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I have a bunch of 10W lamps (like this one) with incorrect power adapters for the emitters - the drivers are meant for a 1x9 configuration (35V, ~300mA), LEDS are 3x3 (~11V, ~900mA).

I am fine with running them at 1/3 power, for now.

Question: what happens to the remaining 2/3 (~23V, 300mA)? Is it radiated off in the driver (and is that a concern, like a short circuit might)? Or does the driver deal with that differently?

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  • \$\begingroup\$ Where's the link for the power supply? \$\endgroup\$
    – Transistor
    Jun 21, 2016 at 8:12
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    \$\begingroup\$ Assuming the LED driver behaves as a current source, it will force a current. It will adjust the voltage so the required current flows. As these are usually switched regulators it will simply switch such that the voltage is lower. Think about what would happen if there was excess power and it would be dissipated in the driver. Would it not get very hot ? So no it is not dissipated in the driver. Same as when using a 12 V, 10 A power supply and loading it with 1 A. There is no remaining 9 A. The adapter only delivers 1 A. \$\endgroup\$ Jun 21, 2016 at 8:18
  • \$\begingroup\$ @transistor It's a cheap waterproof power supply with aluminium and black resin outside. I do not have one at hand. This one shows a range of 20-40V, whatever that means for this issue. Mine shows "34-35V". \$\endgroup\$
    – kaay
    Jun 21, 2016 at 8:24
  • \$\begingroup\$ @fakemoustache Thanks. Thing is, these kind of drivers usually show a RANGE of output voltages, like "6-12V, 900mA". The range is narrow on mine, and I do not understand what using an "unsupported" load will do. \$\endgroup\$
    – kaay
    Jun 21, 2016 at 8:27
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    \$\begingroup\$ And without knowing the design no-one can understand what will happen if you use the driver outside the specified range. With these cheap Chinese drivers my advice would be: don't use them outside their specified range. \$\endgroup\$ Jun 21, 2016 at 8:45

1 Answer 1

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Within the range specified on the driver's output (e.g. "6-12V, 900mA"), the voltage will be adjusted to ensure a constant current. No power wasted.

Outside that range, behavior is undefined.

Thanks, FakeMoustache.

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  • \$\begingroup\$ The only thing wasted is unused power potential in the LED supply which someone paid for in the first place but isn't using. \$\endgroup\$
    – winny
    Jun 21, 2016 at 9:09

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