I'm trying to limit a current of a 5V power supply to 100uA. I first thought of putting in a 50k resistor. That will never allow above the threshold.

enter image description here

Unfortunately, the load requires at least 4.5V and varies in the amount of current in needs. So this is obviously a too simple approach for the problem.

How can I extend this circuit to limit the current and keep the load voltage current independent?

  • \$\begingroup\$ Sounds like a fuse would be just what you need, if 100μA fuses existed. I think the challenging part of this problem is reliably detecting 100μA. That's a pretty small current and you'll need to give some consideration to noise for a robust solution. \$\endgroup\$
    – Phil Frost
    Oct 7, 2016 at 2:44
  • \$\begingroup\$ From 5V? That allows you to only drop 0.5V in the current limiter. See if you can get lucky and find a LDO regulator with adjustable voltage AND current limit. If you can relax the input constraints - say, starting with 9V in - that'll greatly widen your options. Otherwise - Spehro's approach looks good. \$\endgroup\$
    – user16324
    Oct 7, 2016 at 10:13

5 Answers 5


Presumably you are okay with a small drop, provided it is less than 0.5V.

Here is a circuit that will limit at 100uA and will drop less than 100mV before it limits.


simulate this circuit – Schematic created using CircuitLab

The MCP6001 is an inexpensive rail-to-rail input/output op amp that will operate from a 5V supply. The op-amp will saturate at ground until the load current reaches about 98uA nominally (with the values shown). The supply thus 'looks like' 5V with ~1K in series (the MOSFET contributes less than 10 ohms with Vgs =-5V), so it will drop between 0 and 100mV for load resistances of infinity down to 50K.

For lower load resistances the circuit regulates the output current to ~98uA.

The circuit draws about 200uA from the 5V supply, in addition to the load current of 0~100uA.

  • \$\begingroup\$ Are you sure about your calculations? Voltage divider (R2, R3) gives 238mV at non-inverting input of opamp thus it allows 238uA through 1k resistor (R1). \$\endgroup\$ Apr 29, 2019 at 11:58
  • \$\begingroup\$ @Chupacabras Yes, I’m sure. \$\endgroup\$ Apr 29, 2019 at 12:37
  • \$\begingroup\$ You are right, now I see where I made mistakes. One of them is I saw 49.9 ohms there, not kiloohms. My bad. \$\endgroup\$ Apr 29, 2019 at 13:41
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    \$\begingroup\$ Rail-to-rail is relevant because the inputs have to be within the op-amp's common mode range- a part such as LM324/358 will not work well in this application without much more voltage across R1. Input bias current is a direct error on the current. At 1mA any op-amp is okay, at 100nA many are not. Offset voltage and tempco of the Vos (TCVos) are important if you want to minimize the voltage across R1. At 1 volt, any op-amp pretty much is okay- but at 10mV many will contribute a lot of error. \$\endgroup\$ May 28, 2020 at 13:55
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    \$\begingroup\$ @SpehroPefhany oh, sure - thought this was a new question as it was on top, but I visited the wrong SE landing page so it was only on top because of some activity, sorry about that. \$\endgroup\$
    – Arsenal
    Oct 23 at 11:39

The Ohm's Law police will come after you if you attempt to regulate both voltage and current into a fixed load.

Unless you have a very unusual load, the load will draw whatever current it needs if you supply the correct voltage. Any attempt to reduce the current will reduce the applied voltage.

  • \$\begingroup\$ It's for simulating a power supply that can give 4.5V/100uA. The load sometimes has spikes with more than 100uA. There is a decoupling capacitor to handle that but I skipped all that to try to focus on the main goal (getting from 5V/100mA to 4.5V/100uA power supply). You think it's important for the question? \$\endgroup\$ Oct 6, 2016 at 23:30
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    \$\begingroup\$ What kind of supply are you trying to model? A battery limited by internal resistance? A benchtop supply with a foldback circuit? A wall wart with a fuse? For each one you would want to model it a different way. \$\endgroup\$
    – The Photon
    Oct 7, 2016 at 0:28
  • \$\begingroup\$ @ThePhoton I want to simulate a point in the I-V curve of a solar cell. \$\endgroup\$ Oct 7, 2016 at 14:12
  • \$\begingroup\$ So just a quick enquiry. Would you be able to regulate the current if you figured out the impedance of the circuit and finding the correct voltage? \$\endgroup\$ Feb 22, 2021 at 19:57

Generally your current limiting circuit will require a bit of voltage "headroom" to operate in. That means you'll need an unregulated PSU of > 5 V and regulate it down to 5 V while monitoring the current.

It's a while since I've read up on the old LM723 voltage regulator but they offer voltage and current limiting. These were very popular once upon a time so you should find plenty of sample configurations on a web search.

enter image description here

Figure 1. Basic Low Voltage Regulator (VOUT = 2 to 7 Volts). (Figure 4 of datasheet).


It's a bit late but I would use a current regulator diode S-101 from Semitec. S-101T is the SMD version and they are available in a range of currents.


The simplest solution is NCh JFET with gate shorted to source.

The spec is called IDSS with Vgs=0

Here are some choices, which require ESD precautions and reverse voltage failure.


I see these are either not in stock, obsolete or last time buy

Plan B

quick and dirty solution

LM317 cct. for 0.1mA R1 = 1.25V/0.1mA = 12.5k

enter image description here

  • \$\begingroup\$ Can you share a circuit reference for NCh JFET with gate shorted to source? \$\endgroup\$ Oct 6, 2016 at 23:25
  • \$\begingroup\$ I don't have any handy. It's common EE knowledge widely documented on web. Pick any datasheet I linked to read IDSS but also becoming OBSOLETE. Generally we must use a stable Vreg regulator CC design to be a constant current depending on V and I tolerances unknown. Look at LM317 and CC design \$\endgroup\$ Oct 6, 2016 at 23:43
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    \$\begingroup\$ Current regulation isn't the same as current limiting \$\endgroup\$ Oct 7, 2016 at 0:42
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    \$\begingroup\$ ... and neither of these solutions meets the requirement of zero overhead voltage. \$\endgroup\$
    – Dave Tweed
    Oct 7, 2016 at 0:58
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    \$\begingroup\$ The LM317 current regulator definitely won't work at 100uA - it needs some mA, depending, in part, on the input voltage, and it will need to drop several volts (1.25V plus the regulator drop-out voltage) in order to work. \$\endgroup\$ Oct 7, 2016 at 6:36

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