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Which way is better, zener or resistor

Discussion in 'Electronic Basics' started by Allen Bong, Aug 19, 2007.

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  1. Allen Bong

    Allen Bong Guest

    I have a 12V relay and I wanted to operate it with 18V supply. Could
    I use a 6V 1W zener to drop the extra 6V from the supply or is it
    better to use a 150 ohm 1/2W resistor. Which will generate more
    heat? The relay operates with 20mA current.

    Allen
     
  2. In most cases, the resistor would be better.

    As to specifics, you don't say what the coil resistance is.
    But for any given coil resistance, a zener or a resistor
    that drop the same voltage in series with that coil will
    waste exactly the same power. The resistor is probably
    cheaper, and more reliable. It also has a big advantage if
    the input voltage has any variability. If value is chosen
    to drop 6 volts when the coil drops 12, it will drop this
    same fraction of the total voltage if the input voltage
    varies. So if the input voltage rises to, say, 21 volts,
    the resistor will drop 1/3rd of that, leaving 2/3rds of that
    for the coil, so the coil voltage will rise to 14 volts.
    Similarly, if the input voltage drops to 15 volts, the coil
    will get 10 volts.

    If the 6 volt zener is used, it will always drop just about
    6 volts, so a 21 volt input will put 15 volts across the
    coil and a 15 volt input will put 9 volts across the coil.

    All this assumes that the coil resistance varies little as
    the coil current varies. The actual positive coil
    temperature coefficient will reduce the contrast between
    these two methods.

    The really fine way to reduce the voltage would be to use a
    12 volt integrated voltage regulator (LM7812) to throw away
    the excess voltage and regulate the coil voltage to a fixed
    12 volts. Probably will cost about the same as a 1W zener.
     
  3. Nobody

    Nobody Guest

    If it draws [email protected] (i.e. the coil resistance is 600 Ohms), then you need
    a 300 Ohm resistor to drop 6V.

    In either case, [email protected] = 120mW dissipated in the zener or resistor.

    The zener option is simpler if you're using DC, the supply voltage is
    constant, and you don't know the coil resistance or current draw.

    The resistor is likely to be cheaper, and will result in less variation in
    coil voltage if the supply voltage changes. E.g. if the supply voltage
    drops to 15V, the zener will give 15V-6V = 9V across the coil, while the
    resistor will give 15V*2/3 = 10V.
     
  4. John Fields

    John Fields Guest

    ---
    For the same voltage drop across them, with the same current through
    them, the Zener and the resistor will each dissipate the same amount
    of power, so they will each generate the same amount of heat.

    Since you say you have an 18V supply and your relay pulls in with
    12V across it and 20mA through it, to get 12V across the relay
    you've got to drop 6 volts when 20mA is flowing through the relay,
    which means that the series resistance needs to be:

    E 6V
    R = --- = -------- = 300 ohms.
    I 0.02A

    The power which will need to be dissipated will be:


    P = IE = 0.02A * 6V = 0.12 watts


    So, it seems, the most economical choice would be a 300 ohm, 1/4
    watt resistor.
     
  5. Jamie

    Jamie Guest

    I know every one else has given you the answer of 300 ohms which is
    correct how ever, I have a backyard trick that not only gives you
    300 ohms! but it'll also give you a Pilot indicator..! :)

    Get your self a 6 volt 20 ma incandescent lamp and use that as your
    resistor. It'll light up for you and pull the relay in at the same time.

    Of course, if the bulb burns out, your relay won't pull in either! :(

    but either way, you'll know if something is wrong!
     
  6. Eeyore

    Eeyore Guest

    Firstly the resistor won't generate more heat than the zener - see P = IR. Why
    did you think this ?

    Secondly, a resistor is cheaper.

    Thirdly, you calculated the resistor value wrongly. It should be 300 ohms (from
    R = V/I). Use 270 ohms for nearest E12 value.

    Graham
     
  7. Steve Wolfe

    Steve Wolfe Guest

    I have a 12V relay and I wanted to operate it with 18V supply. Could
    Why not just use a regulator? At that low of current, even the small
    to-92 cased regulators would work perfectly well.
     
  8. Allen Bong

    Allen Bong Guest

    Thanks very much for all that replied. That really cleared a lot of
    doubts in me. Why zener is better than resistor? Most probably I
    thought that 1W zener is much smaller than a 1W resistor!

    Allen
     
  9. John Fields

    John Fields Guest

     
  10. John Fields

    John Fields Guest

    ---
    LOL, most of the rest of us have had easy access to 5% resistors for
    years and years, and yet you suggest that the OP should settle for a
    value of 270 ohms.

    Why? Because you relegate him to what you think is an inferior
    status of humans who only have access to relatively primitive
    technology.

    What a provincial cocksucking twit you are.
     
  11. default

    default Guest


    You've already gotten enough responses to your question. The only
    questions that remain: do you care how much power is wasted? Can,
    will you, does the job warrant it, rewind the relay with smaller gauge
    wire?

    Relays, as a general rule, will pull in at a lot less voltage than
    their ratings - since the rating is usually "must pull in by X volts."

    If this is a DC relay - your text implies that - you can do what a lot
    of us have done for years. . . .

    The relay "pull in" voltage and "hold" (in) voltage are different.
    May take 10 volts to pull in and 8 volts to hold in . . .

    You choose a resistor that allows the hold voltage to develop across
    the coil and put it in series with the relay coil. (calculating the
    resistor with ohms law and relay drop and supply voltage).

    TaDa!

    Now you just need to find a capacitor to bypass the relay that will
    allow the pull in current to flow and put it in parallel with the
    resistor (observing polarity since it will need to be an electrolytic
    for large capacity)

    Calculate the resistor using Ohms law - calculate for hold current
    (look up the data sheet) - or just calculate for 12 volts across the
    coil.

    Relay just sits there. Suddenly 18 volts is applied and relay current
    charges the capacitor (which is discharged so appears like a dead
    short) all that current flows into the coil pulling the armature in.
    Now in time, the capacitor reaches some equalizing voltage and passes
    no more current, but the resistor supplies the hold current/voltage.

    Empirical testing? Calculate the resistor based on hold voltage (the
    coil and resistor are in the same voltage divider - 12 volts on the
    relay 6 volts across the resistor - series circuit)

    Try caps until the relay pulls in when 18 volts is applies then double
    it.

    Resistor? 80 milliamps and 6 volts to drop = 75 ohms (ohm's law) (you
    still have the "must pull in" specification for a fudge factor). So
    you could use just a 75 ohm resistor in series with the coil to make
    it work (it would have to be a 1/2 watt resistor or more to keep it
    cool)

    That alone will work. No capacitor, no hassle.

    Using a capacitor allows you to pull in with a lower calculated
    resistor. Let's say it takes 8 volts and 150 ohms to "hold" in.
    That's only 53 milliamps to hold. Then the resistor to do that is 113
    ohms (or 100 for ease of procurement) and .4 watts (still use a half
    watt resistor)

    Try a 200 ufd cap with 20 volt rating across the resistor. Whether it
    works or not depends on the mass and friction in the armature.

    This is only worth doing if you want to save power - like you are
    using batteries to supply the 18 volts.

    Years ago I applied it into a two way radio system in the mobile units
    - yes it saved battery energy - but it also meant the person in the
    field had to run his engine to make the transmitter work when the
    battery got low. No more dead batteries in the field - and saved a
    ton of money - but that wasn't my goal.

    Otherwise the field rep would talk on the radio and kill the battery
    and couldn't restart the engine - just a unintended benefit or
    synergy.
     
  12. They both need the same board area to allow 1 watt to escape
    without damaging the board. In this case, the 6 volt drop
    at 20 mA produces a small fraction ( 0.12) of a watt. So a
    quarter watt resistor versus a quarter watt zener is a
    better comparison.
     
  13. Allen Bong

    Allen Bong Guest

    Jamie,

    I've noted that in my note book. In my application, 99 % of the time
    the relay remains not operated and it only operates to indicate an
    alarm. If it were the other way round round I'd not use a lamp as a
    resistor to drop voltage as the life span of the filament is much
    shorter than a resistor.

    We have alarm panels in our office to indicate status of the working
    of our equipments. The lamps were ON most of the time. Every 6
    months or so there is 1 or 2 bulds that need replacement.

    Allen
     
  14. Rich Grise

    Rich Grise Guest

    For a 6V drop at .02A, my calculator says 300 ohms. R-E/I, right?

    They will both dissipate the same amount of heat - 6V * 0.02A, which
    admittedly isn't a whole lot.

    Use the resistor - the relay will pull in quicker than with the zener.

    Why? Because at first, there's zero current flowing, so there's zero
    voltage drop across the resistor, so the relay sees the full 18V, but
    only for an instant - the current immediately starts to rise, dropping
    volts across the resistor, according to the standard T=LC.

    And then, of course, don't forget the catch diode across the coil. :)

    Hope This Helps!
    Rich
     
  15. Allen Bong

    Allen Bong Guest

    Thanks Rich, As all the people here also think that a resistor is
    better than a zener on this application, I will adapt the idea and
    use a resistor. Thanks for your further analysis which convince me
    more.

    Allen
     
  16. Rich Grise

    Rich Grise Guest

    It's my great pleasure to have the power to explain things in a way
    that beginners can "get" - we were all beginners at least once! ;-)

    Cheers!
    Rich
     
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