I read this question and I have the same issue: acquire and protect an input with 5-24V range. I wonder if nowadays I can use a BSS126 as current source transistor.

It's a depletion mode N-channel, I checked the output characteristics and they seem correct.

I'm afraid about the type of transistor: BFR30 is a field-effect transistor (FET) while BSS126 is a MOSFET (SIPMOS). Is it suitable for this application?


Actually I don't need a constant current to detect the input status in the 5-24V range. But I have very limited space on the PCB board so I'm looking for a minimum-components solution.

I tried with the simplest scheme:


simulate this circuit – Schematic created using CircuitLab

Some notes:

  • I selected a zener with a reverse voltage slightly below the minimum input voltage

  • R2 can be selected to allow about 5 mA through LED and opto-coupler: (4.3V - 2.1V - 1.1V) / 5mA = 860 ohm

  • with 5V at input, R1 should let flow 5mA towards the opto-coupler and a couple more for the zener, say (5V - 4.3V) / 7mA = 100 ohm

  • with 24V, the current flowing in R1 is: (24V - 4.3V) / 100 ohm = 197 mA it's too much (about 4W!)

This is why I was looking for a more efficient way to acquire such a signal with few components.

  • \$\begingroup\$ R2 can be selected to allow about 5 mA through LED and opto-coupler: (4.3V - 2.1V - 1.1V) / 5mA = 860 ohm No: 4.3 V - (2.1V + 1.1V) = 1.1 V / 5 mA = 220 ohm. are you sure the opto coupler LED only drops 1.1 V ? \$\endgroup\$ Feb 11, 2020 at 14:21
  • \$\begingroup\$ Yes, I use ACPL-247-500E. You're right about the calculations, but I'm afraid about R1 and zener. The led and opto are ok with those values. \$\endgroup\$
    – Mark
    Feb 11, 2020 at 14:30

1 Answer 1


I think that in principle you can use the BSS126 (depletion type NMOS) as a crude current source, just like the BFR30 (N-channel FET, looks like a JFET to me)

Both are depletion (normally on) types so both will can conduct a current when Gate and source are shorted (Vgs = 0).

However, the datasheet of the BFR30 tells is that the drain current at Vgs = 0 is between 4 mA and 10 mA. For the BSS126 the range is less tight, there is only a minimum current guaranteed of 7 mA. So the current could be much larger as well, it is not guaranteed to be less than some value.

But don't think that since in the other question the issue (5 to 24 V input range) was solved with a JFET, it now has to be solved like that was well. There are many other ways as well. The JFET was there to keep the current through the optocoupler's LED somewhat constant, is that really needed? Many LEDs will work (light up) with a current between 1 mA to 20 mA. So is there a need to keep that current constant?

If you just need to detect that the 5 to 24 V is present or not, there might be no need to keep the current constant.

Maybe a couple of resistors and a zener diode is all that's needed. If the current does need to be somewhat constant there are other solutions, for example these circuits:

enter image description here

I know it says 6 - 15 V, with some small changes that range can easily be made 5 V to 24 V.

  • \$\begingroup\$ I don't need a constant current, but I wasn't able to find a working solution with the common zener/resistors circuit. I've updated the question with my trial. \$\endgroup\$
    – Mark
    Feb 11, 2020 at 13:52
  • \$\begingroup\$ Your circuit should do the trick after you have replaced the 4.3 V zenerdiode with one that has a slightly higher zener voltage. The 4.3 V might be insufficient to make the two LEDs in series light up. You could also just remove the red LED or give each LED its own series resistor and connect those across the zener diode. \$\endgroup\$ Feb 11, 2020 at 13:56
  • \$\begingroup\$ If I raise the zener voltage I have to lower the value of R1 otherwise there is not enough current at 5V. And thus the things will be even worse at 24V: all the current will flow into the zener. \$\endgroup\$
    – Mark
    Feb 11, 2020 at 14:01

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