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Need LED driving circuit. ( <15uA and >8mA)

Discussion in 'Electronic Design' started by Boki, Apr 21, 2006.

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  1. Boki

    Boki Guest

    Dear All,

    Here is the requirment.

    1. To drive a LED, the current should less than 25uA at turn off state.
    ( The supply voltage now is about 2.8 ~ 3.3V )

    2. At turn on state, the driving current should be larger than 7mA to
    avoid low beta of transistor, in this situation, the supply voltage is
    about 2.4~2.6V.

    The Vf of LED is 1.9 ( somebody's measurment result, no other
    informtion ..)

    Some of my other questions:

    1. Why beta will be changed?
    2. The Vf of LED is normal ?

    Best regards,
  2. Chris

    Chris Guest

    Hi, Boki. You seem to be impervious to advice here.

    Yesterday, Mr. Fields offered you a perfectly good one-chip solution
    for your problem. Looking at the post through Google Groups, the link
    was cut, but it's available if you click Show Options and click "Show
    Original". You can then cut&paste.

    Look at the LTC1440, as suggested by Mr. Fields. With one IC and two
    1% resistors, your problem is solved. Total supply current is less
    than 4uA over temp, plus whatever you need for the resistive voltage
    divider (should be less than 2uA). The device is optimized to source
    current, but will sink 8mA with an output voltage of less than 300mV.

    If you want to solve your problem, try Mr. Fields' solution. Go to the
    Linear website and look at the data sheet. Your circuit is the Typical
    Application on the bottom of page 1. Choose other 1% resistors to get
    the trigger voltage you need.

    Considering your skill level, you're not getting under 15uA off-stste
    with discrete transistors. Linear has already (re)invented the wheel.
    And since you've been casting around here, nobody's offered to do your
    job for you.

    Git R Done
  3. Boki

    Boki Guest

    Hi Chris,

    Chip solution is good, but my project is mass production.

    I think I have to consider discrete solution first..., am I right?


    Best regards,
  4. Chris

    Chris Guest

    If there's real money involved, and the extra dollar is that important,
    there are many good EEs who will be happy to consult with you about
    your problem. Subcontract one.

    It sounds like you're a Pioneer/Ranger campaign contributor who's
    lucked into a really good Homeland Security contract. I wish you well.

    Git R Done
  5. Mac

    Mac Guest

    What transistor?
    You can drive the LED with a current source. You can use a FET in series
    with the whole mess to keep the off-current basically zero.

    Here is a circuit that might work:

    Vcontrol-----+------+ |
    | | |
    | \ V LED
    | / R1 -
    | \ |
    | / |
    | | /
    | +------------|
    | | \>
    | \ |
    | / R2 \
    | \ / R3
    | / \
    | | /
    | | |
    | +--------------+
    | __| d
    | ||
    +-----------------|| N-channel
    ||__ s

    When Vcontrol is low, the FET turns off, and no current flows. When
    Vcontrol is high (around 2.5 V), the FET has a very low resistance (be
    sure to pick a FET which has a low Vth), and can be viewed as a short
    circuit. So now the voltage at the BJT's emitter will be one diode drop
    below the voltage at its base. So, R3 will have an essentially constant
    current through it, which means the LED will have an essentially constant
    current. Note that the BJT is NOT saturated in this application, so you
    can count on fairly high Beta.

    If you choose R1, R2, and R3 optimally, you can probably get a reasonably
    good circuit. Maybe 25 Ohms for R3, then pick R1 and R2 so that the base
    is at around 1 Volt. So, R2 = 1k, and R1 = ~1.5 k might work.

    I would try to keep Vcontrol/(R1+R2) > 10 * 7 mA / Beta_min if you know
    what I mean.

  6. Boki

    Boki Guest

    Mac 寫�:
    It is professional and excellent!

    Best regards,
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