If the minimum load of a sensor(page 7) is 65k Ohm, what does that mean?

Does it mean that the sensor needs at least a 65k Ohm resistor, so a 100K Ohm is good?

Or does it mean the sensor needs at least a certain amount of current drawn off through the resistor, so a 10K Ohm resistor is good?

  • \$\begingroup\$ That font is awesome in the datasheet... \$\endgroup\$ – W5VO Feb 16 '11 at 21:50

Because the diagram on that page:

Figure 10

shows "Voltage Out" (i.e, your ADC input) at the far right, I'm going to assume it's asking for a minimum output current to ensure stability, therefore, you need a resistor of 65k or less resistance (500uA @ 100% humidity) or more current) to get a stable output. To put it another way, 65k is the largest resistor you're allowed to have there. If the resistor was internal to an ADC block, I would interpret it as the minimum resistance/impedance of the ADC.

Typically, power supplies specify a minimum current (or maximum load resistance) to maintain stability, while sensors, opamps, and DACs have a minimum output impedance (or maximum current) required to maintain accuracy. The way I'm reading it,his one seems to be backwards to the convention.

It's interesting that they claim an operating current of 200uA when their minimum load draws 500uA at 3V3. Perhaps this value doesn't include the output current, or a marketer is specifying the device at 0% humidity and 0V out?

In the end, the only way to really know is to call Honeywell and ask for clarification, or to attach the output to an oscilloscope and pot and see whether it's both stable and accurate at various impedances below, at, and above 65k.


My first guess is that this sensor needs a minimum load to be stable. With a high load the output may not be stable, or the output regulation may provide a significant error. I can recall that a voltage regulator like a LM317 also specifies a minimal load current, which is the minimal load you need to provide for the regulation to be stable.

What you could try though is get the sensor and just measure the signal open ended (no load attached except your multimeter voltage probes). If it's all over the place or just not right (like the output voltage is clipping) you may need to add a more significant load (try like 56k first). Your multimeter is typically going to have a load of over 1 M ohm, maybe even more.

  • \$\begingroup\$ An oscilloscope would not only give you a known impedance, it would show "All over the place" for signals which oscillate quickly. I don't think that an unstable signal would be "all over the place" so much that a DMM would pick it up, nor would it have a period in seconds. Of course, it's out of specification, so it could do both of those things. \$\endgroup\$ – Kevin Vermeer Feb 16 '11 at 20:53

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