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Soft Starter Using Pic


pramod

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Hi,

Iam trying to develop a soft starter using PIC by controlling 3-phase SCRs to obtain reduced ph-ph voltage.

I have tried it out using both resistive load and inductive load.It worked.But when i started the 25 hp

induction motor voltage across the delta winding of motor got reduced but the starting current was 2 times

full load. What could be the reason?Is it because the o/p voltage waveform of SCR's is not a sinewave?

Iam also not able to increase the firing angle of SCR's above 100 degree(5.5ms) with inductive load.

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Hello pramod

 

Welcome to the forum.

 

The induction motor has a very low impedance when the slip is high. Typically, an induction motor will draw 6 - 9 times it's rated current during start if it is started with full line voltage.

 

Reducing the start voltage will reduce the start current, the current will always be greater than the rated current of the motor.

 

Best regards,

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Hello pramod

 

Welcome to the forum.

 

The induction motor has a very low impedance when the slip is high. Typically, an induction motor will draw 6 - 9 times it's rated current during start if it is started with full line voltage.

 

Reducing the start voltage will reduce the start current, the current will always be greater than the rated current of the motor.

 

Best regards,

Hello Marke

 

Thanks for the reply.

 

I started the motor with full line voltage(415,ph-ph) and reduced voltage(230,ph-ph) using a auto transformer and i checked the starting current.In both these conditions the starting current was around 100 amps for 1 second and then it got reduced to 10 amps immidiately.

 

I am trying get the same result by controlling SCR's.The controlled o/p voltage is 230V(ph-ph) but its a switching waveform.

 

When i apply reduced voltage using thyristors the current never got reduced from 70amps.Why doesn't the current reduce to 10amps or to the no load current ?

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Hello pramod

 

I started the motor with full line voltage(415,ph-ph) and reduced voltage(230,ph-ph) using a auto transformer and i checked the starting current.In both these conditions the starting current was around 100 amps for 1 second and then it got reduced to 10 amps immidiately.

This is due to the way that you are measuriong the curent.

If the motor is 25HP, it will have a rated current of around 38 amps at 400 volts. It will have a locked rotor current of around 240 - 300 amps.

When you start this motor with full voltage, it will draw Locked Rotor Current, irrespecitve of the shaft load. A very light shaft load will allow the motor to start very quickly, and if you use a slow measuring method, it will appear to draw a lower current, but it must draw LRC at line voltage.

Have a look at http://www.lmphotonics.com/m_control.htm and http://www.lmphotonics.com/m_start.htm for some of the background information.

 

At 230 volt, the current will be reduced to 230/400 of LRC, or to around 132 - 170 Amps. As the voltage is lower, the torque will also be lower and the torque reduces with the square of the voltage reduction. So the torque will be one third of the full voltage torque. With the reduction in start torque, the start time will be longer.

At one third of the torque, the start time will be three times longer, hence the average current as measured could be similar. Use a high speed chart recorder and you will see what I am saying.

 

As you reduce the start voltage, you reduce the start torque by the square of the start current reduction. This limits the current reduction that you can achieve. The motor must produce sufficient torque to overcome the resistance and load torque.

 

Best regards,

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Hello pramod

This is due to the way that you are measuriong the curent.

If the motor is 25HP, it will have a rated current of around 38 amps at 400 volts. It will have a locked rotor current of around 240 - 300 amps.

When you start this motor with full voltage, it will draw Locked Rotor Current, irrespecitve of the shaft load. A very light shaft load will allow the motor to start very quickly, and if you use a slow measuring method, it will appear to draw a lower current, but it must draw LRC at line voltage.

Have a look at http://www.lmphotonics.com/m_control.htm and http://www.lmphotonics.com/m_start.htm for some of the background information.

 

At 230 volt, the current will be reduced to 230/400 of LRC, or to around 132 - 170 Amps. As the voltage is lower, the torque will also be lower and the torque reduces with the square of the voltage reduction. So the torque will be one third of the full voltage torque. With the reduction in start torque, the start time will be longer.

At one third of the torque, the start time will be three times longer, hence the average current as measured could be similar. Use a high speed chart recorder and you will see what I am saying.

 

As you reduce the start voltage, you reduce the start torque by the square of the start current reduction. This limits the current reduction that you can achieve. The motor must produce sufficient torque to overcome the resistance and load torque.

 

Best regards,

Hi marke,

Thanks for the detailed explanation.

 

I made a mistake.I mentioned of starting the motor at full voltage and measuring the starting current.I have not done that.

I have started motor at 230V using autotransformer and measured the current.The starting current(r.m.s value) got reduced to 10amps after one or two seconds.230V was applied for 10 seconds and the current was 7amps during this period.

Iam sorry if iam annoying u asking the same question.

I have used the same scope for measuring starting current using soft starter.The current measured during the period of 10seconds was 70amps .

 

Hello pramod

This is due to the way that you are measuriong the curent.

If the motor is 25HP, it will have a rated current of around 38 amps at 400 volts. It will have a locked rotor current of around 240 - 300 amps.

When you start this motor with full voltage, it will draw Locked Rotor Current, irrespecitve of the shaft load. A very light shaft load will allow the motor to start very quickly, and if you use a slow measuring method, it will appear to draw a lower current, but it must draw LRC at line voltage.

Have a look at http://www.lmphotonics.com/m_control.htm and http://www.lmphotonics.com/m_start.htm for some of the background information.

 

At 230 volt, the current will be reduced to 230/400 of LRC, or to around 132 - 170 Amps. As the voltage is lower, the torque will also be lower and the torque reduces with the square of the voltage reduction. So the torque will be one third of the full voltage torque. With the reduction in start torque, the start time will be longer.

At one third of the torque, the start time will be three times longer, hence the average current as measured could be similar. Use a high speed chart recorder and you will see what I am saying.

 

As you reduce the start voltage, you reduce the start torque by the square of the start current reduction. This limits the current reduction that you can achieve. The motor must produce sufficient torque to overcome the resistance and load torque.

 

Best regards,

Hi marke,

Thanks for the detailed explanation.

 

I made a mistake.I mentioned of starting the motor at full voltage and measuring the starting current.I have not done that.

I have started motor at 230V using autotransformer and measured the current.The starting current(r.m.s value) got reduced to 10amps after one or two seconds.230V was applied for 10 seconds and the current was 7amps during this period.

Iam sorry if iam annoying u asking the same question.

I have used the same scope for measuring starting current using soft starter.The current measured during the period of 10seconds was 70amps .

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Hello pramod

 

using a 230V transformer, I would expect to see some additional voltage drop depending on the internal impedance of the transformer, so 70 amps is probably not unrealistic. Perhaps you should measure the voltage also during the period of the overload to see what it drops to.

 

So this shows that there is an major overload current during start, and this is required to develop the start torque. If you reduce the start current down to 1/6 of full voltage (approx equal to rated motor current) then you will get 1/36 torque which is very low. If the waveform is not a pure sinewave (chopped by Thyristors), then the torque will be even lower, so you will not start the motor at this low level of current.

 

If you really need to start the motor at less than rated current, you must reduce the frequency and voltage as is done in a VSD.

 

Best regards,

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Hello pramod

 

using a 230V transformer, I would expect to see some additional voltage drop depending on the internal impedance of the transformer, so 70 amps is probably not unrealistic. Perhaps you should measure the voltage also during the period of the overload to see what it drops to.

 

So this shows that there is an major overload current during start, and this is required to develop the start torque. If you reduce the start current down to 1/6 of full voltage (approx equal to rated motor current) then you will get 1/36 torque which is very low. If the waveform is not a pure sinewave (chopped by Thyristors), then the torque will be even lower, so you will not start the motor at this low level of current.

 

If you really need to start the motor at less than rated current, you must reduce the frequency and voltage as is done in a VSD.

 

Best regards,

Hi marke,

But there are electronic soft starters that work by applying reduced voltage similar to a star delta starter.Do u know how it works?

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Hello pramod

 

I have never heard of a soft starter that works by applying reduced voltage to a star delta starter. This would totally defeat the purpose and basically be a waste of money.

 

If a delta designed motor is connected in star, it draws about 200% of rated current (1/3 DOL current) and produces 1/3 DOL torque. The torque output is often insufficient to get the motor to full speed, so the transition occurs at part speed making the start current as high as DOL. Using a soft starter in star, would reduce the torque even further and make a bad situation even worse!!

 

Best regards,

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Hello pramod

 

I have never heard of a soft starter that works by applying reduced voltage to a star delta starter. This would totally defeat the purpose and basically be a waste of money.

 

If a delta designed motor is connected in star, it draws about 200% of rated current (1/3 DOL current) and produces 1/3 DOL torque. The torque output is often insufficient to get the motor to full speed, so the transition occurs at part speed making the start current as high as DOL. Using a soft starter in star, would reduce the torque even further and make a bad situation even worse!!

 

Best regards,

Hi marke,

I think u didn't get my question.In a star delta starter reduced voltage is applied across stator windings and the motor starts.I asked whether u have heard about a soft starter that employs reduced voltage method to start a motor.

 

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Hello pramod

Sorry, I misunderstood your question.

 

Yes, all soft starters reduce the voltage applied to the motor as a means of reducing the start current and start torque. This has been common practice for close to 30 years.

 

Simple soft starters use a simple voltage ramp technique, and better soft starters use current feedback to control the output voltage.

 

Best regards,

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Hello pramod

Sorry, I misunderstood your question.

 

Yes, all soft starters reduce the voltage applied to the motor as a means of reducing the start current and start torque. This has been common practice for close to 30 years.

 

Simple soft starters use a simple voltage ramp technique, and better soft starters use current feedback to control the output voltage.

 

Best regards,

Hi marke,

Can u tell me for how long this reduced voltage must be applied to a motor.Since i have just started to develop a starter i have not used the logic of voltage ramp up in my program instead i just reduce the voltage applied to the motor for 10 seconds and then apply the full voltage using bypass contactor.

Will it work or should i change my program.

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