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schematic

simulate this circuit – Schematic created using CircuitLab

Suppose we setup a potentiometer with two power supplies +10V and -10V as shown above. A digital voltmeter is setup to measure the voltage.

If the potentiometer is turned all the way up, a digital voltmeter would read +10V, if the potentiometer is setup all the way down, the digital voltmeter would read -10V.

My question is

When writing the equation for calculating the voltage across the voltmeter, how can we include the +10V and as well as -10V in our potential divider equation and calculate the voltage at the voltmeter?

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How about, V = 20x-10 where x is the potentiometer wiper position as a fraction from 0 to 1.

Other related points (to consolidate those made in the follow-up comments and answers):

  • General form: \$ V_{out} = x(V^+-V^-)+V^-\$
  • Potentiometers (pots) are generally used in low power applications. Make sure to check your part's specs to ensure it will operate properly in your circuit.
  • Pot wiper terminals are typically designed to carry low current. Again, check the datasheet.
  • In response to your question in the comments section: if you intend to test your device with an ohmmeter, be sure to first disconnect the power source in order to obtain accurate measurements.
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    \$\begingroup\$ Or, of course, 10 (2x - 1). \$\endgroup\$ Commented Jul 12, 2017 at 3:22
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    \$\begingroup\$ @Amy if the potentiometer has a linear (and not logarithmic) range: 0.5xR1 = 50 Ω. If you are using a standard DMM, you'll need to disconnect the power before making your measurements. \$\endgroup\$ Commented Jul 12, 2017 at 4:03
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    \$\begingroup\$ @Amy No, for an ideal linear pot, when the wiper is half way through the full rotation, the resistance will be half of the potentiometer's full value. Or in your case, 50 Ω (as stated in the comment above). \$\endgroup\$ Commented Jul 12, 2017 at 5:07
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    \$\begingroup\$ @Amy By the way, 20 Volts across a 100 Ω pot is probably a bad idea. Pots are typically used in low power applications. Unless this is purely academic, make sure your part is rated for the 4 Watts. \$\endgroup\$ Commented Jul 12, 2017 at 5:27
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    \$\begingroup\$ Also, pot wiper terminals are typically designed to carry very, very, little current. Check the datasheet. \$\endgroup\$ Commented Jul 12, 2017 at 14:33
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A more general equation would be:

\$ V_{out} = x(V_1-V_2)+V_2\$

\$ x = \$ a wiper from 0 to 1

I strongly suggest that you do not use 100 ohm potentiometer (use a higher value like 10k). Nor do I suggest that you use a potentiometer to power something else.

There is something called output resistance and input resistance, but that is something that should be answered in another future question for when the time comes.

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