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Current source with low voltage drop.

Discussion in 'Electronic Design' started by joble, Sep 30, 2003.

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  1. joble

    joble Guest

    Hi,

    I'm looking for a 15 mA current source. The application is place
    critical so a discrete circuit is no option. The voltage drop between
    in and out should be as small as possible. The input voltage is 20 to
    24 Vdc.

    I've tried an LM317 (in SOT23 package), but the voltage drop is to
    large.

    Does somaone know an alternative for the LM317, but with a lower
    voltage drop?

    Thanks!
     
  2. Jim Thompson

    Jim Thompson Guest

    Dream on!

    ...Jim Thompson
     
  3. John Fields

    John Fields Guest

     
  4. joble

    joble Guest

    OK,

    The load is 5 leds (about 3,5V forward voltage per led --> 17,5V).

    Maybe a FET current source could do the job. This is just one J-fet
    with a resistor between gate and source.

    Those things also exist in a 2 pin package. But i can't find one at 15
    mA. Does anyone know a manufacturer who makes those???


    Greetz
     

  5. +20-24V---string of leds----------------.
    |
    |
    |
    -------------------'
    | |
    | |
    | |
    | |
    | | low voltage opamp
    | |\| sot-23
    | .-------|+'
    ' | | >---.---.
    .-. | .-|-. | |
    | | | | |/| | |
    | | | | | | '
    '-' | '---|----' .-.
    ---------- | | |
    | | | |47 ohm
    V | '-'
    - | |
    . diode | .
    | | |
    GND -----------------------------------------------
    created by Andy´s ASCII-Circuit v1.21 Beta www.tech-chat.de
     
  6. Fred

    Fred Guest

    [/QUOTE]

    Put a few in parallel.

    Fred
     
  7. Current regulator "diodes" need a lot of voltage (maybe 5V) across
    them to work.

    This has only 4 parts (two resistors and two SOT-23-x) and will work
    down to around 1.3V. As a two terminal device it would drop maybe 1V
    more, still better than the LM317, but not by much.

    ...
    +24
    |
    | V
    | -
    .-. |
    | | V
    | | -
    '-' |
    | |
    | |/ MMBT3904
    +------------|
    | |>
    | / |
    ----- |
    / / \-------------+
    / \ |
    ----- |
    | |
    TLV431 | .-.
    | | |
    | | | 82R
    | '-'
    | |
    | |
    === ===
    GND GND

    You could also use an LM334 (plus a booster circuit) but I think there
    will be more parts and cost that way @ 15mA output.


    Best regards,
    Spehro Pefhany
     
  8. Ban

    Ban Guest

    cheap and only 2mA self consumption (with 1mA through the zener). For the
    reference voltage you can use a 7805L or so or a 6.8V zener with low tempco.
    The current mirrors 1mA to 17mA approximately. Transistors are BCV61(dual)
    or 2N3904, 2222A etc.
    o-----+
    +5V .-.
    reference 3k9
    Voltage | | 25Vmax/17mA
    '-' LED-string
    | |
    +---+ |0.5Vmin
    | | |
    \| | |/
    NPN |-+-| NPN
    <|1V |>
    | |.25V
    .-. .-.
    270R 15R
    | | | |
    '-' '-'
    | |
    +-------+
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    ciao Ban
     
  9. John Fields

    John Fields Guest

    ---
    First things first.

    If your source voltage varies from 20 to 24VDC and your load voltage
    needs to be 17.5V, then the voltage drop across whatever you use as a
    constant current regulator _must_ vary from 2.5V (with a source voltage
    of 20V) to 6.5V with a source voltage of 24V. Period.

    If you can guarantee that your source voltage will never fall below
    20.75 and your load voltage will never exceed 17.5V, then you could use
    an LM317L like this:

    lm117L
    +-------+
    VIN>----|IN OUT|--+--18.75V
    | GND | |
    +---+---+ |
    | [RS]
    | |
    +------+--17.5v
    |
    [RL]
    |
    GND>---------------+

    The output current is defined as

    Vref
    Io = ------ + Iadj
    Rs

    Since the Iadj term is small (100µA max) compared to the load current
    (15mA) we can neglect it and rearrange to solve for Rs:

    Vref 1.25V
    Rs = ------ = -------- = 83.3 ohms
    Io 0.015A

    The closest 5% value is 82 ohms, which would raise your output current
    to 15.2mA +/- 5%, so on the high side you'd be pushing 16mA through the
    LEDs.


    If you need a little higher compliance you could do this:


    +V +V +V
    | | |
    [620] | [LED]
    | | |
    +-----+ +------|+\ C
    | | | | >---[1K]---+-------B 2N4401
    [Z6.8][5.76K]| +---|-/LM358 | E
    | | | | | | |
    | +---+ +---------------------------+
    GND | | | |
    [1.02K] | [0.1µF] [68]
    | | | |
    GND GND GND GND
     
  10. Robert Baer

    Robert Baer Guest

    ....and there are dual NPN transistors in a SOT-23 package; even Zetek
    has a rather hi beta dual in the SOT-23, if i remember correctly.
     
  11. Keith Buck

    Keith Buck Guest

    Hi
    This problem is usually solved using switched mode technology.
    Regards keith buck.
     
  12. Robert Baer

    Robert Baer Guest

    Here is the Zetex (correct spelling) info..
    8-MSOP package:
    ZXT12N20DX 50V, beta(min/max) 300/900
    ZXT12N50DX 100V, beta(min/max) 300/900
    SM-8 package:
    ZDT690 45V, beta(min) 400
    ZDT651 60V, beta(min) 100
    There are a number of others in this package running from Vceo of 15V
    to 120V and betas from 75 to 2000.

    There is a tricky way to make a 2-terminal current source that will
    work down to about 650mV across it:
    R2
    + O--*---- ^------*--/\/\/--*-----------O -
    | \ / | |
    | ---- ---- |
    R1 -/\/\--|-------/ \-------|

    Set R2 so that the I*R2 is about 600mV, where I is the current you
    want.
    Using high beta transistors means R2 can be fairly large, making the
    I*R2 closer to what you want, and R2 making a less of a burden on the
    "current source" result. This can be useable from microamps to a few
    hundred milliamps (depending on dissipation).

    Another way is to use a MOSFET, tie the gate to "ground" (-) terminal,
    and use a current setting resistor from drain to same "ground" (-)
    terminal. This can be useable from nanoamps to at least an amp
    (depending on dissipation). Ther are a lot of MOSFETs in SOT23, SOT-223
    etc packages.

    Both schemes have a temperature dependence that you may or may not
    want; compensation can be done by using an appropiate thermistor for R2
    or the drain to "ground" resistor.

    BTW, "LDO" is used for *voltage* regulators, and is inappropiate and
    inapplicable for *current* regulators.
     
  13. Robert Baer

    Robert Baer Guest

    ********* correction on "schematic":
    R2
    + O--*---- ^------*--/\/\/--*-----------O -
    | \ / | |
    | ---- ---- |
    R1 -/\/\--|-------/ \v------|

    On the MOSFET, use of a logic version will reduce the needed drop
    across the resistor; "ordinary FETs (by IRC) need about 3.5V to produce
    10-100uA, and a logic FET needs about 1.5V for the same current.
    The Vgs versus current plot is an excellent log relationship ofer a
    wide current range, almost as good as the Vbe versus Ie plot.
     
  14. Ian Bell

    Ian Bell Guest

    What kind of LEDs are these. I never heard of ones with this big a
    forward volts drop.


    How critical is the 15mA? A 17.5V drop at 15mA implies an equivalent
    resistance of 1166 ohms. For 24V operation an additional series
    resistance of 430 ohms will achieve a 15mA current. When the volts
    falls to 20V the current falls to 20/(1166 +430) = 13.36mA.

    Ian
     
  15. That's a fairly typical Vf for blue and some other short-wavelength or
    white LEDs (which generally use a blue die).
    LEDs don't behave that way. To a first approximation the voltage drop
    across the string is fixed so long as they are forward biased. To a
    better approximation, it's like a fixed voltage source with a resistor
    in series- a fairly small valued resistor.

    For example, a Panasonic LNG992CFB blue LED has a Vf of 3.5V at 20mA
    and 3.25V at 10mA, so the series resistor is r = 0.25/0.01 = 25 ohms.
    The equivalent Vf is 3.25 - 0.01 * 25 = 3.0V.

    So, if you put a resistor in series with 24V you need R= (24
    -15)/0.015 - 125 = 475 ohms. At 20V, the current will be
    I = (20 - 15)/(475 + 125) ~= 8.3mA

    Best regards,
    Spehro Pefhany
     
  16. R.Legg

    R.Legg Guest

    Spe
    1N5283 - 1N5314

    xx Ir Vf
    83 .22
    84 .24
    85 .27 1v00
    86 .3
    87 .33
    88 .39
    89 .43
    90 .47 1v05
    91 .56
    92 .62
    93 .68
    94 .75
    95 .82 1v20
    96 .91
    97 1.0
    98 1.1
    99 1.2
    5300 1.3 1v45
    01 1.4
    02 1.5
    03 1.6
    04 1.8
    05 2.0 1v85
    06 2.2
    07 2.4
    08 2.7
    09 3.0
    10 3.3 2v35
    11 3.6
    12 3.9
    13 4.3
    14 4.7 2v9

    To make your own, the figures of interest are IDss and Vp.

    Look for JFet with high IDss and low Vp, for a self-biasing fet
    limiter with higher current and lower voltage headroom. The
    gate-source resistor determines current, below the IDss value, with
    ~Vp across it.

    Types like 2N4393 (150mA 3V)are a good bet.

    RL
     
  17. normanstrong

    normanstrong Guest

    You've managed to ask the question without providing any information
    except that it is a 15mA current source.

    1. It needs the lowest burden voltage possible; so what is the
    minimum burden voltage?
    2. What is the range of the output voltage?
    3. How accurate does the current have to be?
    4. What is the minimum output impedance of the current source.
    5. Does this have to be a 2-terminal source, or can it be 3-terminal?
    If 3-terminal, what is the maximum voltage available for the 3rd
    terminal?

    Give us some help here.

    Norm Strong
     
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