7
\$\begingroup\$

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...)

\$\endgroup\$
  • 3
    \$\begingroup\$ Have you heard about Thevenin? \$\endgroup\$ – Eugene Sh. Jun 28 '16 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 '16 at 19:06
5
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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 '16 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 '16 at 16:03
  • \$\begingroup\$ I think I understand why it is like this now. \$\endgroup\$ – JimmyB Jun 29 '16 at 9:48
2
\$\begingroup\$

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

\$\endgroup\$
  • 1
    \$\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 '16 at 18:09
  • 3
    \$\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 '16 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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