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I have two IC chips. On one chip I have one 10K ohm pull-up resistor and on another chip I have three 10K ohm pull-up resistors.

At first I was using four through hole pull-up resistors. Then I moved to an array of SMD resistor.

Could I simplify the design by attaching all four pins of the ICs to the same pull up resistor? Should this be a 40K ohm resistor? Is this OK or is it better to attach each pin to a different resistor?

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  • \$\begingroup\$ It depends on the ICs and the functions of the pins, which you propose to attach together. At least, post the links to the IC datasheets and identify the pins you're writing about. Ideally, post the present version of the schematic with separate pull-ups. \$\endgroup\$ – Nick Alexeev Aug 3 '12 at 23:17
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    \$\begingroup\$ Schematics or it didn't happen. \$\endgroup\$ – user3624 Aug 3 '12 at 23:28
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If these pullups are permanent (ie the only connection is from the input pin to the resistor) then there's no problem. If not, but all of the inputs should have simultaneously the same level, then it's OK too. I assume that's what you intended.

About the resistance, it theoretically should be lower, not higher, because when sharing the resistor the other input pins are acting as other resistors connecting that point to the ground, thus reducing the voltage at that input; but considering that today almost all chips are MOS with a very high input impedance, it makes no practical difference (for the number of pins you mentioned), so you can keep the same 10K ohm. If they were TTL, then maybe you should need to reduce it to make sure the voltage delivered is recognized as '1'.

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It depends on what the pins are doing, since connecting all the pins to the same 10k pullup means all the pins will directly be tied together.

This means that only one pin will be able to drive the line at a time, so if any of the pins are performing independent functions to any of the others, there will be contentions (i.e. one or more pins trying to drive the line at the same time)

If you have e.g. one pin as an output and all the rest as inputs that read from the output, such as in I2C (which uses "shared" pullups) then it would be okay. Another similar bus is a 1-wire bus, in which a number of pins share the same pullup.

If you give us the details of the ICs and pins you wish to share the pullup then a definite answer can be given.

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That depends. If the pins are just inputs you could connect them together and maintain the 10KΩ resistor, but if there is a chance that any of the pins would be an output you will better use an individual resistor per pin to avoid that some pin that eventually turns to be an output drives the others to an undesirable state.

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Alexis, if I can assume all these pins are not actively driven signals and instead just config pins, the limiting factor really is input leakage current vs input voltage thresholds. Each pin has a leakage current typically in the microamp (uA) range that increase over temperature. The maximum values can be found in the datasheet. If a pin draws 100uA into itself and across a 45kohm pullup resistor, you would have a 4.5V drop across that resistor! This is bad because even though you are pulling up to a HIGH voltage, the excessive voltage drop will be detected at the input as a LOW instead of a HIGH. For example, if you pulled up to 5V, the actual pin voltage due to a 100uA leakage would be 5V - (100uA*45kohm) = 0.5V and your chip wouldn't work as you designed. So, you need to look up both the min/max input voltage thresholds and also the maximum input leakage currents and check using ohm's law. The more pins in parallel the more leakage across the shared pullup resistor! This is why the pullup should be smaller as more pins are shared. This principle also applies to pulldown resistors since the leakage current would now be flowing out of the pin instead of into it.

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