A hot-surface electric igniter heats a continuous resistive element. This calculator estimates its electrical input from RMS voltage and an entered hot resistance. The illustration follows current reversal and resistance heating in the continuous element.
Electric Igniter Interactive Calculator
Compare RMS electrical input for a hot-surface igniter using its specified hot resistance. The animation shows current reversal and positive resistance heating.
Equation Used
- Sinusoidal voltage applied across the element.
- Constant specified hot resistance.
- No transient or combustion model.
Fixed hot resistance and sinusoidal AC. Component temperature, warm-up and ignition performance are not calculated.
A continuous resistive heating path
The illustrated ceramic element has a continuous U-shaped path mounted in an insulating support, with two insulated leads and a connector. Its orange zone identifies the heated portion; it is not a calculated temperature map.
For a sinusoidal AC supply and a constant hot resistance, current reverses with voltage. Heating power is positive in both half-cycles because it depends on current squared.
The animation follows one slowed electrical cycle. The element does not mechanically move, and its thermal glow is not made to switch off at every electrical zero crossing.
Use the specified hot operating resistance
The inputs are RMS voltage and hot resistance. The outputs are average electrical power, RMS current and the same power expressed as a thermal energy rate.
A cold resistance measurement is not automatically the hot resistance. Real ceramic elements change resistance with temperature, and appliances may use supply or control arrangements outside this simple model.
The entered ranges are arithmetic comparison ranges. They are not ratings for the pictured style or instructions for selecting a replacement component. The manufacturer reference establishes the recognizable hot-surface component form, not an approval of every calculator setting.
RMS values determine average resistance heating
For RMS voltage V and fixed hot resistance R, average power is P=V²/R and RMS current is I=V/R. The International Table Btu conversion used here is 1 Btu=1055.05585262 J, so the equivalent rate is 3600 P/1055.05585262 Btu/h.
At electrical phase θ, v=√2 Vsinθ and i=√2 Isinθ. Instantaneous power is vi=2 Psin²θ. Its average over a complete cycle is P and its peak is 2 P.
The waveforms use those same equations. The heat-rate conversion is not fuel input, combustion output or delivered appliance heat.
120 V RMS across 20Ω hot resistance
The example inputs give 720 W average input and 6 A RMS. Peak current is approximately 8.49 A and instantaneous power reaches 1440 W, averaging 720 W over the cycle.
At 80 V and 20Ω, the result is 320 W and 4 A RMS. Increasing voltage changes power with its square; changing resistance at fixed voltage changes both current and power inversely.
These are ideal electrical results. They do not assert that a particular appliance igniter can operate at these settings.
Electrical power is not an ignition prediction
The model holds hot resistance constant and assumes a sinusoidal voltage across the element. It omits cold-start behavior, control electronics, heat loss, material aging and transient temperature.
No surface temperature, warm-up duration, flame establishment or ignition threshold is calculated. A component’s usable voltage and resistance must come from its specified system.
This model applies to a hot-surface appliance igniter with a specified hot resistance. Spark-gap and pyrotechnic igniters use different operating principles.
References
- Robertshaw hot-surface igniter product description — ceramic element, insulated leads and appliance application.
- Robertshaw Silicon Carbide HVAC sell sheet — component photograph and construction. Its model ratings are not inferred from this generic calculator.
- NIST Guide to the SI: conversion factors — energy-unit conversion.
Hot-surface igniter questions
Why does heating remain positive when current reverses?
Resistance heating is proportional to i², so reversing current does not reverse heat production.
Why is peak power twice the reported power?
The reported result is the cycle average of a sinusoidal supply; the instantaneous squared-sine waveform peaks at twice that average.
Can I use cold resistance?
Not as an assumed hot operating value unless the component data establishes that equivalence.
Does the orange color give a temperature?
No. It identifies the hot element qualitatively.
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