How to throttle power consumption of a heating coil?

Question

Which is an appropriate but simple electrical circuit to reduce the energy (power) consumed by a heating coil which has an AC feeding without generating an asymmetric load (large DC component, e.g. with a diode)?

I know that this would make it cooler.

Some heating coils consume a lot of power (kilowatts), so a simple resistor in series would be a bad idea (difficult to cool). A capacitor in series is also very easy, but way to expensive for the two months application I want it for.

I think what I need is a kind of dimmer, a thyristors or triac. But I want to have a fixed rating, it doesn’t have to be adjustable. Of course I have no problem if it is adjustable if the electrical network is still easy to understand and to build.

E.g. if my power source is 230 V AC, 50 Hz. The heating coil has a power consumption of 2300 Watts at this voltage but I want it to consume only 1400 Watts, which parts are most recommendable and how would the circuit look like (or what is it’s name, I could then search for it).

PS: This is the second attempt to ask this question. The first one was to specific on my dishwasher and inverter, but also got some interesting answers.

Answer 1

The argument:

If your heating element is rated to dissipate 2300 watts with 230 volts across it, then when it's dissipating 2300 watts the current through it will be:

I = P/E = 2300W/230V = 10 amperes,

and its resistance will be:

R = E/I = 230V/10A = 23 ohms.

In order for the element to dissipate 1400 watts, then, the voltage across it must be decreased to:

E = sqrt (PR) = sqrt (1400W * 23R) ~ 180 volts,

and the current through it decreased to:

I = P/E = 1400W/180V ~ 7.8 amperes

Since the mains voltage is unchangeable at 230 volts, then in order to limit the current through the heating element to 7.8 amperes, a series attenuating element must be introduced into the circuit which will drop the mains' 230 volts down to the 180 volts required by the heater.

That's 50 volts, and it must pass the 7.8 ampere heater current, so its resistance will be:

R = E/I = 50V/7.8A = 6.4 ohms.

The total resistance of the series string will therefore be 29.4 ohms, the sum of the two resistances,.

If we use a capacitor to losslessly drop the voltage to the heater, then we can define the resistance of the string as an impedance,

Z² = R² + Xc², where:

Z is the impedance of the string, R is the heater resistance, and Xc is the reactance of the series capacitor, all in ohms.

Rearranging to solve for Xc we get:

Xc = sqrt (Z² - R²) = sqrt (864.4R - 529R) = 18.3 ohms

and, finally, to get the capacitance,

C = 1/(2pi f Xc) = 1/(6.28 * 50Hz * 18.3R )= 174 microfarads .

Cornell-Dubilier makes a nice line of caps which seem perfect for this application.

Pricey, no doubt, but probably less so - and certainly lighter and smaller - than an inductor or a transformer capable of doing the same job.

The LTspice proof:

Version 4
SHEET 1 1460 680
WIRE 256 64 96 64
WIRE 256 96 256 64
WIRE 96 160 96 64
WIRE 256 192 256 160
WIRE 96 320 96 240
WIRE 256 320 256 272
WIRE 256 320 96 320
WIRE 96 368 96 320
FLAG 96 368 0
SYMBOL voltage 96 144 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 325 50)
SYMBOL cap 240 96 R0
SYMATTR InstName C1
SYMATTR Value 180µ
SYMBOL res 240 176 R0
SYMATTR InstName R1
SYMATTR Value 23
TEXT 104 344 Left 2 !.tran 1

Answer 2

A series resistor would be difficult to cool.

This is true, but let's follow this idea to see where it leads. The heating element is a resistor, and note that it gets hot without damage. This tells us that the kind of resistor you want would be another heating element.

Now, it seems as though you're getting nowhere because you've just moved the power from one heating element to another. But the power decreases if you add resistance. The total resistance is higher, so the current through both is reduced. The voltage (across both) is still the same, so if you wanted to calculate the power, you'd use the same voltage times the new current. (Or V^2/R).

Just as an easy example, if you got two of the same heating elements and wired them in series, they would draw half the power (a quarter going to each).

Note: This is a late answer to an old question, but this explanation may be useful to someone with a similar issue who searched and found your question.

Answer 3

The schematic below shows a pretty simple way to do it, and there's an LTspice simulation here that shows the circuit in action.

In words, U1 is wired as a 4-bit binary down counter which uses the mains for a clock source and is designed to count down to zero from 14 in one instance, and then down to zero from 9 in the next, the plan being to keep the mains connected to the heater for 14 mains cycles, disconnect it for 9, and then to repeat that sequence forever, causing the heater to dissipate 1400 watts over the long term, instead of 2300.

In order to do that, U1's carry out (COUT) is used to toggle U3, which sets U1's broadside load inputs to either 1110 or 1001, which is then loaded into U1 by the high on U1-PE, which SETs or RESETs U1's internal count stages. That causes COUT to go high, enabling the counter and allowing it to count down until it gets to zero, when U3 will toggle and the cycle will begin anew with the new preset.

During the time the counter is counting down from from 14 to zero, U3 is used to turn on the zero-crossing TRIAC driver, U4, which in turn turns on the TRIAC and connects the heater to the mains. Then, when U1 gets to zero, U3 will toggle and the TRIAC will be turned OFF for 9 mains cycles. Then, when U1 counts down to zero, from 9, U3 will toggle again, starting the cycle anew by turning on the TRIAC for 14 mains cycles.

Answer 4

As I need this power reduction only for one or two months (until my 3000 watts inverter is repaired and shipped to me again), I used the cheapest method, which was an electric waffle iron I already had, and which was broken in two parts – but the heating still works.

So I used this waffle iron as a resistor, where the iron itself is the heat sink for the attached heating coil of he waffle iron. It gets quite warm and smells like waffles (because I never cleaned the broken waffle iron) when the dishwasher heats. But it works, and it cost me nothing, because I reused the cables and the screw terminals from the waffle iron.

Answer 5

You can use an SSR to power on and off the load. By varying on-off duty cycles, you have a slow PWM. Unlike with series resistors, the unused power is not wasted - but just unused.

You can get a precise control as described

TRIAC Angle of Firing vs Power Delivered to Load, how calculate?

Answer 6

If you want to limit power then control the voltage or the current. Your dimmer link is fine. How much current/ power. The first thing I thought about was a variac/ auto-transformer. (also in your dimmer link.)