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I'm a newbie to circuit design, so I think I'm either really lost or over thinking this (most likely the former). I need a circuit that will act like a current limiting power supply, which I can design in a specific voltage (20mV) and current limit (100mA) across multiple independent outputs. Each of my outputs will likely be shorted or see very low resistance.

So far, I've looked ICs that can provide a constant current source, but none are in the current range I need. I've looked at voltage regulators, but I'm not sure how I could limit the current from them.

The most promising thing I've seen so far is using a two transistor current limiter, like this:

Current limiter

Source: Wikipedia

Is is likely to work? If I have multiple circuits like this (connected to the same VCC and GND), would it still work as intended? Is there a much better way to approach this problem?

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An assumption need to be made which you may wish to clarify. If you want 20 mV at UP TO 100 mA then none of the circuits will work without both a current control and voltage control function. The following circuits deal with current control. These can be used to feed a standard 20 mV voltage regulator circuit. If Iload is < Icurrent_limit then the VR works as desired. If Iload tries to exceed current limit then VR is starved of current. So ...

Your circuit will work but is relatively low quality - it depends on the Vbe of Q2 being well defined - which it tends not be be.Steven says that the top transistor needs more than 20 mV Vbe, which is true BUT if you set the current of choice and IF the load drops 20 mV at that current then Q1 will assume whatever Vce is required to drop 20 mV across the load.

Original circuit. Not marvellous -

enter image description here

Much better are circuits similar to the one below from here.
I've modified this slightly but left their values in place as this is to give an idea only.
The system turns Q3 on until V_Rsense = Vref.
So I_constsnt-current = Vref/Rsense.
Vref can be divided from some input voltage as shown using a divide ratio to suit. . You can provide a variable voltage from a pot or a microcontroller etc if desired to vary the current. Note that this is a current sink with the load being supplied from some V+ of choice. Need not be the same V+ as Vcc = BPS+

enter image description here

https://www.google.co.nz/search?q=current+source&hl=en&safe=off&prmd=imvns&tbm=isch&tbo=u&source=univ&sa=X&ei=K2P6T_iZEOijiAfU4JjlBg&ved=0CGAQsAQ&biw=1536&bih=864

Similar but with MOSFET and uC drive


This is similar but uses the voltage across R1 to provide a high side current source. From here but he says it's copied from "The art of electronics". Note again that Vref is the HIGH side voltage across R1.
ie Icc = Vref / R2 = [V+ x R1/(R1+R3)] / R2

enter image description here


Some thoughts from Maxim from here.

enter image description here


Many related thoughts from all over

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  • \$\begingroup\$ Thanks for your answer. The thing that's confusing me is how to limit both voltage and current (I realize only one will be limited at a time). If I did something like this, I should be able to adjust the open circuit voltage drop between the voltage regulator I drew and the existing circuit. I can then short the load, and adjust the Vref you drew to get the right current. Then my limits will be set, right? \$\endgroup\$ – flndr Jul 9 '12 at 23:57
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No, the transistor circuit won't work at 20 mV, since it needs at least 0.7 V to overcome the base-emitter junction's voltage drop. It's unlikely you'll find any kind of circuit that will work at this low voltage.

But all is not lost. A current limiter may well work with a higher input voltage. It will be the load that determines the output voltage, per Ohm's Law: voltage = current x resistance. So even with a 5 V supply and a 0.2 Ω load you'll still have only 20 mV across it if you limit the current to 100 mA. So use a supply voltage of a couple of volts with the transistor current limiter and everything is peachy.

Keep an eye on the transistor's dissipation. At 5 V in and 100 mA that's 500 mW and that's often the limit for a general purpose transistor. The BC337's Absolute Maximum Rating is 625 mW, for instance. So a 3 V supply is safer than a 5 V one.

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It is fairly easy to 'program' an 7805 voltage regulator to act like a current source. Look at the image below that I found here:

enter image description here

The voltage across R is known, it is 5V. Therefore if you want a 100mA current R should equal 5V / 100mA = 50Ω. Beware that the resistor dissipates 5V × 100mA = 0.5W so a 1W resistor isn't a bad idea.

Not sure what your intention is with the 20mV requirement. Input voltage should be at least 5V (regulator output) + 3V (required across the regulator) = 8V above your maximum required output voltage.

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  • \$\begingroup\$ The 20mV/100mA are limits. When I short the output, I want to limit the current to 100mA. On an open circuit, the voltage should be "limited" to 20mV. \$\endgroup\$ – flndr Jul 9 '12 at 21:09
  • \$\begingroup\$ 600mV(ish) would be easy to implement with a diode at the output. \$\endgroup\$ – jippie Jul 10 '12 at 16:00

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