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Thread: Bleeder Resistors

  1. #16
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    I haven't had the chance to think much on this, but I've been curious about the thread.

    If I follow the motor workings correctly, the start cap is in the circuit when the motor starts and then drops out when the motor reaches speed. I know the cap is there to provide a second phase by providing a phase shift to a second winding in the motor (this is fact, I have my MSEE) to make the motor rotate. Similar, if not identical, to three phase motor electromagnetic theory.

    I have two thoughts.

    1) After the start cap is out of the circuit, there would be a stored voltage on it from the point in time when the start switch opened. This voltage would be somewhere between 0 and 277V (220VAC peak voltage). This is providing the circuit/switch did not arc and discharge the capacitor when the switch opened. There's nothing we can do about this. The breaking of an inductive circuit while under load causes arcs.

    2) Another option would be if only one leg of the start cap opened when the motor reached speed, the other side may be connected to the main which would result in charging the capacitor while the motor was using the running capacitor.

    But either way we would need to address the likelihood that when the start switch re-engages during shut down that the sudden discharge would be harmful.

    On another note, I would like to ask Aaron to explain how the light bulbs are connected such that it is lit while the motor is running and the start capacitor is no longer connected?

  2. #17
    Quote Originally Posted by Anthony Whitesell View Post
    2) Another option would be if only one leg of the start cap opened when the motor reached speed, the other side may be connected to the main which would result in charging the capacitor while the motor was using the running capacitor.
    *cough* charging a capacitor with only one leg connected? *cough*

  3. #18
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    Anthony ......we are talking AC voltages......capacitors don't store AC voltages.....they filter AC voltages......as in pass them...not charge....not store.....
    Ken

    So much to learn, so little time.....

  4. #19
    Quote Originally Posted by Ken Fitzgerald View Post
    Anthony ......we are talking AC voltages......capacitors don't store AC voltages.....they filter AC voltages......as in pass them...not charge....not store.....
    This gets confusing, but it helps to think about what physically happens on a capacitor. With a DC voltage applied, positive charge will accumulate on one plate and negative charge on the other plate. With AC voltage, the same thing happens, but the positive and negative charges switch sides every cycle. Thus, a capacitor doesn't really "pass" AC current (as no current can physically pass through the insulating layer between the capacitor plates), but the presence of alternating charges makes it appear that current is flowing "through" the capacitor.

    So in the case being discussed, the capacitor could have anywhere between -170 and +170 on it, depending on at what point in the AC cycle the switch opened. But it also depends on the size of the capacitor and the size of the load. Depending on the length of the RC time constant versus the AC line frequency, the cap may not get fully charged, meaning it may never even reach 170, no matter when the switch opens.

  5. #20
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    Dan, I agree with your explanation of the appearance of current flow through caps...it doesn't happen.......

    I'm not sure that I can agree with the rest of your explanation...... while in theory, we would want 0 volts on the capacitor at the time the centrifugal switch closes...... I can't believe we can have enough residual voltage on that capacitor to warrant a bleeder resistor..............

    I'm going to have to do a little research on this one..... what determines the value of capacitance and it's relationship with 60 hz....... I'll get back on this one.
    Ken

    So much to learn, so little time.....

  6. #21
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    Quote Originally Posted by Aaron Rowland View Post
    ...I recently burnt out a very expensive motor on my Euro band saw. The bleeder might have prevented this...
    "might have"...Hmm...

    In 25 years I have never suffered motor meltdown that lack of a bleed resistor could produce.

    Also, I have never had my shop blow up from static discharge could produce in the DC ducting. Am I just LUCKY or what???
    [/SIGPIC]Necessisity is the Mother of Invention, But If it Ain't Broke don't Fix It !!

  7. #22
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    Current most definitely flows through a capacitor. I=C*(dv/dt). If it didn't we wouldn't be able to make a high pass filter, build multi-gigabit switching circuits (gak...no more internet!!!), among other useful things.

    Below is a brief quote from one of the finest books on basic electronics, "The Art of Electronics" by Horowitz and Hill, a book I would highly recommend to anyone with a passing interest in electronics:
    "To a first approximation, capacitors are devices that might be considered simply frequency-dependent resistors. They allow you to make frequency-dependent voltage dividers, for instance. ... So a capacitor is more complicated than a resistor; the current is not simply proportional to the voltage, but rather to the rate of change of the voltage. If you change the voltage across a farad by 1 volt per second, you are supplying an amp. Conversely, if you supply an amp, its voltage changes by 1 volt per second."

    Andy - Newark, CA

  8. #23
    Quote Originally Posted by Chip Lindley View Post
    "might have"...Hmm...

    In 25 years I have never suffered motor meltdown that lack of a bleed resistor could produce.

    Also, I have never had my shop blow up from static discharge could produce in the DC ducting. Am I just LUCKY or what???
    This is perhaps the best indication that this bleeder stuff is BS. Single phase induction motors are ubiquitous - they're everywhere (such as in your clothes washer and dryer) - and yet we don't see these motors dying at a significant rate. Induction motors are extremely reliable and have been for with us since about 1890. If they needed bleeder resistors they would have them by now - we wouldn't be hearing about them for the first time here on this forum.

    Complete, absolute, total BS.

    Mike
    Go into the world and do well. But more importantly, go into the world and do good.

  9. #24
    Not an urban legend. The original fax to me from 1991:
    baldor_sketch (Medium).JPGbaldor_fax_cover.jpg

    I also have a new 1.5 hp Leeson that came with a bleeder resistor installed. My original post stated the size wrong--it's 11W 180 Ohm that was recommended, but my local motor shop had only the size I mentioned. Worked fine. I wouldn't be so quick to disbelieve the OP--I've seen a discharging electrolytic motor start capacitor light a 60W bulb momentarily. Sorry about the scan quality, but the fax is old and faded--kinda like it's keeper! The sketch shows the switching scheme I mentioned in my earlier post. Those of you who still don't believe, I invite you to use your fingers to play bleeder resistor!

    Regrds
    Bob
    Last edited by bob hertle; 03-05-2011 at 7:15 AM.

  10. #25
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    Voltage is voltage, there is no difference between AC volts and DC volts for a given point in time. The difference being that AC would be a different value a split second later. That's the distinction between the two in the first place. If you break a circuit with a capacitor in it the capacitor will stay charged with whatever voltage was on it when the circuit was opened (provided there is no discharge path) and it doesn't matter whether it was connected to AC or DC. Volts are volts.

    The capacitor value is not based on frequency. The impendence or resistance to AC current flow is based on capacitance and is equal as follows: X= 1 / (freq. * Capacitance)

    The voltage on the capacitor will be whatever the voltage was when the switch opened. Based on my emag book, a capacitor will build a charge with only one leg connected to an AC source. The other leg will use the RF impedence of air to ground. Secondly, I tried it just this morning. I found a 4.7uF 350V cap discharged it and hooked one side hot leg of the AC outlet for a second. I tested the voltage on it with DVM. Granted this was a very rudimentary test, but there was well over 90V on the capacitor. I'm not going to say it was charged to peak, but it did not have 0 volts.

  11. #26
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    I didn't sleep well last night wrestling with this instantaneous voltage theory.

    Going through my library, I found in the 41 years since studying motor theory, I'd lost my reference materials in one of my many moves.

    There were some interesting arguments on the internet but they often seemed conflicting.


    Anthony,

    Did you measure it connected 1 leg to the source and then repeat the test with it disconnected from the source?
    Last edited by Ken Fitzgerald; 03-05-2011 at 9:24 AM.
    Ken

    So much to learn, so little time.....

  12. #27
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    Capacitors will hold a charge. When I work on a motor, or any equipment with capacitors, I will always discharge the cap prior to working with the equipment. While there's usually not enough of a stored charge to damage switch contact points, or windings, there's enough to really brighten your day.
    I put bleeder caps on the bank of starter cap's on my homebuilt "RPC". I used a 1 megohm. The cap's will bleed down in a couple of seconds I didn't put resisters on the run/balance caps, because they'll bleed down through the motor windings when power is removed.
    3 phase photo 2.JPG
    The big black cap's are the start cap's. The two small gray cap's are to balance between the phases.
    Jim Davenport
    Reporting from the depths of the Magic Garage

  13. #28
    Quote Originally Posted by Andy Pedler View Post
    Current most definitely flows through a capacitor.
    Yes, of course. What I was pointing out is that current, being the flow of electrons, doesn't flow "through" a capacitor in the sense that many people think. If you could put a red dot on a single electron, you would not see it flow into one leg of the cap and out the other (it can't - there's an insulator between the plates). Current just appears to flow directly through the cap because the charge on the plates alternatives. This is a trivial distinction, though.

  14. #29
    Quote Originally Posted by Anthony Whitesell View Post
    Based on my emag book, a capacitor will build a charge with only one leg connected to an AC source. The other leg will use the RF impedence of air to ground.
    Wait, what?

    First, we aren't talking about RF - this is a 60Hz system. The impedance of air at 60Hz is near infinite.

    Second, you are suggesting that current can flow to charge a capacitor without a complete circuit being present. Clearly, this is impossible (KCL).

    Quote Originally Posted by Anthony Whitesell View Post
    Secondly, I tried it just this morning. I found a 4.7uF 350V cap discharged it and hooked one side hot leg of the AC outlet for a second. I tested the voltage on it with DVM. Granted this was a very rudimentary test, but there was well over 90V on the capacitor. I'm not going to say it was charged to peak, but it did not have 0 volts.
    You did the test wrong. Specifically, I think you didn't disconnect the cap from the source before measurement, and then you had your DVM set to AC volts, not DC.

    Do this:
    1) Short the leads of the cap together to discharge it. Measure the DC voltage of the cap and confirm that it's 0 volts.

    2) Connect one leg of the cap to the line for as long as you like. If I understand your argument, you somehow believe that the impedance of air is such that charge will just drift through the air and find its way back to ground, charging the cap (this is silly, of course - if true, we wouldn't need power cords on our 60Hz appliances - the current could just drift through the air!)

    3) Disconnect the cap, and measure the voltage across it with your voltmeter set to measure DC volts. You will see that it still reads 0 volts.

    What you did wrong was leave the cap connected - the cap has some impedance at 60Hz, and your DVM inserts some parasitics in order to take a measurement - the combination of these factors tricked you in to thinking that there was voltage on the cap.

    Sorry, Anthony, but there is simply no way what you propose is true - if it were, it would be a violation of Kirchhoff's current law! Even at RF, it's not magic - there still has to be a return path for current. If you really believe it is possible, you should look into getting a refund of your grad school tuition, because this is freshman-level circuits 1.

  15. #30
    Motor starting capacitors are usually non-polarized electrolytic caps and have a fairly significant leakage current. They will self discharge in a fairly short time without a bleeder resistor.

    Adding some large wattage resistor across them is just a waste of power.

    Mike
    Last edited by Mike Henderson; 03-05-2011 at 11:56 AM.
    Go into the world and do well. But more importantly, go into the world and do good.

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