EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Pugbutt on April 24, 2016, 04:58:26 pm
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Hi Everybody.
I'm wanting to use a Diodes Incorporated AP3032 LED driver in a design. (http://www.diodes.com/_files/datasheets/AP3032.pdf (http://www.diodes.com/_files/datasheets/AP3032.pdf)) The datasheet says that, by default, you can just hook the OV pin to the output to set the max voltage to 27V. The example schematic though shows a voltage divider there though, and the resistors are unlabeled. This implies a variable setting to me, but there is no more detail given.
A similar (but much more expensive) part is a Linear Technology LT1618. (http://cds.linear.com/docs/en/datasheet/1618fas.pdf (http://cds.linear.com/docs/en/datasheet/1618fas.pdf)) They have a pin serving the exact same function, and they were nice enough to specify the relation between divider ratio and cutoff voltage.
Does anyone see how I can set a specific cutoff voltage on the AP3032? 27V feels way too high for my application. I'd rather set ~5V.
Thanks,
Craig
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If you connect the OV to the output and get 27V, any voltage divider will only increase the maximum voltage.
I guess the only way to know for sure is to see what happens if you connect OV to output directly.
A work-around is to trick the current feedback with a resistor divider and a extra schottky diode, like attached. You do the math for the resistors :).
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27V feels way too high for my application. I'd rather set ~5V.
You're probably looking at the wrong chip then - as mentioned in the datasheet this one is specifically intended for driving 7 white LEDs in series, so around 20-25V output.