voltage regulator as constant current source

Discussion in 'General Electronics Chat' started by okin, May 9, 2010.

  1. okin

    Thread Starter New Member

    May 9, 2010
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    Hi everyone,
    I am trying to make constant 30mA current source. I wanted to use lm317 but it have 1.7V of dropout voltage. can you suggest another one with lower dropout voltage. Load is approximately 130ohm.
    Thank you for advice

    * I didn't say that maximum voltage for input pin is 5V. If I use Lm317 for input voltage I need ~8V. I found these voltage regulator http://www.farnell.com/datasheets/62273.pdf
    If I set current of 30mA with R1 then output voltage is ( 0.8*(1+R2/R1)=4.7V -- R1=130ohm - load ). Now I can use 5V for input pin or not?
     
    Last edited: May 9, 2010
  2. SgtWookie

    Expert

    Jul 17, 2007
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    When used as a current source or current sink, an LM317 has a minimum total dropout of 3v; the 1.7v for input to output differential, and 1.2v to 1.3v for Vref.

    Why don't you post a schematic of what you have, and what your available voltages are?

    [eta]
    Your 130 Ohm load will have 30mA flowing through it when the voltage across it is 3.9v.
    E=IR; I=30mA, R=130 Ohms.
     
  3. okin

    Thread Starter New Member

    May 9, 2010
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    Thank you for replay. I understand that for lm317 (Vin=1.7V+1.2V(Vref)+3.9V(load voltage)).

    Maximum voltage that I can connect to input pin Vin is 5V. Between Vout and ADJ I wanted to put resistor to choose current, and load is between ADJ and ground. Now I see that for my example ( http://www.farnell.com/datasheets/62273.pdf) resistor between ADJ and ground choose current.
    So my load in that example will be between Vout and ADJ.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Your math is off. The LM317 (and LM337 negative version) has 1.25V ± 0.05V output, plus the 1.8V used to power it. Wookie is giving you the straight scoop. I always use 3V as the insertion loss for this constant current source.

    The schematic I use looks more like this...

    [​IMG]

    It will handle 2V max on the output, 10ma minimum, 1.5A max. It can be a sink or a source.

    There are transistor versions that have less insertion loss than I am fond of.

    [​IMG]

    This assumes a regulated voltage input.
     
    Last edited: May 9, 2010
  5. SgtWookie

    Expert

    Jul 17, 2007
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    I'm afraid you're not going to be able to use an LM317 unless you have >7v available, unless you build some kind of DC-DC booster circuit.

    You could use a pair of transistors and a pot as a current mirror.
    The two transistors have to be thermally bonded to each other, or else it won't work properly. The voltage source needs to be constant.
    See the attached.
     
  6. okin

    Thread Starter New Member

    May 9, 2010
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    I didn't said that my schematic is current regulator. I said that I must have 30mA trough my load of 130ohm. So just on that your schematic choose R1=41.6ohm and put my load on right side of R1. Or I don't understand something good?
     
  7. okin

    Thread Starter New Member

    May 9, 2010
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    Thank you for your advice with transistor. I will try. And if I want to use boost DC-DC converter on what I must pay attention.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    Actually a current source is usually a current regulator. Possibly a bad choice of words, but that's what I read.

    There are low insertion voltage regulators. You can also use a transistor, though the specs are a bit rougher. The last schematic is a example of this.

    [​IMG][​IMG]
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    This will work with vcc as low as about 4.5V. Mouser has the op amp in a DIP for 34 cents. They have no minimum order, but you do have to pay shipping.
    As shown, you can use your 5V supply, if it is stable, to create the reference voltage.
    You could use a different op amp, as long as it is rail-to-rail I/O, and will work with a 5V supply.
     
    Last edited: May 10, 2010
  10. okin

    Thread Starter New Member

    May 9, 2010
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    Thank you for advice
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Ron_H correctly pointed out on a related thread that I forgot to include ballast resistors in the emitters. Without the ballast resistors, a small difference in Vbe (manufacturing tolerances) will result in a large difference in current between the two transistors.

    I've attached an updated schematic to correct the oversight.
     
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