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When connecting a 10S LiPo (42V) to my ADC via a voltage divider, I began wondering how to calculate the output impedance of the divider.

schematic

simulate this circuit – Schematic created using CircuitLab

For example, what is the output impedance of the above circuit?

(The ADC's specs recommend a signal impedance of 10kOhm or less, which I simply translated to 0.5mA @ Vcc=5V (worst case) and figured I'd be fine as long as the signal can source/sink more than 0.5mA. But that's not really 'calculating' the impedance...)

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    \$\begingroup\$ Have you heard about Thevenin? \$\endgroup\$
    – Eugene Sh.
    Jun 28, 2016 at 14:22
  • \$\begingroup\$ As indicated the parallel resistance. It only becomes a problem when you have a variable potential divider and it keeps changing as you set the wiper position. from zero to half of full value and back to zero in a parabolic function (if you have a linear taper pot, otherwise things are more complicated). \$\endgroup\$
    – KalleMP
    Jun 28, 2016 at 19:06

2 Answers 2

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The output impedance is the equivalent of the two resistors in parallel.

If you are lazy, you can approximate this by taking just the lower-valued one, in this case the 5k.

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  • \$\begingroup\$ Thank you. I'm thinking this through now and it seems that my difficulty is that I was implicitly trying to factor in the capacitance at the ADC's input. When I do that, things get more complicated... \$\endgroup\$
    – JimmyB
    Jun 28, 2016 at 14:34
  • \$\begingroup\$ How 'more complicated'? Just replace the two resistors by one (and a lower voltage source) and add the capacitor. \$\endgroup\$
    – Wouter van Ooijen
    Jun 28, 2016 at 16:03
  • \$\begingroup\$ I think I understand why it is like this now. \$\endgroup\$
    – JimmyB
    Jun 29, 2016 at 9:48
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Wouter has the right answer, here is the background why.

A voltage source in an ideal sense has zero output resistance. In terms of small signal analysis, this can be replaced by a Ground, grounds have zero resistance and in that case your drawing reverts to:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Maybe I'm being pedantic, but I think the use of the term "ground" can be misleading. More generally you are shorting the source and re-analyzing the circuit. Ground is merely a reference node, and the only real ground is beneath your feet. \$\endgroup\$
    – Brendan Simpson
    Jun 28, 2016 at 18:09
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    \$\begingroup\$ No, You're not being pedantic, a ground has a very particular meaning, it is indeed a reference node, but what you're missing is that a true ground also can source and sink current. In small signal usage this is an important aspect of determining operation at a given bias point. \$\endgroup\$
    – placeholder
    Jun 28, 2016 at 18:12

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