ULN2003A - Where does pin 9 go?

Discussion in 'General Electronics Chat' started by NVergunst, Feb 13, 2007.

  1. NVergunst

    Thread Starter New Member

    Feb 13, 2007
    7
    0
    First off, hello. I have a question that I hope you geniouses can solve really fast!

    Im going to replace all the individual transistors in my circuit with a ULN2003A. But the chip confuses me. All the PDF's I found assume you know how to connect it, and just show a simple in/out driver on the schematic.

    My question lies with this: What is pin #9 used for? This site says "The common diode connection is the Cathode end of all internal diodes connected together. This is usually connected to your supply voltage. EG: 12 volts." Link: http://www.dontronics.com/hints.html

    In my application the load goes through the collector, and then ground is right after the emmiter. The base is switched via a PIC that puts out 5v. As seen here: http://www.kpsec.freeuk.com/trancirc.htm

    So I got that I connect the input pins (pins 1-7) to the PIC, and then the outputs (pins 10-16) are all connected to their individual loads, and then to +12v. The ground (pin8) is ground. But where does Pin 9 go? Does it go to ground, or the +12v? Are all the transistors using the ground (pin8) as their emmiters, or is pin9 the common emmiter, so in my application that would go to ground?

    Thanks in advance.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    The diodes are snubbers. Their purpose is to protect the darlintons from reversed voltage (inductive loads). Connect pin 9 to +12V.
     
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