Blog – Page 13 – AC DRIVES SOLUTION.

AQ: Snubber circuit for IGBT Inverter in high frequency applications

Q:
First i had carried out experiments with a single IGBT (IRGPS40B120UP) (TO-247 package) 40A rating without snubber and connected a resistive load, load current was 25A , 400V DC and kept it on continuously for 20min . Then i switched the same IGBT with 10KHz without snubber and the IGBT failed within 1min. Then i connected an RC snubber across the IGBT (same model )and switched at 10KHZ. The load current was gradually increased and kept at 10A. This time the IGBT didn’t fail . So snubber circuits are essential when we go for higher switching frequency.

What are the general guide lines for snubber circuit design in You are not looking close enough at the whole system. My first observation is that you are using the slowest speed silicon available from IR. Even though 10KHz is not fast, have you calculated/measured your switching losses. The second and bigger observation I have, is that you think your circuit is resistive. If your circuit was only resistive, any snubber would have no effect. The whole purpose of a snubber is to deal with the energy stored in the parasitic inductive elements of your circuit. Without understanding how much inductcance your circuit has, you can’t begin designing a cost and size effective snubber.

To echo one of the thoughts of Felipe, you need to know the exact purpose of the snubber. Is it to slow down the dV/dt on turn off or is it to limit the peak voltage? Depending on which you are trying to minimize and your final switching frequency, will dictate which snubber topology will work best for you. The reason that so many snubber configurations exist, is that different applications will require different solutions. I have used snubbers in various configurations up to 100KHz.

AQ: Earthing conductors size calculation

1. As per IEE/BS7671, The Minimum cross-sectional area of protective conductor in relation to the cross-sectional area of associated line conductor (Say =S mm2) by taking into consideration that are of same material, as follows:

If S ≤ 16 then the Minimum cross-sectional area of the corresponding earth conductor = S mm2

If 16 < S ≤ 35 then the Minimum cross-sectional area of the corresponding earth conductor = 16 mm2

if S > 16 , then the Minimum cross-sectional area of the corresponding earth conductor = S/2 mm2

2. It may be necessary to verify the same by using the following equation
I²t ≤ S²K

I earth fault current and t tripping time.

while the following equation is applicable for bonding conductor

Zs < 50/Ia where Zs = Earth loop impedance and Ia is protective device operating current

AQ: 480volt Solidly grounded system versus HRG system

High Resistance Ground will limit the current to about 5 amps. The good news is that it no longer be necessary to trip on a ground fault. The bad news is that you may not connect any single phase loads to that substation. If the single phase loads are an issue, it may be possible to support all those loads with one or two feeders. In that situation, an isolation transformer is added to create a separately derived ground.

High resistance grounding is an excellent option in systems where continuity of service is important. However it is important to understand that if a ground fault occurs, it needs to located and repaired. This can be at a time convenient to facility operation, but it must be chased down and fixed.

This troubleshooting is accomplished by pulsing the ground fault current between 5 and 8 amps. Then a hand held clamp on ammeter is used to search out the offending feeder. Most HRG manufacturers include this feature into their design.

480volt solid ground system the fault current L-g is limited to the fault rating of the system may be 50KA,and usually have a disastrous consequences during faults and system has to be properly selected and protected accordingly.

Presently for some emergency power systems in power plants there is a 480voltb 3ph supply is made with neutral grounded through a NGT ie to limit the L-g fault current within 10Amp, and this may result in overvoltage of the other two phases too. Overall this becomes a more stable system.

I would be very hesitant to use HRG for a medium voltage system. While a fault is on the system the neutral will not be at earth potential. This is usually not so important because the neutral is not carried to any of the loads on a high resistance grounded system. But the effect of the neutral straying away from ground potential is the two unfaulted phases will have a higher than normal line-to-ground voltage. Low voltage systems have a lot of built in margin in their insulation so it is not a problem with most equipment connected to that system. However MV equipment does not have so much “spare” insulation so the effect of high resistance grounding of a MV system is that it significantly increases the chance of migrating the fault from a single-line-to-ground fault to a double-line-to-ground or line-to-line fault.

Also at the point of the fault, the energy of 5 amps flowing in a 480 volt or 400 volt system may be somewhat dangerous but in most cases will be dissipated easily. On a higher voltage system, whether it is 4.16kV, 6.6kV, or higher, the energy at the point of the fault is higher and is much more likely to damage the insulation on the adjacent conductors and quickly turn into a more severe fault.

So I would resist using a high resistance grounded design on a medium voltage system (over 2000 volts).

AQ: How to connect 3 phase motor?

Making a connection of 3 phase motor the nameplate shows different voltages for delta it is 380-400 volt and 660-690 volt for star, what option should be selected? the supply Line to Line voltage is 380-400.

Each stator winding of the motor can withstand 380-400 V.
Thus, if you connect your motor (the stator of your motor) in delta, it should be connected to 380-400 V line-to-line.

On the other hand, if you connect the stator winding of your motor in Y, you’d be able to connect your motor to line-to-line voltage that is sqrt(3) x 380-400 V = 660-690 V.

The actual output power (for a standard squirrel cage 3-phase AC motor) is not determined by the motor itself, but by the load it is driving. The motor will attempt to run at a speed near its synchronous speed, and to deliver the power required by the driven machinery at that speed. This means that the current taken up by the motor at any given voltage, will be almost the same whether it is star, or delta connected. If you therefore connect the motor in star while supplying it by the voltage it is designed for when delta connected, the current through each winding will be sqrt(3) times the winding is designed for. This again means that the heat dissipation in the winding will be approximately 3 times what it is designed for, and therefore it will burn out if you load the motor with its nominal load.

We should be aware that the motor power as mentioned on its nameplate, in relation with the available power of the MCC panel to which it is connected, are the important factors in choosing the type of starting of the motor. Take into account the fact that starting the motor direct in Delta connection (which is the correct one based on your network voltage) the currents may be up to 8xInomianl of motor and if your MCC doesn’t have the capacity to withstand this current (by decreasing its supply voltage ) you may fail with DOL Delta starting type. Is that why, based on the power of motors, in order to avoid high currents during the starting time, it is recommended the Y/D connection. Limitations in starting currents by Y/D are considerable by decreasing the current first with sqrt3 because the feeding voltage is not 660V (you feed the motor with 380-400V) and the current initially in Y is sqrt3<I delta, so it is 3 times less than Delta DOL. Y/D is not the single one, there are a lot of solution to start AC motors.

AQ: What need to be concerned to start a motor?

First, you need to know power (rated power and rated current) of your power source with whom you will supply your motor. For example, if you want to supply your motor by using low voltage synchronous generator (through high voltage power transformer), you need to know rated power and rated current of synchronous generator and rated power and rated currents of high voltage power transformer. This information is very important because if you don’t have powerful source for supplying your motor, there is possibility that you’ll never reach rated rotational speed during rated time which means that you’ll not start your motor.

Second, you need to know kind of your motor. Is that motor asynchronous motor with cage rotor or is that asynchronous motor with sliding rings? This information is very important because these kinds of asynchronous motors have different values of starting current: for asynchronous motors with cage rotor starting current is 6-8 times higher than rated current of mentioned kind of motor while for asynchronous motors with sliding rings starting current is 3-5 times higher than rated current of mentioned kind of motor. Also, too much higher starting current of your motor could be a reason for unallowed warming of windings of stator what it could lead to dangerous consequences, first all, for people in surrounding of motor and then also for equipment in surrounding of motor.

In relation with start of your motor with lower voltage because you will, on that way, reduce starting current 2 times and starting torque will be 4 times lower than rated torque of your motor. On that way, you will easily start your motor.

AQ: AC induction motor constant power

An AC induction motor is supposed to be a constant power motor, which implies it draws more current on low voltage. Consider a motor running a constant torque load at a particular speed. Suppose now the voltage is reduced, which should cause it to settle down at a lower speed supplying the same torque as per the new torque speed characteristic. If we consider the electrical side, higher slip will cause more current to be drawn that too at higher pf, which should maintain the power justifying the above theory. But on the mechanical side the new output power Torque x speed is supposed to be lesser now as speed is less now. Is it this contradiction?

The following guidelines prove there is no contradiction since your question about Motor under running condition:

1. Torque / Slip characteristic for Induction Motor has three Zones.
a)- Starting Torque @ S=1, selection of this torque depends on the application. The starting should be greater the system torque at time of starting.
b)- Unstable Zone during which acceleration and torque development took place. This zone up to the Max. Torque can be developed.
In this regard, it may be necessary to mention that the seventh harmonics to be considered otherwise crawling / clogging may occur.
c)- Normal Operating Zone. NOZ about which your query raised. NOZ ranged as ” 0 < S< 1″ ie up to the Max. Torque. It is worth mentioning that Max. torque always remain the same regardless to its location of occurrence.

2. The torque is directly proportional to rotor resistance “r2” & varies with slip “S”. hence increase of rotor resistance is the most practical method of changing the torque (ie wound rotor Slip ring Motors). Moreover, the Max torque achieved when rotor Resistance “r2” = The Stator impedance, At starting S=1.

3. Accordingly, the ration r2/x2 gives the location of the max. Torque w.r.t Slip (if the max. torque is required at starting (S=!) then r2/x2 should equal “1”.

4. load being constant. Mechanical output = Electrical input – losses.

5. Tmax Propotional to Sq(v). decrease of 50% of the supply voltage generate a reduction of 20% in the max. running Torque (zone c) , increase in slip and also Full load current and temperature raise increase while the full load speed decrease. the status of the above parameters will be opposite if the voltage increases by 10%.

Based on the above, in all cases since the Motor is running within the operating range will be no issue unless the supply voltage falls behind the above limits (-50%, +110%). Accordingly, variable frequency drives provided by under/Over voltage protection relay to avoid damage to insulation due to Heat/temperature rise that will be generated due to excessive current intend to composite load.

AQ: How to select a drive between motor and machine?

We should select a drive (direct/flexible, chain, flat/vee/ribbed belt, gearbox, soft start). The motor/starter/drive characteristics should match that of the load. Design and factors to be considered in selection.

AQ: VFD replace mechanical gearbox to drive the load

Can an AC drive to replace the mechanical gearbox that used to decrease motor speed in conveyor application i.e to use a motor that will drive the load directly throw a coupling, belt or chain, without gearbox, motor rated up to 18.5 kw.

Theoretical is true as far the speed variation is concerned. Practically is not recommended for your application if the conveyor is required to be used with constant speed, on the other hand the gearbox also used for Torque purposes.

For light conveyors used on packing lines on which rate of production varies in accordance to some industrial parameters (Automation & PID control), direct coupled motor controlled by variable frequency drive may be feasible.

VFD is expensive (capital & running cost) its selectivity should be done carefully among the other available options.

By using a variable frequency drive we can change the speed of an AC Motor, and working for any time on any choosing speed, even in some case we can exceed the speed more than the normal one if the motor can withstand it. Noting that:
1- We should be careful when choosing the type of AC drive that should ne normally done according on the application “Conveyor, Fan, Pump, Compressor, ext ” to determine the torque’s level at running time.
2- In some special case when the motor runs at too low speed comparing by his normal one, maybe we need a forcing cooling for that motor.
3- Each VFD has a value of the Short Circuit’s level that can be withstanded, so, we should be careful of that point.

AQ: MCCB burn out in connection with 22kW motor

125A rated MCCB is connected with 22KW motor. The motor runs in normal condition, no overload. But the contacts of MCCB is burnt out. Why?

When the transition between wye and delta takes place, be very sure that that the mechanical interlocks on the contractors are properly adjusted. If one doesn’t completely open before the other closes, you have a line to line fault. That will eventually take out the upstream breaker. Be careful, this is a very dangerous starter. I have been done this road many times.

“Star/delta switching” is delicate maneuver.
* The engine has almost no force to push something more than itself at startup. When it’s time for mode switching start to delta, may not happen too quickly. The arcs that occurs when the star contactor switch off, needs a few milliseconds (minimum 20 ms). Typical of an MCCB to go off is 50 ms and on is 20 ms.
* If the load is too high during startup, the engine will get overloaded when switching from star to delta and arcs can become very powerful and devastate even the strongest Components. In case the engine does not start at idle, the start method is directly harmful to both motor and power grids. Instead of a soft and comfortable start, we get instead two powerful surges. (In that case it may be even better with only one MCCB).Should this be the case, there are two solutions: Make sure the engine starts without load or replace the Star/Delta switch to a soft starter.
3: Next step is a frequency converter but then it’s about something completely different.

AQ: How to find the KA rating of Circuit breakers?

Before breaker’s selecting for your electrical system, you need to calculate value of expected short circuit current at the place of breaker’s installation. Then you need to calculate value of heat pulse and 1s current (expected value of current during one second). After that you need to calculate power of breaker and finally, after all, you can select appropriate breaker. Values of characteristics of selected breaker need to be higher from calculated values of characteristics of your power system.

1. The fault level of the upstream NW (Source) to be known, normally 500MVA or 250MVA.
2. Upstream impedance (reactance and resistor, capacitor to be ignored for Short cct calculation) can be determined accordingly.
3. The LV System starting from the secondary of the distribution transformer.
4. Short circuit percentage voltage for Transformer is known (normally 4% for 1000kVA and 6% for 1500 kVA) and hence reactance and impedance can also determined.
5. Impedance of Cables also can be determined from manufacturer TD sheet.
6. Subtotal impedance to be determined by conventional way (Submit if are in series/ (Z1+Z2+….Zn)/(Z1XZ2X…Zn) if are in parallel.
7. divide voltage by the Impedance up to the required location, will give you the fault current at that location.
8. Determine Maximum and minimum fault current. By the former you can decide the breaking capacity of CB and by the later the setting can be achieved.
9. verify the thermal constraints of the conductors(cables). ie

I²t ≤ S²K² , I short cct current, t time( < 5 s valid), K cable material Factor and S cable section area.
I²t Known as let through energy. accordingly breaking capacity of CB should be > than Circuit Maximum fault.

The MCB, MCCB, & ACB are all Low Voltage Circuit Breakers, where SF6 is a Non-active gaze used in Medium Voltage Circuit Breakers.

Now, to determine the value of Breaking Capacity of any circuit breaker, we should, by calculation, the Maximum Short Circuit Current Value ” Isc3max ” at the installation point of that circuit breaker, where we can calculate it by assuming a ” Short Circuit between 3 phases at that point “, then after knowing ” Isc3max ” we can determine the Breaking Capacity value that should be ” equal or bigger than Isc3max “.

Further:
1- The value ” 250 … 500MVA ” is the short circuit power at Medium voltage side for up to 36kV.
2- About the Short Circuit Voltage percentage value: we called ” Ucc or Usc ” and the value is ” 4& for up to 630kVA transformers “, and ” 6% for up to 2500kVA transformers “, but in all case, we can read it at the transformer’s name plate.
3- Sorry Mr. Omar, we can’t do, you mentioned, the sum of all Z, as these values aren’t on the same vector, so, we should first calculate ” R & X ” for each component, then do the sum of all R ” R total ” and all X ” X total “, then calculate the ” Z total “.
4- By knowing the Minimum Short Circuit Current value ” Iscmin “, we use it to determine the value of “Setting Value” of “Magnetic Protection or Short-time Protection”.