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i am using an LM317 as a Current Regulator for series LED similar to protecting LEDs From over current ( in time of short circuiting) for my LEDs and Mosfets, similar to this picture:

enter image description here

the problem is that i need bigger voltage source ( up t0 20V) and momre cost fo voltage source and LM317 IC, because of dropping voltage on the LM317 ( in range of 4.5v to 5.5V) so if i found another eleman ( similar to fuse or ...) for protecting LEDs AND MOSFETs at the time of over currenting. so is there any alternative (for protecting LED ANd MOSFETS) and is it OK i use one fuse instead of current regulator ( which one is first to burning : MOSFET or Fuse). Thanks a lot.

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  • \$\begingroup\$ Have you tried another (switching or linear) regulator with a greater supply range yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 24 '15 at 6:57
  • \$\begingroup\$ The LM317 can actually drop max 40V from input to output, so your output can surely exceed 20V. With some additional precautions it can even output more than 37V because it's a floating regulator, see electronics.stackexchange.com/questions/148340/…. The ~5V drop across the IC and sense resistor when used as a current source (which you correctly note) is however pretty wasteful compared to some alternatives. \$\endgroup\$ – SX welcomes ageist gossip Aug 30 '15 at 20:03
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Many current regulator or LED driver ICs in the market. For example: Linear, Texas Instruments, etc.

For your question:

is it OK i use one fuse instead of current regulator ( which one is first to burning : MOSFET or Fuse).

Not predictable.

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  • \$\begingroup\$ Actually it can be predictable enough if know the I^2t rating of both the fuse and load/MOSFET etc. \$\endgroup\$ – SX welcomes ageist gossip Aug 30 '15 at 19:58
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I think using a fuse is not a good idea to protect the LEDs. I have put much time to find a decent current source. LM317 is probably is the easiest solution, but the at least 3V voltage drop is not nice. You can use other alternative constant current circuits like the one below
enter image description here

here you set the current by R_sense=0.7/I .

The voltage drop would be merely 0.7V to 1V depending on the transistor Q1 you use. You can find a huge amount of info on the web and here on stackexchange about this circuit.

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  • \$\begingroup\$ That circuit tarted up has provided yeoman service driving LEDs .In fact it works well with mosfet for Q1.Even better the current drops with Q2 temp which is desirable . \$\endgroup\$ – Autistic Nov 7 '15 at 10:48

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