Emergency Light

[rizl4s]

Hi to all,
i'm Niklas from Austria and this is my first thread here, and as many first threads i'm looking for help

Currently i have ten emergency light (Rilux 20/1NC) stocked in my basement that I want to save from being garbage. I looked at a video on youtube where this guy, transform succesfully an emergency light into a simple light by grounding a single resistance. Of course every circuit is different from others and i'm here to ask you if i can do something similar with my Rilux light.
I'll post here some photos of the internal circuit hope that someone can figure out the necessary steps to accomplish my task.

Thanks in advance for your support

Photos Low:

https://imgur.com/8xz4o6Y

https://imgur.com/Je5K1OY

https://imgur.com/CMqUkeK

Photos HD:

https://ibb.co/kMyZRC1
https://ibb.co/hgjmKYf
https://ibb.co/P6ZJwL5

Youtube video:

 

[alec_t]

Welcome to ETO!

transform succesfully an emergency light into a simple light

What do you mean by 'simple light'? Do you want it to be controlled by a manual switch instead of mains-failure detection? Do you want to power it from the mains instead of the battery? Do you want to eliminate battery charging?

 

[rizl4s]

Welcome to ETO!

Thank you

Do you want it to be controlled by a manual switch instead of mains-failure detection? Do you want to power it from the mains instead of the battery? Do you want to eliminate battery charging?

Yes to all

 

[alec_t]

Then I think the modifications would be a good bit more than tinkering with one resistor, and would require the lamp's schematic.

 

[rjenkinsgb]

The tube driver appears to run directly from the charger/battery connections.
Other than ground, there are only to tracks in to it (the right-hand section looking at the solder side).

One via the fuse and choke, which must be the power feed - and one other track, which comes from a resistor near the centre of the board.
That is presumably the on-off control for the inverter / lamp.

 

[rizl4s]

Then I think the modifications would be a good bit more than tinkering with one resistor, and would require the lamp's schematic.

Of course i think it too, i looked around the web for schematic without luck

One via the fuse and choke, which must be the power feed - and one other track, which comes from a resistor near the centre of the board.
That is presumably the on-off control for the inverter / lamp.

Can you make a mark on the photos?

 

[rjenkinsgb]

 

[rizl4s]

Thank you.. So how can I bypass the normal function?

I forgot to say, the red circle marks 220v connection.

 

[rjenkinsgb]

Measure the voltage on the control trace, relative to battery negative, with normal power applied.

Then lift the lower end of the resistor that feeds that trace (above the red circle) and connect the now free end of it to the opposite battery wire; eg. if the control is near whatever voltage the battery should be, connect the resistor to negative, if near 0V connect it to positive.

That should hopefully turn on the lamp inverter.

Be extremely careful doing anything with power connected. The best thing is to connect an extra wire to the control track at one of the component joints and use that and the battery negative twisted / taped to your multimeter leads.

Or just connect the resistor end to each battery terminal in turn and see if either polarity works to switch the lamp on...

It is possible the battery charger part cannot provide enough power to run the lamp, without a battery connected; it could just be a fairly slow charger.

 

[rizl4s]

Thank you, really appreciate your effort in this! In the weekend I will have some free time and I will give it a try.
I'll be in touch

 

[rizl4s]

Measure the voltage on the control trace, relative to battery negative, with normal power applied.

Then lift the lower end of the resistor that feeds that trace (above the red circle) and connect the now free end of it to the opposite battery wire; eg. if the control is near whatever voltage the battery should be, connect the resistor to negative, if near 0V connect it to positive.

That should hopefully turn on the lamp inverter.

Be extremely careful doing anything with power connected. The best thing is to connect an extra wire to the control track at one of the component joints and use that and the battery negative twisted / taped to your multimeter leads.

Or just connect the resistor end to each battery terminal in turn and see if either polarity works to switch the lamp on...

It is possible the battery charger part cannot provide enough power to run the lamp, without a battery connected; it could just be a fairly slow charger.

Ok if i understood all correctly i should try this way:

• Measure voltage with one end of the multimeter on control trace and the other end on battery negative (What should I see here?)
• Remove the lower end of the resistor and connect it to battery negative. If voltage is near 0V connect it to battery positive.

See attached photo: https://ibb.co/9WR0LL6

Also I suppose that I can ran this test without battery plugged in, right?

Thank you

 

[Pommie]

If you replace the battery with an equivalent power supply it should work as required.

Mike.

 

[rizl4s]

Without connecting 220v i suppose, power supply will act as a battery. It can be a way but I want to try this zero cost way before.

 

[Pommie]

Without a schematic it'll be almost impossible.

Mike.

 

[rjenkinsgb]

Ok if i understood all correctly i should try this way:

• Measure voltage with one end of the multimeter on control trace and the other end on battery negative (What should I see here?)
• Remove the lower end of the resistor and connect it to battery negative. If voltage is near 0V connect it to battery positive.

See attached photo: https://ibb.co/9WR0LL6

Also I suppose that I can ran this test without battery plugged in, right?

All correct; I have no idea what voltage you may see on the control track, that's why you need to find out!

Also measure the voltage across the battery leads, as a reference for its working voltage.