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How much current can you draw all outputs driven from a 74HC595?

Discussion in 'Datasheets, Manuals or Parts' started by blueroomelectronics, Nov 9, 2007.

  1. blueroomelectronics

    blueroomelectronics Well-Known Member

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    The 74HC595 is a popular choice for driving LED signs, it can source 35ma per pin according to the Philips datasheet. But what is the maximum for the package?
    [​IMG]
     
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Bill,
    The datasheet says a maximum dissipation of 500mWatt
     
    Last edited: Jul 7, 2008
  3. Leftyretro

    Leftyretro New Member

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    Well looking over a few data sheets from different manufactures of this chip it seems like you might reconsider the use of this chip unless added drive help is given.

    The +/- 35 ma is a maxium DC spec where going above will damage the chip. Recommended DC drive is +/- 6ma.

    Also the max package total dissipation spec limit is 750mw. So 35ma x 5v = 175mw X 8 = 1.4 watts, well about the 750mw and that does not include power for the chip logic that it consumers. It would have a short if not dramatic life if driven will all outputs on ;)

    Lefty

    Edit: The 750mw rating is for the HC595A version I found in a motorola data sheet, Eric's 500mw is for the non A version.
     
    Last edited: Nov 9, 2007
  4. dave

    Dave New Member

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  5. mneary

    mneary New Member

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    The number you're looking for is in the data sheet: "DC current drain per VCC, GND". It's 75 mA. That's the total that the power/GND pins can carry without damage. Since all drive current passes through Vcc or GND, it then becomes the absolute maximum total drive current.

    As noted, this is a damage threshold and function isn't guaranteed.
     
  6. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The output transistor will have 5V across it only when it is shorted. Driving a resistor the output voltage with 35mA will probably be 1V for its saturation voltage loss. Then it dissipates only 35mW.
     
  7. blueroomelectronics

    blueroomelectronics Well-Known Member

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    I saw the 75ma and that's what I thought it meant. So the design is a poor one. Not sure what you mean audioguru.
    The 2003 darlington drivers have a fair voltage drop if I recall...
    So at 5V VDD, 1.2V per LED and 1.2V (assumed) for the 2003 and a 150ohm resistor I calculate 17.33ma per LED or 130ma which is way over the spec. Seems like a poor design.
     
  8. mneary

    mneary New Member

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    I agree it's poor, and I wouldn't manufacture it, but as an alphanumeric display on the hobby bench, it's not really very far over the edge. You don't want to ever try to light an entire block.

    Same problem on the 2003, whose saturation voltage rises past 1.6V at 500 mA (14 * 40 = 560 mA). The higher sat voltage would starve the segments down to 14 mA or so.

    And, of course, the AT90S2313 is obsolete.
     
  9. blueroomelectronics

    blueroomelectronics Well-Known Member

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    Never thought of that, the old Elektor design used 74LS164 with only 8ma max segment drive (sink 330ohm & PNP drive transistors)
     
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I think the 74HC595 will have an output saturation voltage loss of 0.7V. Red LEDs are 1.8V and the ULN3003 also has an output saturation voltage loss of 0.7V at this low current.
    So the remaining 1.8V across the 150 ohm current-limiting resistors produces an LED current of only 12mA.

    Aren't the LEDs multiplexed?
     
  11. blueroomelectronics

    blueroomelectronics Well-Known Member

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    The LEDs are multiplexed, and at 12ma per LED the 74HC595 is sourcing 96ma and the ULN2003 will need to sink 40x12ma 480ma.
     
  12. mneary

    mneary New Member

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    Good catch; I forgot all about the 74HC595 losses. And the ULN2003 would be driving from 1 up to 40 segments; its saturation voltage would be 0.7V for one, and more like 1.2V for all.

    So you get about 12mA for one segment alone, and 7mA each when they are all on. Lots of flicker.
     
  13. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Bill,
    Have a look at the TPIC6595 S/R.
     
    Last edited: Jul 7, 2008
  14. blueroomelectronics

    blueroomelectronics Well-Known Member

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    Nice part for the column drivers.
     
  15. Suraj143

    Suraj143 Active Member

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    Hi all in the above diagram I'm going to use a 47R instead of 150R.Is my value ok for a good brightness?
     
  16. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The resistors are supposed to be 150 ohms so that the 74HC595 ICs do not get too hot inside.

    For more brightness then the resistor value can be reduced if a betterdriver IC is used that can pass the additional current without getting too hot.
     
  17. Suraj143

    Suraj143 Active Member

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    Hello audioguru I have a problem.

    *My LED needs 20mA current to light up to its maximum brightness.
    *I multiplex 7 LEDs (7 rows) then the duty cycle will be 14.28%.
    *So I have to supply 7X current (100/14.28) to light my LED to maximum brightness.
    *That is 20mA X 7 = 140mA current I have to provide.

    The problem is ULN side it can sink this current.But 595 side it cannot supply this 140mA current because its rated only for 35mA max.

    So my question it is not worth reducing the resister either.Can you tell with this 35mA current for the 7 rows how will be the duty cycle?My figures show it will be 25% duty cycle.

    Give me a good answer .
     
    Last edited: Feb 25, 2010
  18. audioguru

    audioguru Well-Known Member Most Helpful Member

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    35mA/7= 5mA. The brightness will not be 1/7th because brightness is logarithmic. 10mA is only a little dimmer than 20mA and 5mA is only a little dimmer than 10mA.
     

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