• State Resolved
  • Locked Locked
  • Replies 6 replies
  • Subscribers 19 subscribers
  • Views 512 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

CD40107BM NAND Gate

Justin Forrester
Expert 1015 points


Hello,

I am working with the CD40107BM NAND gate chip and I am curious if there is a maximum resistance that can be used on the output (and likewise, input) pull-up resistors, or what possible considerations must be made which determining the resistance to be used.  These lines will not be used for high-speed applications, if that makes a difference.

Thanks!

  • 0
    TI__Mastermind 46480 points

    There are several thing to consider.

    first this part is a push pull output so it does not need a pullup.

    If you use a pullup on the output it should not be stronger than the part can drive. Ioh/Iol

    Too strong of an pullup will increase your Vol wich may violate the next part in line's Vil.

    The input pullup will be determined by the part driving that input.

    So if the system does not need pullups dont use them. If it does then usually between 1k to 10Kwill work 

     

  • 0 in reply to Chris Cockrill
    Expert 1015 points

    The data sheet states that to obtain a solid '1' output (VDD) a pull-up resistor is required, otherwise it is essentially a 'Z' state.  I confirmed this in my prototype because it did not obtain the necessary voltages until I included the pull-ups.   I wish to obtain minimum current draw in this circuit, so ideally I would go with the maximum resistance possible on the pull-ups.  As such, having too strong of a pull-up is not an issue.   That said, what would be the downside to using a very large resistance, such as in the 1M ohm range?

  • 0 in reply to Justin Forrester
    TI__Mastermind 46480 points

    Sorry about that I must have looked at the wrong datasheet.

    The part does have open drain outputs. The resistors can be choosen acording to the drive you will need for the next device. the resistor will produce the high drive current.

    The device itself will produce the low drive.  1Mohm will only give you 5ua of high drive at 5V Vdd , so it depends on what you will be driving. A 1Mohm resistor will also make your riseng edge very slow. the next device in line may be sensitive to that.

  • 0 in reply to Chris Cockrill
    Expert 1015 points

    Would 5uA of current on high be sufficient to drive one of the inputs on these NAND gates? The output of one NAND is connected to the input of the other to form a desired unit of combinatorial logic for my circuit.  Again, this is not a high speed application, so the rise/fall times are not an issue assuming outputs can stabilize within a couple seconds.

  • 0 in reply to Justin Forrester
    TI__Mastermind 46480 points

    The number to look at here is Iin. It should have enough drive  overcome this. Iin on this part is 1ua. Unless you are trying to save energy (running off abattery) I think I would decrease the resistor a little. It should work at 5ua though.

  • 0 in reply to Chris Cockrill
    Expert 1015 points

    Yes, this will be operating from a battery.  Thanks for the help.  You have answered all of my questions.