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I've come across two variations on the voltage divider formula, when solved for \$x_2\$: $$R_2={R_1V_{out}\over V_{in}-V_{out}}$$ $$R_2={R_1\over {V_{in} \over V_{out}} - 1}$$

I solved the one equation I can find that most sources agree on for \$R_2\$ $$V_{out}={V_{in}R_2 \over R_1+R_2}$$ and got the first equation. This question seems to agree with that result. However, Wikipedia and this calculator have another version (use a test set of values to check the calculator).

I graphed both, to the best of my ability assigning constants to the voltages, and they are not equal. Which equation is right, and if it's the second how, how do you get there from the base equation?

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  • \$\begingroup\$ What is V1 in the 1st equation? Vout (mistyped)? Or the voltage across R1? \$\endgroup\$ – M.Ferru Feb 28 '17 at 21:49
  • \$\begingroup\$ These two are the same, if in the first equation V1 is Vout \$\endgroup\$ – user28910 Feb 28 '17 at 21:51
  • \$\begingroup\$ @user28910 How are they the same? would you mind posting a proof as an answer? I've tested values and gotten different answers, though I may have messed up somewhere \$\endgroup\$ – cat40 Feb 28 '17 at 21:53
  • \$\begingroup\$ If in the first equation V1 is a typo and is supposed to be Vout, then factoring out Vout from the numerator and denominator yields the second equation. \$\endgroup\$ – Envidia Feb 28 '17 at 21:53
  • \$\begingroup\$ V1 is supposed to be Vout. Just fixed it \$\endgroup\$ – cat40 Feb 28 '17 at 21:56
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I'm assuming that the first equation is a typo and is really $$R_2 = \frac{R_1V_{out}}{V_{in} - V_{out}}$$

We can factor out \$V_{out}\$ which yields: $$R_2 =\frac{V_{out}}{V_{out}} \frac{R_1}{\frac{V_{in}}{V_{out}} - 1} = \frac{R_1}{\frac{V_{in}}{V_{out}} - 1}$$

Which is the second equation.

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Well, both equations are the same...

Vin - Vout = Vout ( Vin / Vout - 1 )

So you can remove Vout. Simple math

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