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Transistor Help.

Discussion in 'Electronic Basics' started by Andy Ladouceur, May 20, 2005.

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  1. Hi everyone,

    I took a basic electronics course a few years back, otherwise only know
    what I've found online, and am having one hell of a time trying to work
    out the right voltages / resistances for a project I'm working on.

    I have 2.4V @ 24mA coming off an output from an IC, that I'd like to
    use to power an LED. The LED needs 3.3V, 20mA.

    I'm able to vary the 24mA output from the IC in 16 1.5mA steps. (From
    1.5mA -> 24mA)

    I could be mistaken, but I'm quite sure what I'm looking for here is a
    NPN transistor. I just don't know what values of.. -anything-, I'd
    need. I had it calculated as:

    NPN Transistor, Beta 50.

    Power to collector: 12V, with a 420 Ohm resistor.
    Power to base: 2.4V
    Power to emitter: 0V (Ground), with an 80 Ohm resistor.

    This would give me 3.25V at 20mA coming out of the collector. (Er..
    right?) Which would work perfectly, if I didn't plan on varying the
    current. But, I'm not sure how to calculate what the current going to
    the LED would be when the base current is dropped to 1.5V. What I'd
    ultimately like is to be able to vary the current and have 1.5mA
    pretty much be 'off' for the LED, and 24mA at its brightest.

    With my current setup, is this how it would work, or am I right in
    assuming most of its current comes from the 12V collector?

    Could someone please advise me of a setup that'd let me do the above,
    if my current(sorry) one doesn't allow for it? Thanks in advance for
    any help you might provide. I'd like to be able to understand the
    workings of a transistor enough to be able to figure this out myself,
    but it's been two hours, and it seems that's not happening, right now.

  2. Rich Grise

    Rich Grise Guest

    Well, as you seem to have already found out, 2.4V isn't enough to power
    a LED that needs 3.3V.

    Exactly what IC is this? We need to know this before we go any further.

    The rest of the post shows only that you're getting yourself all tied
    in knots trying to do design principles that aren't even applicable
    Please, what IC is this? It's now sounding like an analog IC. Please,
    please, tell us the part number of the IC - you don't need anything
    anywhere near what you're using here.

    Can you make the output of your IC zero? Then that would be your "off,"
    and then you just drive a transistor as a switch.

    But first, we have to know exactly what this "2.4V @ 24mA...[variable]
    from 1.5 mA -> 24 mA" is coming from.

  3. Chris

    Chris Guest

    Hi, Andy. I'm not entirely sure you're interpreting the data sheet
    properly. But the data sheet isn't too clear, either.

    The data sheet for the MAX6956 isn't very clear on output pin voltage
    while sinking LED current, if you set it for current drive rather than
    voltage drive. It does mention on page 21 of the data sheet that the
    expected minimum voltage at an output pin set to sink current as an LED
    driver will be about 0.6V. If you've got 3.3V V(f) white LEDs, you
    might want to start thinking about using a 5V supply, and using series
    33 ohm resistors to limit total package power dissipation (if you're
    lighting up most or all 28 outputs with LEDs). A 33 ohm series
    resistor will drop .66V, which will add to the 3.3V with the LED to
    leave 1V remaining (20mW) dropped across the IC output. That would be
    560mW due to voltage drop if you're using all 28 port pins as LED
    drivers, which keeps you in safe limits for all packages at reasonable

    | |
    | White LED
    | |
    .----o-----. V~
    | | -~
    | | |
    | | o1.7V
    | | |
    | | .-.
    | |33 ohm| | |
    | | | | |
    | | '-' |
    | MAX6956 | | V
    | | | 20mA
    | | |
    | o-------o 1.0V
    | |
    | |
    (created by AACircuit v1.28.5 beta 02/06/05

    This is a good thing, because you can't get 66mW (3.3V * 20mA) out of
    57mW (2.4V * 24mA) anyway. Just won't work. TAANSTAAFL and all that.

    If this is confusing, or you're stuck with a supply less than 4V, look
    at Maxim app note 2197 for more information and how to interface the IC
    to transistors:

    If this doesn't clear things up, please provide further information
    with your response:

    * What's your power supply for the MAX6956? If it's less than 4V,
    would you be willing to look at a DC-DC converter to increase the power
    supply voltage to the IC?

    * How many LEDs are you driving? Anything else besides LEDs? The
    trick to using this chip successfully is total power dissipation.

    * Which IC do you have? You have to be more careful with the SSOP
    packages than the PDIP on power dissipation. The complete IC part
    number would have been helpful.

    I hope this has been of help. Again, feel free to respond, but with
    more information, please.

    Good luck with the programming
  4. Ban

    Ban Guest

    Andy, if you want to use the current adjustment possibility of your IC, it
    would be better to supply +5V to it, then you can control the LEDs without
    any external circuitry in 16 brightness steps. If the other ICs work on 3.3V
    you can use a simple level translator to fit to the 5V logic, but they can
    probably be interfaced just straight.
    In the Datasheet is written that output level is 0.7*Vcc, which translates
    to 4.7V to get 3.3V out. There are also tolerances of the LED forward
    voltage, so 5V will give some margin. With 20 LEDs full on simultaneously
    the dissipation in the IC will be almost 1W, so it will need a small
    glued-on heatsink and/or a large plain copper area on the PCB to radiate
    that heat. If only part of the LEDs are on or the current is set to lower
    values, you do not need this.
  5. Chris

    Chris Guest

    I had a chance to look at the data sheet for the MAX6956 again. It
    notes on page three that

    Port Drive Logic Sink Current, Port Configured as LED Driver
    V+ = 5.5V, VOUT = 0.6V at maximum sink current = 24mA typ.

    So the data sheet does guarantee a saturation voltage for the pin when
    configured as a current sink of 0.6V (in a backwards kind of way).

    Let's try putting 3.3V white LEDs as a load on all the pins of your
    MAX6956 (5VDC power supply) without series resistors. By the
    definition of a current sink, once the voltages across the load are
    taken care of, the rest of the voltage will be across the driver. That
    means each pin will have about 1.7V across it. Accounting for the
    power, that means 1.7V * .02A = .034 watts will be dissipated by the
    driver at each pin. If you've got 28 pins, that would mean 28 * .034
    Watts = .952 Watts will be generated by the outputs alone (this doesn't
    include power dissipation from the rest of the IC). The graph on page
    4 suggests the IC itself will take up about 10mA by itself with all
    outputs set as current sinks, all outputs on, and no load on any of the
    pins. That would mean another 5V * .01A = .05 Watts. So, worst case
    (all LEDs on), that would mean 1 watt of power would be dissipated by
    the chip.

    We're getting pretty close to the dreaded abs-max (absolute maximum
    ratings, usually listed on the top of the data sheet, do not exceed or
    you'll be sorry). It depends on your IC package. The DIP package is
    supposed to be rated for 1.667 watts, and I'm assuming that's the IC
    you've got here. But if you have an SSOP IC, you may already be over
    abs-max. Either way, that's probably an uncomfortable amount of power

    But all of this depends on the details. You could be using some of the
    pins as inputs. You could be using some as outputs, but not to drive
    LEDs. You might say that, except for a few moments for a reset, all of
    the LEDs will never be on. Any of these would change what's going on
    here, and affect the basic question of whether you need series
    resistors. The thing is, much of the advantage of the chip is based on
    not needing them.

    If you're driving this with a PIC, you might want to look into using an
    adjustable regulator to set the power supply voltage to around 4.2V or
    so. That's an oddball but perfectly permissible solution that will
    reduce the power dissipation of the chip by almost half.

    All of this is fairly simple hardware stuff. Your main issues will be
    the programming for that chip, which is not trivial, and the cost of
    the IC itself.

    Good luck
  6. Rich Grise

    Rich Grise Guest

    Thanks, Ban and Chris, this is pretty much what I was going to say, but
    you have done so much better than I could have. :)

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