Snubber circuit

  • Thread starter H012
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  • #1
H012
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I have been doing a lot of reading online concerning this circuit but having little luck in my dilemma. I work in the bowling industry the machines use 1/3 HP 115 v single phase induction motors. FLA on these motors is 6 Amp. These motors use capacitors 480-520 MF for braking. This braking circuit is achieved by using 2 NC contacts The motor run circuit uses 2 NO contacts The relay is of course 4PST My problem is: The NC contacts used for braking burn off at least 50% compared to the NO contacts. This is a common problem for years across this industry. The concern has been addressed with the manufacturer to no avail. I guess selling more relays is more important! Does anyone have a snubber/ suppression circuit that would help with this problem? Sure hope someone can help me
 

Answers and Replies

  • #2
Svein
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I had that problem once. I solved it by inserting an 8Ω 50W resistor in series with the relay contacts.
 
  • #3
anorlunda
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[PLAIN]https://en.wikipedia.org/wiki/Contact_protection said:
[/PLAIN] [Broken]
Protection[edit]
The degradation of the contacts can be limited by including various contact protection methods.

One method is to add electronic components such as: capacitors, snubbers, diodes, Zener diodes, transient voltage suppressors, varistors, in-rush current limiters PTC resistors, NTC resistors, voltage-dependent resistors.[3] However, this is the least effective method as these do neither significantly influence the creation nor suppress the arc between the contacts of the electromechanical power switches, relays and contactors.[4]

A slightly more effective method is to make the contacts themselves larger, i.e., a contactor.[5]

A similar method to increasing contact size is to make the contacts out of more durable metals or metal alloys such as tungsten.[2]

The most effective methods are to employ arc suppression circuitry including arc suppressors, solid state relays, hybrid power relays, mercury displacement relays and hybrid power contactors.[6][7][8][9]
 
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  • #4
berkeman
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Do these circuits typically use zero-crossing detection for voltage (closing the contacts) and current (opening the contacts)?
 
  • #5
H012
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I tried a solid state relay they didn't last long at all. I you have ever gone bowling you would see that the motors are cycling on and off a lot!. These machines on a busy day will run 10 hrs. straight. A compact arc suppression circuit is what I need. Using a rated TVS in the circuit might also help.What do you guys think.
 
  • #7
H012
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20160715_163238.jpg
As you can see the 2 middle NC contacts are shot
 
  • #8
jim hardy
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These motors use capacitors 480-520 MF for braking. This braking circuit is achieved by using 2 NC contacts The motor run circuit uses 2 NO contacts


Can you show that braking scheme ? Does it try to reverse the motor ?
Can you tolerate say a half second delay before applying the brake?

Are you free to experiment ? Perhaps a couple amps of DC through the motor would be an equally effective brake.
 
  • #9
H012
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Yes I have a schematic will try my camera on my android. I will get back .Thanks
 
  • #10
H012
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I added as a thumbnail. Blue is the braking circuit Red is the run circuit Let me know if picture is ok
 

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  • #11
H012
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Solid state relay was only one of the arc suppression ways mentioned in the Wiki.

These devices are another. http://www.mouser.com/ds/2/88/Q-QRL-27503.pdf
Type Q/QRL (Quencharc ) Arc Suppressor/Snubber Network
I looked this Quencharc up.Interesting. I'm not sure exactly what they mean " put it across the contacts"
 
  • #12
H012
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I looked this Quencharc up.Interesting. I'm not sure exactly what they mean " put it across the contacts"
Like this?
 

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  • #13
jim hardy
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I added as a thumbnail. Blue is the braking circuit Red is the run circuit Let me know if picture is ok

are these the contacts burning up? i only count half enough on the relay picture to agree with the drawing....
upload_2016-7-15_21-30-23.png


is it safe to assume the NO contacts labelled "T" (above and left of this snippet)
and the NC contacts labelled "T" (in this snippet)
are on same relay? The one you pictured?

Interesting, i never saw this scheme before - i wonder how it works....
...are they switching the start caps from series to parallel and connecting them across the motor run coil, to turn the motor into an induction generator so it'll slow itself down? The two resistors providing electrical load? They must be low ohm... Do they ever get hot?

Or maybe i dont understand the scheme at all.....

If i do have it right, then here's a hypothesis for you to talk over with your guys
........
On a start sequence
after CS inside the motor opens(Centrifugal Switch, opens when motor reaches speed)
it is the job of those two resistors to discharge those two big starting capacitors.
>>>>Were one of those resistors to fail open <<<< the capacitor no longer has a discharge path..
...the first symptom would be motor doesn't brake as quickly as it should
...second symptom would be relay T gets burnt contacts 3 and 8(lower right) but probably not contacts 5 and 10
reason is
because as soon as relay T closes its NC to brake,
any charge that got trapped on the capacitors when CS opened,
discharges immediately through those two bottom right contacts , and probably at hundreds of amps !
If you've ever put a screwdriver across a 500uf capacitor that has any voltage on it you probably melted a big chunk out of your screwdriver. I know for i have done that and the flash and noise of the arc is, well, positively shocking..(:rolleyes:,:sorry:, sorry, just couldn't resist)

To test the hypothesis
How about placing a voltmeter across the capacitors, from TS20 to 115volt return
and watch voltage? It should drop to zero almost immediately after the motor reaches speed
then move meter to TS12
it should read zero until relay T applies the brakes , at which point it should jump up to probably 115volts and fade back to zero as motor slows down.
i would use an analog meter so the DMM doesn't get confused trying to autorange.
If voltage stays high after start your resistors are not discharging the caps
you'll have to try several start-stop cycles because it's random where in the line cycle CS opens, and by Murphy's Law on first few tries it'll open at just the instant there's no voltage on the caps..

If you find an open resistor
then of course replace it,
and I'd try placing a 12 volt 50 watt or so lamp in series with each of those resistors. Every time the motor brakes you should see the lamp light up ever so briefly while motor decelerates.
If it fails to light that tells you the resistor has failed open and your capacitors are abusing your contacts.
Those little Halogen lamps for accent lighting might be an easy fit . Experiment with different wattage to get a noticeable glow, you don't want a bright flash..
That gives you a visible indication that your maintenance guys can watch for. "Let there be light" seems easy to remember....

What do you think ?

What did i miss ?
(Oh wow, what if CS doesn't open ? Voltmeter test described will show that)

old jim
 
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  • #15
jim hardy
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Hmmmmm...
Does that motor ever get "jogged", that is, switched on briefly and then back off before it gets up enough RPM's for CS to open ?
If the caps don't have time to discharge , it assures the NC contact will destroyed by the mechanism i described above. And in that scenario they aren't even offered a chance to discharge.

What controls relay T ? Can it "jog" the motor, switching it back off before it gets up to speed ?

If that sort of operation is possible, try it and watch for outrageous arcing on those contacts when switched off too soon after start.

I'm beginning to think that's more likely your culprit. Don't know why, just a feeling.
It would explain why the failure is industry-wide, though.

old jim
 
  • #16
jim hardy
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if you find that exaggerated arcing on jog
and you have the freedom to try something

Are T's contacts break-before-make ?
if so , this should work
...
lift and tape wire that goes between TM-1 and TMP-Z
put a 250 uf motor start capacitor CTM-3 as shown, with a bleeder resistor, maybe 10K 5 watt
add jumper around lower right NC contacts, C1-32E to 115 volt return C1-34P
bowlingalley1.jpg


in other words fix it so CTM1 and CTM2 don't get charged during motor start. That way they can't wreck the relay when motor gets jogged..

I know it costs you a lot more for the man-hours to replace those relays than for the relays themselves....
 
  • #17
H012
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are these the contacts burning up? i only count half enough on the relay picture to agree with the drawing....
View attachment 103296

is it safe to assume the NO contacts labelled "T" (above and left of this snippet)
and the NC contacts labelled "T" (in this snippet)
are on same relay? The one you pictured?

Interesting, i never saw this scheme before - i wonder how it works....
...are they switching the start caps from series to parallel and connecting them across the motor run coil, to turn the motor into an induction generator so it'll slow itself down? The two resistors providing electrical load? They must be low ohm... Do they ever get hot?

Or maybe i dont understand the scheme at all.....

If i do have it right, then here's a hypothesis for you to talk over with your guys
........
On a start sequence
after CS inside the motor opens(Centrifugal Switch, opens when motor reaches speed)
it is the job of those two resistors to discharge those two big starting capacitors.
>>>>Were one of those resistors to fail open <<<< the capacitor no longer has a discharge path..
...the first symptom would be motor doesn't brake as quickly as it should
...second symptom would be relay T gets burnt contacts 3 and 8(lower right) but probably not contacts 5 and 10
reason is
because as soon as relay T closes its NC to brake,
any charge that got trapped on the capacitors when CS opened,
discharges immediately through those two bottom right contacts , and probably at hundreds of amps !
If you've ever put a screwdriver across a 500uf capacitor that has any voltage on it you probably melted a big chunk out of your screwdriver. I know for i have done that and the flash and noise of the arc is, well, positively shocking..(:rolleyes:,:sorry:, sorry, just couldn't resist)

To test the hypothesis
How about placing a voltmeter across the capacitors, from TS20 to 115volt return
and watch voltage? It should drop to zero almost immediately after the motor reaches speed
then move meter to TS12
it should read zero until relay T applies the brakes , at which point it should jump up to probably 115volts and fade back to zero as motor slows down.
i would use an analog meter so the DMM doesn't get confused trying to autorange.
If voltage stays high after start your resistors are not discharging the caps
you'll have to try several start-stop cycles because it's random where in the line cycle CS opens, and by Murphy's Law on first few tries it'll open at just the instant there's no voltage on the caps..

If you find an open resistor
then of course replace it,
and I'd try placing a 12 volt 50 watt or so lamp in series with each of those resistors. Every time the motor brakes you should see the lamp light up ever so briefly while motor decelerates.
If it fails to light that tells you the resistor has failed open and your capacitors are abusing your contacts.
Those little Halogen lamps for accent lighting might be an easy fit . Experiment with different wattage to get a noticeable glow, you don't want a bright flash..
That gives you a visible indication that your maintenance guys can watch for. "Let there be light" seems easy to remember....

What do you think ?

What did i miss ?
(Oh wow, what if CS doesn't open ? Voltmeter test described will show that)

old jim
Thanks for getting back. I work nights so if I don't get back don't worry I will. Anyway the T stands for the machine TABLE motor
There is also a S which stands for the machine SWEEP motor {Not shown] The relays are 4 pole 2 NO & 2 NC The braking scheme is exactly
as you said. The resistor is a 3 terminal center tap 625 ohm. In my opinion
using 2 480-520 caps is way over kill. These 1/3 motors only require a 250-320 MFD cap. The schematic is odd because the motor plug has 5 terminals V X Y and Z } V is return[neutral] X is 115 V} Y & Z are the start circuit. This problem with the NC contacts burning off is a problem with all Bowling machines it's a flaw of the design I think. That's why I'm here on this forum. I will try your suggestion tonight. If I could a 12 DC injection for braking makes more sense.
if you find that exaggerated arcing on jog
and you have the freedom to try something

Are T's contacts break-before-make ?
if so , this should work
...
lift and tape wire that goes between TM-1 and TMP-Z
put a 250 uf motor start capacitor CTM-3 as shown, with a bleeder resistor, maybe 10K 5 watt
add jumper around lower right NC contacts, C1-32E to 115 volt return C1-34P
View attachment 103303

in other words fix it so CTM1 and CTM2 don't get charged during motor start. That way they can't wreck the relay when motor gets jogged..

I know it costs you a lot more for the man-hours to replace those relays than for the relays themselves....
Motors never jog. Once bowler throws a ball a DATA SENSOR sends a signal to start the machine cycle. Once machine has completed a cycle the machine returns to HOME posistion
 
  • #18
H012
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For everyone helping Thanks! I wanted to add just for clarification, The braking circuits main job is to maintain the timing of the T & S motors
If any of you have bowled you would see the two main components The sweep which sweeps away the fallen pins then there is the
table [T] which picks the pins up and down. There for the two components must stay in unison.
 
  • #19
jim hardy
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using 2 480-520 caps is way over kill. These 1/3 motors only require a 250-320 MFD cap.

When they're in series for start , two 500's make a 250... In parallel for braking they make 1000, surely enough to overcompensate the induction motor and turn it into an induction generator.
Wow 1000 uf at 60 hz is only 2.65 ohms ?
It's important that relay T's NO contacts open before the NC ones close else line voltage is applied briefly across 2 ohms of capacitance. Relay races are awful troublemakers.

That voltmeter test will be interesting.
500uf X 312.5 ohms is a time constant of 16 milliseconds, one line cycle
so after VS opens, voltage at TM-1 should decay to zero in around 1/10 second
i'd hook a dc meter there on maybe 250 volt scale and watch a few machine cycles
if you see dc more than just a blip, there's trouble in River City alright..
a DATA SENSOR sends a signal to start the machine cycle.


oh - a computer ? see my signature.. Look for occasional big sparks while there's a meter across those caps.
I'll be happy to be wrong,
old troubleshooter's maxim: " Before you figure out what it is, you'll figure out a lot of what it ain't".

old jim
 
  • #20
H012
153
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Ah - I did not read your post carefully enough. So there is a pair of NC contacts in the braking circuit. That takes it into an entirely different area. Breaking a circuit is much harder than you think, especially in this context. You need circuit breakers, not relays. Here is a link to some: http://new.abb.com/low-voltage/products/circuit-breakers/emax2.
Braking not breaking

When they're in series for start , two 500's make a 250... In parallel for braking they make 1000, surely enough to overcompensate the induction motor and turn it into an induction generator.
Wow 1000 uf at 60 hz is only 2.65 ohms ?
It's important that relay T's NO contacts open before the NC ones close else line voltage is applied briefly across 2 ohms of capacitance. Relay races are awful troublemakers.

That voltmeter test will be interesting.
500uf X 312.5 ohms is a time constant of 16 milliseconds, one line cycle
so after VS opens, voltage at TM-1 should decay to zero in around 1/10 second
i'd hook a dc meter there on maybe 250 volt scale and watch a few machine cycles
if you see dc more than just a blip, there's trouble in River City alright..



oh - a computer ? see my signature.. Look for occasional big sparks while there's a meter across those caps.
I'll be happy to be wrong,
old troubleshooter's maxim: " Before you figure out what it is, you'll figure out a lot of what it ain't".

old jim

Thanks I'll check tonight. I'm glad you're working with me on this. I believe what you're saying is what's ocurring. I never understood after so many years why even the new machine use AC motors. at 115 volt
If they must use AC why not 230 volt. My opinion!
 
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  • #21
Svein
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Braking not breaking
Yes, I know. What I said was that you used NC (normally closed) contacts for braking. That means (in normal usage) that the contacts are closed when not braking. Which again means that you open the contacts when you want to brake - or, in other words, you want to break the circuit when you want to brake.

The problem with using relay contacts to break a circuit is that lots of things can happen that relays are not designed to handle. Relay contacts cannot hinder arcing and are prone to contact welding. That is why I suggested Circuit Breakers (https://en.wikipedia.org/wiki/Circuit_breaker).
 
  • #22
H012
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Thanks that makes sense.
 
  • #23
H012
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this is how it's described in the manual.
I did a amp draw on the wire that feeds a voltage to the fields when braking and for a
.250 of a second the amps jump to 29 amps
let me know if you can read the photo
 

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  • #24
jim hardy
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let me know if you can read the photo
Great, thanks. Works just like we thought.


I did a amp draw on the wire that feeds a voltage to the fields when braking and for a
.250 of a second the amps jump to 29 amps
Only a quarter second ? Does the motor very nearly stop in that time ?

29amps.... .. Digital meter?AC or DC ?

...machines use 1/3 HP 115 v single phase induction motors. FLA on these motors is 6 Amp.

5x FLA is okay for starting current, if that's what it is.. can you say which wire on your post #10 diagram ?

how about that voltage measurement TMP-Z to 115 return ? Should be zero DC and AC just before brake and stay there,
Voltage from centertap of resistors to 115 return should be zero DC and AC before brake, but maybe 120-ish AC while braking...
 
  • #25
H012
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Great, thanks. Works just like we thought.



Only a quarter second ? Does the motor very nearly stop in that time ?

29amps.... .. Digital meter?AC or DC ?



5x FLA is okay for starting current, if that's what it is.. can you say which wire on your post #10 diagram ?

how about that voltage measurement TMP-Z to 115 return ? Should be zero DC and AC just before brake and stay there,
Voltage from centertap of resistors to 115 return should be zero DC and AC before brake, but maybe 120-ish AC while braking...
I was using a digital amp clamp on. I got to get my analog meter out of the closet. Just like you guys said. Digital doesnt like quick changes in readings. I was reading the return at C1 21. which goes to one side of the NC contact when it closes. I will check tmp z to return Tues when I return to work
This system has been employed by these machines since 1964. technology has come
a long way. Just hoping to find a fix that doesn't involve reinventing the wheel. Know what I mean? Effiecentcy wasn't as inportant as it is today. I consider myself pretty smart
in most things But not so much in this matter
 
  • #26
jim hardy
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I was reading the return at C1 21. which goes to one side of the NC contact when it closes.

click 'copy image'
paste into Paint
draw a red circle around C1-21
save as jpg with a name
click "upload" ?


i see four NC contacts on relay T. which ones burn up ?
bowlingalley2.jpg
 
  • #27
H012
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Sorry the place I checked was C1 31. I was looking at the sweep motor schematic which is C1 21.
 

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  • #28
Svein
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this is how it's described in the manual.
I did a amp draw on the wire that feeds a voltage to the fields when braking and for a
.250 of a second the amps jump to 29 amps
let me know if you can read the photo
This is exactly the situation i had with a motor I was controlling with a relay. Inserting an 8Ω 50W resistor in series with the relay contacts dropped the current spike to acceptable levels and did not interfere with normal operations.
 
  • #29
jim hardy
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Thanks !

In the spirit of not re-inventing the wheel

are you free to experiment with this thing? Simple stuff?
Why i ask is this
induction motor establishes a field in the rotor
which, upon removal of external power, dies out with the rotor's natural L/R time constant
on big motors that can be a goodly part of a second
and during that time that rotor field will induce voltage in the stator, which decreases in both magnitude and frequency as the field dies off and the rotor slows down.
The job of those huge capacitors is to extend the amount of time the rotor's field persists
which begs the question
by how much did they really need to extend rotor decay time? In other words, how much capacitance do they need?

We could embark on a research project, observe voltage decay on a motor and estimate rotor time constant that way
see http://scholarworks.boisestate.edu/cgi/viewcontent.cgi?article=1050&context=electrical_facpubs&sei-redir=1&referer=http://www.bing.com/search?q=induction+motor+rotor+time+constant&qs=n&form=QBLH&pq=induction+motor+rotor+time+constant&sc=1-35&sp=-1&sk=&cvid=B2E246CDE9E84E089C19605491FC6377#search="induction motor rotor time constant"
but one experiment would be a lot easier.

At the instant brakes are applied
current will be :
(voltage made at motor terminals by the rotor field spinning inside the armature ) divided by (sum of impedances) ,
and sum of impedance is Zmotor + Z external
and our best measurement so far is your 29 amps though i suspect it's higher because of meter response time. We'll use it anyway.
If Vmotor is at that instant let's just guess 130 volts,
130/29 = 14.4 ohms is sum of motor and external impedance
We know the capacitors are only 1/2πfc = ##\frac{1}{377 X 1000uf}## = 2.65 ohms,
and motor winding resistance is typically only a couple ohms, throw in a little more for interconnecting wires
suggesting motor impedance dominates and it'll be inductive.
In fact since inductive and capacitive impedances subtract, motor Z might be around 16 ohms.

Do we reallly need all that capacitance to extend motor time constant? I don't know, the more capacitance you have the longer it can maintain terminal volts as it slows down and frequency drops off..
The calculations are complex
but you have access to a test bed !

What If

we lifted this wire and observed whether it significantly affects coastdown ?




bowlingalley3.jpg


That leaves the capacitors in series,
if your 29 amps for ¼ second gets larger then we are moving closer to series resonance between motor and capacitors
if it gets smaller then motor impedance is way less than my ~14 to 16 ohms

but the real question is how does it affect coastdown ? Does the mechanism stop at the right place?

To try and explain (to myself as much as to you)
thinking in simplest terms
Given that the objective is to remove energy from the rotating parts so the mechanism doesn't overshoot its resting place,,, and
Capacitance does not remove energy, resistance does,,, and
The only resistance in the circuit is motor winding resistance (well plus Rs of the capacitors where we'd rather not deposit energy for it heats them)
maybe there's a different balance of capacitance and resistance we could use that avoids this high current.

Remember your power theorem, maximum transfer when Zload = conjugate of Zsource
if Zmotor = 2+j13 , maximum slowdown will be when Zload = 2-j13
and our load is now roughly 0-j2.65
lifting that lead leaves the caps in series making our load 0-j10.6 at 60 hz

so the current might get really big.
If it does, that's a clue

do you think we might extract energy faster with some resistance external to the motor?
Maybe a 300 watt incandescent lamp here ?
bowlingalley4.jpg


Trial and error, but based on fundamentals...
every piece of data improves our understanding of the fundamentals.

old jim
 
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  • #30
H012
153
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Thanks !

In the spirit of not re-inventing the wheel

are you free to experiment with this thing? Simple stuff?
Why i ask is this
induction motor establishes a field in the rotor
which, upon removal of external power, dies out with the rotor's natural L/R time constant
on big motors that can be a goodly part of a second
and during that time that rotor field will induce voltage in the stator, which decreases in both magnitude and frequency as the field dies off and the rotor slows down.
The job of those huge capacitors is to extend the amount of time the rotor's field persists
which begs the question
by how much did they really need to extend rotor decay time? In other words, how much capacitance do they need?

We could embark on a research project, observe voltage decay on a motor and estimate rotor time constant that way
see http://scholarworks.boisestate.edu/cgi/viewcontent.cgi?article=1050&context=electrical_facpubs&sei-redir=1&referer=http://www.bing.com/search?q=induction+motor+rotor+time+constant&qs=n&form=QBLH&pq=induction+motor+rotor+time+constant&sc=1-35&sp=-1&sk=&cvid=B2E246CDE9E84E089C19605491FC6377#search="induction motor rotor time constant"
but one experiment would be a lot easier.

At the instant brakes are applied
current will be :
(voltage made at motor terminals by the rotor field spinning inside the armature ) divided by (sum of impedances) ,
and sum of impedance is Zmotor + Z external
and our best measurement so far is your 29 amps though i suspect it's higher because of meter response time. We'll use it anyway.
If Vmotor is at that instant let's just guess 130 volts,
130/29 = 14.4 ohms is sum of motor and external impedance
We know the capacitors are only 1/2πfc = ##]frac{1}{377 X 1000uf}## = 2.65 ohms,
and motor winding resistance is typically only a couple ohms, throw in a little more for interconnecting wires
suggesting motor impedance dominates and it'll be inductive.
In fact since inductive and capacitive impedances subtract, motor Z might be around 16 ohms.

Do we reallly need all that capacitance to extend motor time constant? I don't know, the more capacitance you have the longer it can maintain terminal volts as it slows down and frequency drops off..
The calculations are complex
but you have access to a test bed !

What If

we lifted this wire and observed whether it significantly affects coastdown ?




View attachment 103374


That leaves the capacitors in series,
if your 29 amps for ¼ second gets larger then we are moving closer to series resonance between motor and capacitors
if it gets smaller then motor impedance is way less than my ~14 to 16 ohms

but the real question is how does it affect coastdown ? Does the mechanism stop at the right place?

To try and explain (to myself as much as to you)
thinking in simplest terms
Given that the objective is to remove energy from the rotating parts so the mechanism doesn't overshoot its resting place,,, and
Capacitance does not remove energy, resistance does,,, and
The only resistance in the circuit is motor winding resistance (well plus Rs of the capacitors where we'd rather not deposit energy for it heats them)
maybe there's a different balance of capacitance and resistance we could use that avoids this high current.

Remember your power theorem, maximum transfer when Zload = conjugate of Zsource
if Zmotor = 2+j13 , maximum slowdown will be when Zload = 2-j13
and our load is now roughly 0-j2.65
lifting that lead leaves the caps in series making our load 0-j10.6 at 60 hz

so the current might get really big.
If it does, that's a clue

do you think we might extract energy faster with some resistance external to the motor?
Maybe a 300 watt incandescent lamp here ?
View attachment 103375

Trial and error, but based on fundamentals...
every piece of data improves our understanding of the fundamentals.

old jim
I can lift this wire Tues when I return to work. If you remember earlier in this thread I had talked about the capacitance of this circuit
I agree with the time constant that the fields are energized is a question. By the way just for clarification The relays are turned on an off by means of micro switches riding along a cam faces. The signal to the relay 24vac coils for S & T is done via a PCB. This in basic terms
 
  • #31
jim hardy
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The relays are turned on an off by means of micro switches riding along a cam faces.
Great - that's something you can see!

The signal to the relay 24vac coils for S & T is done via a PCB.
?? Just a passive pcb that conducts current to the coils, or an active one with circuitry ?

If you remember earlier in this thread I had talked about the capacitance of this circuit

Yes i do remember .
I sure wonder how the design guys arrived at their capacitance values.

That's a fun part of maintenance, figuring out why the designers did things the way they did.
Is the gear mechanism some sort of "geneva wheel" gear mechanism where modest motor coastdown shouldn't matter very much?
genevawheel.jpg
 
  • #32
Tom.G
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are these the contacts burning up? i only count half enough on the relay picture to agree with the drawing....
i see four NC contacts on relay T. which ones burn up ?

Well?
 
  • #33
H012
153
7
Great - that's something you can see!


?? Just a passive pcb that conducts current to the coils, or an active one with circuitry ?



Yes i do remember .
I sure wonder how the design guys arrived at their capacitance values.

That's a fun part of maintenance, figuring out why the designers did things the way they did.
Is the gear mechanism some sort of "geneva wheel" gear mechanism where modest motor coastdown shouldn't matter very much?
View attachment 103380
It's a active board The main function of the board is scoring. The other part of the circuitry is for the start timing for when the [T] table and [S} sweep start. As I stated the two components work in unison. That's the problem of the NC burning up If the Sweep motor coasts to far then it out of time with the Table motor. When the PCB sends it signal to start Sweep and Table the machine goes into a whats called interlock The motor just mounts to a right angle gear head.
 
  • #34
H012
153
7
Well?
both NC's burn up but never at the same time. The NO's when I take the relay apart always have plenty of contact surface left. This is always the case.
 
  • #35
Tom.G
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I'll add my vote to the approaches by @Svein and @jim hardy , it sure seems the simplest and most likely to work. Although I have my reservations about an incadescent lamp in that vibration environment.

I solved it by inserting an 8Ω 50W resistor in series with the relay contacts.

Inserting an 8Ω 50W resistor in series with the relay contacts dropped the current spike to acceptable levels

Maybe a 300 watt incandescent lamp here ?
bowlingalley4-jpg.103375.jpg

@jim hardy In your post #29 calculating motor impedance, would it be more appropriate to use Line Voltage and LRA (Locked Rotor Amps)?
Hmm... Maybe not. I just measured the inductance of a couple motors here w/ similiar FLA rating and got 10mH and 8.5mH for an induction motor and one with brushes (Skil Saw, compound wound?)
 

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