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A bit confused about: ULN2803, open collector, and nPn?

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  • lcdpublishing
    Hi guys, While looking over the ULN2803 for my latest circuit, I got myself good and confused (as usual). After some digging, I think I have my layout good
    Message 1 of 4 , Aug 3, 2008
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      Hi guys,

      While looking over the ULN2803 for my latest circuit, I got myself
      good and confused (as usual). After some digging, I think I have my
      layout good and will certainly have to test things before making a
      PCB, however, this is a great chance for me to learn something at a
      deeper level.

      The ULN2803 uses nPn transistors (in pairs). It was (is) my
      understanding that you use an nPn transistor to switch the high side
      of a load. However, this chip is an open collector output (I think)
      which I interpret as "Switching to ground" and as such, this would be
      on the low side of the load. As this seems to be in "contradiction",
      I am not sure how to interpret this.

      The second area of that is a tiny bit confusing is the current to
      turn the output on. If I am reading the data sheet correct(which I
      probably ain't), I need 1ma at 2.7 volts to switch on (approx). I
      will be switching this with a Atmega2560. I would like to limit the
      current at that device to a safe margin below 10mA. Will I need a
      series resistor between the two? I am thinking that I won't because
      the ULN2803 should only draw 1mA of current to switch.

      Sorry for the questions like this....

      Chris
    • Leon
      ... From: lcdpublishing To: Sent: Sunday, August 03, 2008 2:22 PM Subject: [Electronics_101] A
      Message 2 of 4 , Aug 3, 2008
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        ----- Original Message -----
        From: "lcdpublishing" <lcdpublishing@...>
        To: <Electronics_101@yahoogroups.com>
        Sent: Sunday, August 03, 2008 2:22 PM
        Subject: [Electronics_101] A bit confused about: ULN2803, open collector,
        and nPn?


        > Hi guys,
        >
        > While looking over the ULN2803 for my latest circuit, I got myself
        > good and confused (as usual). After some digging, I think I have my
        > layout good and will certainly have to test things before making a
        > PCB, however, this is a great chance for me to learn something at a
        > deeper level.
        >
        > The ULN2803 uses nPn transistors (in pairs). It was (is) my
        > understanding that you use an nPn transistor to switch the high side
        > of a load. However, this chip is an open collector output (I think)
        > which I interpret as "Switching to ground" and as such, this would be
        > on the low side of the load. As this seems to be in "contradiction",
        > I am not sure how to interpret this.

        "High-side" means that the switch is connected to the positive supply and
        sources current to the load.

        Leon
        --
        Leon Heller
        Amateur radio call-sign G1HSM
        Yaesu FT-817ND and FT-857D transceivers
        Suzuki SV1000S motorcycle
        leon355@...
        http://www.geocities.com/leon_heller
      • John Popelish
        lcdpublishing wrote: (snip) ... One of those contradictory ideas is wrong. ;-) NPN transistors are most easily used to switch the negative aide of loads since
        Message 3 of 4 , Aug 3, 2008
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          lcdpublishing wrote:
          (snip)
          > The ULN2803 uses nPn transistors (in pairs). It was (is) my
          > understanding that you use an nPn transistor to switch the high side
          > of a load. However, this chip is an open collector output (I think)
          > which I interpret as "Switching to ground" and as such, this would be
          > on the low side of the load. As this seems to be in "contradiction",
          > I am not sure how to interpret this.

          One of those contradictory ideas is wrong. ;-)

          NPN transistors are most easily used to switch the negative
          aide of loads since the emitter is common to both the input
          and output currents and must be the most negative terminal
          of the 3 when the transistor is on.

          > The second area of that is a tiny bit confusing is the current to
          > turn the output on. If I am reading the data sheet correct(which I
          > probably ain't), I need 1ma at 2.7 volts to switch on (approx). I
          > will be switching this with a Atmega2560. I would like to limit the
          > current at that device to a safe margin below 10mA. Will I need a
          > series resistor between the two? I am thinking that I won't because
          > the ULN2803 should only draw 1mA of current to switch.

          Take a look at the internal schematic on the data sheet:
          http://www.ottomat.hu/linkek/ULN2803-D.pdf
          (bottom of page 4. There is already an internal 2.7k input
          series resistor for each stage. The graph above the
          schematic shows the input current versus input voltage.

          --
          Regards,

          John Popelish
        • rtstofer
          ... I usually think of PNP transistors as switching the high side with the emitter connected to VCC and pulling the base down to turn on the transistor.
          Message 4 of 4 , Aug 3, 2008
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            --- In Electronics_101@yahoogroups.com, "lcdpublishing"
            <lcdpublishing@...> wrote:
            >
            > The ULN2803 uses nPn transistors (in pairs). It was (is) my
            > understanding that you use an nPn transistor to switch the high side
            > of a load. However, this chip is an open collector output (I think)
            > which I interpret as "Switching to ground" and as such, this would be
            > on the low side of the load. As this seems to be in "contradiction",
            > I am not sure how to interpret this.

            I usually think of PNP transistors as switching the high side with the
            emitter connected to VCC and pulling the base down to turn on the
            transistor. Remember, PNP points in - the emitter arrow points inward.

            >
            > The second area of that is a tiny bit confusing is the current to
            > turn the output on. If I am reading the data sheet correct(which I
            > probably ain't), I need 1ma at 2.7 volts to switch on (approx). I
            > will be switching this with a Atmega2560. I would like to limit the
            > current at that device to a safe margin below 10mA. Will I need a
            > series resistor between the two? I am thinking that I won't because
            > the ULN2803 should only draw 1mA of current to switch.

            I spent a bunch of time wrangling with Ohm's law and coming up with
            conflicting results. THEN I read the 2d paragraph of the Description:
            "the ULN2803A has a 2.7kohm resistor for 5V TTL and CMOS". So, no,
            you don't need a series resistor.

            Just check that Vce goes to 1.1V when the device is on. That's about
            the lowest drop you can expect and anything higher means the device
            isn't fully on and is dissipating more heat than necessary.

            The problem with Darlington's is that Vce(sat) is so high. Your relay
            will never see more than 3.9V. That may be adequate.

            A simple transistor will have a Vce(sat) of about 0.2V. The gain
            won't be as high and a base resistor will be required but the Vce
            voltage drop will be a LOT less. You can get some pretty small SMD
            transistors.

            Richard
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