Questions About JFET 100mA Constant current source.

Discussion in 'General Electronics Chat' started by BrightDiode, Nov 1, 2012.

  1. BrightDiode

    Thread Starter New Member

    Nov 1, 2012
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    First post here, I've etched and built many circuits but I'm self taught and learn as I go, so please be gentle.

    My goal is to cheaply and simply use a 3.7V Li-Ion cell to power a 3.0Vf LED with 100mA constant current through the voltage range of the cell, 4.2V-2.8V.

    Here are a couple of the articles I have been reading:

    http://socrates.berkeley.edu/~phylab...Files/bsc4.pdf

    http://www.vishay.com/docs/70596/70596.pdf

    So I think I have the basic understanding down. The negative voltage developed over the source resistor is fed back to the gate to "close" the JFET more as current increases, as current decreases the voltage will rise closer to zero, which will open the JFET to more current flow.

    I have found some JFETs with an Idd of 200mA:
    http://www.fairchildsemi.com/ds/J1/J105.pdf

    Using the formula provided for basic source biasing on the first page of the vishay document:

    -4V Gs(off)
    0.1A Id
    0.2A Idd
    2.0K

    I end up with 11.7 ohms for the source resistor.

    This however, doesn't make sense to me, since no current flows through the gate of the JFET, it all must go through the 11.7 ohm resistor, but if I calculate what resistor I'd need to run the LED with no JFET I end up with:

    3.7V supply, 3Vf drop, and 100mA = 7ohms.

    So I can't be using more resistance on the JFET circuit, doesn't make sense to me, where am I misunderstanding? Is my battery voltage just not high enough to make this work properly? If that's the case should I look into cascading Jfets or running two with a lower Gs(off) in parallel?

    Or maybe there's a much more simple way to achieve a 100mA current source from a 3.7V Li-Ion cell?

    Thanks a bunch for any help offered. I'm ordering some JFETs to play around with but any insight is much appreciated.
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You're going to need a buck-boost regulator, like this. There are others out there.
    Caveat: I haven't used this one, or any switching regulator, for that matter.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    Obviously you can't operate a 3V LED from a 2.8V battery without boosting the voltage. That's why Ron suggested a buck-boost converter, which is what you would need.
     
  4. BrightDiode

    Thread Starter New Member

    Nov 1, 2012
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    Sorry, I should have been more specific, I'm not worried about the 0.2V, I just want most of the range of the cell, 4.2V-3V is fine, or even 3.2V say, to make up for any losses.

    This needs to be cheap and simple, a buck-boost regulator is far too expensive and complex for this project, I will be making about 20 pieces.

    Anybody used a JFET for 100mA current source before at these voltages? This seems like the most simple and effective way to do this, if it will work.

    A low voltage drop linear current regulator like the AMC7135 would be perfect, except it's 350mA, and I can't just run 3 LEDs in parallel from one, because they will all be separate pieces.

    http://pdf1.alldatasheet.com/datasheet-pdf/view/202788/ADDTEK/AMC7135.html
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You are not going to be able to make a JFET current source with the low voltage compliance that you need.See the first two plots under Typical Characteristics in the Fairchild datasheet. Keep in mind they are typical, and subject to wide tolerances. Notice that Vds has to be a couple of volts before the current is reasonably constant, even with Vgs(off)=-0.7V (second plot). In order to get 100mA, you will need J105 or J106, which limits you to the first plot on the left side. Notice that the Vds has to be many volts before the current is reasonably constant. You need 0V<Vds<1.2V (approximately). It just won't work.
    I have a circuit that will work, but it requires an op amp, a MOSFET, a diode, and 4 resistors.
     
  6. BrightDiode

    Thread Starter New Member

    Nov 1, 2012
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    Thanks for your help Ron H.

    I'd definitely be interested in the circuit you have in mind that will work.
     
  7. Ron H

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    Can you work with surface mount devices? I'm almost certain that the op amp is only available in a tiny package. Still searching, though.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    Don't keep the poor guy (and the rest of us) in suspense, Ron. Post your circuit. ;)
     
  9. Ron H

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    I'm still looking for a better op amp, both for performance and for package choices. This configuration seems to work in simulation. If temperature stability is an issue, the diode can be changed to a 1.25V reference, with appropriate changes in the resistors around it.

    Obviously, it's the standard feedback current sink circuit, optimized for low voltage operation.
     
  10. BrightDiode

    Thread Starter New Member

    Nov 1, 2012
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    Thanks Ron!

    I can definitely work with small SMT stuff. I'm going to order these parts today and I'll let you know how it turns out.

    Edit: I'm just adding up the parts now and I didn't realize but 200M resistors are almost a dollar each! Possible to use something more like 10M for R1?
     
    Last edited: Nov 5, 2012
  11. Audioguru

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    Dec 20, 2007
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    The circuit uses a 200m resistor that is 200 milli-ohms (0.2 ohm). A 0.22 ohm resistor is more common.
    200M is 200 million ohms and is completely different and might not be available.
     
  12. Ron H

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    What Audioguru said.:)
     
  13. tubeguy

    Well-Known Member

    Nov 3, 2012
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    I think R1 is a 200 milliohm or 0.2 ohm resistor. (small m)
     
  14. Ron H

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    It's unanimous! (Except for the OP.):D
     
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