# Simple bidirectional current limiter?

Discussion in 'General Electronics Chat' started by RichardO, Mar 30, 2016.

1. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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I need a current limiter that works up to about 20 volts at the input and limits to a few milliamps.

I have simulated a circuit using a J-FET and a couple of diodes and it seems to work good enough. However, I have this feeling that I have overlooked something. What do you think?

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2. ### ian field Distinguished Member

Oct 27, 2012
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The usual method is to short gate and source, but JFETs are notorious for large parameter spreads.

Grounding the gate directly and grounding the source through a current setting resistor gives more repeatable results.

3. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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I need the diodes to get the bidirectional current limiting so I can't directly short the gate and source. The diodes do increase the limiting current, however.

I don't have a feel for how much variation I will see from device to device. I am guessing about 2 to 1 since transconductance varies that much. Vgs(off) varies about 10:1 but I don't think that matters here.

I could put the 100 ohm resistor in the source circuit to help stabilize the current. Unfortunately, the voltage across the 100 ohm resistor is small -- less than 100 mV. I don't think that small a voltage will have much effect on the current.

4. ### crutschow Expert

Mar 14, 2008
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Note that, according to the data sheet for the 2N5484, the zero gate-bias drain current can vary from 1mA to 5mA.
The 2N5485 varies from 4mA to 10mA.
Are these wide variations acceptable for your application.

Edit: Note that your circuit depends upon the two diodes having identical leakage characteristics (which the simulated diodes do, of course) but real diodes may have enough difference in them to change the forward and reverse limiting current symmetry.
Something to be aware of.

Last edited: Mar 30, 2016
5. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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The absolute current is not critical. It has to be over 1 mill1amp and less than a few milliamps. I may have to calibrate the circuit's gain. The high-end current of the 2N5485 is probably too much -- especially when the diodes are added to the circuit.

6. ### ian field Distinguished Member

Oct 27, 2012
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I've never seen your original configuration anywhere else, and I doubt it will do what you claim.

There are bidirectional current sources on the web, but all use more than one active device. Search under medical electronics.

As a quick cheat; you can enclose the standard single JFET current limiter inside a bridge rectifier - but you knock 2x Vf off your voltage headroom.

7. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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But it works great in simulation. Seriously, I agree with you but I am not sure why it won't work. I don't have time to try it with a real FET right now.

Good idea. Thanks.

Unfortunately, my signal is only 100 mV. Severely limits my options -- that and my cheapness.

8. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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I tried a few old 2N3819 J-FET's in the circuit and got the following results:
4.3, 4.3, 7.8 and 11.7 ma. The positive and negative currents were nearly the same.

The circuit has a slow risetime but that is not an issue in my case -- tens to hundreds of microseconds, I think. I did not bother measuring it.
( I keep calling the circuit a current limitter but the goal is really to protect the 100 ohm resistor from too much power dissipation.)

9. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Then can't you simply omit the FET and connect another resistor in series with the 100Ω?

10. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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Unfortunately, no. This is an input stage for measuring small resistances (<150 ohms).

The requirements are:
Maximum voltage across the 100 ohm resistor less than 100 mV.
Current through the 100 ohm resistor less than 1 mA.
Voltage across 100 ohm resistor in parallel with 150 ohms greater than 50 mV.

11. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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I was thinking the same thing so I shunted one diode and then the other with a high value resistor and I did not see any significant change in limiting currents.

Thanks to your note, I will redo the test in simulation to make sure I did not delude myself. If it works in sim., I will try the same experiment with real parts.

I have now tried a six 2N3819 J-FET's in the circuit and got the following results:
4.3, 4.3, 7.5, 7.8, 11.7 and 14.9 ma.
The positive and negative currents are nearly the same. I have not tried different diodes or added "leakage" resistors in the tests.

Edit: I don't think the diodes need to be matched for leakage. One diode is always on while the other is off. I think what is important is that the diode leakage is greater than the gate leakage of the FET. I am guessing that this is always the case but I need to look at diode and FET spec's in detail to be sure.

12. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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I earlier said that the signal is less than 100 mV so the source resistor will have little effect. Oops. I forgot that the fault condition is much higher current so putting the resistor in the source of FET _will_ reduce the limiting current. Of course, there have to be 2 resistors -- one in the source and one in the drain -- to have the currents of both polarities match.

I measured the rounding (not rise and fall times) of the corners of the square wave of the 2N3819 test circuit.
There is about 1us rounding on the positive corner and 0.5us on the negative corner. These times are _not_ seen in the simulation. :-(

I also added another current limiter to the simulation that uses two J-FET's. I would expect this circuit to be more trustworthy in the real world than the 2-diode version. The trade off is higher cost and larger RDSon of the two FET's in series.

13. ### ian field Distinguished Member

Oct 27, 2012
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Some JFETs are symmetrical - many aren't.

Shottky barrier diodes might work better in your circuit.

14. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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All the FET's (such as the 2N3819) I have happened to play with are symmetrical.

I wonder, if one manufacturer's part is symmetrical will then all manufacturers of that part number will be symmetrical? Probably, but a dangerous assumption. This is making me tend toward the two-FET circuit.

I had a similar thought about Schottky diodes. In simulation, I could not see any difference. One advantage of Schottky diodes is that they are a lot leakier than silicon diodes. See my comment above about the diode leakage needing to be greater the the FET gate leakage.

15. ### ian field Distinguished Member

Oct 27, 2012
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If you want leaky................seek out some NOS germanium diodes.

SB diodes certainly have significant leakage, but it doesn't seem to be very consequential on the application of working voltage.

But then the voltage ratings start at only 20V - anything over 60V starts getting pricey.

16. ### RichardO Thread Starter Well-Known Member

May 4, 2013
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From the spec sheets:
1N914 peak reverse current at 20 volts is 25 nA max.
2N5484 gate reverse current at 20 volts is 1 nA max.

It looks like the silicon diode is leaky enough.
It does not look like there is anything to be gained by "exotic" germanium or Schottky diodes in this circuit.