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

enter image description here

However, the load requires at leas 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?

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  • \$\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 '16 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\$ – Brian Drummond Oct 7 '16 at 10:13
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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.

schematic

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.

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  • \$\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\$ – Chupacabras Apr 29 '19 at 11:58
  • \$\begingroup\$ @Chupacabras Yes, I’m sure. \$\endgroup\$ – Spehro Pefhany Apr 29 '19 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\$ – Chupacabras Apr 29 '19 at 13:41
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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.

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  • \$\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\$ – TheMeaningfulEngineer Oct 6 '16 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 '16 at 0:28
  • \$\begingroup\$ @ThePhoton I want to simulate a point in the I-V curve of a solar cell. \$\endgroup\$ – TheMeaningfulEngineer Oct 7 '16 at 14:12
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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).

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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.

http://www.digikey.com/product-search/en/discrete-semiconductor-products/transistors-jfets/1377093?k=&pkeyword=&pv609=99&pv609=59&FV=fff40015%2Cfff80345&mnonly=0&newproducts=0&ColumnSort=609&page=1&quantity=0&ptm=0&fid=0&pageSize=500

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

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  • \$\begingroup\$ Can you share a circuit reference for NCh JFET with gate shorted to source? \$\endgroup\$ – TheMeaningfulEngineer Oct 6 '16 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\$ – Tony Stewart Sunnyskyguy EE75 Oct 6 '16 at 23:43
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    \$\begingroup\$ Current regulation isn't the same as current limiting \$\endgroup\$ – Scott Seidman Oct 7 '16 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 '16 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\$ – Spehro Pefhany Oct 7 '16 at 6:36

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