Connect with us

overheating of the variable resistors

Discussion in 'General Electronics Discussion' started by di1026, Jul 24, 2013.

Scroll to continue with content
  1. di1026

    di1026

    7
    0
    Jul 24, 2013
    Hi I am developing a very simple circuit in which I have a power supply (can work either in cc or cv modes) working in cv mode to give voltage to R1+r1 and R2+r2 in parallel. R1 and R2 are variable resistors that range from 0 to 1000ohm and r1=0.8ohm while r2=0.4ohm (two coils). So it is just R1 is in series with r1 and parallel with R2, which is in series with r2. The power supply can give a voltage between 0 and 30v. As I increased the voltage to around 5v both of the variable resistors got burned. I think it is an overheating problem but I do not know how to solve it. Thanks for any suggestions.
     
  2. john monks

    john monks

    693
    2
    Mar 9, 2012
    What is cc, cv, R1, r1, R2, and r2?
    Can you post a schematic?
     
  3. di1026

    di1026

    7
    0
    Jul 24, 2013
    Hi thanks cc mode is when the power supply only gives a fixed value of current while cv mode is giving a fixed value of voltage and i am attaching a scheme...thanks for any advice
     

    Attached Files:

  4. duke37

    duke37

    5,364
    771
    Jan 9, 2011
    Show a circuit diagram. If the resistors burt out, they onerheated !

    What are you trying to do?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Solve it by not turning the variable resistors down anywhere near 0 ohms.

    Even better, place a fixed resistor in series so as to limit the current through the variable resistor to the maximum they can safely handle.

    For a pot R ohms rated at W Watts, and a voltage of V volts, the series resistor should be V*sqrt(R)/sqrt(W).

    So, for 5V, 1000 ohms, 1W, the fixed resistor is 5 * 31.6 / 1 = 158 ohms (for 30V it would be around 1k).

    But what are you trying to do?
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Alternatively, set the maximum current to the minimum value of sqrt(W/R) for both pots.

    (for a 1k 1W pot, that would be about 31mA)
     
  7. di1026

    di1026

    7
    0
    Jul 24, 2013
    Hi thanks a lot for your reply. what i am trying to do is to get a certain range of tunable current for i1 and i2 so that i can generate a tunable magnetic field in the coils to be used in other experiments.where did you get the formular you used in your reply? thanks again :)
     
  8. di1026

    di1026

    7
    0
    Jul 24, 2013
    i think the problem is that i do not want to limit the current, say that, i want i1 to be tunable between 0-2.5A and i2 to be tunable between 0-0.5A, r1=r2=0.8ohm so i need two variable resistors in each branch to control the voltage across the coils r1 and r2. so i can calculate the range of variable resistors i need to use but i think i should look for potentiometers that can dissipate heat at that certain rate, right?
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    The formulas are all derivations of V = IR and P = VI

    A potentiometer rated at 1W is designed to dissipate 1W over the entire resistance element. This is effectively a current limit. Calculating that is pretty straightforward, but it pointless in this case because using potentiometers to control current directly is wasteful, expensive and inelegant.

    It would be far better to use 2 power supplies in CC mode and adjust the current directly. Presumably the power supplies will be up to this task and your life will be simpler.

    I'd also be looking at having coils with far more turns of finer wire so that their DC resistance is higher. This will assist you in 2 ways. Firstly you can get the same magnetic field at a lower current, and secondly control by voltage may be practicable.

    What are you actually trying to do?
     
  10. di1026

    di1026

    7
    0
    Jul 24, 2013
    Hi thanks for your reply. What I am trying to do is that for cold atoms experiment we have a lot of magnetic coils and right now each of them is using a single power supply and it costs a lot of room so we are thinking to use just one power supply but constructing a circuit so that we can have different currents output for different coils. but i also think it would be better to be tunable and that's why potentiometers came into the plan :)
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Well, potentiometers are NOT the way to do it (not directly anyway)

    There are various circuits which allow you to create a variable current source, but I'm not sure that you'll save an awful lot more space over a small power supply.

    There are options if you're willing to go switch-mode, but that would have more ripple and thus more variation of the magnetic field. This may not be suited to your experiment.

    If you're driving your 0.8 ohm load at 2.5A, it will drop 2 volts. If your constant current source is operated from 5V, then the peak dissipation is only 7.5W which is easily handled.

    This page shows the basic circuit for a constant current sink. In this case the constant current flows between +12V and the drain of the mosfet. (He's not using it for that purpose exactly). Note also that the circuit is powered by a split power rail (+12/-5V). The current is given by the voltage at the non-inverting input of the 741 divided by the sense resistor.

    In your case, the sense resistor could also be the coil! In any case, I would place a reverse biased diode across the coil to remove any chance of problems due to its inductive nature.

    The Mosfet will need a heatsink, and it would be preferable to operate it from a lower voltage (say +5V)

    Note that this means you'll need 3 voltage rails, +12, -5, and +5. (the -5 could be anything between -5 and -12 without causing problems) Only the supply to the drain (+5 in this case) needs to supply significant current, but the regulation of the +12V supply needs to be good as this is also a reference for the current limit. If you wish to monitor the current, do so in the drain of the mosfet.
     
  12. di1026

    di1026

    7
    0
    Jul 24, 2013
    Hi I am interested in what you said in the 4th paragraph, is it easy to find a resistor that can deal with a heating rate of 7.5w, just as an example of your calculation? I mean is that a better way to use the lowest possible voltage in the power supply, something around the voltage required for the coil that has the maximum voltage drop? I need to calculate that power and see if it is acceptable for the resistor specifications right? and then to find the specific resistor that can stand this amount of power. But let's say I have 0.8ohm for the coil and 2.5A current flowing through then I let the power supply give out 2.6v then I only need a resistor that can stand 0.25w heating rate right? That would not be hard? At this point I need a resistor of 0.04ohm and let's say another end of current I need is 0.1A then the voltage drop across the coil is 0.08v and I still have 2.6v output then I need a resistor of 25.2ohm (heating rate in this case is 0.252w). Thus I need a variable resistor that varies between at least 0.04ohm and 25.2ohm while can stand the heating rate at least of 0.252w. Is this correct? If it is, I can try this tomorrow in the lab. Thanks!
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Well, you could add a 1 ohm resistor in series with the coil.

    Then you set the voltage of your power supply equal to the current you require in amps.

    But that requires 1 power supply per coil and is just a less efficient way doing what I suggested in an earlier post (CC mode on power supplies).

    Getting resistors with values under 1 ohm is not hard, under 0.1 ohm is a little harder, but they come in preferred values (0.1, 0.15, 0.22, 0.27, 0.33, 0.39, etc.). You may not be able to find the *exact* value you want.

    You can also get resistors that can handle 10, 20, 50, or even 100W without a lot of trouble. But expecting to find exactly the right low value resistor in the power rating you need may be a bit of an ask.
     
  14. di1026

    di1026

    7
    0
    Jul 24, 2013
    Thank you so much Steve, I'll think about this more and get back if I am still confused. Many thanks! :)
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    Another approach would be to use a single power supply in constant current mode, connect all the electromagnets in series, and put a low resistance in parallel with each electromagnet, as necessary, to pass some of the current, leaving less for the electromagnet. You would need fairly low resistances, and probably rotary switches to select between different values. I'm not sure if this would be better than the other approaches suggested, but it might be worth considering.

    FWIW I think your best bet is as Steve suggested, use more turns of thinner wire in your electromagnets, so for a given magnetic field strength, less current is required.
     
  16. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    What you should make is an adjustable current source, using a power transistor as regulating element. You can then use a pot to control the current, but now on the lowcurrent side of the construction.

    If you want to continue with the potensiometer idea, you need to use a rheostat, a wirewound adjustable resistor for higher power. Due to the wirewound structure, it will have a step like effect on the resistance, and resulting current.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-