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WARNING! To others reading this question: This circuit uses high voltage. DO NOT mess with 220V/110V unless you know what you're doing and know the risks (electrocution and death, mainly)

Now that that's out of way...

Hello everyone, I'll try to be as brief as possible: I have a heating element taken from a 220V hot plate and connected like so:

Original circuit

Components:
Vs: 220V (RMS) 50Hz sine AC (from wall outlet)
U1: Bog standard DIAC-TRIC dimmer
R1: The heating element, 100 Ohms in room temperature (assumed more when hot)
R2: A very small incandescent light bulb, also taken from the hot plate, I assume its current is negligible

I need to slightly reduce the current of the heating element. I thought about a non polarized 100V 100uF capacitor but I don't have one. I do have 50V 1000uF polarized capacitors, so, my first question: If identical capacitors in series have their max allowable voltages added up but their capacity divided by the number of capacitors, AND, if 2 identical polarized capacitors connected in series with reverse polarity make a non polarized capacitor with half the capacity (reference), then, will connecting eight 50V 1000uF polarized capacitors back to back create the equivalent of a single 200V 125uF bi-polar capacitor?

The new schematic:

"Improved" circuit

New components:
C1 - C8: Radial aluminum electrolytic polarized 50V 1000uF capacitors

So if my calculations are correct, when the dimmer is at full and the element is at room temperature, the capacitors will measure 63V at peak amplitude and will reduce the RMS current by 447mA (2.2A without the capacitors, 1.753A with them).

Second question: Is everything I wrote correct? Or did I accidentally build the equivalent of a hydrogen bomb that will destroy my house, neighborhood, and everyone I've ever known and loved??

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  • \$\begingroup\$ You know you'll be adding a lot of impedance to the circuit by introducing all those capacitors don't you? \$\endgroup\$ – MCG Mar 27 '18 at 10:19
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    \$\begingroup\$ Better to just control the dimmer range, surely? \$\endgroup\$ – Brian Drummond Mar 27 '18 at 10:42
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    \$\begingroup\$ Why do you need to reduce the current of the heating element, when it's already connected to a device for reducing its power? Is it necessary, or will your capacitors be sufficient, for the problem you actually have, rather than the one you state? \$\endgroup\$ – Neil_UK Mar 27 '18 at 10:46
  • \$\begingroup\$ MCG: Is there a downside to this impedance? Brian Drummond: The dimmer is a built product, not a circuit by me and I don't want to modify it. Neil_UK: I want to reduce the whole range, I.E. I want a lower current at both the minimum and the maximum values of the dimmer. I hope I managed to explain it correctly now. \$\endgroup\$ – C4lculated Mar 27 '18 at 11:29
  • \$\begingroup\$ Why was my question downvoted? Did I do something wrong? Please explain my mistakes to me so that I can avoid making them again. \$\endgroup\$ – C4lculated Mar 27 '18 at 11:43
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Capacitors in series need balancing resistors to ensure that one capacitor doesn't get over-supplied with voltage. This problem is normally one that is associated with DC circuits and series capacitors used in power supplies but, can also be seen on AC supplies.

A problem I see is that most electrolytic capacitors have a wide tolerance range and this will inevitably mean that one (or some) of the capacitors (that have the lowest capacitance value) will receive an unequal share of the voltage and possibly cause a breakdown.

Another problem is that relying on an electrolytic to turn into a diode when reverse polarized is a bit hit and miss. So, if you choose capacitors that are fairly well balanced in value and maybe consider using reverse diodes across them en bloc it might be OK.

Alternatively, get a lamp that is low voltage and wire it in series with the element in order to drop the voltage to the element.

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  • \$\begingroup\$ Thanks Andy, you have given me some good ideas, maybe I'll use an Arduino capacitance meter to measure the capacitors tolerances. Unfortunately, I managed to somehow sprain my finger so I'll have to let it heal before I can try anything. \$\endgroup\$ – C4lculated Mar 28 '18 at 5:20

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