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Determining Rectifier Size

Discussion in 'General Electronics Discussion' started by Gravimetric_Chemistry, Apr 11, 2012.

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  1. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Yeah, Im making a circuit to run an electromagnet off my 15a wall plug and am trying to get a better idea of what is going on as it is not working properly. Im using a 600v 40a rectifier and want to make sure the problem is not that a rectifier hooked up to sufficient wattage is going to automatically pull that wattage (like a photon hitting a photocell) and that it is simply because my load, the electromagnet, needs more turnings to increase resistance such that it lowers the feedback more appropriately. My calculus is pretty bad so I cant measure anything here except that it blows my circuit breaker when I turn it on so thanks for any help.
     
  2. timothy48342

    timothy48342

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    Nov 28, 2011
    I would think it is the load.
    600V 40a just means don't apply more than 600V to the rectifier and don't let more than 40a run through it. It doesn't "pull" anything. However it might allow unlimited current to run through it till it fails. If you hook it to a very low resistance load (or short the terminals with a screwdriver) it would allow more than 40a to run through it then who knows what might happen, but I don't think your exceding 40a, because it is driven by only 15a.

    Now the 15a wall plug. Sort of the same thing. If you hook it to something that can draw more than 15a it would fail in some way. (If your lucky it would fail to provide the appropriate voltage, but it could also catch fire or load the mains. I think in this case the 15a power supply is loading the mains.)

    The mains work a little different. The available power (or amperage) is huge, but there is a circuit breaker. As long as the current stays below the breaker's amperage threashold, they work as if their is seemingly unlimited power. When the current is too high, they cut it.

    The point is that the rectifier doesn't "pull" power thought it. It just "provides" power to the load. If you have a small number of turns in your coil. That is what is going to excede the limits of the other components.

    So the weakest link is going to fail. (In this case the circuit breaker. It tripped, which is a good thing.)

    Adding more turns to the coil might work, but even better would be to determine the resistance of the coil and do a little math, (No calculas required. Volts divided by resistance is amps and it better be no where near 15a.)

    You said, "cant measure anything." You don't have any kind of ohm-meter or multimeter available? You might be able to get an estimate of the coils resistance based on the gague of the wire, the number of turns and the diamater of the turns. Still no calculas, but a little algebra.

    One more thing. 15 amps?? Are you sure? That wall plug is HUGE. (I think... maybe I'm off on that. I don't know.) Can you post the data printed on the plug? Hopefully it has the details. Like "voltage in," "voltage out," "amps in," "amps out." Those detail would be nice.

    -tim
     
  3. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Ok, thanks for the help. Thats a good point. In a photocell the cell does not "pull" the current, the photon hits it providing its magnetic energy to it. My problem must be that Im using a super low load. Im going to have to sit down and really think about that. Im still winding my magnet so Ill let you know how it goes.

    But yeah, the wall plug. The power plant puts out a super amount of electricity that flows down huge cables. Along the way throughout the city Im assuming it gets split into parallel circuits that cuts the amperage down. Im not sure but considering that they have "high voltage" sections thats what I would say opposed to it all being one series circuit. I would have to think about that a little more. But anyway, out on my telephone pole is a transformer that converts the volts from whatever it is in the road line to the 110 volts that my house circuit runs on. Again Im not a certified electrician and that cant be the case because it gets divided into what must be parallel circuits in the circuit breaker, so the transformers must be custom to every house? Anyway, yeah my main circuit is definately 15a 110v.

    The calculus: I was talking about using that to calculate the magnetic reluctance of the iron core of my magnet. I am assuming the reluctance (magnetic force) gets converted into increased resistance in my copper wire coil. You dont NEED to use calculus just as long as you can test the wattage of your first winding. Cont. (out of space)?
     
  4. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Then you use that wattage reading to convert it into watts per inch by dividing the watts by the length of the first pass of winding. Otherwise you have to use calculus to determine it by number of turns. Again Im assuming. Im not a magneto physics major either.
     
  5. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    In my case the circuit breaker trips it off before I can do any measurements while its running.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
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    Jan 21, 2010
    Or more accurately before you start a fire.

    Your understanding of things (based on your last few posts) is very poor.

    You are doing a slightly more sophisticated version of sticking a fork into a power point.

    edit: A warning that I'm ashamed that I didn't write earlier...

     
    Last edited: Apr 15, 2012
  7. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Yes super moderator steve, a slightly more sophisticated version of sticking a fork into a power point, and maybe the "In my case the circuit breaker trips it off before I can do any measurements while its running", but "very" poor? If you would please induldge me a little.
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Nothing magnetic is involved. Depending on the type of detector, electrons freed by the photons may generate an electric field, or a small current, or similar.

    It sounds to me that you're using a super high load. Super low resistance maybe.

    Do you understand ohms law?

    No. It actually is distributed as a high voltage and low current and transformers convert it to a lower voltage and (therefore) higher current.

    Nothing actually "cuts the amperage down", in any sense of those words that make sense to me.

    That's correct, it's certainly not one series circuit.

    No, the transformer isn't in any way custom to your house. It takes some distribution voltage (say 415V, or 3kV, or 2.2kV) and steps it down to 110V (It is actually a little more complex than that, but this explanation will do). The transformer can handle a certain power, and your house (and the other houses on that transformer) must not place a load on it that is higher than it can handle.

    Yes, the circuits in your house are in parallel, but there is no "dividing of current" in any real sense. Each circuit gets what it demands up to the point at which the supply can no longer supply it, or a circuit breaker breaks.

    And you main circuitry is definitely NOT 110V 15A, it's 110V with a 15A rated maximum load (and presumably a 15 or 20A breaker or something similar)

    I would start with calculating the resistance of your wire. Sure the inductance may have an effect, but let's assume the load is resistive at first. Note that 110V at 15A is over 1500W, and your wire had better be able to radiate that 1500W of power without melting.

    This really makes no sense at all, unless you're saying that you will use this method to determine the power dissipation per unit length of the wire. Remember that if you wind it in a coil, the inner portion will not be able to radiate heat as fast as the outer portion.

    Ohms law would allow you to calculate all of this before you connect it up. If inductance is an issue (and we need to know how many turns and what sort of core, and whether the core saturates) then even measuring the current wouldn't help you as it's out of phase with the voltage.

    Which should tell you that you're drawing massively more than the breaker is rated at.

    And in that way, it's because what you're doing is placing a heavy load (essentially a short circuit) to the mains. The same thing that happens if you stick a fork in a power point.
     
  9. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Steve, buddy. Youre acting like Im not studying right now for a masters in photochemistry. Im going to read more after this post but based on your first two statements Im seeing hot air. Photons are called electro-magnetic waves for a very strong reason. Have a good one, and thanks for the help.
     
  10. alfa88

    alfa88

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    Dec 1, 2010
    Might I suggest you use a battery and scale things down for your experiment.
    As you have it now. 120V, rectified = 60V

    26 [email protected] = 61.215 Ohms
    60V/61.215Ohms=0.98A (current capacity of 26 guage wire is 2.2A)

    That's alot of wire! But you wont trip a breaker. By the way you want to use inslated wire.
     
    Last edited: Apr 12, 2012
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
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    Jan 21, 2010
    No, I'm acting like you don't understand electrical engineering, which is probably something you're not studying.

    You make assumptions that are incorrect, and them base decisions on them. If you want me to go back and simply pretend that all your assumptions are right then I will.

    In that case the answer to your problem is

    However, good sense suggests that you shouldn't do that.
     
  12. CocaCola

    CocaCola

    3,635
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    Apr 7, 2012
    Small correction or maybe just some clarification, and of course simplified... The US 'residential' service is supplied to the house/unit at 220 volts, with a center tap neutral from the transformer... Your 'circuit breaker box' in your house divides the 220 volts supplied into two half phases @ 110 volts and a neutral, or uses the entire phase to supply 220 volt sockets... Commercial service brings in three phases and generally 440 volts...

    This is very important to note, your house is supplied with a BOAT load of available Amps!!! Most new circuit panels have a main fuse that limits the in house power to a total of 100-200 Amps @ 220v, 40-60 Amps was common a few decades ago or for small structures like house trailers... The fuse box further limits each run of electricity leaving the circuit box even further, to usually 15 or 20 Amps @ 110v or 15-60 Amps at 220v... But trust me the lines coming to your house will supply A LOT more Amps before they become a large light bulb filament...

    *Edit look, there someone hooked me up with an avatar :)
     
    Last edited: Apr 12, 2012
  13. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    0
    Apr 11, 2012
    *******************************************************************************************************
    Correction: It appears now that Im down one multimeter that the rectifier changed the amperage. I have a different load on it now but Im guessing that the change in amperage was just too high for my circuit breaker.

    Im going to add two more passes of wire on my magnet. Right now Im estimating its about 1100w with 8 passes of 14 gauge copper wire around a 1.5' iron rebar core bent in a U shape. I wasnt planning on it taking quite that much wire. Looks pretty impressive though. Right now 350' of wire. Glad I saved the money on the 500' wire. Now to make the transformer and buy another multimeter and run this properly.

    @Alfa88- yeah, thats a good idea. I hooked just the magnet up to a car battery charger. I didnt do any measurement other than the fact that it didnt get any stronger as I increased the power so thats how I estimated it was using only around 24w.
     
  14. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Oh yeah. Steve- Im TOTALLY paying attention. Keep up the FABULOUS work.
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
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    Jan 21, 2010
    You've got 8 turns of copper wire around an iron bar and you seriously expect that to limit the current?

    From here, I see the resistance is around 0.00297 ohms per foot. Let's call that 0.003.

    A rough guess is that each turn requires about 6 inches of wire, so you have 8 x 6 x 0.003 / 12 ohms of resistance in your coil.

    So 0.012 ohms. Let's round that up to 0.02 ohms to allow for some leads on the end of that coil.

    At 110 VRMS, you would expect a current of about 5500 amps. That exceeds the rating of your mains circuit, and also of the wire itself, but let's look at instantaneous power anyway...

    5500 amps from 110V gives us 605kW. Perhaps a tad more than 1100W...

    I'm not sure of the exact resistance of a fork, but it's probably not much different to your coil.

    I'm still not totally convinced you're an expert at this.

    And 2 more turns... Nah, not significant.

    My guess is that you should start with about 7000 turns. I'm not totally convinced that the coil would last long without overheating, but it would almost certainly guarantee a current less than 10A (if the assumptions about the length of wire required for each turn are correct)

    And the wire is insulated, right?
     
  16. davenn

    davenn Moderator

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    Sep 5, 2009
    he's saying 8 passes ...whatever that means ??? maybe 8 layers ?
    he's said he's used 350ft of 14 gauge wire and the 1.5 inch presumably refers to the diameter of the rebar core ??

    bit of confusion going on here it seems ;)

    Gravimetric_Chemistry exactly how many turns have you got on the rebar with that 350 ft of wire ?
    and when you say passes do you really mean layers ?

    Dave
     
  17. BobK

    BobK

    7,680
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    Jan 5, 2010
    My Galvanic Chemistry,

    Just as a sanity check, let's say you create an electromagnet that runs on 110V and draws the 15 A you are talking about. That is 1650W. Go to your local big box store and get a 1500W heater. Your electromagnet will produce a little more heat than that. Are you ready to handle that?

    Bob
     
  18. alfa88

    alfa88

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    Dec 1, 2010
    He said he was using a rectifier. That basically cuts the voltage in half.
     
  19. Gravimetric_Chemistry

    Gravimetric_Chemistry

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    Apr 11, 2012
    Ok. Alfa88: I hooked a smaller rectifier up but it fried. Was it too big? Can I hook up a much smaller rectifier so that I wont need to increase my load? Im really not too thrilled about wrapping another coil.
     
  20. BertMan

    BertMan

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    Dec 15, 2011
    Whoa

    I'm sorry, but I have to chime in here before someone gets killed. You sir, should not be playing with 110V with your understanding of electronics. 110V will easily kill, and this is no area for a novice to be testing circuits in. My suggestion would be to start smaller. A lot smaller. Get a wall wart that will bump the voltage down to 9 or 12V DC, and start playing around with different coil setups. You will find that wire thickness and the number of turns play a huge part in the resistence of the coil and the amount of inductance. But to cut your teeth on a 110v electromagic without the proper tools or education is insane.

    If you must insist on tying this, here is my advice. If you want to stop popping circuit breakers, then you have to increase resistance of your coil. Simple Ohms law is no calculus. 110V AC rectified to 60V DC with a 4 ohm coil will draw 15A. Get an ohm meter if you dont have one and check resistance. You can buy a cheap one for under $20. And if you plan on drawing 15A get prepared for some serious heat. If this thing doesn't pop a cicuit breaker, then you will probably end up melting the insulation off the wire and catching a fire. Good luck to you.
     
    Last edited by a moderator: Apr 14, 2012
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