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DS8867 info (found in the MSI/88e)

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  • erd_6502
    I got a response to a request for specs on the DS8867... DS8867 is an 18-pin DIP 8-segment constant-current driver, designed to be driven from MOS circuits
    Message 1 of 5 , Mar 5, 2003
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      I got a response to a request for specs on the DS8867...


      DS8867 is an 18-pin DIP 8-segment constant-current driver, designed to
      be driven from MOS circuits operating at 8V +/-10% minimum, and will
      supply 14mA typical current to an LED display. Output current is not
      sensitive to Vcc variations.

      Features:
      internal current control - no external resistors
      100% efficient - no standby power
      operates in 3- and 4-cell battery systems
      inputs and outputs grouped for easy PC placement

      Absolute maximum ratings:
      Vcc 7V
      input voltage 10V
      output voltage 10V

      Operating conditions:
      Vcc min 3.3V max 6.0V
      ambient temperature min 0deg C max 70deg C
      logical '1' input typ 4.9V max 5.4V
      logical '0' input typ 0.1uA max 10uA
      logical '1' output min -8mA typ -14mA max -18mA
      logical '0' output typ -0.5uA max -10uA

      Pinout is straightforward, pins 1..8 are inputs 1..8, pin 9 is ground,
      pins 10..17 are outputs 8..1 (ie the outputs are almost opposite the
      inputs), pin 18 is Vcc.
    • J.C.Wren
      I was always under the impression that a diode drop was worth 0.6 volts. --John ... From: Lee Hart [mailto:leeahart@earthlink.net] Sent: Wednesday, March 05,
      Message 2 of 5 , Mar 5, 2003
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        I was always under the impression that a diode drop was worth 0.6 volts.

        --John
        -----Original Message-----
        From: Lee Hart [mailto:leeahart@...]
        Sent: Wednesday, March 05, 2003 17:55
        To: cosmacelf@yahoogroups.com
        Subject: Re: [cosmacelf] DS8867 info (found in the MSI/88e)


        erd_6502 wrote:
        > I got a response to a request for specs on the DS8867...

        Great! I've got the full data sheet. Basically, it is 8 copies of this
        circuit (view with fixed width font):

        VCC (pin 18)
        |
        | Collector
        base |/
        input___/\/\/\________| NPN transistor
        R1 | |
        2.3k _|_ |\ Emitter
        _\_/_ |
        | > R2
        2 diodes _|_ > 50 ohms
        _\_/_ >
        |______|____output

        It clearly doesn't need 8v +/-10% on the inputs. If the input is being
        driven by a CMOS chip at 5v, the output will be roughly 2 diode drops
        less, or 3.8v. This is still plenty for an LED and darlington driver to
        ground (2v + 1.5v = 3.5v).
        --
        Lee A. Hart Ring the bells that still can ring
        814 8th Ave. N. Forget your perfect offering
        Sartell, MN 56377 USA There is a crack in everything
        leeahart_at_earthlink.net That's how the light gets in - Leonard Cohen


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        [Non-text portions of this message have been removed]
      • J.C.Wren
        D oh! I missed the word LESS. I though you meant it dropped BY 3.8V. Nurse! More coffee! --John ... From: J.C.Wren [mailto:jcwren@jcwren.com] Sent:
        Message 3 of 5 , Mar 5, 2003
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          D'oh! I missed the word LESS. I though you meant it dropped BY 3.8V.
          Nurse! More coffee!

          --John

          -----Original Message-----
          From: J.C.Wren [mailto:jcwren@...]
          Sent: Wednesday, March 05, 2003 16:12
          To: cosmacelf@yahoogroups.com
          Subject: RE: [cosmacelf] DS8867 info (found in the MSI/88e)


          I was always under the impression that a diode drop was worth 0.6 volts.

          --John
          -----Original Message-----
          From: Lee Hart [mailto:leeahart@...]
          Sent: Wednesday, March 05, 2003 17:55
          To: cosmacelf@yahoogroups.com
          Subject: Re: [cosmacelf] DS8867 info (found in the MSI/88e)


          erd_6502 wrote:
          > I got a response to a request for specs on the DS8867...

          Great! I've got the full data sheet. Basically, it is 8 copies of this
          circuit (view with fixed width font):

          VCC (pin 18)
          |
          | Collector
          base |/
          input___/\/\/\________| NPN transistor
          R1 | |
          2.3k _|_ |\ Emitter
          _\_/_ |
          | > R2
          2 diodes _|_ > 50 ohms
          _\_/_ >
          |______|____output

          It clearly doesn't need 8v +/-10% on the inputs. If the input is being
          driven by a CMOS chip at 5v, the output will be roughly 2 diode drops
          less, or 3.8v. This is still plenty for an LED and darlington driver to
          ground (2v + 1.5v = 3.5v).
          --
          Lee A. Hart Ring the bells that still can ring
          814 8th Ave. N. Forget your perfect offering
          Sartell, MN 56377 USA There is a crack in everything
          leeahart_at_earthlink.net That's how the light gets in - Leonard Cohen


          Yahoo! Groups Sponsor
          ADVERTISEMENT




          ========================================================
          Visit the COSMAC ELF website at http://www.cosmacelf.com, or view the
          Wiki/FAQ at http://1802.bitting.com/

          To unsubscribe from this group, send an email to:
          cosmacelf-unsubscribe@yahoogroups.com



          Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.


          [Non-text portions of this message have been removed]


          Yahoo! Groups Sponsor
          ADVERTISEMENT




          ========================================================
          Visit the COSMAC ELF website at http://www.cosmacelf.com, or view the
          Wiki/FAQ at http://1802.bitting.com/

          To unsubscribe from this group, send an email to:
          cosmacelf-unsubscribe@yahoogroups.com



          Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.


          [Non-text portions of this message have been removed]
        • Lee Hart
          ... Great! I ve got the full data sheet. Basically, it is 8 copies of this circuit (view with fixed width font): VCC (pin 18) ... base |/
          Message 4 of 5 , Mar 5, 2003
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            erd_6502 wrote:
            > I got a response to a request for specs on the DS8867...

            Great! I've got the full data sheet. Basically, it is 8 copies of this
            circuit (view with fixed width font):

            VCC (pin 18)
            |
            | Collector
            base |/
            input___/\/\/\________| NPN transistor
            R1 | |
            2.3k _|_ |\ Emitter
            _\_/_ |
            | > R2
            2 diodes _|_ > 50 ohms
            _\_/_ >
            |______|____output

            It clearly doesn't need 8v +/-10% on the inputs. If the input is being
            driven by a CMOS chip at 5v, the output will be roughly 2 diode drops
            less, or 3.8v. This is still plenty for an LED and darlington driver to
            ground (2v + 1.5v = 3.5v).
            --
            Lee A. Hart Ring the bells that still can ring
            814 8th Ave. N. Forget your perfect offering
            Sartell, MN 56377 USA There is a crack in everything
            leeahart_at_earthlink.net That's how the light gets in - Leonard Cohen
          • Lee Hart
            ... It is. There are 2 diodes in series, so the basic drop from input to output is 1.2 volts. The 2.3k resistor doesn t matter much because the transistor s
            Message 5 of 5 , Mar 5, 2003
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              J.C.Wren wrote:
              > I was always under the impression that a diode drop was worth 0.6
              > volts.

              It is. There are 2 diodes in series, so the basic drop from input to
              output is 1.2 volts. The 2.3k resistor doesn't matter much because the
              transistor's base current is so low.

              > Basically, it is 8 copies of this circuit (view with fixed width font):
              >
              > VCC (pin 18)
              > |
              > | Collector
              > base |/
              > input___/\/\/\________| NPN transistor
              > R1 | |
              > 2.3k _|_ |\ Emitter
              > _\_/_ |
              > | > R2
              > 2 diodes _|_ > 50 ohms
              > _\_/_ >
              > |______|____output
              >
              > It clearly doesn't need 8v +/-10% on the inputs. If the input is being
              > driven by a CMOS chip at 5v, the output will be roughly 2 diode drops
              > less, or 3.8v. This is still plenty for an LED and darlington driver to
              > ground (2v + 1.5v = 3.5v).
              --
              Lee A. Hart Ring the bells that still can ring
              814 8th Ave. N. Forget your perfect offering
              Sartell, MN 56377 USA There is a crack in everything
              leeahart_at_earthlink.net That's how the light gets in - Leonard Cohen
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