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ac motor braking system
Engineering and workplace issues. topic
Posted by ravi chandran on 2 January, 2002 - 9:54 am
Dear sirs,

Currently i am working on a project. I have a ac induction motor 11kw which is operated by star delta contactor switching. This motor does not have any mechanical brake. The problem is i want to stop the motor from free run when the power is cut off from the motor. I am aware that there is a method by injection the dc voltage to the ac motor coil will brake the motor. i do not have any knowledge on this and i could not find any books on this either. Please advise how could i stop the motor by injecting the dc voltage.

for your info, my motor is rated 11kw, 3 phase, 4 pole, 415 VAC, 50 Hz.

Please advise me

Thanks in advance
ravi chandran


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Posted by Aquilino Rodriguez on 3 January, 2002 - 9:19 am
Hello:

Time ago I worked on an Engineering where we made washing machines for automotive industry.
We used this method to stop some AC motors.

When we wanted to stop an AC motor, the process used was:

1- disconnect AC supply (switch off contactor)
2- connect the DC circuit (standard DC power supply) to one of the coils of the motor. The magnetic field generated by the coil will try to
brake the shaft in each revolution. 3- wait until the motor stops or the rpm decreases to a safe level.

I do not remember now the rule to determine the rating of the DC intensity (depends on the braking time: shorter time -> larger Intensity,
maybe you will need to make it empirically)
Do not forget accessory protection elements for DC circuitry and interlocking of DC and AC circuits!

I hope this information will be useful.

Best regards.

Aquilino


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Posted by Phil Corso on 4 January, 2002 - 1:52 pm
Responding to Aquilino Rodriguez's Thu, Jan 03, 9: 40am, query:

Rules of thumb, repeat, rules of thumb, are:

a) Star winding - connect dc excitation (+) to phase A and (-) to B or C. Set voltage so that Idc = 1.225 times rated Iac.

b) Star winding - connect dc excitation (+) to phase A and (-) to B and C. Set voltage so that Idc = 1.414 times rated Iac.

c) Delta winding - connect dc excitation (+) to phase A and (-) to B or C. Set voltage so that Idc = 1.50 times rated Iac.

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by ianingram on 3 January, 2002 - 9:22 am
Easy fix.

Step 1; Change motor wiring to Delta.

Step 2; Replace star/delta contactors with an inverter that allows for DC braking. I can't recommend any particular model, there are many
inverter manufactures & models that will do this task.

email: sgi@tpg.com.au


Posted by John watford on 3 January, 2002 - 11:21 am
Allen-Bradley and of course others have several models on the market that will do exactly what you want. Good Luck...


Posted by Johan Bengtsson on 3 January, 2002 - 1:33 pm
One problem you will have to look into is that either way you want to electrically stop the motor you need to have some power availiable.

An AC drive with regenerative breaking could perhaps manage to be powered by the regernerated energy from the motor, in all other cases you probably need a battery if you should be able to electrically stop the motor when power is cut off. This is of curse unless there can be another voltage source that doesn't get cut off.



/Johan Bengtsson

Do you need education in the area of automation?
----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 Hässleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
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Posted by Phil Corso on 3 January, 2002 - 3:57 pm
Welcome Johan:

Sorry, but I must disagree with you one more time. Effective regeneration is only possible if the load "overhauls" the motor.

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by James Bouchard [CPCCA] on 4 January, 2002 - 12:39 pm
In a braking application where you want to bring the motor to a stop the load will always overhaul the motor since you have removed line power and the only thing keeping the motor turning is the inertia of the load. If you use a deceleration ramp that is faster than the coast to rest ramp then the load will always overhaul and you will get regeneration. If you use a deceleration ramp that is slower than the coast to rest ramp then the load will not overhaul and you will actually have to provide power to keep the motor going. The trick is to know what is the coasting time of the motor and load under the various operating conditions.

James Bouchard


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Posted by Phil Corso on 4 January, 2002 - 1:41 pm
James, I agree wholeheartily with the conclusion you presented regarding regeneration braking. However, the theoretical fact that overhauling is aways present does not necessarily mean that "braking" is effective.

Certainly not in the three cases presented, i.e. 3/4 Hp radial saw, the 11 kW grinder, and the 55 kW grinder. Unless the residual magnetic field is substantial, the generated voltage will decay in milli-seconds. In fact, the "technique" discussed in my Wed, Jan 02, 2:33 pm thread illustrated the point.

It pleases me that interest exists about a subject that isn't key-board related. If you would like to to know what plugging current is in per-unit of rated-current, based on usual motor parameters, please let me know.

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by Ed Oleen on 28 January, 2002 - 12:07 pm
Hey!!! Just a minute there? What is the solution for the 3/4 hp radial saw????? I've got a 1 hp (120v) table saw ("contractor saw") that I would like to retro-fit with a brake... The thing takes too bloody long to coast to a stop, and I get a bit impatient sometimes, and I've only got 1 set of fingers... I am looking for a device to do this. I have no luck finding such a device on the
web - everybody seems to be talking about multi-horsepower, high voltage, multi-phase stuff, with the braking as part of the overall speed control.

If possible, please contact me directly: eoleen@earthlink.net



Posted by Johan Bengtsson on 4 January, 2002 - 12:57 pm
Sorry, but I don't understand what you disagree about.

If you tell the drive to stop the motor as fast as possible (ie break with maximum torque) a lot of energy will be drawn from the motor and thereby charge the capacitors inside the drive. This energy are then either burned away in a high power resistor or in some cases delivred back to the grid. This is done by using some of the energy stored in this capacitor that is right, but this energy is not all used up before it is filled again - and more than replaced. What I tried to say was that the extra energy you get when breaking should be more than enough to keep the drive an all the electronics in it alive until the motor is stopped. It would work in theory, if it works practically depends on how the drive is made internally.

Most of this energy does of course come from the load, if that is what you mean but it does not really change my point. The point is that you need electrical energy to electrically stop a motor. If you do this in some way regenerating electrical power there should be more than enough, in all other cases you will need an external power source since all the energy will become heat immediately instead.


/Johan Bengtsson


----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
Internet: http://www.pol.se/
----------------------------------------
Do you need education in the area of automation?


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Posted by Phil Corso on 4 January, 2002 - 2:28 pm
Johan, see my reply to James Bouchard.

Regards,
Phil Corso, PE
(Boca Raton, FL)


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Posted by Phil Corso, PE on 6 January, 2002 - 2:09 pm
1) Re my Fri, Jan 4, 12.46pm reply to Aquilino Rodriguez:
a) There is an error in the dc excitation factor for the Delta-winding case. It should be the same as the 3-terminal Wye-case. Hence, distribution becomes 100% forward current in one winding and 50% reverse current in the other two. The resultant multiplier should be 1.225, not 1.500.
b) Some additional insight. The rule of thumb factors are based on providing a dc current that will develop the same MMF magnitude as ac
current.
c) Please note that, although the resultant braking currents are not excessive, brake torque could be. But that is for another thread.
d) In closing, the material I presented is to enlighten the uninformed, and not as an endorsement of a specific method.

2) Re my Fri, Jan 4, 1:59pm reply to Johan Bengtsson referring to my reply to James Bouchard. Johan's answer expounds on much more than was evident in the original and subsequent threads.
a) Please note that the the original questions referred to ordinary induction motors with conventional supply connections. Johan, I believe, refers to electronically augmented drives.
b) Induction motors need a source of excitation to operate as a generator. I responded to the case of power loss to just one motor. For the motor sizes presented, I seriously doubt that residual magnetism is of sufficient duration to produce effectual brake torque.
c) The consequence is quite different for a group of motors connected to a mutual supply bus that suddenly loses power. For this case, sufficient excitation may be available, but, only for the time that the motors are connected to the bus.
d) There was no mention of capacitors in the 3-phase grinder cases. There was a 0.25uF cap noted in the 3/4 Hp radial saw case, but it had
nothing to do with the original braking circuit.

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by Johan Bengtsson on 8 January, 2002 - 12:23 pm
Ok, I still don't understand what the disagreement is about really

I'll try to clarify my point again:


To electrically break a motor you need power (right?)


If the power grid is disconnected for any reason from you this above statement is still true (right?)


This means you need a power source, for example a battery availiable that can give this power (right?)


An alternate power source could be the energy stored in the inertia of the load (+ motor). This is however availiable only if you actually break the motor by regenerating power, in all other cases the only thing you get is heat.


My point was (but obviously it got lost somewhere) that if power fails and you want to electrically break the motor you need a battery in all cases except perhaps if you use the regenerating method, in that case the power stored by the rotating inertia could (teoretically) be that
power source.
A mecanical break - on the other hand - could work by just removing the power to it.


/Johan Bengtsson

Do you need education in the area of automation?
----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
Internet: http://www.pol.se/
----------------------------------------


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Posted by Phil Corso on 10 January, 2002 - 9:19 am
Attn: Johan Bengtsson, reur Tue, Jan 8,12:21 reply:

My disagreement is still with your explanation of how a disconnected motor, theoretically, provides "adequate" braking torque. Yes, upon
interruption of power the load inertia is providing rotational torque. I also agree that there is residual stator iron magnetism to generate "some" rotor MMF. But, it decays so rapidly that its effect is nil. Hence, additional energy input is necessary. This, then, was the basis of the original query... ac or dc?

The experiment noted in my Jan 02 message essentially proved this point. The analogy I like to use is that of a chicken eating its own droppings for protein. So its eat-drop-eat-drop,,, etc. Sooner, or later it will run out of protein. Simply put, think of residual magnetism being protein.

In closing, I strongly agree with your suggetion about an electrically-operated mechanical brake for the situaton described above, It just took a long time to get you there.

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by Johan Bengtsson on 16 January, 2002 - 2:40 pm
Ok, now I understand - yes it the motor is disconnected from the drive you won't get any breaking torque - I agree completely about that.

It was, however, not what I meant. What I was talking about was that if the motor is still connected to the drive, but rather when the drive gets disconnected from the power grid (or if the power grid fails).

The drive would not die completely immediately (since it have capacitors holding a considerable amount of energy) and if it would try to stop the motor in case of a power failure it could very well succed with that because of regeneration without any external power availiable.

A too long delay would - of course - kill the drive and after that there would be no hope of electrically breaking since electrical power - as you say - need to be availiable.


In order for it to work the drive have to be connected to the motor all of the time, and operational. It could (theoretically at lest) stay operational on the power regenerated for as long as it take to stop the motor since there would be more energy regenerated than is needed by the drive. (why would you otherwise need a breaking resistor).

I hope this clears it up!


/Johan Bengtsson

Do you need education in the area of automation?
----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
Internet: http://www.pol.se/
----------------------------------------


Posted by Donald Pittendrigh on 18 January, 2002 - 3:03 pm
Hi All

First things first....... Break means to damage as in destroy, damage,
cause (irreversible) negative change, I don't think this is what you mean, try Brake, which is to slow down/stop.


> Ok, now I understand - yes it the motor is disconnected from the drive
> you won't get any breaking torque - I agree completely about that.


If the motor is disonnected from the drive, presumably his does not mean it is connected to anything else in opposition, it will do exatly what is asked to do i.e. nothing!!!!


> It was, however, not what I meant. What I was talking about was that
> if the motor is still connected to the drive, but rather when the
> drive gets disconnected from the power grid (or if the power grid
> fails).

The drive may or may not be disconnected from the power grid, the real question is where if anywhere will the energy be dissapated, if there are Braking resistors the power may be converted to heat in the resistors, i.e. there is still an electrical path present which will disharge the accumulated energy. If the power grid is disconnected the energy may still be discharged in your neighbours toaster/airconditioner/HiFi system, or anything else which is locally connected.


> The drive would not die completely immediately (since it have
> capacitors holding a considerable amount of energy) and if it would
> try to stop the motor in case of a power failure it could very well
> succed with that because of regeneration without any external power
> availiable.


The amount of energy the capacitors store and are thus able to discharge is immaterial, what is of coincern is can they absorb the Kinetic energy you want to change to (potential) some other form of energy and thus dissapate as heat.

> A too long delay would - of course - kill the drive and after that
> there would be no hope of electrically breaking since electrical power
> - as you say - need to be availiable.


Kill the drive ??? what does this mean, inanimate objets are not to good at dying, the issue here is, is there a discharge path for the energy.


> In order for it to work the drive have to be connected to the motor
> all of the time, and operational. It could (theoretically at lest)
> stay operational on the power regenerated for as long as it take to
> stop the motor since there would be more energy regenerated than is
> needed by the drive. (why would you otherwise need a breaking
> resistor).


It does not (necessarily) take power to stop the motor, it takes conversion of the stored kinetic energy to some manageable form, and subsequent dissapation of this energy, it does not/should not matter if the drive is "powered" up as the motor generates the power, the DC link if it is powered up should keep the electronics going, and the transistors/thyristors firing, and allow the dissapation of the above mentioned energy to proceed, by the time there is no more energy left to realise this, it does not neccessarily matter anymore as the motor is stationary.

> I hope this clears it up!

Clear as a bell i am shoor.

Bye
Donald Pittendrigh



Posted by Johan Bengtsson on 22 January, 2002 - 9:59 am
>First things first....... Break means to damage as in destroy, damage,
>cause (irreversible) negative change, I don't think this is what you
>mean, try Brake, which is to slow down/stop.

OOPS! sorry! of course!


>> It was, however, not what I meant. What I was talking about was that
>> if the motor is still connected to the drive, but rather when the
>> drive gets disconnected from the power grid (or if the power grid
>> fails).

>The drive may or may not be disconnected from the power grid, the real
>question is where if anywhere will the energy be dissapated, if there
>are Braking resistors the power may be converted to heat in the
>resistors, i.e. there is still an electrical path present which will
>disharge the accumulated energy. If the power grid is disconnected the
>energy may still be discharged in your neighbours
>toaster/airconditioner/HiFi system, or anything else which is locally
>connected.

Exactly


>> The drive would not die completely immediately (since it have
>> capacitors holding a considerable amount of energy) and if it would
>> try to stop the motor in case of a power failure it could very well
>> succed with that because of regeneration without any external power
>> availiable.
>
>
>The amount of energy the capacitors store and are thus able to
>discharge is immaterial, what is of coincern is can they absorb the
>Kinetic energy you want to change to (potential) some other form of
>energy and thus dissapate as heat.

Partly true, they will need to hold enough energy for the drive to begin stopping the motor, but since that don't take very long time the energy needed isn't that high. But if you remove them completely it won't work.


>> A too long delay would - of course - kill the drive and after that
>> there would be no hope of electrically breaking since electrical
>> power - as you say - need to be availiable.
>
>Kill the drive ??? what does this mean, inanimate objets are not to
>good at dying, the issue here is, is there a discharge path for the
>energy.

What I mean if of course that it needs energy to work, no energy = no work done = not able to stop the motor. Ok, I am not a language expert (I have never pretended beeing one either <grin>) but of course a thing like that don't die but isn't it quite common to call it dead when it doesn't do what it is expected to do?


>> In order for it to work the drive have to be connected to the motor
>> all of the time, and operational. It could (theoretically at lest)
>> stay operational on the power regenerated for as long as it take to
>> stop the motor since there would be more energy regenerated than is
>> needed by the drive. (why would you otherwise need a breaking
>> resistor).
>
>
>It does not (necessarily) take power to stop the motor, it takes
>conversion of the stored kinetic energy to some manageable form, and
>subsequent dissapation of this energy, it does not/should not matter if
>the drive is "powered" up as the motor generates the power, the DC link
>if it is powered up should keep the electronics going, and the
>transistors/thyristors firing, and allow the dissapation of the above
>mentioned energy to proceed, by the time there is no more energy left
>to realise this, it does not neccessarily matter anymore as the motor
>is stationary.

You definitely need power to electrically stop an asynchronous motor, that is what Phil Corso tryed to say all the time, and he is completely right about that. The power is availiable since it is possible to pull it from the kinetic energy stored in the motor, but not unless you have some amount of electrical energy to begin with.

As opposed to brake with a DC current trru one winding, that will also stop the motor, but you need to feed electrical energy into the motor until it does. All this energy are then transfered to heat inside the motor.

The matter is completely different with a syncronous motor or a DC motor since they work as generators all by themself (if they are rotating of
course) and just short circuiting them make them stop rather quickly (too quickly perhaps ... but that is another story)


>Bye
>Donald Pittendrigh

/Johan Bengtsson

Do you need education in the area of automation?
----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
Internet: http://www.pol.se/
----------------------------------------


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Posted by Phil Corso on 18 January, 2002 - 3:49 pm
Johan,

Once again you responded with a discourse on the regeneration method using an electronically augmented drive. If a VSD, then your repy has merit, If a soft-start, then the bypass contactor would preclude energy available from the drive.

Regards,
Phil Corso, PE
(Boca Raton, FL)


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Posted by Johan Bengtsson on 18 January, 2002 - 3:50 pm
Of course, but then a soft starter can't break with regeneration at all anyway.

/Johan Bengtsson

----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
Internet: http://www.pol.se/
----------------------------------------
Do you need education in the area of automation?


Posted by Yogesh on 7 January, 2002 - 7:03 pm
ravi,
u and me travelling on same boat. I have 55kw motor for which I require breaking . I have few comments on it pl give me mail on
Yogesh.Tapaswi@saint-gobain.co.in.
regards,
yogesh


Posted by Ravindra on 19 January, 2002 - 12:58 pm
Hai,

Do you really require DC braking .
Is it position control problem?
then do the following . When ever you are going to stop the motor , the contactor will de enerzise in the Motor electrical circuit. You take one NO contact from the Contactor , and connected to the Air operated solenoid valve and pneumatic cylinder with arm type brake.

Once you are trying to start te motor the contactor will enerzise , and will relese the pnematic cylinder , because the power to the
solenoid cutoff

This the way we are doing , where ever we need position control of the equipment

regards
mullapudir


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Posted by Phil Corso, PE on 22 January, 2002 - 2:32 pm
In the interest of technical correctness, the following comments are intended to clear up some inconsistencies:

(1) My Wed, Jan 2, 10:44 pm post.
Plugging Method. Although I pointed out that the current will not be excessive, I want to caution readers that net-torque could be. This results from the fact that net-torque is the difference between the torque required by the load and the counter-rotational torque developed by the motor.

(2) My Fri, Jan 4, 1:46 pm post.
Dynamic Braking Method. The rule-of-thumb equivalent dc-current factors for the two star-connected cases were inadvertently crossed. The factors should be 1.225 x FLA for the 3-terminal (3-windings energized) case, and 1.414 x FLA for the two-terminal (2-windings energized) case. The factor for the delta-connected winding is still 1.225 x FLA.

(3) Johan's Fri, Jan 18, 8:05 pm post.
Your statement that all of the energy is transferred as heat, is in error. Of the total power transferred across the air-gap, part is converted to heat as rotor loss, and part is converted to mechanical power (torque times rotational speed!)

BTW, to the List... unless you subscribe to the philosophy that "what's fur free ain't worth nuthin" lets hear some feedback!

Regards,
Phil Corso, PE
(Boca Raton, FL)


Posted by Johan Bengtsson on 23 January, 2002 - 12:59 pm
That would be true - if it were not for the fact that torque times rotational speed, with the correct signs where they should be, you get a negative power when you reduce the speed, ie the the energy are not transfered to the load as an increased speed, but rather the oposite. This means that the mechanical energy (rotational speed times inertia) does have to go somewhere (there is a law against destroying energy) since it is not only hard to brake that law but it is supposed to be impossible the energy simply becomes heat - if you don't do anything else with it - such as regenerate it to electrical power or something.

So by electrically braking without regeneration all the energy you feed to the motor + all mecanical energy stored in the motor and load will end up as heat. And will do so inside the motor (apart from the small amount lost in feeding cables, friction elsewhere, cooling fan and so on - of course)


Btw, if the law actually could be broken it would be possible to create energy too - a very good idea but unfortunately not something anyone know how to do.


/Johan Bengtsson

Do you need education in the area of automation?
----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: johan.bengtsson@pol.se
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----------------------------------------


Posted by schmooot on 21 December, 2011 - 3:53 pm
Can this same theory be used to apply braking to a single phase 220V ac motor?


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Posted by Phil Corso on 21 December, 2011 - 8:13 pm
Schmooot... If you are referring to the 04-Jun-02 post, the answer is yes. Do you need details?

Regards, Phil Corso


Posted by schmooot on 22 December, 2011 - 4:10 pm
> Schmooot... If you are referring to the 04-Jun-02 post, the answer is yes. Do you need details?

Yes I would like details please.

I've got a 3-phase machine that I converted to single phase by changing the motor and minor wiring. The electronics are happy on 220V single phase. And my single phase 220V motor runs the machine just fine.

I would however like to utilize the brake system that is built in. It consists of the start contactor opening (disconnecting the 2 lines)and the brake contactor closing (connecting one line and the + lead of a rectifier into the one winding.)

If I can utilize that brake contactor, and/or the rectifier as well it would be nice. Or if I have to use an external brake of some sort I can run it off the brake contactor as well.

It is a farm duty motor with about 8 wires coming out of it.....the diagram shows most of them either not used or joined together


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Posted by Phil Corso on 22 December, 2011 - 9:04 pm
Schmooot... two questions:

1) Is the motor also reversible?

2) Can you send me the Schematic, so that I can modify it to include the "Brake" circuit?

Phil Corso Cepsicon[at]AOL[dot]com)


Posted by schmooot on 23 December, 2011 - 3:44 pm
www.badschmocustoms.com/schmooot/temp/acmotordiagram.jpg

Here is the schematic off the motor housing. It is reversible.

Also shown is where I attached it to my machine...couldn't be bothered to change the picture so just ignore that

I have tried swapping the wires to reverse the motor while it was still spinning (hoping for a brake and subsequent direction change) but it just spooled back up again in the same direction.


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Posted by Phil Corso on 23 December, 2011 - 11:46 pm
Schmoot... please send me the motor's nameplate data and your your email address. Phil


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Posted by Phil Corso on 25 December, 2011 - 10:35 am
Schmooot... regarding the comments you posted:

1) Can't Reverse Motor While Running.
A single-phase fractional motor can be reversed by switching the power supply leads to either the 'run' winding or the 'start' winding. But, while operating at normal speed the 'start' winding is disconnected from the supply by a centrifugal switch. If reversal is attemped, the 'start' winding effect is negated until the motor slows down enough to re-close the switch. But, at that point the available torque could be too low!

2) Dynamic Braking Problem.
The nameplate reveals the motor is a Dual-Value Capacitor motor; having both a 'Run' and 'Start' capacitor. Hence, braking by DC-current injection, while doable, is more complicated than for the typical motor. The reason is the 'Run' capacitor must be isolated and any residual charge "bled" off.

I will contact you off-list with additional information.

Regards, Phil Corso


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Posted by Phil Corso on 30 December, 2011 - 2:18 pm
Follow up to Schmooot... Attached is a schematic circuit showing how to alter the Dynamic Braking Circuit (originally provided for a 3-ph induction motor) so that it can accommodate the replacement single-phase, 2-Value Capacitor (Start and Run) Fractional Hp motor! Modifications to the original Schematic are shown in red. Following is a description of major components:

o C1, Run Contactor (original.)
o C2, Brake Contactor. The original, 3xN/O contacts, are designed for AC-duty and are not suitable for DC-duty. I suggest replacement with one having 2xN/0 and 2xN/C contacts, or 2xSPDT contacts.
o CS, Start-Capacitor (original.)
o CR, Run-Capacitor (original.)
o Rs, Discharge-Resistor* for CS-cap (new.)
o Rr, Discharge-Resistor* for CR-cap (new.)
o RB1, and RB2, DC-Injection Resistors* for deceleration-time adjustment and to limit RUN-winding current (new.)
* Type, Size, and Wattage to follow.

Regards, Phil Corso


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Posted by Hesham on 22 February, 2012 - 4:22 am
Hello Phil Corso,

I cant see the attachment. can u send it to me on
hesham.marzouk@live.com??


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Posted by Phil Corso on 22 February, 2012 - 1:47 pm
Hesham... please contact me off-forum.

Regards, Phil Corso (cepsicon[at]AOL[dot]com)


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Posted by Phil Corso on 22 February, 2012 - 2:23 pm
Hesham… The reason for contacting me off-forum is that I need detail about your application.

The schematic in Schmooot's case covered a 1.5 Hp, single-phase, dual-voltage, start-cap, run-cap, motor, which had been substituted for a 3-phase motor. The conversion was more involved because of the very large run-cap.

An aside… follow-up to the Schmooot case: The conversion was successful! Originally, the motor took about 30 seconds to stop. The modification resulted in a stopping time of less than about 4 seconds, without exceeding FLA!

Phil Corso

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