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HIH-4000 sensor - weird output value

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nickagian

Member
Hi,

I have connected a HIH-4000 sensor on my breadboard:
  • Pin +ve to 5 V
  • Pin -ve to GND
  • Pin OUT over a 10k to GND
I then measure the OUT pin with a voltmeter and it shows 0.860V. According to the formula and the diagram in the datasheet, this corresponds to almost 0 %RH!!! Does that mean the sensor is dead or there is something else wrong?

Thanx
 

JimB

Super Moderator
Most Helpful Member
Pin OUT over a 10k to GND
Look at the connection diagram in the datasheet, it says 80k Ohm minimum load.
You have 10k, try increasing it to 100k.

JimB
 

nickagian

Member
Thanx for your reply.

Yes I'm aware of that. I just don't think the value is so important that it will not work at all... Perhaps it will not be stable, but showing 0 %RH is, according to my feeling, almost impossible. I will try though to find a bigger resistor to use.
 

ronsimpson

Well-Known Member
Most Helpful Member
I agree with JimB.
I think the data sheet wants a resistor larger than 80k. So 100k or 220k or even open.
I think the part will drive current in a 80k resistor but will not drive a 10k resistor because that is 8x too much.
Try no resistor.
 

Les Jones

Well-Known Member
Most Helpful Member
I interpreted the data sheet in the same way that nickagian did. Taking it to mean that it required a greater load than provided by an 80K resistor.

Les.
 

spec

Well-Known Member
Most Helpful Member
These sensors are sensitive to light as well as humidity- try shading the sensor from light. Although the load statement is confusing, 80K would be a minimum resistance load, so it would appear that as the others have said a 10K load would be way to low.

Incidentally the sensor output is ratiometric with supply voltage.

Is the sensor in an ambient where the RH is zero? Try breathing on the sensor and see if the output voltage changes.:)

spec
 

ronsimpson

Well-Known Member
Most Helpful Member
I interpreted the data sheet in the same way that nickagian did. Taking it to mean that it required a greater load than provided by an 80K resistor.
At first I also thought it wants less then 80k. So 10k should work. BUT 1k is less than 80k and 10 ohms is less than 80k and 0.1 ohms is less than 80k so it should also work.
We know 0.1 ohm will not work.
So it must be the other direction. 100k is greater than 80k
If this is true then an open circuit will work. Real simple to test.
 

Les Jones

Well-Known Member
Most Helpful Member
Good point Ron, I did not think how silly it was taking it literally if the resistor was taken to a very low value. It's nice to know that someone else (At least initially.) interpreted it the same way as me.

Les.
 

spec

Well-Known Member
Most Helpful Member
Good point Ron, I did not think how silly it was taking it literally if the resistor was taken to a very low value. It's nice to know that someone else (At least initially.) interpreted it the same way as me.

Les.
I did too Les.:D

spec

PS: the confusion arises because the data sheet is confusing
 
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