Does this circuit make sense?

Discussion in 'General Electronics Chat' started by Scalpel78, Nov 19, 2014.

  1. Scalpel78

    Thread Starter Member

    Aug 11, 2013
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    Hi, I'm working on a breakout board for the MC33926 motor driver.
    One of the output pins, SF (Statusflag), is an active-low which will go low when there is a problem with the driver.
    In the datasheet it says that the SF should be tied to VDD via a pull-up resistor.
    In addition, I want to add a red LED to indicate when there is a problem (when SF goes low).
    Here is a screenshot of that part of the circuit:
    [​IMG]

    This is probably really basic, but do I really need the left pull-up resistor? Will the Led and right resistor also function as a pull-up resistor?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Possibly not. With only the LED and series resistor, the pullup voltage at SF will likely only be somewhere around 5V minus the LED forward voltage and that may not be sufficient to register as a logic high by the /SF input.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Just use one 470 Ohm or 680 Ohm resistor and a 20mA Indicator LED. The second, parallel resistor is not required. Use a LED with a Vf of 1.8V or so...
     
  4. ian field

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    Oct 27, 2012
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    Assuming that's an open collector output - its just possible that any slight leakage could produce a faint residual glow in the LED.

    The circuit appears to be technically correct, but if there's no residual glow problem, you could save the cost of a resistor.

    If its for production - I'd leave the extra resistor in so future tolerance spreads don't cause embarrassment.
     
  5. MikeML

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    I'll bet that residual leakage spec is on Motor Driver data sheet, and I'll bet it is less than 10uA over the entire voltage and temperature range for the part. That will not cause a "faint residual glow".

    Ps Here is the data:

    clip.gif
     
    Last edited: Nov 19, 2014
  6. ian field

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    Just possible - which is a fair bit short of even likely.

    The circuit the OP posted is technically correct - and advisable on a production run.
     
  7. Scalpel78

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    Aug 11, 2013
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    Thanks for great replies.
    This is just for personal hobby, so I'll try removing the second resistor, and I'll use a red 0603 20mA LED with Vf 1.95V in series with a 680 ohm resistor.
     
  8. MikeML

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    SF is an output...
     
  9. Lestraveled

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    May 19, 2014
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    Wait a Minute. Not only is SF an output, but the "maximum permissible load current" is 0.5 ma. 500 micro-amps!! You need a buffer to drive that 20 ma. LED.
     
  10. MikeML

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    So use a Ultra-bright Red LED. They are more than bright enough with only 1/2mA (even in sunlight).
     
  11. crutschow

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    When I saw pull-up I was thinking logic input but of course, it makes more sense that it's an output. :oops:
     
    Last edited: Nov 20, 2014
  12. ian field

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    Just a suggestion - don't use an ultra-efficient LED, some types reach full brightness at only 2mA, even though you'd only get such a faint glow as only visible in darkness, if it was my project that would irritate me.
     
  13. Scalpel78

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    Aug 11, 2013
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    What would such a "buffer" look like? A transistor?
     
  14. MikeML

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    If you are so anal that you need to worry about the 5uA leakage through the LED with SF_ high, then add the 100K resistor across the LED as shown in the first circuit. I believe the first circuit will work fine as an indicator with or without the 100K. Try it both ways, and let us know.

    89a.gif


    If you think you need a higher current through the LED, then the second circuit will do it.


    89b.gif
     
    Last edited: Nov 21, 2014
  15. Scalpel78

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    Aug 11, 2013
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  16. MikeML

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    Try it on a breadboard or with some clip leads. I'll bet that it is plenty bright enough to determine the state of the logic pin at a glance...
     
    Last edited: Nov 21, 2014
  17. ian field

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    Genius! - that's exactly what the OP's first post said in the first place.
     
  18. MikeML

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    Not quite, his proposed solution was a resistor in parallel with the LED and its current-limiting resistor. Either method would supply the leakage current, although we can make the case that the way I did it, the resistor can be a higher value, so wastes less power.

    I went a tried a high-brightness 20mA Red LED at 5V through 100KΩ. That would be about 30uA, six times more than 5uA leakage you are worried about. Not even a detectable glow. However, with ~500uA through it (5V through 6.8K) , it is bright enough.
     
  19. ian field

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    High brightness and ultra efficient are 2 different types.
     
  20. Scalpel78

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    Aug 11, 2013
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