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An electrical house radiator stop heating, so I opened it and start trying to understand how the control board is working.

I was very surprised to discover a very very low cost board with a power supply without transformer (If I'm right, the supply is provided only with 1 diode + 2 power resistors + 1 capacitor):

  • PCB components face
  • PCB traces face
  • PCB with drawings
  • Scheme with PCB disposition
  • Deducted Scheme

I doesn't find any zener diode as in some capacitive supply scheme I found on the web. (WebSite1,WebSite2,WebSite3)

I de-soldered the 470uF (16V rated) capacitor and found it's value wasn't 470uF anymore (370uF instead) so I replaced it. After that the display leds start working again but the heating led never start and the radiator doesn't become warm. I measured about 8.3V on the capacitor (after replacement)(8.3V with multi-meter on DC or 17.6V multi-meter on AC) On a working radiator, I measured 8.1V on the capacitor (8.1V with multi-meter on DC or 16.9V multi-meter on AC)

The two power resistors marking colors are now gone, maybe because of years of heating. A working radiator doesn't have the exact same value as the broken one: About 7.65(*2) KOhm for a working one vs about 7.5(*2) KOhm for this broken unit.

I didn't find any datasheet for the reference of the big chip UC (131DCI007A). I think it is a multiple channels voltage comparator to do temperature comparison and user selection with the two potentiometers.

Questions :

  1. Is Zener diode optional in capacitive supply? But maybe it isn't a capacitive supply?

  2. I know that the precision of this power resistors is low, but is it possible that with the time the resistors values decrease and cause the capacitor to start to die?

  3. How can I calculate the initial power resistors values? (the colors are gone)

  4. How can I calculate what voltage the supply is supposed to deliver? (I didn't find the UC datasheet)

  5. I wanted to simulated this circuit but didn't success : Simulator What did I do wrong?

  6. What could be the black CMS components I didn't identify? They pass (0 Ohm) in both size while testing onboard (still soldered) with multi-meter. Are they fuse? Or simply strap?

  7. Is this design safe and norm compliant? I don't see any security capacitor (X1/X2/Y1/Y2) nor varistance. The ground is also not connected.

  8. Is this power supply a very bad choice? I think the two power resistors are heating all time the breaker is on (In summer some radiators breaker have to still be on).

  9. Is there a simple way to make a simple adaptation to the board to improve this power supply conception by adding some components? I would like to find a way to stop wasting power in summer. A other simple solution could be to add a small switching power supply with transformer like some small USB charger or light bulb board (no need to high output current I think). But how to be sure of the safety of this circuit? In both cases, How to be sure this modification could handle the radiator temperature?

  10. What is the aim of the T4148 diode? To use the other alternance of 230VAC power? So not the same voltage supply powering?

  11. What else should I test to understand why the radiator isn't heating anymore? I measured onboard the bottom temperature sensor and found the same value as on a working radiator.

Thanks for having read (and sorry for my poor English level)

ROUGEXIII
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  • Welcome to EE.SE! This appears to be a reverse engineering, modification, or repair question. Please be aware that such questions must involve specific troubleshooting steps and demonstrate a good understanding of the underlying design of the device being discussed, so that you can ask specific, focused questions that can be answered concisely. Otherwise, the question is far too broad. More information can be found here: [Is asking how to fix a faulty circuit on topic?](http://meta.electronics.stackexchange.com/q/2478/11683). – Dave Tweed Oct 12 '19 at 18:52

1 Answers1

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Is Zener diode optional in capacitive supply?

A Zener is usually required to prevent the voltage from rise to dangerous levels if the load draws less current than 'normal' or if the mains voltage spikes. But it depends on the particular design. If it doesn't have a Zener then you know it works without it. If it does, it needs it.

I know that the precision of this power resistors is low, but is it possible that with the time the resistors values decrease and cause the capacitor to die?

Unlikely. Power resistors usually increase in resistance or go open circuit with age. The 2% difference between your values is probably just manufacturing tolerance - or perhaps a slightly different nominal value used in different production runs.

How can I calculate the initial power resistors values? (the colors are gone)

Don't bother. If the measured values are reasonable they are probably correct. (BTW the colors aren't gone, they have just changed. If you know how the colors change with heat you can figure out what they used to be)

How can I calculate what voltage the supply is supposed to deliver? (I didn't find the UC datasheet)

First trace the circuit, then apply the usual formulae. Or simulate it.

I wanted to simulated this circuit but didn't success : ... What did I do wrong?

That simulator is a joke. Use LTspice.

What could be the black CMS components I didn't identify? They pass (0 Ohm) in both size while testing onboard (still soldered) with multi-meter. Are they fuse? Or simply strap?

Black as in Ferrite? Might be small inductors used to reduce EMI.

Is this design safe and norm compliant? I don't see any security capacitor (X1/X2/Y1/Y2) nor varistance. The ground is also not connected.

If everything is well insulated then it doesn't need a ground. Capacitor only has to be X rated if connected directly across the mains, and only some circuits need one. It looks safe enough to me...

Is this power supply a very bad choice? I think the two power resistors are heating all time the breaker is on (In summer some radiators breaker have to still be on).

It's not a great design, but a better one would cost more. How much is the saved electricity worth?

Is there a simple way to make a simple adaptation to the board to improve this power supply conception by adding some components?

Perhaps, but unless you know exactly what you are doing it's very dangerous. The 'simple adaptation' could be more trouble that it's worth, and you could be held liable if anything goes wrong (even if it wasn't your fault).

A other simple solution could be to add a small switching power supply with transformer like some small USB charger or light bulb board...

USB charger, light bulb board? No, just don't. Repairing it with equivalent (perhaps higher rated) components is OK, random modifications with unknown consequences is not. If you are not happy with the design then buy a better heater.

I would like to find a way to stop wasting power in summer.

Turn off the power to it when not in use?

Bruce Abbott
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  • Thank you for your big answer. I didn't success to use LTspice simulation, I don't understand if possible to see live voltage? I can't turn the breaker off in summer cause some other equipments are on the same breaker. I make some corrections, add voltages measurements and add questions 11 and 12. – ROUGEXIII Nov 09 '19 at 16:05
  • "After that the display leds start working again but the heating led never start and the radiator doesn't become warm." - some other component must be damaged, but which one? In cases like this (where I have replaced several parts and it still doesn't work) I often find it is better to just buy a new unit. If you want to turn the radiator off in summer then wire an on/off switch into it - much safer than trying to modify the power supply circuit! – Bruce Abbott Nov 09 '19 at 19:45