Ti = Maximum permissible ambient
heat to the air coefficient of 5,5 Wm-2 ºC-1. Only the top, temperature inside the enclosure
the front and two sides of the enclosure are (ºC) and
Calculate the minimum required unobstructed
free to dissipate heat.
surface area Ae for the enclosure as follows: k = ratio of po / pi
The minimum required unobstructed surface
P area Ae for the enclosure is as follows Where:
Ae = (3)
k(Ti - Tamb) po = Air pressure at sea level and
P
Ae = pi = Air pressure at the installation.
where: k(Ti - Tamb)
Typically, a factor of 1,2 to 1,3 can be used to
Ae = Unobstructed surface area (m²) 430
Ae = allow for pressure drops in dirty air filters.
P = Power dissipated by all heat 5,5 (40-30)
Example
sources in the enclosure (W) Ae = 7,8 m2
To calculate the size of an enclosure to
Tamb = Maximum expected ambient
accommodate the following:
temperature outside the
If we select an enclosure with a height (H) of
enclosure (°C)  Three Mitsubishi 15 kW drives
2 m, a depth (D) of 0,6 m, and a minimum
width W  EMC filter for each drive
Ti = Maximum permissible ambient
temperature inside the  Braking resistors mounted outside the
Dissipating surfaces > 7,8 m²
enclosure (°C) enclosure
Top + Front > 7,8 m²  Maximum ambient temperature inside
k = Heat transmission coefficient
(Wmin x 0.6)+(Wmin x 2)+(2 x 0.6 x 2) the enclosure 40 ºC
of the enclosure material
> 7,8 m2  Maximum ambient temperature outside
(Wm-2 ºC-1).
the enclosure 30 ºC
Wmin > (7,8 – 2,4)/2,6
Example Maximum dissipation of each drive
> 2,1 m = 570 W
To calculate the size of an enclosure to

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Maximum dissipation of each EMC filter Typical fuse recommendations for three- where the individual phase conductors are
= 60 W phase AC Inverters are given in Table 2. surrounded by a further insulating medium
before being covered by the screen. For
To t a l d i s s i p a t i o n = 3 x ( 5 7 0 + 6 0 ) Motor and inverter protection
other values of capacitance the length
= 1890 W
Apart from the VSD’s built-in motor overload limits are approximately in inverse to the
Then the minimum required volume of protection, internal heat, short circuit and capacitance.
ventilating air is given by: ground fault protection is provided. The
Note that individual separate phase
electronic overload protection for the motor
Electronic overload relays conductors give lower capacitance, but may
also takes into account the reduced cooling
not be acceptable for EMC reasons. Cables
It is important to recognise that the non- when the motor is operated at low speeds
without forced ventilation. where the screen is laid over the phase
sinusoidal waveforms, and variable frequency,
conductors and mineral insulated cables are
associated with Inverters, invalidate the basis
Long motor cables known to have much higher capacitance
for protection afforded by most electronic
The use of chokes to facilitate the application and should be treated with caution.
overload relays. The use of such devices on
the mains supply of an inverter is also invalid. of long motor cables is rare. It is usually Maximum cable lengths at a specific supply
Please consult the manufacturer before using simpler and more elegant to use a larger voltage and switching frequency
any of these devices. inverter rating
The above cable lengths are for Mitsubishi
Fusing Because of charging currents, long motor FR-A500 series VSDs. Please consult with the
cable runs can cause a reduction in the inverter manufacturer before installing.
Fuses should not be seen as an overload
torque available from the inverter. This could
protection device; inverters regulate the For other switching frequencies:
also cause the inverter to trip due to excessive
current flowing in the system and fusing needs
current normally experienced with smaller Maximum cable length =
to be designed to cater for catastrophic
capacity inverters.
failure within the drive or a short circuit maximum cable length at 3 kHz x
between cables. High rupturing capacity The following guidelines give suggested (3 kHz / switching frequency) (5)
fuses (HRC) act as clearing devices for limits for standard shielded cables. The
Output chokes
sustained high currents and are consequently capacitance of the cable is approximately
well suited to this type of duty, and is 300 pF/m measured from the three phases Calculation of the necessary inductance of
commonly recommended by most drive together to the sheath. This is typical of steel the choke is complex. However the following
manufacturers. wire armoured cable (SWA) and similar, guidelines might be helpful.

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Inverter rating Recommended Inverter Rating Cable length (m) at where:

400 V AC. supply
Power (kW) Current (A) Motor cable Fuse rating (A) voltage and 3 kHz ƒo(max) = Maximum output
(mm²) switching frequency frequency
0,75 2,1 1,5 6 0,75 kW 70 If the loss exceeds this limit, and
1,5 3,8 2,5 10 1,5 kW 100 it is not possible to reduce the
2,2 5,6 2,5 10 2,2 kW 130 switching frequency, then a resistor
4 9,5 2,5 16 4 kW 200 should be connected in parallel
5,5 12 2,5 16 with each choke to extract some
5,5 kW – 11 kW 250
7,5 16 4 20 of the power. The resistor value is
15 kW – 90 kW 300
given by:
11 25 4 35
Table 3: Long motor cables. (11)
15 34 6 40
18,5 40 10 50
22 46 10 60 The value is not critical and variations of
will be rather lower than the
30 60 16 70 ±50% are acceptable. The power rating of
specified 50Hz inductance. the resistor should be at least 0,8P. Provision
37 70 25 80 A good rule of thumb is to must be made for the resistor to dissipate this
45 96 35 100 specify an inductance of power without overheating itself or nearby
55 124 35 125 twice that determined by this equipment. Values of 100 W per phase are
calculation. not uncommon
75 156 50 160
90 180 70 200 The acceptable voltage drop Supply impedance
at the working frequency
110 202 95 250 High quality commercial VSDs are designed
determines the maximum value
to operate from typical industrial power
Table 2: Fuse ratings. of inductance. Calculate this
distribution systems with a maximum fault
from the following expression:
Estimate the cable capacitance (C), from level of ten to twenty times the inverter’s rated
one line to all others. (8) power. Problems can also occur if a drive
system is installed close to the main power
Typical values
supply or power factor correction capacitors,
Screened/armoured cables where there is a where: both of which present low supply impedance
plastic sheath between the phase and the to the drive. AC inverters fitted with DC link
x = Acceptable volt drop over the
screen: 130 pF/m chokes are in general unaffected by low
choke
supply impedance.
Screened cables with no plastic sheath
between cores and screen, mineral insulated VAC = Motor voltage rating (line to line A good rule of thumb is to ensure a total
cables: 300 pF/m RMS) supply impedance of approximately 4%
f0 = Maximum inverter output reactance. Where information about the
Add an allowance for the motor capacitance:
1 nF is a reasonable estimate supply is not known, it is good practice to fit
frequency
line reactors of 2%.
Decide on the available charging current
There is no easy solution for high supply
(6) If 2Lmin < Lmax then any value between these impedance as power lines carrying drive
limits can be used. current need to be oversized, as do
Where: transformers used in order to minimize the
If 2Lmin > Lmax then the inverter cannot operate impedance. This over sizing may need
In = Nominal rated r.m.s output current
with this length of cable and a higher rated to be as high as five times that normally
k = Acceptable short term overload inverter must be used. considerate adequate.
factor
Consideration must be given to the high Brake resistor selection
Factor 1,41 = Maximum DC link frequency losses in the chokes. This can be
Firstly we need to understand what happens
voltage estimated from the following expression:
when a motor decelerates to stop under
Factor 2 = Inverter instant trip current P = 0,8fsCVDC (9) a high inertia load. When the motor starts
to nominal output current to decelerate and the load keeps on
where:
rotating the motor, the motor starts to act
Most inverters are rated at 150%, ƒs = Switching frequency as a generator sending energy back to the
i.e. k = 1,5 VSD. With energy from the main supply and
The factor 0,8 is a rough estimate of the
Maximum DC link voltage can be calculated from the motor, now acting as a generator,
fraction of the total losses dissipated in the
from the highest RMS AC supply voltage the DC Bus voltage level starts to rise from
choke. Note that the loss is proportional to the
times 1,41 approximately 560 V DC to levels that will
switching frequency so the lowest acceptable
activate a DC Bus over voltage trip. Most
Now calculate the minimum inductance per frequency should be selected. inverters can sustain levels up to 780 V DC
phase from the following expression It is now necessary to decide whether the before tripping. Some inverters equipped
with internal braking circuits and resistors will

[ ] (7) choke is able to tolerate this loss. This is a

2C Vdc control this rise in DC Bus voltage but only
Lmin = difficult judgment. As a crude rule, the loss
3 Ich for a short duration. If a lengthy or heavy-
should not exceed 0,1 VA in the choke at
duty brake cycle is required, it is best to fit
If using standard iron-cored chokes, the maximum speed, i.e.:
external units that could cope with the extra
inductance at the high frequencies involved P ≤ 0,2� f0(max) LIn3 (10) energy safely.

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Name of Application Load torque as percentage of full load driving

To reduce the cost and physical size of VSDs,
torque and the fact that very few applications require
DC Bus braking, manufacturers only provide
Breakaway Accelerating Peak running
units with limited DC Bus braking capabilities.
Users should consult with manufacturers as
Liquid 100 100 100 to the capability of the units before installing
Agitators them in these applications.
Slurry 150 100 100
Two important factors to consider are the
Valve closed 30 50 40 required braking torque and the duty cycle
Blowers, centrifugal of the application.
Valve open 40 110 100

Blowers, positive-displacement, rotary, bypassed 40 40 100 Kinetic energy of the motor and load
= 0,5 J ω2 (12)
Centrifuges (extractors) 40 60 125
where:
Compressors, axial-vane, loaded 40 100 100
J = Total inertia (kg m²) of the motor,
Conveyors, shaker-type (vibrating) 150 150 75
transmission and driven machine
Escalators, stairways (starting unloaded) 50 75 100
ω = angular velocity
Valve closed 25 60 50
Fans, centrifugal, ambient
Valve open 25 110 100 If there is gearing between the motor and
the driven machine, J is the value reflected
Valve closed 25 60 100
Fans, centrifugal, hot gases at the motor shaft.
Valve open 25 200 175
As the energy regenerated is proportional to
Fans, propellers axial-flow 40 110 100
the square of the angular velocity, most of the
Frames, spinning, textile 50 125 100 energy in the system is concentrated at the
Grinders, metal 25 50 100 higher operating speed, and is delivered to
the resistor at the start of the deceleration.
Machines, buffing, automatic 50 75 100

Machines, cinder-block, vibrating 150 150 70 Example:

Machines, key seating 25 50 100 A system inertia of 10 kg m 2 is to be

Machines, polishing 50 75 100 decelerated from full load speed to rest. We
need to find the braking resistor value and
Mills, flour, grinding 50 75 100
maximum braking torque
Mixers, chemical 175 75 100
Drive rating = 30 kW
Mixers, concrete 40 50 100
Motor rating = 30 kW
Mixers, dough 175 125 100

Mixers, liquid 100 100 100 Motor full load speed = 1475 rpm

Mixers, sand, centrifugal 50 100 100 Motor nominal torque rating = 191 Nm
Mixers, sand, screw 175 100 100 Repeat cycle time = 30 seconds
Mixers, slurry 150 125 100
Resistor operating voltage = 660 V
Mixers, solid 175 125 175
Our first step is to determine the minimum
Pumps, adjustable-blade, vertical 50 40 125
deceleration time tb
Pumps, centrifugal, discharge open 40 100 100
Maximum braking torque Mb is identical to
Pumps, oil-field, flywheel 150 200 200 maximum accelerating torque
Pumps, oil, lubricating 40 150 150
ω � n
Pumps, oil fuel 40 150 150 Mb = Ja = J = J NM
tb 30 tb
Pumps, propeller 40 100 100 (13)
�Jn
tb =
Pumps, turbine, centrifugal, deep-well 50 100 100 30Mb
Pumps, vacuum (paper-mill service) 60 100 100

Pumps, vacuum (other applications) 40 60 100 But maximum deceleration occurs at 150%
Pumps, vacuum, reciprocating 150 60 150 of motor nominal torque. The value to apply
for Mb is therefore
Rolls, crushing (sugar cane) 50 110 125

Rolls, flaking 30 50 100

1,5 x 191 = 286,5 Nm

Screens, centrifugal (centrifuges) 40 60 125 So the actual deceleration time tb is:

Screens, vibrating 50 150 70 10�x1475
tb = seconds
Separators, air (fan-type) 40 100 100
30x286,5

Washers, laundry 25 75 100 = 5,39 seconds

Table 4: Application load torque percentages This is the minimum deceleration time. For

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APPLICATION
example, let us decide on a more practical Pb 35 = 17,5 kW
Pr = =
deceleration timex of 7seconds. We can O / L Factor 2 (15)
now calculate the maximum braking torque
For practical purposes, it can be assumed
to decelerate the load in 7 seconds: that 15% to 20% of energy dissipated
� n 10�x1475 during the regenerative braking is due to
Mb = J Mb = Nm
30 tb 30x7 electrical losses in the motor and inverter, and
mechanical losses in the motor and load, all
= 220,64 Nm of which assist the braking. In practice, using
the recommended resistor value will result in
Braking Power is:
extra braking torque available. However, the
rate of energy feedback from the load inertia
n Mb n � x 224,38x1475 is determined by the rate of deceleration.
Pb = Nm =
30x103 30x1000
A braking resistor must be installed in
= 34,65 kW accordance with instructions provided by its
supplier or manufacturer. The braking resistor
Resistor resistance value (Ω): should incorporate a thermal tripping device,
The motor is capable of delivering up which should be connected, to a trip release
to 150% of its full load rating for up to mechanism to stop the VSD.
60 seconds maximum, i.e.: Resistors intended for braking duty should be
capable of tolerating thermal shock. “Pulse
1,5 x 30 = 45 kW.
rated” resistors are recommended.
It is therefore equally capable of regenerating
Application selection
the same short time power. As 45 kW is in
excess of 35 kW required for the application, The following table is a guide to torque
the value of the resistor is: requirements in certain applications.

V2 6602 (14) References

R= = = 12,45Ω
Pb 35x1000 [1] P. Sala and F. Molinelli:”UFS Braking units”

Resistor power rating Pr Static Control Systems, Revision 5, pp 1-21,

December 2001
As the braking operates intermittently, the
[2] ”Industrial Automation and Control
resistor can be selected from a range offering
Equipment” Mitsubishi Electric, page 26,
“intermittent” rather than “continuous” power January 1997
absorption. Advantage can also be taken of
[3] G.R. Jones, M.A. Laughton and M.G. Say:
the overload rating of the resistor by applying
”Electrical engineer ’s reference book”
an overload factor, which will be derived
1993
from a set of cooling curves obtained from
[4] “FATEC Inverter application course”
the manufacturer or supplier of the resistor. In
Mitsubishi Electric, 2000
this example, deceleration time is taken as 7
seconds, repeat cycle time 30 seconds. From Contact R G van der Merwe, Motiontronix,
rudi@motiontronix.co.za;
typical data the overload factor is 2.
C Hoogendoorn, Circuit Breaker Industries,
The power rating of the chosen resistor is: cbi@cbi.co.za 

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