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To avoid an address conflict, I need to connect an SHT31's address pin to high. The datasheet, doesn't specify if I need any resistors for this or if it has to go directly to VDD. I'm concerned about power consumption since it is a battery powered device. Is it ok to connect it directly and if not, what value of resistor must be used to obtain the high address while consuming little power?

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Since the datasheet does not specify any requirement, it can be directly connected to VDD.

So a resistor is not necessary. However, nothing prevents putting a resistor to pull the pin up. It does not make much difference, as long as the resistor has a reasonable value, like between 0 ohms and 100 kohms.

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  • \$\begingroup\$ Can I assume that no extra power will be consumed than if there was a resistor? \$\endgroup\$
    – mmvgm1
    Sep 5 at 11:47
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    \$\begingroup\$ The datasheet does not indicate anything about what kind of leakage currents would be. It is possible to use a resistor just as well, it does not matter much. If there is very little leakage current, there would be very little voltage drop over a resistor. \$\endgroup\$
    – Justme
    Sep 5 at 12:09
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Logic low is GND and logic high is VCC. No resistor needed.

If you would add a resistor (with a large value) to VCC, a small current would lower the voltage at the pin and it would no longer be at logic high.

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  • \$\begingroup\$ I understand there would be a drop due to the resistor, but I'm concerned of the current through the pin if I don't limit it with the resistor \$\endgroup\$
    – mmvgm1
    Sep 5 at 12:04
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    \$\begingroup\$ I think the rationale is, if there was a significant current, and you limited it with a resistor, then the pin wouldn't be high any more. Therefore it's very unlikely that the chip would be designed that way, especially after boasting its microamp-range idle current. \$\endgroup\$
    – DamienD
    Sep 5 at 12:10
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    \$\begingroup\$ @mmvgm1 The input impedance will probably be in the M\$\Omega\$-range. So the current will be very small already. Even adding some k\$\Omega\$-resistor will not change the current much. But a to large resistor will change your pulled-up pin to logic low. \$\endgroup\$
    – mais
    Sep 5 at 14:01
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Using a resistor to pull up the pin, will make the pull-up weaker (both adding the resistor and the need of longer traces) and the pin will be vulnerable to noise (for example if a antenna radiates nearby, it will disrupt the pull up, and it may cause the pin to read 0 instead of 1).

And yes, it will draw a little bit of current, but usually the pin is already some Mohms of input impedance, so a little current is drawn anyway.

So you have to trade between power consumption (if you do not add a resistor) and noise tolerance, higher price, less space (if you add a resistor)

If your IC has this option, pull down the pin instead of High. It will solve all the above issues.

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  • \$\begingroup\$ How would the use of resistor or no resistor will affect the power consumption - in any significant way that is? How would it have any effect if it is a pull-up or pull-down to solve any issues (assuming there are any issues)? \$\endgroup\$
    – Justme
    Sep 5 at 12:28
  • \$\begingroup\$ It will not change the power consumption in any significant way. \$\endgroup\$ Sep 5 at 12:50

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