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voltage drop from 555 / 556

Discussion in 'Electronic Design' started by Lyle, Jun 3, 2005.

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  1. Lyle

    Lyle Guest

    I'm playing with designing an IR tracking system using IR LED's flashed from
    a 556. I want to run it off of a 9 v battery and was trying to calculate
    the right resistor for high output but I can't find specs for output voltage
    drop under load. I may just have to manually graph it out but by chance,
    does anyone know what kind of voltage drop below Vcc I can expect out of a
    555 or 556 at 100 - 150 mA max load?
    thanks
    Lyle
     
  2. Ban

    Ban Guest

    http://www.national.com/ds.cgi/LM/LM555.pdf
    page 2,3
    with 9V the droop will be typical around 1V with 100mA, but can be as high
    as 2V worst case. You can see the 4 lower diagrams on page 3 which refer to
    a supply voltage of 5V, 10V and 15V. These are the typical values, the max
    values are specd for 15V only on page2.
     
  3. If you read the data sheet, it says Voh for sourcing 100mA on a 15V
    supply, 13.3V typical, 12.75V minimum. At 200mA, 12.5V typical.

    So, one guess would be 1.7V drop at 100mA and 2.1V at 150mA. Assuming
    you're talking about the bipolar type. Be sure to bypass the battery
    well.

    If you use a transistor such as a 2N4401 wth 15mA of base drive, the
    drop will likely be less than a couple hundred mV.


    Best regards,
    Spehro Pefhany
     
  4. I assume he's talking about sourcing current, rather than sinking,
    since he mentions "voltage drop below Vcc", but I could be wroing
    (sic).


    Best regards,
    Spehro Pefhany
     
  5. MSC

    MSC Guest

    Build it, try it...

    Sometimes we get so bogged down in designing things on paper and in
    software that we forget that occasionally the best method is just to
    get the old breadboard out and try it!

    Mike
     
  6. Ban

    Ban Guest

    Well, the droop is pretty symmetrical sourcing or sinking, even if the
    graphics and the tables are not giving the same values.
     
  7. Ban

    Ban Guest

    Well, the droop is pretty symmetrical sourcing or sinking, even if the
    graphics and the tables are not giving the same values.
     
  8. Ban

    Ban Guest

    Well, the droop is pretty symmetrical sourcing or sinking, even if the
    graphics and the tables are not giving the same values.
     
  9. Ban

    Ban Guest

    Well, the droop is pretty symmetrical sourcing or sinking, even if the
    graphics and the tables are not giving the same values.
     
  10. Ban

    Ban Guest

    Well, the droop is pretty symmetrical sourcing or sinking, even if the
    graphics and the tables are not giving the same values.
     
  11. Typical values @25°C:

    Source Sink 2N4401 ZTX1051A
    NE555 NE555 Ib=Ic/10 Ib=Ic/100
    1mA 1.3V 0.015V 0.006V 0.05V
    10mA 1.4V 0.1V 0.005V 0.05V
    100mA 1.6V 1.5V 0.1V 0.05V
    200mA 2.5V 2.5V 0.15V 0.06V




    Best regards,
    Spehro Pefhany
     
  12. Lyle

    Lyle Guest

    thanks all! gives me a good place to start and I'll bread board it and test
    from there.
    thanks
    Lyle
     
  13. Lyle

    Lyle Guest

    LOL, yup, I just wanted to get a good starting idea as I'm trying to get as
    much out of the IR LED's as is reasonable but their voltage drop is about
    half that of a standard LED so its easy to fry them, very easy.
    Lyle
     
  14. Bob Monsen

    Bob Monsen Guest

    gesundheit!
     
  15. Lyle

    Lyle Guest

    I see it now, thanks for pointing it out. What do you mean bypass the
    battery well? I was just going to use the output from the 556 in this
    version until I need higher currents, do I need a transistor driver? (I'm
    just a hobbyist)
    Lyle
     
  16. The battery will have some internal resistance. It will get worse as
    the battery runs down. You should put a capacitor across the battery
    that will supply most of the current *during* your high-pulse with
    little droop. The battery can recharge the capacitor between pulses.

    For example, suppose you wanted to pulse at 150mA for 1msec and 200mV
    of drop was considered okay.

    You'd need C = 0.15A * 0.001 sec/0.2V = 750uF (use 1,000uF/10V).
    Put 0.1uF ceramic in parallel.


    Best regards,
    Spehro Pefhany
     
  17. Pooh Bear

    Pooh Bear Guest

    That's why you use a series resistor. To set the current. The variation in Vf
    for the diode ( or the margin in Voh for the 556 ) will make very little
    difference to the current.

    Best not to design right on the edge anyway.

    Graham
     
  18. Pooh Bear

    Pooh Bear Guest

    Put a big capacitor across it ! The current drawn by the IR led(s) will see the
    full battery impedance otherwise and the supply volts will sag.
    A transistor driven by the 556 would give a better saturation voltage. On a 9V
    supply that's not really a big issue. A bipolar transistor would also 'waste'
    the base current thus reducing battery life.

    Graham
     
  19. Fred Bloggs

    Fred Bloggs Guest

    Do you have the data sheet for IRLED you're using? Some of them can
    really take a wallop, go ohmic, and develop 5V drop. But you need the
    datasheet to determine maximum current, maximum permissible pulse width,
    and allowable average power, as well as heat sinking techniques using
    short leads to large traces/planes. 9V Type N are not particularly
    famous for high capacity, and in alkaline will spend most of their
    useful life in the 7.2-8.5V range, so that the resistor drop method is
    not the best way to drive the LED.
     
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