What resistor size?

Elementary, Watson. What resistor, if any, should be used to protect a Honeywell HIH Series 4000 humidity sensor hooked up to 5V? The data sheet says max supply current is 0.5 mA at this voltage. If someone could tell me if one is required and, if so, how to figure it, It will be appreciated. Thanks.

Matt

Sensor data sheet:

V = I*R

10K

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Matt

I have a degree in electronics (but it was mostly vacuum tubes then!) and I use this calculator often in place of pencil and paper.

Great little calculatorâ€¦ Covers all sorts of topics.

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@ willgeorge - Neat little app. Thanks for sharing.

I believe thatâ€™s the maximum ammount of current it should draw under normal operation so you can have a basis for power consumption of the device.
I couldnâ€™t find any official in-circuit application note for it, but some google search shows that people use it connected straight to Vcc and GND, sometimes also adding low pass filter on its output(good practice).

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Reading the datasheet, I should say: just apply 5V without resitor and the sensor will draw 500ÂµA. No need for Ohms-law here .

Imagine the sensor needs 4.95V, that would be 0.05V over the resistor, which would limit the current to 5ÂµA

Makes sense. Thank you!

Thanks for the input.

The formula and the calcs give 10k reistance with 5V and 0.5mA. Simple enoughâ€“I assume thatâ€™s in a series w/ the sensor.

But is it needed or not? How does one know?
WouterH says no, just hook it up without resistor.
Farsa, is that what you are saying, too?â€“just hook it to VCC (which would be 3V for a PandaII so that might not be good).

WouterH, I donâ€™t understand about imagining the sensor uses 4.95V, the resistor .05V and the current 5 microamps. Why not imagine it uses 4.8V or 5.0V, or whaterver?

Yes, I agree with WouterH - donâ€™t use a resistor.

You should connect the sensor + to 5V and - to GND. Also, you must use a voltage divider on itâ€™s output because it can go higher than 3.3V, which is bad for the Panda ADC. Example:

OUTPUT
|
[11k]
|
|
[20k]
|
GND

F.e. take a 3V/5mA LED that you power from a 5V source: the voltage over the serie resistor will be 5V - 3V = 2V. So to calculate the serie resistor value you do 2V / 5mA = 400 Ohm.

In your example the sensor needs 5V and you will power it with 5V, which means that the voltage over the serie resistor would be 0V. And 0V / 0.5mA = 0 Ohm. And a wire is the closest to 0 Ohm you can get

Correct, but if the sensor output comes from a variating resistor, then a simple voltage divider will change the linearity of the output. If you see this, you can compensate it in software or add a R2R buffer amplifier before the divider. (or divide with the amplifier).

Thanks Wouter. So if the sensor is 5v, there is no need to worry about a resistor. And its corollary: there is no need to protect a sensor from too much current since it draws what it draws. I think i can grok that. For now.

No I think Wouter is saying that if the sensor needs to be powered by 5v, just power it by 5v because you already have that voltage available. If you have 5v and 3v3 available and you need to provide say 3v0, then you can use a resistor to give you that (remeber that a resistor is a heat producing device when doing this).

And youâ€™re dead right on the current for a sensor as it will draw what it needs. (thatâ€™s not the same for an LED though for anyone reading this)