Shop OBEX P1 Docs P2 Docs Learn Events
embarrassed again trying to use a darlington array — Parallax Forums

embarrassed again trying to use a darlington array

BadgerBadger Posts: 184
edited 2008-12-05 16:49 in General Discussion
Hello


I cant find or dont understand how to hook up a darlington array. I would like to know ware to hook the power up to it. I know pin 9 is ground and 10 is the common is common the general power ?? Can you all point me to a nuts and volts or a web site that explains this. I dont quit understand the data sheet. So I think I am asking to be spoon fed but I guess I am asking. So if someone would be willing to help me that would be great .. As I said I am embarrassed. I am still in the beginning stages of beginner electronics.


Badger

Comments

  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-12-04 23:38
    The "COM" pin isn't a positive power pin.· It's a common for the integral free-wheeling diodes.·

    The ULN2803 isn't so much like logic gates: it always sinks current, opening/closing the path to ground; it doesn't source voltage.·

    In the attached drawing are examples of circuits for resistive (non-inductive) and inductive loads.
    2803_2ex.JPG
    746 x 414 - 54K
  • Mike GreenMike Green Posts: 23,101
    edited 2008-12-04 23:39
    Have you looked at:

    1) The datasheet for the device you want to use

    2) Nuts and Volts Column #6
  • Peter JakackiPeter Jakacki Posts: 10,193
    edited 2008-12-04 23:42
    Your subject line has already answered your question. It is a "darlington" array, that is an array of darlington transistors on a common substrate. Where do you connect the power on a transistor if the emitter is grounded? You don't because the collector normally goes to the load which connects to the power.

    However, due to the fact that the array is fabricated on a common substrate and is not really in fact independent transistors there is one little thing you need to factor in depending upon what you are driving.
    If you are driving inductive loads such as relays etc then the spikes from these can couple back through the substrate and affect the other transistors or damage the device as a whole. I have never really looked into the failure mechanism but in the datasheet you will see a diode common which is in fact part of the process of integrating multiple transistors onto a common substrate. Typically you would connect the diode common to the power supply of the load and these diodes clamp the inductive "spikes" if they are connected to the power supply.

    I'm trying to keep this as simple as possible so just use them as a bunch of common-emitter, open-collector darlington transistors and always connect the diode common to the load supply to protect the device.

    *Peter*
  • BadgerBadger Posts: 184
    edited 2008-12-04 23:45
    hello

    Thanks PJ Allen the drawing Helped a lot. I was hoping it was that simple.


    Mike I read the data sheet from several sources and I did not quit understand it.



    Thanks for the info it really helped me. I am getting ready to build that circuit at the moment



    Thanks again

    Badger
  • BadgerBadger Posts: 184
    edited 2008-12-04 23:50
    Peter


    to address your post. If I understand you. You would hook the vss to the comm and of course the ground to the vdd. then the pins to the input pins on the array with the matching output pin to the LED's I am trying to use. PJ Allen showed me a pic of just using the pins from the stamp to the input pins of the darlington array and the output pins to the LED's. I hope this is correct


    Badger
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-12-04 23:55
    No, No, No!
    Pin 9 goes to stamp ground (Vss), like my drawing shows.
  • BadgerBadger Posts: 184
    edited 2008-12-05 00:02
    Hello


    opps did I get that the labels wrong again ???


    Badger
  • Peter JakackiPeter Jakacki Posts: 10,193
    edited 2008-12-05 01:08
    Yes, you have the labels wrong and I don't know how.

    Look at the attachment as it shows the detail for each driver and as you can see it is basically a darlington transistor with the needed resistors plus the diodes which are part of the substrate.

    GND goes to your common ground
    COMMON (not to be confused with common ground) if it is connected as in inductive loads MUST go to the load supply which is highest (in case you mix low voltage and high voltage loads)
    INPUT of course comes from the I/O pin
    OUTPUT connects to the negative side of the load with the positive side of the load connect to the required power supply

    *Peter*

    P.S. RTM (READ THE MANUAL or datasheet)
    ULN2803 transistor.png
    777 x 427 - 12K
  • BadgerBadger Posts: 184
    edited 2008-12-05 16:49
    Hello

    I just want to sincerely thank the all who answered my question on the darrlington array. You all were a big help I have it working now. I dont have and 560 ohm resisters but I do have enough 1K ohm resisters. It seems to work fine. Just a bit dimmer.

    Have a merry Christmas and a Joly New Year.


    Badger
Sign In or Register to comment.