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220V solder iron under 110V

Discussion in 'General Electronics Discussion' started by HellasTechn, Jan 17, 2017.

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  1. HellasTechn


    Apr 14, 2013
    Hi !

    I have this silly quiestion. Most ships i work on have 110V-120V mains power (two supply lines of 60 volts 180 deg out of phase). I also have a solder iron that works on 220V and gives 40W. This solder iron is consisted of the heating element and a 1N4007 diode in series.
    If i remove the diode then i should be able to connect it to 110V but will it give the same 40 W ?

    this is my way of thinking:

    At 50 hz and with the diode on the heating element receives 25 pulses of 110V per second and produces 40W.

    At 100V without the diode it will receive around 60 pulses of 60 V per second.

    So wattage should be about the same or slightly higher right ?

    Am i thinking correctly ?
  2. duke37


    Jan 9, 2011
    A resistor run on half voltage will dissipate one quarter the power.
    A resistor in series with a diode (AC) will dissipate one half the power since it conducts half the time.

    I do not think that taking the diode out will give the same heat on half the voltage.

    This seems to be an 80W iron run through a diode. Taking out the diode and running on half voltage will dissipate 20W. This may be enough, depending on the job.

    Does the iron have a thermostat? The diode may be there to ease the work of a thermostat contact.
    Last edited: Jan 17, 2017
  3. HellasTechn


    Apr 14, 2013
    No there is no thermostat. Just the heating element and the 1n4007 in series.
    It is a cheap (probably chineese, i am not sure) solder iron. It actually runs quite well on 220V.

    the heating element is probably a 40W and the diode is there to keep it from burnig up i assume. I am definetly not sure though.

    Can you explain a little more why ?
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    The effective voltage of a rectified sine with Veff = 220 V (not recitified) is 155 V (rectified)
    When you operate the iron without diode it will see 110 V (eff).
    Assuming the resistance of the hot iron is approx. the same for 110 V or 155 V, the power will drop from 40 W to 20 W, confirming Duke's reasoning.
  5. duke37


    Jan 9, 2011
    As Harald says, this assumes that the resistance stays constant.

    The power dissipated is V*V/R so if V is half, P = V/2 *V/2 /R = V*V/R/4
    If a diode is used, then the voltage is applied for half the time so the average power is half.

    The iron is a 80W iron with a diode so dissipates 40W as rated.
    Half the voltage on an 80V iron will dissipate a quarter as much, so 20W.

    If the resistance is not constant, then there will be some compensation since resistance goes up with temperature.
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