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I have a 3.3V circuit with an ADC chip that reads values between 0V and 2.1V, that means anything above 2.1V will return 2047. If I use a voltage divider with the Vcc power supply, I would lose a lot of range. What is the best solution to measure all values? Do I have to buy a 2.1V regulator? Or can I supply 2.1V with another voltage divider?

The variable resistor in that provides the voltage to measure would be a potentiometer or a light sensor (photoresistor).

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    \$\begingroup\$ The normal setup for a photoresistor/potentiometer setup is a potential divider, so I'm confused by "would lose a lot of range". You can place a resistor of suitable size towards the positive side of the supply, which is going to constrain the maximum voltage output. \$\endgroup\$ – pjc50 Jul 12 '13 at 13:24
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For a photoresistor that varies (for example) from 10 MOhms with no light, to 100 Ohms with bright light, the following voltage divider arrangement would provide a 0-2.1 Volt range as desired, with a 3.3 Volt supply:

schematic

simulate this circuit – Schematic created using CircuitLab

At pitch dark, this would yield 2.1 Volts, while at maximum light, it would be approximately 0 Volts.

In order to take better advantage of say the 100 Ohm to 1 MOhm range of the photoresistor and consider anything above 1 MOhm as "too dark", R1+R2 can be chosen to total 570 kOhms, instead of the 5.7 MOhms shown.

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