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Simple Q Ohms law

Discussion in 'Electronic Design' started by [email protected], Jun 15, 2006.

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  1. Guest

    I have a 4 W DC power supply. I am getting -200 V out of it. When I
    connect 2K resistor across it I get something like -25 Volts across the
    resistor. However when I connect 200K resistor across it I get -200V
    across it and the resistor gets very hot and smells like somethings
    burning. I dont understand why?

    I though smaller resistor would allow large current to flow through it
    and larger R would inhibit current and hence smaller resistance should
    get hot.

    Second thing is, shouldnt the votage acorss the resistor always -200V
    no matter what resistor I choose?
    Thanks for all replies
     
  2. Tim Wescott

    Tim Wescott Guest

    sci.electronics.basic is the correct group for this sort of question.

    The power supply is probably current limited, with 'foldback', which
    means that if you put a way-too-small resistance on it the current goes
    lower than the maximum to keep the power supply internal parts from
    toasting.

    The power relation is P = I*E. Combine this with ohms law and you get P
    = V^2/R. 25V * 25V / 2000ohm = 0.31W, so that's not bad at all.
    Strangely, 200V * 200V / 200kohm = 0.2W, which should be better.

    Four watts into a 5 watt resistor would get things hot, but you won't
    get four watts out of that power supply unless you load it with a 10k
    ohm resistor. A 20k ohm would give you 2W, which will generally make
    things pretty hot unless you're using a 10W resistor or something
    similarly big -- are you sure you're not using a 20k resistor?

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/

    "Applied Control Theory for Embedded Systems" came out in April.
    See details at http://www.wescottdesign.com/actfes/actfes.html
     
  3. Tom Bruhns

    Tom Bruhns Guest

    200V across 2Kohms would be 0.1A and 20 watts. Your supply is rated at
    4W, apparently, and likely has some current limiting built in to
    prevent you from damaging the supply to too high a load current.

    200V across 200kohms is 1mA, and 200 milliwatts dissipation. That is
    well within the rating of your supply, so the current limit is not
    activated. I'm surprised you say it gets "very hot"; it must be
    physically small, or perhaps not really 200k ohms.

    200V at 4W is 20mA. 200V at 20mA represents a 10k ohm load. 4 watts
    dissipated in a small volume should be expected to make it quite warm;
    the heat must go somewhere or the temperature will continue to
    increase. (That's a whole 'nuther resistance question--thermal
    resistance. Temperature rise equals thermal resistance times
    dissipated power. If the part gets rid of heat just by conduction,
    it's a pretty linear relationship, but if by radiation, it's not linear
    at all.)

    I trust that wasn't a homework problem...

    Cheers,
    Tom
     
  4. Jon

    Jon Guest

    With a 4 W supply, the lowest value resistance load you can have is
    10K. The supply is probably going into current limiting with the 2K
    load. Regarding the smell; you are only dissipating 0.2W with the
    200K load. Are you sure it's 200K?
     
  5. ian field

    ian field Guest

    The OP said the burning smell is with a 2k resistor - and the burning smell
    is most likely the PSU!
     
  6. ehsjr

    ehsjr Guest

    You need to recognize that any real world source has
    an internal resistance:


    .----Internal resistance-------.
    | |
    - -
    |S| |L|
    |o| |o|
    |u| |a|
    |r| |d|
    |c| -
    |e| |
    - |
    | |
    . .
    ------------------------------

    The "internal resistance" could be an active current limiter
    or simple resistance. If your numbers are right, then it has
    to include an active limiter. A useful exercise for you is to
    solve the circuit for I and P with first the 2K load
    resistance, then the 200K load resistance, then solve
    for Vload. You will see why the limiting must be active,
    not merely a resistance.

    The power dissipated in the load resistor is E^2/R, or
    ..2 watts. A 1/4 200K resistor should get hot, but not
    cause a burning smell. What size resistor did you use?

    As to the load resistor always having 200 volts across
    it, look at the impact of the source internal resistance.
    Ed
     
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