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I just installed a dimmer for a set of (non-dimmable) MR16 bulbs wired in parallel. The dimmer sits between a 12V power supply and the bulbs.

My observation when trying to dim the non-dimmable bulbs is that the brightness stays the same until the input voltage to the bulbs drops to below roughly 1V. Then the brightness drops rapidly.

I suspected that the bulbs were drawing more current in order to compensate for the drop in input voltage in order to keep the amount of power drawn the same. So I measured it, they weren't.

So therefore I know that as input voltage drops from 12V to 6V the power drawn by each bulb drops, but I also know the apparent brightness of the bulbs does not. I can only conclude that the extra power drawn from a bulb run at 12V must be dissipated as heat.

My question is: is my conclusion correct? Is there a reason to run my bulbs at 12V or is 3V good enough? Will running the bulbs at a lower voltage lengthen their life? Is there a simple (UK A-Level / US High-School-Diploma) circuit diagram that can explain the behaviour and be used to calculate how much power is converted into heat?

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    \$\begingroup\$ How did you measure the current? How does your dimmer work? I am pretty sure the mistake you've got is the current draw measurement. Current will increase as the voltage decreases. Each bulb has most likely got a constant current source in it, which pushes the required current through the bulb. This keeps energy wastage low. \$\endgroup\$
    – Puffafish
    Jan 24 '19 at 11:38
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    \$\begingroup\$ There might be a good reason why they are called "non-dimmable bulbs". \$\endgroup\$
    – Oldfart
    Jan 24 '19 at 12:40
  • \$\begingroup\$ I measured the current by connecting my multimeter in series with the dimmer. \$\endgroup\$
    – Resonance
    Jan 24 '19 at 12:59
  • \$\begingroup\$ You can't measure real power by just measuring current and multiplying by the assumed voltage. You have calculated apparent power. You need to know the power factor of the load to get the real power. \$\endgroup\$
    – Elliot Alderson
    Jan 24 '19 at 13:03
  • \$\begingroup\$ @Puffafish. I do not know how my dimmer works. I do know that at low dimming values the 240V -> 12V power supply starts to buzz. \$\endgroup\$
    – Resonance
    Jan 24 '19 at 13:10
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LEDs are better driven by current sources than by voltage sources. This means that LED bulbs tend to have a small current regulator embedded in them.

This current regulator can be something sophisticated like a buck regulator to something as simple as a capacitive dropper.

These regulators will provide the current even when you drop the voltage (up until they cannot create the voltage needed to light up the LEDs). And because they are active (in the case of switch-mode power supplies) or reactive (like the capacitive dropper) they won't create excess heat like a simple resistor in series would.

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  • \$\begingroup\$ But in that case the input current would change with the input voltage, otherwise a switched supply couldn't keep the output current the same. \$\endgroup\$
    – jusaca
    Jan 24 '19 at 13:31
  • \$\begingroup\$ Inside the bulb the regulator provides a constant current (say for example, 350 mA) To achieve that it has to regulate voltage (also for example but it's common value) around 3V. But it can be 3.2 or 2.8, the regulator knows how to adjust the voltage so that current remains at 350 mA. (Current through leds varies with voltage). Power will be in this case +-1W. If a higher voltage is applied to the regulator, the regulator will still convert to the +-3V no matter the input voltage. \$\endgroup\$
    – Fredled
    Jan 25 '19 at 2:05
  • \$\begingroup\$ It works as long as the input voltage is higher than 3V. If input is 12V, idealy it would take 87 mA to generate 350 mA at 3V. But because the conversion process wastes power it's a litte bit more than that. 100 or 120 mA. So instead of 1W, it will read +-1.2W. \$\endgroup\$
    – Fredled
    Jan 25 '19 at 2:05
  • \$\begingroup\$ Thank you for your answer, reading up on buck regulators and capacitive droppers has kept me busy, although I quickly felt out of my depth. I guess I must have measured the current incorrectly, although I'm not sure how. \$\endgroup\$
    – Resonance
    Jan 25 '19 at 12:14
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As stated above, there is a current regulator in the bulb which prevent the light to go dim. If there is no difference in light output and you can measure less consumption at lower voltages, it means that the bulb is compatible from 3V to 12V but that it takes considerably less power to convert from 12V to 3V than from 3V to 3V.

The LED diodes won't live longer if you apply less voltage but the regulating circuit may. There is no guarantee to that effect.

You could power the bulbs with less voltage but bear in mind that at lower voltages, wires must be larger because it draws more amperes. If your wires are too thin they will heat and even melt at 3V. Hence it may be necessary to apply 12V between the power supply and the bulbs.

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