Bleeder Resistors

[Rick Christopherson]

It's baloney alright. Dan nailed the reasons in his earlier post. Plus, the cap is across AC as Ken pointed out, not DC, so you're not going to have much of a residual charge on the cap when the starting switch kicks out at startup.

I'm just amazed that these "urban legends" get passed along without anyone questioning whether they make any sense or not.

Mike

Anthony ......we are talking AC voltages......capacitors don't store AC voltages.....they filter AC voltages......as in pass them...not charge....not store.....

Capacitors do in fact store a significant charge from an AC line. How much will depend on where in the sinusoidal wave the circuit was broken. Anyone that has built/balanced their own Rotary Phase Converter knows this all too well....it scares the heck out of you when you reconnect the caps differently.

The capacitor will have a charge on it that depends on where in the sinusoidal signal it was disconnected. The problem isn't simply reconnecting it to power while it is charged, but the 50:50 chance that it will be charged with an opposite polarity to the instantaneous position of the sine wave.

There will always be some arcing when the capacitor is reconnected because the odds of it having the same voltage as the existing sine wave is low. How much arcing depends on how far apart these voltages are. The greatest arcing occurs when the polarities are reversed. By discharging the capacitor, you minimize the maximum voltage difference between the capacitor and the incoming sine wave.

That being said, I certainly am not going to run down to my shop and monkey around with bleed resistors on such a trivial issue. The contacts on the switch are rated for the use.

[Ken Fitzgerald]

If they do store AC voltage, it's only because of the action of the switch which could be providing an instantaneous rectifying action as a diode or some switching type device would provide....and therefore...it's the ac has become a dc potential.

[Dan Friedrichs]

If they do store AC voltage, it's only because of the action of the switch which could be providing an instantaneous rectifying action as a diode or some switching type device would provide....and therefore...it's the ac has become a dc potential.

As time goes by, the voltage provided by an AC source goes from some negative value to zero to some positive value back to zero, etc...

If you have a cap connected to an AC source, and then disconnect it, the voltage on it is going to be whatever voltage the source was supplying at the instant you disconnected it. That could be negative, zero, or positive.

If you hooked an oscilloscope up to the AC line, and set it to capture from 1us to 1.01us, what would you see? Probably something that just appears to be a flat DC voltage. Again, depending on when you triggered it, it may be negative, positive, or close to zero.

[Tom Stenzel]

The residual DC voltage across the motor start capacitor will be random as it depends on when in the AC cycle the switch opens. But that voltage will always be limited by the start winding that is in series with it.

Also the start winding has inductance which will limit the instantaneous current flow when a DC voltage is applied to it.

It's been my experience that contacts have a lot more problems with arcing when they open that when they close.

Compared to the stresses and currents involved when the the switch opens on motor startup, the maximum potential current surge that occurs on motor shutdown is inconsequential. It's much ado about nothing.

-Tom Stenzel

[Dan Hintz]

When was the last time you've seen a motor fail because the centrifugal switch welded itself shut from arcing?

I don't really do much power design. I leave that to the guys who specialize in that field... my specialization is in signal processing (but I end up doing a lot of hardware design at the microprocessor level, which includes industrial controls). I will say that relay contact arcing/welding is a serious issue, especially in the industrial sector. Even at low voltages, contacts can weld themselves together if it's not accounted for. It requires a little bit of external circuitry to prevent such an occurrence, including reverse diodes and resistors, sometimes caps if you want a pull-in delay. The Oneida rep specifically mentioned the centrifugal relay contacts as being the problem child in multiple on/off cycles. While I have never really worked with centrifugal relays or large motors, there's no leap of faith whatsoever to recognize the relay contacts will act the same, regardless of purpose.

I cannot verify the starter winding is of higher resistance... I thought it was part of the standard coil that was simply wrapped differently for a few turns, but I'll take your word for it. If it is higher resistance, than I would agree that it is a weak link. But I cannot agree that it is the only problem, regardless, and I lean more heavily towards the relay being the major cause of failure.

[Dan Friedrichs]

The Oneida rep specifically mentioned the centrifugal relay contacts as being the problem child in multiple on/off cycles. While I have never really worked with centrifugal relays or large motors, there's no leap of faith whatsoever to recognize the relay contacts will act the same, regardless of purpose.

I agree that it can happen, but in that case, it would be dependent on TOTAL number of on/off cycles, not the number of on/off cycles in a short time. Even opening and closing of the contacts may cause a small arc, but in a centrifugal switch with large metal contacts, certainly the little bit of heat generated would be quickly dissipated, meaning that repeated cycling in a short time interval is no more detrimental to the contacts than repeated cycling over a long time interval.

I stand by my assertion that heating of the starter winding is the ONLY reason not to repeatedly cycle an induction motor over a short time period. Obviously there are other effects (such as wear on the magnetic starter, wear on the centrifugal switch, etc), but those are not specific to repeated cycling over a short time interval - they will be the same if you turn the motor on twice, 3 hours apart, or twice, 5 minutes apart.

[Anthony Whitesell]

I did #1, 2, and 3. The DVM was set to DC volts. As it has been pointed out AC voltage cannot be stored. That's like trying to store a breeze. You can store high air pressure, but not a moving wind. Same thing, you can store DC but not AC.

The DVM spiked to over 90 volts and bleed to zero because the meter input impedence is not infinite. This one is probably lower than others as it is a less expensive DVM (not cheap, but also not a Fluke).

If you place a hose with a hole in over a bucket (that doesn't leak) does the bucket fill up? Yes. Same thing. Just the voltage (or the level in the bucker) will not be predicitable as the other leg of the capacitor has no ground reference. There's no violation of KCL, but no one can tell you what the voltage will be.

[bob hertle]

All very interesting! I'd like to point out that I did not make any assertions about reduced motor life. I have replaced several sets of contacts on single phase motor starter switches and it's relatively easy and cheap! Biggest hassle is removing the motor from the machine. Frankly I couldn't care less if I had to replace the contacts every 20 years or so--just part of the game. My reason for adding bleed resistors (specifically to 2 Baldor motors) was to get rid of the very annoying shudder, or shake when the still charged starter cap discharges as the centrifugal switch closes on spindown (~75% of motor base speed). It was a problem that was well known to Baldor, as evidenced by their faxed response to my call at the time. Since then, I have run across the problem twice more while helping friends. Solved it with the same way--bleeder resistor--and never really looked back--til now! LOL!

Regards

Bob

[Anthony Whitesell]

There is nothing "theoretical" about this. AC voltage cycles from -120 to +120 60 times per second. The voltage across/through the capacitor is doing the same thing. When you break the circuit, the capacitor remains charged with the voltage it had when the circuit broke. The caveat here is...IF the switch contacts did not arc when the circuit was broken. IF the contacts arced then chances are the capacitor has discharged (hence the cause of the arc). If the contacts ALWAYS arc, then the cap is always discharged when the start cap is removed from the circuit and we have been either discussing the wrong problem or the wrong solution. So we would need a resistance across the capacitor that will discharge the capacitor before the arc is generated. That would get into the dielectric constant and impedence of air at the time the arc occurs and then we need a resistor smaller than that value. I think I'll take my chances or wait for the motor manufacturers to solve it for me.


Try this experiment. I did in high school.

AC Hot->Pole 1 SPDT Switch
SPDT Swicth Common->Cap->AC Neutral and DVM Ground
Pole 2 SPDT Switch->DVM High side

Set the DVM for 200VDC.

Set the switch to Cap side for a second, then to the DVM. Notice the voltage. Switch it again. Notice the voltage, it will probably be different (might not be if you manage to time it right). Generally every time you set the switch to the cap and then back to the DVM, you get a different reading. If you DVM is like mine, the reading doesn't hang around long as the DVM is acting as a discharge path for the capacitor. The variation in the reading is due to the 60 cycles per second and where the capacitor is removed from the circuit during the "charge" cycle.

[Anthony Whitesell]

You're making me wonder if that's what is going on with my jointer. My table saw doesn't do it, nor my bandsaw. But the jointer makes one heck of a thud when shutting down. It also makes a shudder at startup, so I thought it was the belt. Based on comments from here, I have played with the tension but can't get rid of it. It is much worse during the shutdown than the start up.

[Dan Friedrichs]

There is nothing "theoretical" about this. AC voltage cycles from -120 to +120 60 times per second.

No, it goes from -169 to +169. 120 is the RMS, not peak.

[Dan Friedrichs]

I did #1, 2, and 3.

Are you sure the cap was disconnected from the line before you measured? I just went and repeated your experiment, and got zero volts. I also SPICE'd it, and got zero volts.

If you place a hose with a hole in over a bucket (that doesn't leak) does the bucket fill up? Yes. Same thing. Just the voltage (or the level in the bucker) will not be predicitable as the other leg of the capacitor has no ground reference. There's no violation of KCL, but no one can tell you what the voltage will be.

Yes, it is a violation of KCL. You are implying that electrons are piling onto a plate of a capacitor - where are they coming from? There must be an circuit path for them to flow from.

I think if you draw this out, you'll see how ridiculous it sounds - a voltage source, connected to one side of a capacitor, and the other leg of the cap floating. You are telling me that current goes into one leg of the cap, and doesn't come out the other? That is a prima facia violation of KCL!

Your water analogy is entirely incorrect. Please see the section on "Capacitor" here: http://en.wikipedia.org/wiki/Hydraulic_analogy

[Anthony Whitesell]

Are you sure the cap was disconnected from the line before you measured? I just went and repeated your experiment, and got zero volts. I also SPICE'd it, and got zero volts.

Yes Dan it was disconnected.

As for the spice model, that's confirms the statement in my emag book. It can't be resolved to an exact number.

I did what I did. The cap was charged by sticking one leg into the electrical outlet. The DVM was set to 200VDC. I just did it again and came out with 70-something volts to double check the procedure, even though it was the same methodology as last night and as you have asked.

The pipe analogy from the wiki website isn't applicable. The more appropriate analogy would be static charging. By placing a higher potential on one side of an insulator than the the other with no discharge path.

[Dan Friedrichs]

The pipe analogy from the wiki website isn't applicable. The more appropriate analogy would be static charging. By placing a higher potential on one side of an insulator than the the other with no discharge path.

WHY is it not applicable? What do you mean by "static charging"? How can you place a potential on an insulator?

[Bill Huber]

I have not read all the replies but I do know this, I would lover to have a bleeding resistor in my shop. Just the other day I cut my finger on the sharp edge of a package I was opening and it would would have been nice to have that bleeding resistor at that time.

I just looked at the title again it was bleeder resistor, not bleeding resistor, sorry about that.