1
\$\begingroup\$

I'm trying to use a HIH4030 humidity sensor, where the output value is provided to a XBee.

Considering that output sensor max output voltage is around 4V and the XBee ADC only supports 1,2V, I took some time to choose the correct resistors for a voltage divider to get a decent ratio. I end up choosing a 120 ohm and a 270 ohm resistors to get a ratio of 4V to 1V.

And in this case I wasn't getting any values on my XBee. When I measured the signal, I had about 31mV going out the voltage devider when I should have around 800mV.

Now, correct me if I'm wrong (and there's a good chance that's the case), but the reason for these low values are due to the low total resistance value (R1+R2 = 390ohm) and the low current value going out the sensor. The HIH4030 specs say that the current supply varies from 200uA to 500uA, so I'm assuming that from this sensor I won't get higher currents than 500uA. Is this a correct assumption?

Anyway, taking this assumption as the correct one, through ohms law,

V=RI. V=390*0.0005, thus V=195mV which is hardly enought.

I was considering getting higher resistance values, at least to a total of R=1,2/0.0005.

Am I proceeding correctly?

\$\endgroup\$
2
\$\begingroup\$

Yes, you're on the right track. The values in your voltage divider should be driven mainly by the characteristics of the ADC on the XBee.

Typically, the ADC on a microcontroller requires a source impedance no higher than about 10 kΩ. You could use 27 kΩ and 12 kΩ in your divider; the resulting source impedance would be about 8.3 kΩ, well within spec.

The load on the sensor would be 39 kΩ, drawing just over 100 µA at 4V.

\$\endgroup\$
  • 3
    \$\begingroup\$ It's the parallel combination of the two divider resistors: $$\frac{1}{1/12k + 1/27k} = 8.3k$$ \$\endgroup\$ – Dave Tweed May 1 '14 at 1:05
  • 1
    \$\begingroup\$ This probably is dumb question, but... why the parallel combination? Aren't the resistors on voltage divider in series? \$\endgroup\$ – cvicente May 1 '14 at 8:21
  • 2
    \$\begingroup\$ They are in series from the point of view of the sensor, which is why it sees a 39k load. However, as long as we stay within its current limits, we can treat the sensor as a voltage source, which means that its output impedance is low. This means that the effective source impedance "seen" by the ADC input is the two resistors in parallel. \$\endgroup\$ – Dave Tweed May 1 '14 at 11:17
  • 1
    \$\begingroup\$ Don't confuse the input impedance of the XBee (the load that it presents to the source) with the souce impedance that it can tolerate while still meeting its perforance specifications; these are two different numbers. Note that the input impedance shifts from 10M to 1M when taking a sample, a delta of 9M; this implies that your source impedance needs to be less than 9K if you want this shift to induce less than a 1% error. \$\endgroup\$ – Dave Tweed May 1 '14 at 16:26
  • 1
    \$\begingroup\$ That sounds more like a drive specification, applicable when those pins are used as outputs. Probably irrelevant to your usage. \$\endgroup\$ – Dave Tweed May 2 '14 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.