I was looking at the AD8403 A1 Digital potentiometer to use in a circuit where the voltage on the brush would be 12V and the voltage on the extremities would be 11.9V. Whilst this is way outside of the specification of 5V, since the difference across the pot is only 0.1V can it still be used?
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\$\begingroup\$ If your supply was referenced to a virtual ground at say 11V and you could be sure input and output currents were limited when power rails came up and went down you could possibly make it work. You would have to shift all your digital IO as well. \$\endgroup\$– KalleMPCommented Mar 7, 2017 at 23:23
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2 Answers
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The "absolute maximum ratings" table in the datasheet is the place to look. It indicates that Va, Vb, and Vw must all remain between 0V and Vdd. Since Vdd cannot exceed 8V, this chip will not work for you.
To see why, consider that - while the "pontentiometer" nodes may see only a small differential voltage - there are switches internal to the chip which span between the potentiometer nodes and the chips supply / ground. The voltage limitations of these switches is what prevents you from bringing the whole pot up to a higher common-mode voltage.
Fortunately, there are other digital pot products that should meet your needs!
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Generally, for any IC, the voltages on the signal pins cannot be more than 0.3 volts, or so, outside the power supply pins. (There are some exceptions, such as some open-collector TTL parts - but this would be clearly stated in the datasheet.)
You should always read and understand the Absolute Maximum Ratings table in the datasheet.
Short answer: NO.
answered Mar 7, 2017 at 20:51