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What the title says. I'm curious how exactly a digital potentiometer works. Is it something like a pair of pass transistors like below?

schematic

simulate this circuit – Schematic created using CircuitLab

Or is it more like a series of switched resistors like below?

schematic

simulate this circuit

Or is it something else entirely, that I can't think of?

I looked around, and found exactly one source suggesting that it is the latter case, but I couldn't find anything else talking about the internals of a digital pot at all. Maybe there's more out there, though, and I'm just really bad at finding things.

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Second option is a good representation. You can fabricate well matched resistors in CMOS processes, and also well matched analog switches. As ever, the absolute value tolerance for passives in CMOS is high, but the matching is excellent. Have a look at the datasheet for the PGA2311, which is a audio potentiometer.

The first option would be a bit grim, as you've alluded to. The MOSFETs are non-linear, and you have issues with the bulk connection with changing \$V_\mathrm{DS}\$.

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  • \$\begingroup\$ Yes, the first option would be a nightmare to design I'm sure. The bulk connection I don't see being a problem though, as it would likely not be connected to source in this case (but that's the only MOSFET symbol in the schematic editor, so it had to do). \$\endgroup\$ – Hearth May 14 '18 at 20:30
  • \$\begingroup\$ @Felthry True, if your input voltage (that could be + in your schematic) is constant, but not in the general case (unfortunately!). \$\endgroup\$ – awjlogan May 14 '18 at 20:31
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I don't understand why this question is posted. A 10-second Google search brings one of the leaders in digital potentiometer design, Maxim/Dallas Semiconductor. Almost every datasheet clearly shows the design idea. For example, the DS2890 block diagram:

enter image description here

Or MAX5460 functional diagram: enter image description here

It is quite clear that the IC implementations follow the "idea #2" in OP question.

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  • \$\begingroup\$ I did say I looked around and could only find one source saying anything either way. \$\endgroup\$ – Hearth May 14 '18 at 22:01

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