I plan to make a module for my motorbike to replace fuse by current limiters and overload detection system. I've heard nice things to do with J-FET. It should work up to 12V 30AMP. The detection should return the signal to a PIC Microcontroler.

I have found the explaination below in litterature. It says that when it is overload the current falls to 8mA. Is there any ways to measure it and send a return signal to microcontroler? Thanks in advance!

JFET current Limiter

enter image description here

JFET current limiting circuit is shown in figure. Almost all the supply voltage therefore appears across the load. When the load current tries to increase to an excessive level (may be due to short-circuit or any other reason), the excessive load current forces the JFET into active region, where it limits the current to 8 mA. The JFET now acts as a current source and prevents excessive load current.

A manufacturer can tie the gate to the source and package the JFET as a two terminal device. This is how constant-current diodes are made. Such diodes are also called current-regulator diodes.

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    \$\begingroup\$ Sorry to break the news, but lots of luck finding a 30 amp FET. \$\endgroup\$ Commented Apr 8, 2016 at 18:05
  • \$\begingroup\$ not looked far have you @WhatRoughBeast \$\endgroup\$
    – user16222
    Commented Apr 8, 2016 at 19:09
  • \$\begingroup\$ And not even SiC and from a distributor no a die manu mouser.co.uk/ProductDetail/GeneSiC-Semiconductor/GA05JT01-46/… \$\endgroup\$
    – user16222
    Commented Apr 8, 2016 at 19:11

3 Answers 3


That won't work for multiple reasons.

1) You can't get JFETs that handle much more than 10 mA

2) If you current limit 30 A overload on a 12 V system, you'll dissipate 360 W of power and quickly fail your circuit

3) Most systems like this limit current for a short time (< 1 ms), then turn off completely, and perhaps retry at intervals. If you turn off too quickly, the voltage spike from the inductance of the will will also damage your circuit.

Take a look at this - http://www.nxp.com/files/training/doc/dwf/DWF13_AMF_AUT_T1023.pdf and look for eXtreme Switches to see if any of those would work for you.

  • \$\begingroup\$ "much more than 10mA" ORLY... infineon.com/dgdl/… \$\endgroup\$
    – user16222
    Commented Apr 8, 2016 at 19:08
  • \$\begingroup\$ and not a die manu: mouser.co.uk/ProductDetail/GeneSiC-Semiconductor/GA05JT01-46/… \$\endgroup\$
    – user16222
    Commented Apr 8, 2016 at 19:11
  • \$\begingroup\$ It looks that it Works!!! google.fr/url?sa=t&source=web&rct=j&url=http://… \$\endgroup\$ Commented Apr 8, 2016 at 21:42
  • \$\begingroup\$ that SiC device has 100mohm on resistance and is not really suitable for this application -- its current rating is only 26 A (only with a wonderful heatsink), and will still fail if you actually have a short circuit. \$\endgroup\$
    – jp314
    Commented Apr 9, 2016 at 1:24
  • \$\begingroup\$ Yes i understand jp314 but my idea was to cut the circuit with a microcontroler before i have the short circuit by measuring the current level and more precisely an image of the current like in section 2.2. What do you think. \$\endgroup\$ Commented Apr 9, 2016 at 5:17

I realize this is an old question. But the correct answer to this question is "you don't want to do this." Fuses are specifically there to prevent wires from overheating and smoking the insulation or even starting a fire. They are simple and reliable. You cannot replace a fuse with an active circuit incorporating mosfet's and whatnot, because the active circuit failure modes may be unsafe. For example mosfet's usually fail with drain and source shorted together. So this is just a terrible bad and unsafe and misguided idea. It may be OK to put active current limiters between the existing fuse and load. But not to REPLACE the fuse as this question says.


I've tried this w/ the classic 2-terminal [Doubt patent] and it is only a "runaway resistor"; there is a 3 transistor version that works. I've Spice simulated & built all of the above, but the best option is a big honking power MOSFET w/ a big heatsink. If you use the "analog mode", you will dissipate a lot of power. The thing I haven't tried is a SMPS approach; "smart control" is unnecessary, as a proper regulator will inherently work or burn up.

OK, to turn this to an answer: JFET, NO, MOSFET, YES. It will be large, I've done this. Bipolar BJTs are current devices and don't turn off as well as the voltage device FET types. JFETs are not made in the power ranges that would work. But MOSFETs are available in "brick" packages. My goal was 100A and I used MOS to get there. The MOSFET bricks and a heatsink can dissipate the reject power. I'm talking 1200V/100A devices used in megawatt UPS's that I worked on.

To get away from the high power dissipation, a SMPS [Switch Mode Power Supply] method would be needed. This makes use of a true switching OFF mode rather than imposing a dissipation. However, conduction reject heat in the ON mode would still need to go somewhere, so big likely does not go away.

By "proper regulator" I mean a circuit that regulates w/out an external "decision maker" [CPU]. A lot of such burn up w/ thermal runaway when pushed beyond a SOA [Safe Operating Area] restriction, such as the Doubt 2-terminal BJT [Bipolar junction Transistor] circuit. The Doubt circuit [Pat US3769572] looks promising but is best left to 20 mA or so applications. I've tried them @ >1000X.

Another problem is that these limiters use a series resistor as a sense element. To get high currents, this sense R goes into the milliohm region. You can, or will have to, make your own shunts by using a tapped heavy gage wire, i.e. the drain or emitter lead itself. You will need a milliohm or microhmmeter for any high current shunt trimming.


  • 1
    \$\begingroup\$ This is more of a comment than an answer, as answers should answer the question. when you have enough rep you can comment \$\endgroup\$
    – Voltage Spike
    Commented Mar 21, 2019 at 20:24

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