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IC to NPN transistor

Discussion in 'Electronic Design' started by [email protected], Oct 22, 2006.

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  1. Guest

    Hey I need some help calculating some resister values, I cant seem to
    find the formula anywhere. I'm trying to trigger a NPN transistor to
    switch on a set of leds when my IC's output goes HIGH. Heres my values:

    NPN transistor:
    Typical Hfe - 200
    Max Vce - 30v
    Ic - 800mA
    Power dissipation - 1.8w

    IC:
    Lowoutput - 8mA, 0.35v
    Highoutput - 0.4mA, 3.5v
    Power Source - 5v

    Leds:
    4.5v x 6
    20mA x 6


    What I need to happen is when the output goes high is to turns on 6
    leds (parralle) operating at 20ma each from a 9 volts source. I already
    know I need a 39 ohm resistor for the leds, what I really need to know
    is how to calculate the resistor value I need between the IC and the
    transistor.
     
  2. Johnny Boy

    Johnny Boy Guest

    It looks like you might be pushing it a bit from the figures you've
    given. (6*20mA)[LEDs] / 200 [Hfe] = 0.6mA, but the IC can only supply 0.4mA
    @ 3.5V. What voltage will the IC's output drop to when sourcing 0.6mA?
    What is the IC and what is the transistor?
    .... Johnny
     
  3. This thread should be in sci.electronics.basics, I think. Adding it
    there, with followups set to .basics.

    Johnny, the transistor's beta is 200 only if it is kept out of
    saturation and even then only if not at too high currents, often. So
    the situation is probably still worse than just the 600uA versus 400uA
    would otherwise indicate.

    I think this kind of thing does call for two transistors, to relieve
    the load on the IC output.

    : +4.5V
    : |
    : |
    : \
    : / R2 +4.5V
    : \ 4.5k |
    : / |
    : | |<e Q2
    : +----------| PNP
    : | |\c
    : |/c Q1 |
    : IN---| NPN |
    : |>e +-----+-----+-----+-----+-----,
    :3.5V | | | | | | |
    :400uA | \ R \ R \ R \ R \ R \ R
    :max \ / 39 / 39 / 39 / 39 / 39 / 39
    : / R1 \ \ \ \ \ \
    : \ 1k | | | | | |
    : / v v v v v v LEDs
    : | - - - - - -
    : | | | | | | |
    : gnd gnd gnd gnd gnd gnd gnd

    But this would mean a PNP capable of the loading.

    Q1 gets close to saturation in this case because Q1's collector
    probably has to get dragged down some 800-900mV. That places it very
    close to the Q1 base voltage, when on. So if the 3.5V drive is a
    little higher up than 3.5V (and it might very well be) then the base
    drive current will rise a lot. So really this isn't so good, even
    though it really limits the drive current -- if the Q1 base voltage is
    kept low enough to keep Q1 out of saturation, but not so well
    otherwise.

    So perhaps a better circuit using another resistor is:

    : +4.5V
    : |
    : |
    : \
    : / R2 +4.5V
    : \ 15k |
    : / |
    : | R3 |<e Q2
    : +---/\/\---| PNP
    : | 1k |\c
    : R1 |/c Q1 |
    : IN---/\/\---| NPN |
    : 15k |>e +-----+-----+-----+-----+-----,
    :3.5V | | | | | | |
    :400uA | \ R \ R \ R \ R \ R \ R
    :max gnd / 39 / 39 / 39 / 39 / 39 / 39
    : \ \ \ \ \ \
    : | | | | | |
    : v v v v v v LEDs
    : - - - - - -
    : | | | | | |
    : gnd gnd gnd gnd gnd gnd

    That will load the IC output more, as Q1 is also saturated now, but it
    will deal with an input voltage that spans the likely range and the
    worst case current is about half the IC output max spec of 400uA.

    Jon
     
  4. Tim Wescott

    Tim Wescott Guest

    * Use the _minimum_ Hfe, if you want the circuit to be successful all the
    time. If I'm driving a transistor into saturation I'll calculate a
    base current assuming Hfe/10. So assume that your transistor
    type has a minimum Hfe of 100, that gives you an Hfe of 10.
    * With an Hfe of 10, and 120mA, you'll need 12mA into the base of the
    transistor -- that's bigger than what your microprocessor can supply,
    so you'll want a Darlington pair, or other current boost topology.
    With a Darlington, you'll need 120uA into the first transistor, but
    I'd go with everything your processor can deliver.
    * Oh -- you don't want one 39 ohm resistor -- you'll be much better off
    with one R per LED, so take R = 220. With one resistor you have a
    much greater chance that the LED's will be at significantly different
    brightnesses.

    So I'd do something like this:


    +9V
    ---
    |
    .-----------o------o------.
    | | | |
    | .-. .-. .-.
    | 220| | 220| | 220| |
    | | | | | | |
    | '-' '-' '-'
    | | | |
    | | | |
    | V -> V -> V ->
    ___ 1.3V |/ - - -
    o---|___|--o--| | | |
    10K | |> o------o------'
    | | |
    .-. | |/
    | | '-------o-|
    12K| | | |>
    '-' | |
    | .-. |
    | | | |
    | 1K| | |
    | '-' |
    | | |
    | | |
    === === ===
    GND GND GND
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/

    "Applied Control Theory for Embedded Systems" came out in April.
    See details at http://www.wescottdesign.com/actfes/actfes.html
     
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