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Discussion in 'General Electronics Discussion' started by Anders Lund, Jun 16, 2015.

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  1. Anders Lund

    Anders Lund

    3
    0
    Jun 16, 2015
    Hallo

    Apparently I have lost all knowledge of electronics, so I hoped you would be so kind to assist me.

    I am planning this current containing to resistors (heating elements) and one component which requires 12V.

    I am trying to figure out if my two resistors are enough to bring down the voltage, or I will need additional resistors.

    I have my input of 230V
    Two identical serial connected resistors of 1.27 x 10^-7Ω (R1 = 1.27 x 10^-7Ω and R2 = 1.27 x 10^-7Ω)
    This is where I get stuck. I would appreciate if someone could help me back on track.

    I have made a quick drawing of the circuit.

    Regards
    Anders
     

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  2. Martaine2005

    Martaine2005

    3,075
    847
    May 12, 2015
    Hi Anders Lund,
    Math, formulas and equations is certainly where I FAIL..
    Your math may be correct but probably the wrong approach. (notice probably).

    It would be better to have a small transformer of 230v to 12v rather than resistors.
    Is this by any chance for a heated floor element and controller?
    Martin.
     
  3. Martaine2005

    Martaine2005

    3,075
    847
    May 12, 2015
    Also, is the 12 volt AC or DC?
     
  4. Anders Lund

    Anders Lund

    3
    0
    Jun 16, 2015
    Hallo

    Thanks for your prompt reply.
    Unfortunately, the heating elements is a crucial part, the 12 Volt is for a DC motor,

    The idea is to create a two-piece heating element which are revolving

    Regards

    Anders
     
  5. Martaine2005

    Martaine2005

    3,075
    847
    May 12, 2015
    OK, and thanks.
    I still think the best way is to use a 230v to 12v transformer for the motor.
    You might have to rectify the DC for the too.
    Martin
     
  6. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    Hi Anders
    Sorry but what is the actual question? Bring down what voltage? What do you want to work out?
    Adam
     
  7. davenn

    davenn Moderator

    13,713
    1,911
    Sep 5, 2009

    where's the heating element and motor in your circuit ??

    you need to give a lot more specifics for what you are trying to do
    And as has been suggested, using resistors to drop from 230V to 12V is a really bad and unsafe idea

    you have shown no rectifier ... 2 resistors are not going to provide 12V DC from a 230VAC source


    Dave
     
    Last edited: Jun 17, 2015
  8. hevans1944

    hevans1944 Hop - AC8NS

    4,547
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    Jun 21, 2012
    Welcome to Electronics Point @Anders Lund!

    I think we may have a communications problem.

    None of what you posted makes any sense. For example, you state that the two resistors are 1.27 x 10^-7 ohms heating elements. This is nonsense, as that would be 0.000000127 ohms, or 0.127 μΩ. The current in these two series-connected elements would be I = E/R = 230/(2.54 x 10^-7) = 905 mega-amperes! Moreover, if the resistors in series are supposed to be connected to a load, across which you want to develop 12 volts from a supply of 230 volts, then the load resistance must be(12/218)(2.54 x 10^-7) = 13.98 nΩ. This is an absurdly low resistance for a motor.

    But let's be somewhat charitable and say you meant to have an exponent of +7 instead of -7 for the power-of-ten multipliers, so the two resistors have a value of 1.27 x 10^7 ohms. If these are "heater" resistances, that value is absurdly large instead of absurdly small. But, if accurate, it would require a load resistance of "only" (12/218)(2.54 x 10^7) = 1.398 megohms. This is an absurdly high resistance for a motor.

    If you are trying to build a voltage divider, you need a resistor, R1, in series with another resistor, R2, and this series combination connected across the voltage, Vs, you want to divide. The decreased or divided voltage, Vd, appears across R2. From Ohm's Law, the current in the voltage divider is I = Vs / (R1+R2). This current also flows through R2, producing a decreased voltage Vd = I (R2) = Vs (R2 / (R1 + R2). Solve this equation for R2 = R1 Vd / (Vs - Vd). Plug and chug to find the value for R2, given some reasonable value for R1.

    So... please tell us what you are really trying to DO. Do you have a 230 volt supply connected to two heaters that are normally in series across the supply? Is this an AC or DC supply? Is it your desire to also run a 12 volt DC motor from this supply? How much current do the two heaters in series draw from the 230 volt supply? How much current does the 12 volt DC motor require? It might be helpful to express the heater resistance as a decimal number without the power-of-ten scientific notation.
     
  9. Anders Lund

    Anders Lund

    3
    0
    Jun 16, 2015
    Hallo, and thank you all, for taking the time explaining this to me.

    The basic idea is to create a two revolving heating elements (not the idea but imagine a kettle, where the heating elements are rotating).

    The power source will be normal danish wall plug, which is supplying 230 Volts. It is then connected to a slip-ring, surrounding the heating elements (allowing it to rotate), then serial connected to another heating element.
    As I need these heating elements to rotate, it will require a motor. The most common motors I have found with the required RPM, is requiring 12 Volts.

    I got inspired by a hair dryer, where they are using the heating elements as a resister to drop the voltage to the motor, 6 min into the video.
    I didn't take the transformer from AC to DC into consideration, thanks for reminding me.

    As heating elements, I have chosen Nichrome rods. Nichrome is a common heating element in electronic devices and has a Resistivities of 0.000001 Ω * meter.
    I will need the rods to have a diameter of 1 cm and a length of 10 cm.
    Cross Sectional Area of a Cylinder = π x r^2 = 0.785

    As the Resistance = resistivity * length/area
    Resistance = (0.000001 [Ω*m] * 0.1 [m])/0.0000785 [m^2]
    Resistance = 0.001273Ω I did miscalculate this in the first try

    Using ohms law as @hevans1944 said

    I = Vs / (R1+R2) = 23V/0.002546Ω = 9033.77848 amperes

    My question is:
    Am I able to drop the voltage from the 230V to 12V, using my two heating elements or will I need to add additional resistors? And do you have any experience with which types of AC-DC transformers I should use?


    (I am also considering if it would be more beneficial to do a parallel connection ?? )

    I understand, that it is an atypical question and the explanation is a bit confusing if i need to reformulate some of it, please tell.

    Best regards

    Anders Lund
     
  10. Martaine2005

    Martaine2005

    3,075
    847
    May 12, 2015
    Hi Anders Lund.
    The answer is YES.
    But safely NO.
    And I believe you would be foolish to try your idea!!
     
  11. KJ6EAD

    KJ6EAD

    1,114
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    Aug 13, 2011
    Anders, there are numerous difficulties with your approach to this problem. First, your motor probably won't run on AC current so simply dropping the voltage through the resistance of your heaters may not be adequate. Second, even if the motor would run on AC, it would probably spin too fast for your purposes without gear reduction. Third, unless you were very clever or lucky, the amount of heat generated by the heaters when they're dropping the motor voltage correctly will probably not be the amount of heat you require. Fourth, the current requirements for slip rings will probably make them more expensive than you like.

    Perhaps you could restate your original premise in terms of system requirements such as this hypothetical example:

    I intend to construct a novel marshmallow toaster as part of an automatic s'more making apparatus. What I think I need is a unidirectional radiant heat source of approximately 100W to be directed towards a cylindrical object of approximately 4cm diameter and 4cm length (the marshmallow before toasting). I need the heat source to rotate circumferentially around the target at a radius of approximately 100cm from the target center. What can I use to do this, motor, heater, slip rings, etc.
     
  12. BobK

    BobK

    7,682
    1,686
    Jan 5, 2010
    If you apply 230V to a heating element that is 0.001273Ω you will do nothing more than blow a fuse.

    Bob
     
    davenn likes this.
  13. hevans1944

    hevans1944 Hop - AC8NS

    4,547
    2,122
    Jun 21, 2012
    Assuming your calculations are correct (I haven't checked them), where are you going to get more than 9000 amperes at 230 volts to feed into your nichrome heater rods? That's somewhere north of two million watts! It sure isn't going to come from a normal Danish wall plug! And where are you going to find an approximately 0.001 ohm resistor for that 9000 amps to flow through, so it develops 12 volts of alternating current (AC) that you will need to rectify to operate a 12 volt direct current (DC) motor? Just asking...

    Rotating heater rods sounds like a good way to heat all sorts of things that need evenly applied heat around a cylindrical axis. Either that, or mount a whole bunch of smaller, non-rotating, heater rods in parallel around a cylindrical axis. I would go with that approach and save the cost of a motor, bearings, and whatnot. Especially the whatnot. You are going to need one huge whatnot.

    How many watts of heat do you need for your application? And how did you arrive at that figure? Is this design a rectal extraction? I ask because I see no evidence of engineering yet.
     
    BobK, Arouse1973 and davenn like this.
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