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Starting Current wth Soft Starter
8

Starting Current wth Soft Starter

Starting Current wth Soft Starter

(OP)
I'm in a disagreement with an Engineer with a major electrical equipment manufacturer over the basics of a thyristor controlled soft starter. I am trying to calculate the voltage drop profile on the supply system during starting.

The Engineer's analysis shows the initial starting voltage at 30% and the motor current at 330A with 20%PF and ramps up to 598A as the voltage is raised. That's the range of current values I say needs to be used in the voltage drop calc. The Engineer says the current to use is the current on the source side of the starter, which would be 30% of the current on the motor side at start and would rise to equal the motor side current as the voltage is ramped up.

I say he's wrong, that the current is virtually the same on either side of the starter.

RE: Starting Current wth Soft Starter

Power is virtually the same on either side of the starter.  There is no law of conservation of current across any device that can change voltage.

RE: Starting Current wth Soft Starter

The engineer and davidbeach are correct.

Think of the reduced voltage starter as a varibale voltage transformer. Input V*I = Output V*I

The source side voltage is not changing only output(motor) side is, so obviously the current will be different on the two sides when voltages are different to deliver a given power.

RE: Starting Current wth Soft Starter

Huh???

There is NO transformer action on a solid state starter! Maybe you were thinking of an Auto-Transformer starter. In a solid state starter, voltage is not transformed, it is reduced by using partial waveforms. So really it can be considered line potential for smaller time slices, hence a lower RMS voltage, which in combination with the motor impendence, reduces the current consumption.

Line current and motor current are the SAME in a solid state soft starter!

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

jraef, I wasn't trying to imply a transformer action, and maybe only the phase angle of the current changes, but if you have a different rms voltage on each side of the starter, and power is conserved (accounting for starter losses), there has to be some difference in the currents.  If that difference is only phase angle, it might be hard to detect with a ammeter, and the magnitude might be the same, but it isn't the same current.

RE: Starting Current wth Soft Starter

(OP)
I agree there might be some slight difference in the line and motor, but that difference is not defined by Input V*I = Output V*I.

It's easy to show how a conversion device like a transformer changes voltage and current on either side of the device and that Input V*I = Output V*I, but a solid state soft starter is not a conversion device, so I'd like Mr davidbeach, or Mr rbulsara to explain to me how a soft starter changes the current as suggested in their comments.

RE: Starting Current wth Soft Starter

If output V (phasor) differs from input V, such as being at a lower rms value then output I will have to differ from input I.

Assume for the moment that the starter is lossless.  One method that can be used is to turn on the thyristors for a portion of each half cycle.  V in is a sinusoid, V out is non-sinusoidal and has a much lower rms value.  Power out equals power in, therefore since V in and V out are not equal, it is not possible for I in and I out to be equal.  It is entirely possible for I (scalar quantity, magnitude only) in to equal I out.

RE: Starting Current wth Soft Starter

I am open to learning.

I would be interested to know if the voltage does not change how does the motor see "reduced" voltage? and if does, then why V.I in will not be V.I out? Less the losses in the electronics?

You possibly can't have 480V in and 144V (30%) out and not have different currents. If so then the solid state starter will have losses equal to a primary resistor or reactor starter. Is that true?





RE: Starting Current wth Soft Starter

jwilson3;

Quote:

The Engineer says the current to use is the current on the source side of the starter, which would be 30% of the current on the motor side at start...

But of course, since right at start the motor is not being sent  full voltage.  The motor side will have the power consumption defined by that (partial) starting voltage and the current.  The same amount of 'power' will be drawn by the starter but at Vsource.  The result is you will need less current on the start side than the current of the motor side.  30% in this case.

The key is "constant power" as there is no energy "storage" or consumption in the starter. (except losses)

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Starting Current wth Soft Starter

There seems to be some confusion here. A conventional solid state starter has three thyristor switches, one in each line. The switch has only 2 possible states, closed or open. When its closed the line current is the same as the motor current and when its open  the line and motor current are zero.

RE: Starting Current wth Soft Starter

First let me start off by saying that after 15 years of providing technical support for a soft starter manufacturers (3 of them), I no longer surprised by the number of very knowledgeable and intelligent engineers out there who have no experience with them and therefore don't understand how they work. So please do not construe any of this as condescending, I am just trying to illuminate the problems of misconception out there. Before I took that first job, I too was confused. I had to see it (and deconstruct it) to truly understand, and I accept that not everyone has that opportunity.

And for the moment, trust me when I say that if you put a CT on the input side or the output side of an RVSS starter, the current measures almost exactly the same (1.5W or heat per amp per phase is the only power loss).

Stop thinking of power "conversion", be it by rectification, as in a VFD, or by induction via transformation or by restriction via resistance/impedance. An RVSS is "restricting", but not with resistance/impedance. It plays with time, not magnetism or chemical interaction, to NOT ALLOW full power through the devices. To understand what I mean, go back to basics for a moment. What is a Volt? The potential that makes 1 Amp flow through a conductor and dissipate 1 watt. What is an Amp? 1 coulomb of charge per second through a conductor.  What an RVSS is doing is altering the time domain of that basic equation. By only allowing power flow for a restricted time, the current, and by definition the voltage, is lowered. So even though you can measure the RMS voltage on the load side as being lower than the line side, it is different by virtue of a reduced net amount of current flowing through the circuit. But from an Amperage standpoint, the current is the current, regardless of which side of the SCR you measure it on.

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

And since I didn't expressly say it jwilson3, the "engineer" in your original post is wrong. You were right to question him on that minor point, not that it ultimately matters in the application you were considering..

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

I think we are missing the fact that the waves are no longer 100% sinusoidal.
The current "in" must be the same as the current "out". . However the incoming current is zero for an extended time until the SCRS fire on. On the output the current is the same but the voltage is chopped (not reduced as for a transformer). The product “VxI” is the same on both sides.
See figure below.


RE: Starting Current wth Soft Starter

(OP)
Thanks to JRaef, for a fundamental explanation. I was beginning to doubt my own understanding of electrical theory,since so many knowledgeable people were disagreeing with me.

The hardest thing to overcome is the concept that full voltage on the line side times the current passing thru to the motor side doesn't equal the motor side voltage times the current. If you consider that the voltage on the line side is only "driving" current during the part of the cycle that the thyristors are firing, and you disregard the voltage of the non-conducting part of the cycle, then the applicable line side and the motor side RMS voltages are the same. Then the fundamental equations that we grew up on remain valid. Everyone can then be happy.

RE: Starting Current wth Soft Starter

aoldale/Jraef:

Good picture. OK. I buy that. Learn a thing everyday.

RE: Starting Current wth Soft Starter

Well it would be a good picture if the current waveform was correct.

RE: Starting Current wth Soft Starter

Thank you for the advice cbarn24050. If you have a picture from an oscilloscope I will appreciate it. I kind of guess-deducted the current feeding an inductive load, just for a quick illustration.

RE: Starting Current wth Soft Starter

Oh you drew it yourself, I just assumed you got it from somewhere. It shouldn't be too hard to find it.

RE: Starting Current wth Soft Starter

Hi Itsmoked. I think we are talking of a "soft starter" not a "VFD".

RE: Starting Current wth Soft Starter

Howdy aolalde!  You're right!  I was in too big a hurry.  I thought Saftronics only did softstarters and that was their "here's how they work page"..

Here's the pics from Siemens "starters" page.





These seem to be showing that they're switching at nonzero voltage points - on AND off.  ???

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Starting Current wth Soft Starter

The magnitude of the available voltage at the input is imaterial during that part of the cycle that the current control devices are not conducting.
The input voltage waveform during the time that the SCRs are conducting, will be the same as the output voltage wave form. When the switch is closed, the voltage is the same on both sides of it.
If the effective value of this waveform is used instead of the effective value of the input voltage sin wave, the equations will balance.  
Think of a flashing light that is on 50% of the time. The output voltage equals the input voltage 50% of the time.
Similarly, the soft starter input voltage equals the output voltage only during the time that the SCRs are conducting.
respectfully

RE: Starting Current wth Soft Starter

(OP)
Thanks, waross. You stated it better than I did.

RE: Starting Current wth Soft Starter

A few deserving stars.

Can I blame this temporary blockage of my understanding on my personal termoils going on in my house?

smile Cheers

RE: Starting Current wth Soft Starter

Sorry to get in so late. I had an interesting experience in May this year. A large fan for a boiler, 1200 kW, had a failure in its Siemens soft starter.

I made some recordings. This is what it looked like when we finally got it working:



Voltage in upper trace and current in lower trace. We are at about 50 percent of end speed. Current scale is 3000 A/division.

It may be interesting to see what happened when we tried to start with the fault still in the controller (one firing pulse missing). It looked like this:




The scale factor is now 6000 A/division. As you can see, the missing firing pulse produces a DC component, which not only drives motor current high - it also saturated the transformer feeding the system so that it had a very high overcurrent which made a protective relay fail. It did that after about ten minutes.  There were probably other damages to the system as well. But I left when the soft starter was OK. So I do not know what else had happened.


Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Starting Current wth Soft Starter

Lotta amps there.. Whew!  You do put a lot of time on your data recorders don't you Gunnar..

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Starting Current wth Soft Starter

Ya know what? My customers pay me to do that. And as a side effect, I get some RL data for you.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Starting Current wth Soft Starter

Perhaps it would help to state that real and reactive power are conserved quantities, but that distortion power is not. And in general, apparent power is not either. I confess I'm not entirely sure if reactive power is conserved in the general case (help me out if you know), but I thinks it's fair to say that distortion and apparent power are not.

RE: Starting Current wth Soft Starter

Introducing concepts like distorsion power is more confusing than helpful. It obscures the fact that the same current is flowing in the supply wires as in the motor wires after the soft-starter.

Distortion power was used as a way of explaining what harmonics do to the apparent power. Remember the 3D Pythagoras rule used by some educators and salespeople that actually had no idea what they were talking about? Nor had their listeners, I think. They just got confused. That's why you do not see this concept any more.

The soft-starter just lets a smaller part of the sine voltage through. The smaller the part, the smaller the RMS voltage - and hence the current. It is no more complicated than that.

jwilson3 is right. And so are the guys supporting his view. I jumped in only because I thought that my recordings are illustrative. The soft-starter has a very limited ability to manipulate the voltage (Can only fire thyristors, has to take into consideration that the phase/phases returning the current must be "on" when thyristor is fired. It must also take the inductive nature of the load into account. As you can see, the thyristor conducts long after zero voltage crossing)  All these circumstances complicate things considerably. But it does not change the fact that Current In = Current Out.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Starting Current wth Soft Starter

Keith,
You are batting 1000 buddy!
That 2nd waveform posting is not a wave form for a true 3 phase controller. Siemens (and a few others) have a "low cost" version of their soft starters that only control 2 of the 3 phases in an attempt to reduce the component count. One phase is just a piece of bus bar. While this works OK on low inertia loads, it is somewhat controversial and IMHO should not be used where long acceleration times are expected. That waveform shown is for their 3RW - 2 phase controller only, and even then it is a little "simplistic" to say the least. They publish this sort of stuff because people who don't understand it get excited from comments from competitors about negative sequence currents and DC components imposed on the sine wave which can theoretically damage the motor windings. They are trying to simplify the concept for the masses.

Most of the "waveforms" published for soft starter outputs are simplified graphical representations of one phase, partly because of exactly what Gunnar's post shows; the complexity of a scope readout. Remember, a soft starter is never controlling just one phase at a time, it is controlling 6 SCRs which are almost carrying current from more than one line phase 2/3 of the time. The effect on scope traces of the output voltage waveform in any one phase becomes very complex to interpret.

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

By the way waross, nice addition to what I had said earlier, I need to remember that way of stating it.

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

I wasn't aware that distortion power was a dated concept. It's in my latest IEEE dictionary. Made sense to me. Just thought it might help those thinking input V*I=output V*I understand that this only true for sinusoidal quantities.

RE: Starting Current wth Soft Starter

The way it is (or was) abused is not seen any more. Of course, it can be used - but only where it makes sense.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Starting Current wth Soft Starter

Yeah jraef... bomb  Teach me to be superficially helpful!

Skog's scope pictures are quite a bit better.  Nasty complex especially in the one phase of three aspect.

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Starting Current wth Soft Starter

Hey guys, again Like a few others I have leapt into this a little late, But I want to point out to itsmoked something he missed in his post containing the Waveforms and included comments about the non zero Voltage switching.
With SCR's the switching is done at zero current, not zero Voltage, and from the manufacturers diagram of the current waveform, you can see that indeed it is switching at zero AMps.IF you look at the opeation of a SCR Rectifyer operating in the regeneration region, the switching is always done at quite high voltages, both on and off.

Like Gunnar, I spend enormous amounts of my customers money investigating problems with these things. Have literally CD's full of high speed recordings of all sorts of things, except, sadly, Thyristor starters.


Stars to jreff and others for some very good points

 

RE: Starting Current wth Soft Starter

Thanks for the comments jraef and jwilson3;
I have a lot of respect for the knowledge and experience of both of you. Your remarks are more flattering than quite a few LPSs.
Thanks
Respectfully

RE: Starting Current wth Soft Starter

Just came across this thread, and I agree with the conclusions made. After many years experience in designing and working with soft starters I can confirm that the current waveform is a double humped waveform as shown above with the forst hump being the current flowing from phase A to phase C and the second hump being the current flowing from phase A to phase B. As you reduce the conduction angle, the OFF time increases and the humps become more prominant. As the conduction time is increased, the humps blend into a single hump and at 180 degree conduction angle we have the traditional sinewave we are so familiar with.
When the SCR is conducting, we have the same voltage input and output minus tha voltage drop across the SCRs which is typically around 1 - 1.2 volts per SCR. The current is the same, so at each instant that the SCR is ON, the VA input equals the VA output minus the conduction losses of the SCR. For each instant that the SCR is OFF, the VA input equals the VA output, so the laws of conservation etc are fully met.
The difficulty comes when considering discontinuous waveforms. At that time, you can not take the average voltage times the average current to get the average power. This is a stunt commonly pulled by the promoters of energy savers!!

Unfortunately, much of the technical literature about SCR control of induction motos shows waveforms that are true for an SCR based controller with a resistive load. This shows the current waverform to be very different from reality and the commutation occuring at 0 and 180 degrees.
The reality is that the current is smoothly rounded and takes on the double hump, and it commutates at an angle which reflects the power factor of the motor, hence the commutation angle changes with slip.

During the non conduction period, the voltage on the motor terminals does not fall to zero as logic would suggest. The spinning motor acts as a generator and creates it's own voltage profile to fill in the gap. The actual voltage seen is dependant on the voltage generated, but can make the waveform "applied" to the motor seem prety wicked at times. The rality is that the waveform applied to the motor is only the portion where the SCRs are conducting.

So in calculating power and VA, we need to take the integral of the products of the instanteous values to get meaningful results.
Returning to the OP, for voltage drop calculations, is is usually sufficient to consider the average start current to get reasonable results. The maximum start current flows around the voltage zero crossing at zero shaft speed and increases higher up the sinewave as the motor reaches full speed. This is the same for a standard induction motor starter. The major difference is that there is a non conduction period before the voltage crest that can at times include the voltage crest.
So treat the soft starter like a full voltage starter with a reduced average start current. - somewhat like a primary resistance starter with a very small power loss.

Best regards,

Mark Empson
http://www.lmphotonics.com

RE: Starting Current wth Soft Starter

TomG33 and Mark thanks for the further clarification.

I have learned a truck-load on this one.

Looking at a truly 3ph system by looking at only one of the ph's really adds to the confusion.

Rather like looking at an elephant thru a soda straw at night.

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Starting Current wth Soft Starter

Why would you do that, Smoked? It's dark at night, you know.

I think that MarkE's comment is the final word. He is THE Soft-Starter Authority. He, and Jeff, I meant to say.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Starting Current wth Soft Starter

And a lot of what I have learned in the past few years has come from Marke! wink

Mark, you need to get the Google tool bar spell checker!

JRaef.com
"Engineers like to solve problems.  If there are no problems handily available, they will create their own problems."   Scott Adams  
For the best use of Eng-Tips, please click here -> FAQ731-376

RE: Starting Current wth Soft Starter

jraef,

'Forst' is a word where I come from: forst, sicond, thord, forth...

This has been an interesting thread - thanks to all for making me get my notes out and look some stuff up. Sometimes I forget howe much I used to know!

----------------------------------
  Sometimes I only open my mouth to swap feet...

RE: Starting Current wth Soft Starter

Well done Marke! a correct explaination.

RE: Starting Current wth Soft Starter

We can avoid the whole distortion power question by simply saying that active power is always conserved, while non-active power in general, is not.

Still hanging tough davidbeach? Kind of alone out there.

RE: Starting Current wth Soft Starter

I don't know.  If all the soft starter can do is switch the current, then current in and current out have to be the same thing; I think I was trying to use input voltage as the reference and carrying the same phase angle over to voltage out, but looking at the various plots above, it becomes clear (to me anyway) that the voltage phase angle has to change across the starter, therefore the phase angle between current and voltage changes between the two sides.  What ever the details, I will certainly stand behind my first statement that power doesn't change.

RE: Starting Current wth Soft Starter

Hello davidbeach

The soft starter is literally three solid state AC switches in series with the three phases to the motor.
The switches (SCRs) can only be turned ON or OFF and not change to the phase angle occurs when current is flowing.
The voltage patterns get confusing because of the voltage generated by the motor which fills in some of the OFF time.

Best regards,

Mark Empson
http://www.lmphotonics.com

RE: Starting Current wth Soft Starter

(OP)
To davidbeach,
You are correct that power doesn't change as you cross thru the starter. However, you can't use full voltage on the source side in the calculation. The "effective" value of that portion of the voltage waveform that's being "utilized" by the motor while the solid state devices are "firing" is what's used in calculating power in. And it's the same as power out. Current in is the same as current out, and voltage in is the same as voltage out.

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