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I have a constant current led driver which gives 180ma @24volts. But I need 120ma to drive a led strip at same volatge. Is there any way to lower the current from this driver? I tried shunting led driver with a 500ohm 1w resistor and managed to get 130ma but the resistor gets little hot. Is there any other solution? Any help is appreciated.

Thanks David

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  • \$\begingroup\$ What is the voltage across your LED strip? Is it the whole 24 V? \$\endgroup\$ – The Photon Oct 12 '17 at 4:44
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    \$\begingroup\$ There is probably a sense resistor in the module, so if you increase the value of that resistor in the ratio 1.5:1 it should regulate at the lower current. Look for a low value resistor, possibly of a larger size, maybe in the low ohm or sub ohm range, it may be connected directly to one side of the load. \$\endgroup\$ – Spehro Pefhany Oct 12 '17 at 6:34
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    \$\begingroup\$ Could you please post the part number, picture, datasheet of your constant current driver? Is an IC, or a finished product? In the second case, is it sealed (you have no access to the internal circuitry) or can you open it easily? \$\endgroup\$ – next-hack Oct 12 '17 at 6:56
  • \$\begingroup\$ Hi, Thanks for the info. Led driver IC used is FT886A-RT. Sense resistor used was 2.7ohm (for 180ma). I replaced it with 4.7ohm and now I get about 130ma . Guess it should ok to work with. \$\endgroup\$ – David Oct 12 '17 at 18:08
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    \$\begingroup\$ @SpehroPefhany Your comment is better than the answers given below, IMO. Would you care turning it into a proper answer? \$\endgroup\$ – Lorenzo Donati -- Codidact.com Oct 12 '17 at 19:14
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Since the OP has provided some information on the chip I will extend the below comment to include the specific case, but it should apply to most modules of this type:

There is probably a sense resistor in the module, so if you increase the value of that resistor in the ratio 1.5:1 it should regulate at the lower current.

Look for a low value resistor, possibly of a larger size, maybe in the low ohm or sub ohm range, it may be connected directly to one side of the load.


In this particular case the sense resistor is 2.7 ohms, so the current can be reduced by increasing the resistor to something like 4.7 ohms. In this case, it is not connected in series with the load, but on the low side of the switch (connected between the source of the N-channel MOSFET switch and the supply negative rail). The chip is an FT886A-RT from Fremont, and the datasheet is rather sketchy, however here is a typical application circuit showing the sense resistor:

enter image description here

The comparison voltage is 500mV and they appear to assume ripple in the input so a lower average current than you are observing with a DC input to the chip. If you greatly decreased the current it might be necessary to increase the inductor value, especially if the input voltage was expected to be relatively high compared to the module design maximum. The chip is advertised as having protection against inductor saturation so even that would not be expected to kill the chip, just affect the operation.


If there is a lot of ripple, due to poor input or output filtering, remember that the perceived visible light output (in the normal working range) is roughly proportional to the average LED current, but the maximum LED current is proportional to the RMS LED current, so the further you stray from smooth DC the less light output you can safely expect.

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Many drivers are fixed output, but some can be programmable, use dip-switches, has multiple outputs at different current or use some kind of LEDset/Rset. Check the datasheet for your driver to se if it can be regulated.

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  • \$\begingroup\$ Hi, I solved it by changing sense resistor. (Increased it from 2.7 to 4.7ohm). Thanks \$\endgroup\$ – David Oct 12 '17 at 18:10
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The target sense resistor value is 500/0.120 = 4.166

Use a 4.22Ω sense resistor. This will give you 118mA.

Fremont FT886xx Datasheet

DATASHEET LINK: Datasheet Fremont FT886xx

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