Specifically, I am trying to find an efficient and cost effective method for minimizing required length of a Kanthal A1 heating wire element, but at the same time keeping the diameter large enough to allow for a relatively low surface load (w/cm^2) - to increase the service life of the element.

I would like to run the heating element off 120v-240v AC, with 200-400 watts. For 120v @ 200 Watts = 120^2/200 = 72 Ohms - After calculating, it appears I have a choice between too much wire for the available space, or too much surface load on a thin&shorter wire. Excess surface load is what causes the the wire to prematurely oxidize and burn up, or even melt.

For the purposes of surface load, can I effectively halve the load by placing a rectifying diode in series? If yes, what would the best diode be for this application (assume the circuit would be continuously open, and there would be minimal heat dissipation for the diode beyond it's inherent rating.) How do you calculate the heat produced/dissipated by a rectifying Diode? Would these work? 10A10

Are there better/additional ways to efficiently lower the power in the circuit? My understanding is that a resistor would dissipate power equal to it's reduction, so that's not an option. In case it influences the answer, any solution up to $50 parts cost would be worth considering, but under $10 is ideal.

  • \$\begingroup\$ Do you need 120v can you design around an old atx computer power supply? \$\endgroup\$
    – sstobbe
    Commented May 21, 2017 at 22:26
  • \$\begingroup\$ For the diode, simply work out the current and multiply by the forward voltage at that current. If it's rectifying AC, so only doing a 50% duty cycle, divide by 2. You'll need to get the forward voltage at the particular current from the graph on its datasheet. \$\endgroup\$
    – Ian Bland
    Commented May 21, 2017 at 23:30
  • \$\begingroup\$ Do also note that the same resistance at double the voltage will dissipate four times the power. I = V / R; P = V * I = V * V/R; <-- V counts twice, so doubling it will quadruple P. \$\endgroup\$
    – Asmyldof
    Commented May 21, 2017 at 23:49
  • \$\begingroup\$ Start with your acceptable measurable specs, Pd, ΔTmax , ΔV , Rmax, minimum efficiency ε and medium (air , oil, water, metal tank etc) !!! These are all critical to answer. \$\endgroup\$ Commented May 22, 2017 at 2:15
  • \$\begingroup\$ Half-wave rectification will cause a horrible power factor. I'd suggest a transformer (~ 1:0.7) instead, and perhaps/if needed a largish capacitor (if in doubt, use X-rated) to counteract the inductive load of the transformer. \$\endgroup\$
    – Oskar Skog
    Commented May 22, 2017 at 4:58

1 Answer 1


The diode will cut the power in half, although you may want to check on the thermal capacity of the wire, will it heat up fully in the 8.3ms that is a single half cycle of AC? If so then the surface loading will still be to high and most likely cause premature failure.

As for the diode, you say a 400W system max.

\$P=400W, P/V=I\$ So \$400\div120=3.3A\$

Now cut that value in half since the diode is blocking half the cycle.


Assume the voltage drop across a general silicon diode is 0.7V. The power dissipated can be found by:


You should be able to find a diode online easily that meets these requirements. I would generally double it and find something meeting 3A and 2W of thermal dissipation at a minimum.


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