Why would a digipot require the voltages on its terminals to be between ground and VCC?
Why would a digipot require the voltages on its terminals to be between ground and VCC?
I'm not sure if other digipots are like this, but with the MCP4151, which I'm using, the datasheet says that the "Resistor Terminal Input Voltage Range (Terminals A, B and W)" has a min voltage of Vss and max of Vdd.
I was skeptical about that since internally it's just a resistor ladder, but when I fed an audio signal through it slightly outside of Vss and Vdd, there was a bunch of staticky noise not present when I use a mechanical pot instead. I don't understand why it matters what the input voltage is as long as it doesn't melt the internal resistors.
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Because a digipot isn't actually a pot? It's not made from an adjustable resistor. It simulates one, using transistors and stuff. (or just transistors)
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– immibis
Sep 13 '18 at 5:00
1 Answer
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This device is far more than "just a resistor ladder". It includes a great deal of digital logic that controls the serial interface and allows you to program the resistance.
The resistors are connected to the "wiper" by analog switches. These switches are made of transistors. There are parasitic PN junctions between the transistor terminals and the power and ground terminals. These PN junctions remain reverse-biased, as desired, only as long as the internal voltages remain between Vss and Vdd.
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You are likely hearing shot-noise thru diode-junctions. The current noise will be sqrt( 2 * I * q) where "q" is charge on the electron, and "I" is the about-to-turn-on-even-more current thru the isolation diodes, as discussed in"Elliot Alderson" answer. Assume 1uA current. The sqrt( 2 * 1uA *1.6e-19) is sqrt(3.3e-25) or sqrt(33e-26) or 6e-13 or 0.6 picoAmperes noise current. If you have 100,000 ohm resistors, you'll have 0.06uV noise.
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– analogsystemsrf
Sep 13 '18 at 4:47