This is a follow-up question to my previous one, What does Low/High mean on the connections of a chip?.

To create a HIGH connection, you often just connect the pin to VDD, which is often above VIH (and therefore sets it to HIGH). However, some people choose to add resistors to those pins, as shown below:

enter image description here

Credit: Wolfson

In this case, DVDD is 3.3V and the VIH is 0.7×DVDD. Looking at that, it seems that the resistors aren't necessary. What are these resistors for, and do I even need them?

  • 3
    \$\begingroup\$ It seems you're thinking the resistors are there to drop voltage - they're not. They're there for various reasons; to limit current into the pin (may help under fault/ESD conditions, sometimes saves power, sometimes stops bizarre bugs like back-feeding power from a battery backup etc...) and also to give designers/manufacturers the chance to remove/no-fit the resistor(s) for testing, bug fixing, version setting, board ID setting, etc. \$\endgroup\$ – John U Jul 8 '14 at 16:38
  • \$\begingroup\$ @JohnU: Those are possible reasons for series resistors, but not for the example shown. \$\endgroup\$ – Ben Voigt Jul 9 '14 at 2:20

These are called pull up resistors. These are used to ensure that inputs to logic systems settle at expected logic levels if external devices are disconnected or high-impedance is introduced. Usually used with open-collector or open drain outputs.


If you are positive you will never connect the input to anything but "high" and if the input can only be an input, there is little need for them on modern chips (antediluvian logic sometimes recommended a pullup rather than direct connection, but that's history).

If you might want to probe around and change the logic state, having the resistor rather than a copper trace that you'd have to cut makes sense.

If the chip can make it input an output under some conditions then having the resistor there limits the current that the chip will have to sink, and could save it from overheating or damage.


In the schematic you show, the connection isn't just to the pin, but also off-page to some other circuitry. (look where NON_AUDIO, TRANS_ERR etc. end up).

So the pin isn't meant to be high all the time, but U1 sometimes pulls those outputs low. The resistors are there because U1 has open-collector outputs, or floating outputs while in reset, or something similar.

RESETB is an input to U1, so it might be driven by an external signal. It's fairly common to leave reset lines unconnected, in which case the resistor pulls the chip out of reset. Often there will be a capacitor to ground, to allow the supply to settle a bit before the chip comes out of reset.


It is not clear what you want to do with the resistors. If you remove them and make direct connections with DVDD, you have just shorted all outputs and input! If you just remove them, the outputs and input will "float" (can't rely on what state they will be in). Without a pull up resistor, the input and output lines will experience either a 0 (ground), or ? (float), instead of either a 0 (ground) or 1 (> 2.5v). In addition, "floating" lines are more susceptible to "electrical noise." Your circuit could experience "random" resets.


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