1
\$\begingroup\$

I am not very good at understanding current, so I was hoping someone could help calrify the difference between pulling up multiple digital IC pins to Vdd with either a single resistor OR one resistor per pin?

if you are connecting multiple pins to the same resistor, would you just use a higher value resistor? Ie. instead of 4 10k resistors connected to Vdd, pull-up everything with one 40k resistor? Or perhaps the other way around?

Example:

enter image description here

VERSUS

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ 40K would be too weak probably. Where is 40K coming from anyways? Parallel resistors results in a LOWER resistance, not higher. In any case, if these are CMOS pins you don't need to worry about current limiting at all so you would either use 10K shared to all pins, or 10K for each pin. If you did have to worry about current limiting (BJT) then you would use 10K per pin or 2.5K for all pins, but I think it is risky to share pullups with BJTs since the base-emitter voltage drops may not be the same and you end up with an imbalance of current in the base-emitter with the lowest voltage drop. \$\endgroup\$
    – DKNguyen
    Jan 25, 2020 at 1:28
  • \$\begingroup\$ But on the other hand, all the BJTs are on the same die which makes them very well matched so maybe it isn't an issue. \$\endgroup\$
    – DKNguyen
    Jan 25, 2020 at 1:30
  • \$\begingroup\$ Are you ok with with possibly shorting out local circuits? \$\endgroup\$
    – user105652
    Jan 25, 2020 at 4:36
  • \$\begingroup\$ @DKNguyen if it's bipolar TTL the inouts have internal pull-ups, if it's CMOS 100K is plenty. \$\endgroup\$ Jan 25, 2020 at 5:21
  • \$\begingroup\$ @Jasen I just know that I've had trouble with pull-ups higher than 10K in the past that were fixed when the pull-up was reduced to 10K or lower. And it still doesn't change the backwards OP's backwards logic of turning replacing four parallel 10K resistors with one 40K resistor. \$\endgroup\$
    – DKNguyen
    Jan 25, 2020 at 5:32

1 Answer 1

0
\$\begingroup\$

Assuming DC loading isn't a factor here (generally not with modern CMOS devices), one advantage of the first (individual resistors) is that it makes it easier to use one of the unused inputs just by connecting a wire/signal to that pin. If the second option were being used, you'd have to isolate the desired pin from the shared pullup trace first.

Another reason for individual resistors is that it makes routing easier on the PWB.

Many times you'll see a hybrid, where you might have one pull up resistor for a given IC. Or one resistor for a common signal, such as an enable for a bunch of differential line drivers.

\$\endgroup\$
2
  • \$\begingroup\$ well in my case, I just have these four interrupt input pins which I do not need, but I DO need to pull them all high so the IC functions normally. At least thats what is says in the datasheet. I would like to just use one resistor to save board space / reduce part count. What would you recommend? \$\endgroup\$
    – scottc11
    Jan 25, 2020 at 1:19
  • 1
    \$\begingroup\$ @scottc11 If I don't need them and know I will never need them, I just tie them directly to Vdd. The resistor is for older BJT structures where you had to limit the current into the pin but CMOS doesn't need that (with high impedance MOSFET gates and all). But if you think you may need to mess with the pins for debugging then pull them with individual resistors. That way you can at least tap into the connection to pull them low as required in debugging. If you just tie them all with one resistor it's not very useful for debugging since the state of one is the state of all. \$\endgroup\$
    – DKNguyen
    Jan 25, 2020 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.