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Just as the title implies... I'm curious whether given a heater is on for exactly 1/2 of a period of time does the frequency of the heater turning on and off impact the net result of heat created? For instance if it was on for 10 seconds and off for 10 would that have the same net effect as turning it on every 1 second then off the next repeatedly over the same length of time?

Inc1982
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  • Assuming a heater with no tempco, it is the same. Unfortunately heating elements have tempcos and you would have to calculate this and likely the higher frequency one consumes more power – PlasmaHH Apr 06 '16 at 06:34
  • I'm thinking largely of a resistive heater with no temp control. – Inc1982 Apr 06 '16 at 06:53
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    tempco is temperature coefficient of the resistor. – PlasmaHH Apr 06 '16 at 08:28
  • If you need better heater control, the usual go-to is an SCR. These operate in either phase-control or zero-cross modes - for resistive heaters usually a zero-cross produces less electrical noise. Here you want a 50% duty cycle; a zero-cross SCR would do this by applying power every other cycle of the mains voltage (ie: 50-60Hz). A phase control SCR would apply power for half of the voltage cycle, either the leading or trailing edge of the sine wave depending on type. The abrupt cutoff in phase control is what can generate the electrical noise. – J... Apr 06 '16 at 13:18

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Your premise is largely correct but there are a few things to take into consideration:

  • Temperature coefficient of resistance causes the heater resistance to increase with temperature. The primary observable effect of this is that on switch-on the current is initially somewhat higher but decreases as the heater warms up. This is most famously observed in the incandescent bulb whose turn-on current can be five to ten times higher than its steady-state current. The effect on your heater would be to increase the average power slightly.
  • Thermal response time is the rate at which the load temperature reacts to the power turning on and off. In most industrial applications a steady temperature is required / preferred but on/off proportional power control is used (as in your question) for simplicity and reliability. The solution is to switch with a cycle time that is short relative to the response time of the system. e.g. If the system is at operating temperature and turning on the power results in a temperature rise of 10°C / minute and we want to hold the temperature to within 1°C then I would expect that running at a 2 or 3 s cycle would manage that.

Mathematically, you are putting energy into the system and this can be calculated by \$ E = \int P dt \$. i.e., the area under the power-curve.

Transistor
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Yes but the actual effect depends on the heating and cooling rates in the system. The warm up time and cool down of the system is critical here; if it is a perfect system with no warm up or cool down time then the frequency will have no effect. Real systems will have a warm up time and so at any point below the critical frequency (which we can define as the frequency at which the system warms up during the on cycle as much as it cools down during the off cycle) will not appreciably warm up. After the critical point the on cycle will add more heat to the system and so warm up. This assumes that the system heats and cools at the same rate which is unlikely. If the system heats and cools at different rates the you may need to have an asymmetric duty cycle to get the same effect.

Once you have the warm up and cool down rates of your system then you could build your duty cycle and frequency accordingly.

AndyW
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  • "... at any point below the critical frequency will not appreciably warm up." This can't be true as there is energy entering the system. What's happening the energy otherwise? – Transistor Apr 06 '16 at 08:11
  • Assuming a closed system it will cause the closed system you are correct. I was thinking more of anti-condensation heaters that I have on large marine motors for pumps. The ACH is used to keep the motor above ambient temperature when the motors are not turning. In this case the heaters must first overcome the heat loss externally before it can raise the internal temperature in the same way that pumping water into a vessel will only raise the water level one the inlet flow exceeds the rate at which it flow out. I should have been clearer in my definition of the system boundaries, sorry. – AndyW Apr 06 '16 at 08:35