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Dummy Load

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Nov 20, 2022
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I would like some advice/help on how to add a dummy load to a circuit when changing from 230v incandescent lamps to LED, on the visual display it has a flashing icon showing lamp failure when using the replacement LED lamp, replace with the original working 4w 230v T5 lamp and the icon cancels. Looking for something similar to a "canbus load resister" on a 12v vehicle when changing to LED lamps but for 230v ac.
 

kellys_eye

Jun 25, 2010
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Connect a resistor across the LED of a value sufficient to match the current flow in the original lamp circuit. This assumes your LED is a 'modular' one i.e. is in a package that works when supplied directly from 230VAC.

i.e original lamp (4W @230V) draws 17mA. Your LED draws ???

Your resistor should be 230/.017 LESS the current drawn by the LED - so if the LED draws 5mA then the calculation would be 230/(0.017-0.005). In this example the resistor would be 19,166 ohms (nearest practical value 18k).

The resistor wattage would be around 3watts (use a 5W resistor to be safe).

Adjust the values given above according to your own situation.
 

Harald Kapp

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What's the use of using an energy efficient LED lamp when you add a resistor to "burn" energy?
What kind of circuit is it that behaves as described? Could this circuit possibly be modified to accomodate a low-energy LED lamp?
 

kellys_eye

Jun 25, 2010
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I see this as a system that detects lamp failure therefore monitors lamp circuit current. Replacing a filament lamp with an LED reduces the current such that a 'fault' is detected. If this detection is part of software it would be difficult to 'fix'.
 

Keonte45

Aug 29, 2022
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I see this as a system that detects lamp failure therefore monitors lamp circuit current. Replacing a filament lamp with an LED reduces the current such that a 'fault' is detected. If this detection is part of software it would be difficult to 'fix'
Okay...
 

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Connect a resistor across the LED of a value sufficient to match the current flow in the original lamp circuit. This assumes your LED is a 'modular' one i.e. is in a package that works when supplied directly from 230VAC.

i.e original lamp (4W @230V) draws 17mA. Your LED draws ???

Your resistor should be 230/.017 LESS the current drawn by the LED - so if the LED draws 5mA then the calculation would be 230/(0.017-0.005). In this example the resistor would be 19,166 ohms (nearest practical value 18k).

The resistor wattage would be around 3watts (use a 5W resistor to be safe).

Adjust the values given above according to your own situation.
Thank you for your reply and solution, by the way, the other posters, seem to think this is for energy saving, but the reason for changing to LED is for longevity (50,000 hrs) the T5 UV tubes only last about 1 year. I have ordered 5w 18k ceramic resistors and will give it a try, you were about right the LED draws 5mA +/- I thought that this was the way to go but not sure of the calculation. kind Regards.
 

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Thank you for your reply and solution, by the way, the other posters, seem to think this is for energy saving, but the reason for changing to LED is for longevity (50,000 hrs) the T5 UV tubes only last about 1 year. I have ordered 5w 18k ceramic resistors and will give it a try, you were about right the LED draws 5mA +/- I thought that this was the way to go but not sure of the calculation. kind Regards.
Just to get this correct, do I connect the resister across the 230v feed (Live and Neutral) to the LED or in parallel on just the Live side?
 
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