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I want to connect Sharp GP2Y1014AU0F Dust Sensor with an ESP32. According to the datasheet, the VCC can be from 0 to +7V (5V recommended). With Vcc = 5V, the sensor output voltage varies between 0.6 to ~3.6V. enter image description here

I suppose if I power the sensor at 3.3 the output graph won't be valid and I need to calibrate the sensor myself at 3.3V supply. So I want to keep the Vcc at 5V. Although I don't expect the output to > 3v all the time but still want to keep it safe for ESP32 (i.e. <3.3V)? My initial idea was using a voltage divider but wonder if it will affect the ADC measurements. Is there is a better to achieve the same objective?

Sensor Datasheet

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    \$\begingroup\$ "According to the datasheet, the VCC can be from 0 to +7V (5V recommended)" - No, that's not what the datasheet tells you. "Absolute maximum ratings" are not an operating spec, they're the range over which the device won't (likely) be destroyed. If you want it to behave 'normally' then refer to the "Operating Supply Voltage" which clearly tells you that you need to supply 4.5V-5.5V. \$\endgroup\$
    – brhans
    Sep 12, 2021 at 13:36
  • \$\begingroup\$ You are right. I interpreted the full range to be usable. \$\endgroup\$
    – Zaffresky
    Sep 12, 2021 at 17:14
  • \$\begingroup\$ There are a few discussion around the module keyestudio Ks0196 which is a GP2Y1014AU0F with resistors —but 3.3V = rubbishy low value is the unanimous consensus... so it's a common hope! (Commenting solely to drop term Ks0196 for folk googling 3.3V and whether one could get away without a level converter as the plot (which is easily missed in the dataset) answer it (=no) —The KS manual subtracts .1 which is the min at 0 g/m3, but the spec says mean=.6, and multiplies by mystery slope .170 mg/g3·V, so this plot is helps!) \$\endgroup\$ Feb 3 at 18:06

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The sensor is not specified to work on 3.3V supply. It is specified to work on 5V +/- 0.5V supply, and therefore it is not required to operate in any way outside that specification.

The curve you are looking only provides typical example output. The datasheet only guarantees that the output is at least 3.4V or more at 4k7 ohm load, but since it does not give a maximum, it could go up to the 5V supply voltage.

If it can work with 4k7 ohm load, surely that can be split into for example 2k2 and 3k3 resistors to have a divided output for ADC measurement.

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  • \$\begingroup\$ I misunderstood the Vcc range. Where is the 4.7k resistor/load that you are referring to? I only saw one 150R resistor in the suggested circuit in the datasheet. \$\endgroup\$
    – Zaffresky
    Sep 12, 2021 at 17:20
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    \$\begingroup\$ @Zaffresky that 150R resistor is for driving the LED. In the "Electro-optical Characteristics" table on the same page, the spec for "Output voltage range" lists "RL=4.7k" as the Condition. \$\endgroup\$
    – brhans
    Sep 12, 2021 at 17:26
  • \$\begingroup\$ Yep I see it now :) thanks for pointing it out. \$\endgroup\$
    – Zaffresky
    Sep 12, 2021 at 19:43
  • \$\begingroup\$ @Justme Following your answer to split the RL, how can it be done. I suppose I need to modify the internal circuitry or is there some other way? \$\endgroup\$
    – Zaffresky
    Sep 12, 2021 at 19:52
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    \$\begingroup\$ The RL is an external load that the sensor is capable of driving, it has been tested with such a load. It does not exist in the sensor, it is a load you put to the sensor output, and it's most likely the maximum load. You can always use higher impedances. \$\endgroup\$
    – Justme
    Sep 12, 2021 at 20:20

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