Controlling the power through a heating element?

Question

I need a 25mm x 30mm 120V 200 watt band heater.

Unfortunately, the only 25mm x 30mm 120V band heaters I have found are 300 watt.

What sort of circuit could I construct or purchase that could regulate the current flowing through a large heating element like that?

Ideally looking for something off the shelf.

I'm used to limiting current (say, through an LED) with a small resistor. I have no clue how to design for something high current like this.

Doesn't have to be adjustable, literally all I want to do is ensure the heating element dissipates 200 watts, not 300. I'm guessing the element is resistive.

My power supply is 250 watts, so I'm not trying to dissipate any power in the control circuit.

Answer 1

AC: use a lamp dimmer

If your power source happens to be somewhere in the range of normal grid voltage and AC:

A resistive heater is indistiguishable from an incandescent light bulb (if you close your eyes).

So, a dimmer for halogen lamps will work. (internally, that's a phase-fired controller, but really, for your purposes: it only turns on the heaters for a percentage of time).

They can be had for a couple of € at furniture stores and construction / home improvement stores.

DC: follow piojo's advice

... and just use PWM with a MOSFET as switch (no need for overly large flyback protection diodes, can probably use one with an integrated one), setting your duty cycle to whatever makes the device output what you need.

There's ready-to-use PWM controllers for that, including FET gate drivers.

Answer 2

You can't get the power down to 200 W with resistors unless those resistors can handle their share of the power. In other words, the resistors would need to be more heaters. If your heaters are 300 W, putting two in series would double the resistance and halve the power.

Or you can find another resistor (a power resistor or heater) which will bring you to exactly the power you need: at 120 V, your target current will be 1.66 A. The resistance that will give you this is 72 Ω. Your current heater is 48 Ω, if the given wattage is accurate. So you need to add an extra 24 Ω of resistors/heaters that can handle 1.66 A. The power will be I²*R, and if you split it over two resistors, metal power resistors should be able to handle that.

Another way to arrive at the exact power output you need is to cut out part of the waveform with a solid state AC dimmer (I think something based on a triac). The peak current output will still correspond to a higher power than your power source is rated for, but the average power can be tuned to be the power you want.

I'm not sure whether AC can be dimmed with PWM, but if so, that's also an option analogous to using the triac.

Answer 3

Assuming you mean 120 VAC, there are two simple ways to reduce power.

• A phase-control device such as a light dimmer can easily handle 200 W, but calibrating to exactly that power might be an issue, as is insuring that in the event of failure of the control, power would never exceed 200 W.

• A transformer can be used to buck the AC line voltage down to 98 VAC. The 300 W heater has a warm resistance of 48 ohms, so would dissipate ~200 W at the reduced voltage. Since the cold resistance might be a trifle less, it would initially draw more current, so a 3 amp secondary rating should be safe. For example, either Hammond 167M20, which is rated 20 VAC @ 3 A, or Hammond 166M24-BULK, at 24 VAC @ 3 A, would probably be within your required tolerance. Be careful to check the output voltage, first open circuit, to be sure the secondary is phased to buck line voltage, and under load, to be sure it's close to 98 VAC to the heater.

Phase control is easy to implement and uses a small device, but needs calibration and might go out of adjustment, while the transformer is heavy and bulky, but reliable.

N.B. In any case, use a thermal fuse to prevent overheating.

Answer 4

150W is probably close enough. Just use a diode.

^{simulate this circuit – Schematic created using CircuitLab}