Iacdrive|Automation Solution

Variable frequency drive Constant Torque/Variable Torque

A typical variable torque application would be a centrifugal pump. A typical constant torque application would be a conveyor, and there are positive displacement pumps that are also constant torque. Have a talk with a mechanical engineer, get them to show you curves and explain.

DBR stands for Dynamic braking resistor. Regeneration will happen when the motor rotates a speed higher than the speed which corresponds to the frequency setpoint ie.. the rotor speed is more than the speed of the rotating magnetic field.
Regeneration feeds back energy to the drive which results in DC bus overvoltage. To prevent the drive from tripping due to DC bus overvoltage the DBRs are used. The regenerative energy is discharged in the resistor as heat.

Regenerative Breaking – we used to have VFD on a vehicle rolling road. So when the car is travelling faster than the VFD, the VFD generate back into the power supply – causing a break effect. If you had a large mass- large inertia that you want to stop quickly, you need to break the load- you can do that with regenerative breaking. Otherwise, disconnecting the variable frequency drive, will mean your load just freely rotates, and that can mean it will take 30 minute to come to a stop for a large inertia.

Active Front end- I first came across this term with ABB. It is all to do with how to mitigate harmonics from VFDs. You can use phase shift transformers, but with modern electronics, you can use a opposite phase current to counter act the harmonics generated from the VFD. So the overall impact on the network is small.
In active front end technology the rectifier is basically an inverter with IGBTs.
The main advantage are:
1) Low current THD <5 %
2) It is basically a four quadrant rectifier .Referring my last post please note that you will not require a DBR with AFE. The increase in voltage of DC Bus due to regeneration can be fed back to the input AC supply in the form of energy. So you don’t require a DBR.
3) AFE drives have very good immunity to input voltage fluctuations.

Just an advice. Please go through variable frequency drive literatures (available in plenty) to have a good understanding of the different VFD technologies.
Selection of VFD requires proper understanding of the VFDs and the overall electrical system. There are lots of marketing gimmicks in the world of VFD. Always be careful before selecting a VFD specially higher KW drives.

For large drives, you need to speak with supplier to configure your machine correctly. There are many options, but yes active front ends are available. But there are other solutions; ASI Robicon use a current driven VFD, so harmonics are lessened in the first place, so an active front end is not the right terminology. It is a different solution. I used a 10MW version of that type of ac drive. I think Siemens have bought the company since.

VFD PWM and PAM definition

PWM is shorted for Pulse Width Modulation, it’s a variable frequency drive (VFD) regulate way to change the pulse width according to certain rules to adjust the output volume and waveform.

PAM is shorted for Pulse Amplitude Modulation, it’s to change the pulse amplitude according to certain rules pulse amplitude pulse train to adjust the variable frequency drive output volume and waveform.

Change transformer vector group

Transformer nameplate vector group is YNd1. However, the nature of connection on both its primary and secondary side is such that:
Generator phase A = Transformer phase c
Generator phase B = Transformer phase b
Generator phase C = Transformer phase a

Also, on transformer HV (secondary connected to grid),
Transformer phase A = Grid phase C
Transformer phase B = Grid phase B
Transformer phase C = Grid phase A

The questions are:

1. How does this affect the vector group (YNd1) of the transformer? Will it be changed to YNd11?
2. Will it make any difference as far as the vector group is concerned if instead of phase A and C, phase B and C were swapped on both ends of the transformer?
3. The transformer protection relay is configured for YNd1 group, and it is reading negative phase sequence current (ACB instead of ABC). Changing the vector group configuration will solve the problem?
4. Relay is used for differential protection (percentage differential) of the transformer.
Will this negative phase sequence affect normal operation of the transformer in any way?

1. How does this affect the vector group (YNd1) of the transformer? Will it be changed to YNd11?

Yes, the name plate vector group of a transformer is only valid for a standard phase rotation ABC. for a phase rotation ACB the apparent vector group will be YNd11.

2. Will it make any difference as far as the vector group is concerned if instead of phase A and C, phase B and C were swapped on both ends of the transformer?

No, by swapping any two phases the rotation becomes no standard and the apparent vector group will become YNd1

3. The transformer protection relay is configured for YNd1 group, and it is reading negative phase sequence current (ACB instead of ABC). Changing the vector group configuration will solve the problem?

I think the way the relay is configured at the moment will give you problems, if I’m correct you should be able to see differential current when the transformer is loaded, and it is likely to trip on the first through fault (can you confirm this). To resolve this issue you have two options.
i) Set the vector group to YNd11 in the relay, this will remove the differential current but will mean the relays see’s 100% NPS current and 0% PPS current, this may give you problem if you have any NPS elements enabled in the relay ( inter turn fault detection, directional elements etc)
ii)Set the vector group to YNd1 and the phase rotation setting to non standard ACB this will get rid of the NPS currents and the differential current, so this is probably the best solution.

4. Relay is used for differential protection (percentage differential) of the transformer.
Will this negative phase sequence affect normal operation of the transformer in any way?

No, there will be no problem with the transformer itself just the relay protecting it.

As i said previously if I’m understanding the problem correctly, you should be able to see differential current at the moment when the transformer is loaded, is this correct?

Why industrial induction motor star point not grounded?

In any electrical system, we limit the neutral grounding to 1 or 2 locations at the power source, eg, the star-points of generators or transformers. By keeping the grounded neutrals at the power source, earth leakage current will be flowing radially from the power source to the point of short-circuit at downstream. In this way the direction of earth fault current flow can be easily identified and the earth fault protection relays in the distribution system can easily be coordinated.

Grounding a motor star point will create an earth path for earth leakage current to flow through that motor’s star point. If there are 10 motors in a process plant and their star points are all grounded, there are 10 additional paths for earth leakage currents to flow through.
If all the motors’ star points are grounded in this way the earth fault current detections by the protection relays will be complicated and likely they will trip at the incorrect locations because earth fault currents are flowing in many directions toward multiple grounded neutral points.

Therefore the electrical consumers (ie the load, including the capacitor banks), even if they are star connected, are not to be grounded.

Grounding of neutral point is not being decided base on the presence of unbalance loads. It is decided for safety reason and for earth fault protection requirement. Unbalance 3-phase load will result in some current flowing through the neutral conductor but it doesn’t result in a (residual) current flowing through the neutral-ground connection.

Motor is a balanced 3-phase load, this I agree. However when the system supply voltage is unbalanced caused by unbalanced loads somewhere else or due to network conductors problem, the motor operating under unbalance voltage will result in unbalance current in the 3 windings. The same is true for the generator windings under that condition. The design engineer may then decide that individual machines should be fixed with negative phase sequence current protection.

Even if there is a neutral voltage shift in the induction motor, we should not ground the motor’s neutral point. If you ground it, it may create nuisance trip of earth fault protection relays (the motor’s EF relay, upstream EF relays, or the EF relay connected to transformer’s neutral-ground CT).

I am sure in reality, there is some neutral voltage shift in motor’s star point. However, there is no harm with that.

If you ground the star point, you still will not get rid of the unbalance current/voltage from the motor windings. There the negative sequence current is still present in the motor winding.
If you think an unbalance voltage supply is causing problem to the motors, you should solve the unbalance voltage problem elsewhere, not by grounding the motor’s star point.

Neutral current is less than phase current?

In a balanced 3-phase system with pure sine waves, the neutral current is zero, ideally.
If there is phase imbalance, it shows up in the neutral, so check for imbalance.

The other major cause of high neutral currents is full wave rectification, where the current of each phase is flowing only at its peak voltage. In this case, the neutral current can be as high as three times the phase currents, theoretically.

If you can see the frequency of the neutral current, line frequency currents indicate imbalance. Current due to full wave rectification is high in third harmonics, so it may show mostly 3 x line frequency, or be a ratty square wave at 3 x line frequency.

High neutral currents, and some resulting fires, are largely responsible for the adoption of power factor correction requirements. If your loads are balanced and pfc corrected, you should not have neutral currents.

The neutral current (In) is summation of the phase currents. And obviously, the three phases are decoupled now; and not loading Y makes Iy=0.
So In = Ir + Ib (vectorial sum). Now depending on the amount of loading, nature of loads and their respective power factors, a variety of possibilities (for neutral current magnitude and phase) arise; which may include the case of In being higher.
The statement “neutral current is usually less than phase currents” is naive and not universal.

Nonlinear loads (i.e. rectifiers as Ed mentioned above) draw significant harmonic current. In many cases the current Total Harmonic Distortion (THD) is >100%. In a 3-phase, 4-wire system, the triplen harmonic currents (3, 9, 15, 21…) sum in the neutral wire because they are all in-phase. This is why the neutral current can be much higher than the phase currents even on an otherwise balanced load application. If you can put a current probe on the neutral and look at the waveform – you can see how much fundamental vs. harmonic current there is.

What is the surge impedance load

The surge impedance loading (SIL) of a line is the power load at which the net reactive power is zero. So, if your transmission line wants to “absorb” reactive power, the SIL is the amount of reactive power you would have to produce to balance it out to zero. You can calculate it by dividing the square of the line-to-line voltage by the line’s characteristic impedance.

Transmission lines can be considered as, a small inductance in series and a small capacitance to earth, – a very large number of this combinations, in series. Whatever voltage drop occurs due to inductance gets compensated by capacitance. If this compensation is exact, you have surge impedance loading and no voltage drop occurs for an infinite length or, a finite length terminated by impedance of this value (SIL load). (Loss-less line assumed!). Impedance of this line can be proved to be sqrt (L/C). If capacitive compensation is more than required, which may happen on an unloaded EHV line, then you have voltage rise at the other end, the ferranti effect. Although given in many books, it continues to remain an interesting discussion always.

The capacitive reactive power associated with a transmission line increases directly as the square of the voltage and is proportional to line capacitance and length.

Capacitance has two effects:

1 Ferranti effect
2 rise in the voltage resulting from capacitive current of the line flowing through the source impedances at the terminations of the line.

SIL is Surge Impedance Loading and is calculated as (KV x KV) / Zs their units are megawatts.

Where Zs is the surge impedance….be aware…one thing is the surge impedance and other very different is the surge impedance loading.

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.

AC motor maximum torque

As per Torque/Slip characteristic for AC Motor, the value of the Max. Torque can be developed is constant while the Starting Torque occurs @ S=.1, (T proportional to r2 and S also proportional to r2 where r2 is the rotor resistance, the ratio r2/x2 when equal to 1 gives the max. Torque w.r.t Slip at Starting. Wound rotor motors are suitable and recommended for application for MV drive where it is required to be started on load such as ID. Fans, S.D Fans, Drill, etc.

As you aware the torque is directly proportional to the rotor resistance “r2” & varies with slip “S”, hence injection of resistance into the rotor via Slip Rings, High Starting Torque can be got while the Speed, efficiency and Starting current will be reduced. Therefore resistance is the most practical method of changing the torque (i.e. wound rotor Slip ring Motors). Moreover, the Max torque can be achieved at starting when rotor Resistance “r2” = The Stator impedance, at starting S=1.

On the other hand, the slip of the Induction Motor (speed) can be changed by “extracting” electrical power from rotor circuit, more extraction increases the slip. By using thyristorized Slip-Recovery Scheme “ i.e Kramer Scheme” feedback of Power from rotor circuit to supply circuit which also known as “the slip Power recovery scheme”. The scheme is simply consists of rectifier and an inverter connected between slip-rings and the A.C Supply circuit. The Slip Rings voltage is rectified by the rectifier and again inverted to AC by the inverter and feedback to supply via a suitable Transformer. Such arrangement gives good efficiency with high cost due to Rectifier and Invertor.

System configuration of grounding

There are different types of system configuration for grounding like TT,IT,TN-C etc. How do we decide which configuration is suitable for the particular inverter (string or central). What are the factors that help us to decide the configurations?

One of the main concerns in a system is to avoid large low impedance ground loops.
These are created by the return signal path connected to the chassis (metal work) at multiple points. The large current loop allows noise currents to radiate H fields and hence couple into other electronics. The antenna effect will be proportional to loop area.

Single point grounding of the return path to chassis prevents this. However single point grounding conflicts with good RF practice where you want to ground to chassis at the sending and receiving ends of a signal path. There is therefore no universal best practice.

In my field, spacecraft, the standard practice has all primary power electronics galvanically isolated from the spacecraft chassis. Individual modules must maintain the isolation with transformer coupled DC/DC converters. The centre tap of each PSU secondary output is then single point grounded to the module metalwork. We talk about primary side and secondary side electronics where only secondary side is grounded to the metalwork.

Anything powered directly from the primary bus must be isolated with a maximum capacitance to chassis of 50nF to avoid excess HF loop currents forming.

In general it depends on country specific law and standards required by Power Supply Operators. From design point of view it all depends on which point of grid you are going to connect and what type of inverter is used.

Starter of SAG Mills with rotor resistance

Q: For now I am working on a mining project which involves starting two SAG mills, the method of starting these mills is by rotor resistance and likewise we are using an energy recovery system (SER), could someone tell me how this system works SER? Each mills have two motors of 8000 kW at 13.8 kV.

A: For large mills requiring variable speed, the wound rotor motor and SER drive are economical for a total rating of approximately 2MW to 16MW. Above 16MW, the gearless drive (cyclo-converter) is typically used because gearboxes and pinion gears reach their present limit in size. Around 2MW and below, the squirrel cage/VVVF drive is simple and cost effective.

Advantages of the wound rotor/SER drive are:
1. If the SER converter drive fails, the drive can be switched to fixed speed bypass – starting the usual way with the LRS.
2. The converter only needs to be sized for 15-20% of the total motor rating with associated reduction in floor space, air-conditioning etc. The converter is only sized for the feedback energy which is proportional to the speed difference from synchronous speed. The drives are typically set up to run between about 85% to 110% of synchronous speed for an optimized arrangement.
3. Relatively low capital cost when all things considered – including spare motor cost etc.

Brush/slip ring maintenance is one issue. However, when the brushes are specified correctly for the load, the wear is manageable. Once the maintenance program is set up for shutdowns, it is not a major issue.
I expect that this type of drive would be the most common large mill variable speed drive in the world’s minerals processing industry for the range mentioned above for the last 15 years (approximately).

The SER drive converter controls the voltage in the rotor. Motor speed is proportional to rotor voltage. Resistance in the rotor indirectly achieves the same thing (with a different torque curve shape), but energy is lost in the resistors which is very inefficient. The SER drive via a feedback transformer feeds energy back into the power supply. This returned energy is proportional to the speed difference from synchronous speed. So at say 85% speed, 15% of the motor rated power is returned from the rotor to the supply. At a hypersynchronous speed, the SER drive feeds power into the motor rotor allowing it to run faster than synchronous speed. So for a fixed torque and higher speed, the power obtained from the motor is higher than the motor nameplate rating.

Gearbox ratio is best set up to allow the speed range to be covered using the SER drive’s hyper-synchronous capability.