-2
\$\begingroup\$

I need precision variable resistors for a programmable analog computer that I'm trying to construct.

An analog computer requires variable resistors that can be set with precision.

I don't want to use digital components because the have discrete steps and are not pure resistors (have capacitance)

My first thought was to connect a dc motor via reduction gears to a precision rheostat -- wire that to a digital ohm meter and build a control circuit that dials the rheostat to the desired value.

Are there off the shelf components that I can buy or will I have to construct my own.

Notes:

  1. On using a MOSFET as a variable resistor -- based on this post, it looks like it will be difficult to deal with the nonlinear behavior -- plus the ohomic range seems narrow.

  2. I looked at digitally controlled rheostats but they have several limitations including the fact that they are active solid-state components and that they have discreet steps (typically 256 and 1024 on the high end.

\$\endgroup\$
15
  • \$\begingroup\$ Is solid state really not an option? There are analog solid state solutions. \$\endgroup\$ Commented Sep 21, 2016 at 2:07
  • 1
    \$\begingroup\$ Make up your mind -- in one paragraph, you say you don't want anything with discrete steps, but in another, you say you'll use a digital ohmmeter. How exactly do you intend to specify the desired value of the resistance? \$\endgroup\$
    – Dave Tweed
    Commented Sep 21, 2016 at 2:07
  • 1
    \$\begingroup\$ A MOSFET is basically a voltage controlled resistor. All you need to do know is combat the odd curve. \$\endgroup\$
    – Bradman175
    Commented Sep 21, 2016 at 2:10
  • 1
    \$\begingroup\$ Sounds to me that this is just a dream project from someone who hasn't tried to do the math and found out what kind of precision and accuracy you need. \$\endgroup\$
    – pipe
    Commented Sep 21, 2016 at 2:23
  • 2
    \$\begingroup\$ A JFET and an op amp. \$\endgroup\$ Commented Sep 21, 2016 at 2:50

1 Answer 1

1
\$\begingroup\$

There is indeed - this motorized potentiometer. Note that it has two separate resistive tracks. Or had you not realized that you have to disconnect a resistor from its' circuit to measure it?

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.