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Leszek Luchowski
Posted: Thu Nov 05, 2009 11:47 am    Post subject: Simplest circuit to reduce AC power in a resistor? Reply with quote

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Hello;

I have a 2000 W heating radiator in my bathroom (basically, it's a big resistor that gets orange hot and radiates most of the heat out). The problem is that when this device is cold, its resistance is much lower than at its intended operating temperature, and so it blows the fuse almost every time I turn it on.

However, once I have reset the fuse and got the radiator working, i can switch it off for quite a few minutes and safely turn it on again - it's still hot enough to have a resistance the fuse can live with.

I tried the obvious solution using a double switch and a diode:



In this way, I can switch it on at reduced power first (left switch), but it still does blow the fuse, although a bit less often.

I would like to replace the diode with some kind of thyristor circuit that would cut the "reduced power" even more than the diode does. Preferably a circuit that is small enough to fit into the wall-mounted switch housing like the diode did, and without a need to connect to the other lead of the mains (the top wire in my schematic) so I don't have to ruin the paintwork on the wall.

I am not sure how to get at it. I did study electronics a quarter of a century ago, but it was medical electronics so I didn't learn very much about power devices
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Harold14370
Posted: Thu Nov 05, 2009 3:55 pm    Post subject: Reply with quote

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Maybe you could get a different fuse - one with the same current rating but a little slower acting so that it will ride through the inrush.
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Bender
Posted: Fri Nov 06, 2009 1:07 am    Post subject: Reply with quote

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I'm not really a specialist, but some things that come to mind are:
1) a chopper or a dimmer, if you can find one for that power
2) replace the diode by a similar resistance (another radiator Wink)
3) I think replacing the diode with an inductance could work too. You'd get a filter, which in this case could reduce the amplitude of the 50 Hz signal.
The cutoff frequency is
f_c = \frac{R}{2\pi L}
If you pick L to make f_c = 50 Hz, you'd half the amplitude of the current. The nice thing could be that, since R starts smaller, the cutoff frequency will start small and increase while the radiator heats up (i.e. low current amplitude at first, which increases when the radiator heats up). All theory, though, I've never tried anything like that in practice, and I'm not taking responsibility. I don't know if you can find the right inductance, and it might also heat up.

Harolds idea might actually be the best. Be careful when switching fuses though, since your electric circuit might not be designed for other fuses. Replacing a quick acting with a slow acting is usually quite safe (but don't make it slower than necessary), but I'll repeat his advice to not take a higher current rating!
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Harold14370
Posted: Fri Nov 06, 2009 10:47 am    Post subject: Reply with quote

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Assuming this heater is on its own circuit, there shouldn't be any need to trip it on instantaneous high current, which seems like what is happening. Is it really a fuse or is it a circuit breaker? If it's something you can reset, it must be a circuit breaker.

The current rating of your circuit breaker or fuse in the service entrance panel is designed to protect your house wiring. With a steady state current of 2000watts/230 volts = 8.7 amps, you shouldn't be having a problem. There aren't any other loads on the circuit are there?

If your breaker is just protecting the house wiring from overheating, a slower blowing fuse should work fine. The wire will be cold initially when you turn on the heater, so it will take some time to heat up to the point where it would damage the insulation.

I don't know what the wiring codes are like in Poland, but in the USA you wouldn't have any house wiring less than 14AWG which is good for 15 amps. A 15 amp fuse or breaker should easily carry 8.7 amps without tripping.

There are devices like negative thermal coefficient thermistors that are used to limit inrush to things like computer power supplies. I wouldn't put that or anything else that dissipates heat into your wall switch. Inside the heater enclosure, maybe.
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Leszek Luchowski
Posted: Fri Nov 06, 2009 1:02 pm    Post subject: Reply with quote

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Yes Harold it's a circuit breaker, sorry for using the wrong word.

All the wiring in my home is recently laid 2.5mm^2 copper wire (sorry, no idea how this relates to your AWG). The circuit breaker is rated 16 A. I have yet to find any data about its reaction time. And I am not sure if any outlets aren't connected to the same circuit breaker.

I am reluctant to put any self-made circuitry in or near the radiator - it is hanging almost directly above the bath (tub). This is why I chose a model that is safety rated IP56, i.e. certified for use on open decks of seagoing vessels. I wouldn't trust a home-made 230V device in a place where, hopefully, my children will soon be splashing around in water.

I will try finding a less nervous 16A circuit breaker or replacing the diode with an integrated power regulator (plus the due resistors and capacitors).

Thanks to you and to all others who replied,
Leszek.
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Harold14370
Posted: Fri Nov 06, 2009 1:53 pm    Post subject: Reply with quote

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Leszek Luchowski wrote:
And I am not sure if any outlets aren't connected to the same circuit breaker.


That's the first thing to look into. The answer may be as simple as plugging the other loads, if any, into a different circuit. Trip the breaker off and see what else in the house doesn't work.

The other thing you could try is to swap out the breaker with a spare breaker of the same type. It could be that that particular breaker is too sensitive.
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Leszek Luchowski
Posted: Mon Feb 08, 2010 10:09 am    Post subject: Reply with quote

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At long last, I have found the time to shop for and instal a new circuit breaker. For whatever reason, they aren't labelled "slow" and "fast" in any human language, just "B" or "C", and I first had to find out that "B" is fast and "C" is slow.

After replacing a 16A "B" with a 16A "C" the problem is solved. Thanks to all who replied, Harold in particular for proposing a simple and effective solution. Saved me a lot of time.

Cheers - L.
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Harold14370
Posted: Mon Feb 08, 2010 5:35 pm    Post subject: Reply with quote

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Glad it worked out for you.
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