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current limiting in phototransistors

Discussion in 'Electronic Basics' started by Jeff Lawlis, Aug 15, 2003.

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  1. Jeff Lawlis

    Jeff Lawlis Guest

    I am using 5 infrared phototransistors in parallel (Radio Shack part #
    276-142). The emitters have a rating of 5V reverse voltage and 1.7V
    max forward voltage. How can I calculate which resistance to use (in
    series with the battery I presume) to limit the current when using a
    9V battery? Do I wire the emitters so that the cathode is connected
    to the positive end of the battery? I realize that these are novice
    questions, but I would appreciate the help. (I am not using the
    detectors included with the emitters).

    Jeff
     
  2. You will be using the emitters in the forward direction (that is the
    only way that emit anything). You can wire all 5 in series and put a
    small resistor in series with that to limit the current to the desired
    amount, but the current will fall rapidly as the battery voltage
    sags. But this is definitely the most efficient way to drive them,
    because you get to use the same battery current to run all of them.
    The 1.7 volt spec may be a maximum value, instead of typical, so a
    resistor between 50 and 100 ohms may work okay.

    Are you at all interested in battery life, or is maximum output your
    only concern? If life is no concern, you could drive them separately
    or in smaller series groups with larger resistances in series with
    each (or each group.) For instance, you might put a pair in series
    with a 560 ohm and a triplet in series with 390 ohms to drive all 5
    with about .01 amps (10 milliamps).
     
  3. (snip)

    The cathode of the emitter is the more negative terminal. You add up
    all the forward voltages (1.7 volts, each, in this case) of the
    devices you will operate in series, and subtract that total from the
    available voltage (9 volts, in this case). Then you divide that
    excess voltage by the current you want through the while series string
    to solve for the resistance needed to limit the current ot that amount
    while it burns up the excess voltage.

    For instance, to drive a single emitter, the excess voltage is
    9-1.7=7.3 volts. If you want ot drive the emitter with .01 amp (a
    conservative amount for many emitters), you divide that voltage by .01
    to get a resistance of 730 ohms. Standard 5% values near that are 750
    ohms and 680. If you drive all 5 emitters this way, the total battery
    drain will be 5 times .01 amp or 50 milliamps.
     
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