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I am struggling to understand how to compute a specific, simple, equation relative to potentiometers. Given the image below:

enter image description here

I do not understand how to get to the second equation, i.e., the case in which the potentiometer is not an open load.

I managed to easily verify the OL case. In the second scenario, I applied the first Kirchhoff's law to the node between R1 and R2, but I ended up with a Vout = ƒ(Vin) much more complicated (and, presumably, wrong) than the one reported in the figure, which assume is true, as confirmed by multiple online sources.

As I did not find a demonstration about how to get to that simple equation, I am asking for some help.

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  • \$\begingroup\$ You just find the equivalent resistance of R2||R3 and redraw the schematic with that. Or more generally, just do KVL on the entire circuit without trying to simplify resistor combinations since real problems won't always simplify. \$\endgroup\$
    – DKNguyen
    Apr 10, 2023 at 18:18

2 Answers 2

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Fundamentally, both cases are voltage dividers with a difference in the division ratio due to the changing \$R_3\$ (presumably the potentiometer mentioned). The second case is the same as the first, except that \$R_2\$ is replaced with

$$R_2 \parallel R_3 = \frac{R_2 R_3}{R_2 + R_3}$$

since you can no longer ignore the non-infinite \$R_3\$ (the two resistances are in parallel, which is what the \$\parallel\$ symbol means).

The equation is therefore:

$$V_\text{out} = V_\text{in}\frac{R_2 \parallel R_3}{R_1 + R_2 \parallel R_3} = V_\text{in}\frac{\frac{R_2 R_3}{R_2 + R_3}}{R_1 + \frac{R_2 R_3}{R_2 + R_3}}$$

which can be simplified as:

$$V_\text{out} = V_\text{in}\frac{R_2 R_3}{R_1(R_2+R_3) + R_2 R_3} = V_\text{in}\frac{R_2}{R_1\left(\frac{R_2}{R_3}+1\right) + R_2}$$

as given. The advantage of rewriting the equation this way (so \$R_3\$ only appears once) is that it is easier to see what happens as \$R_3\$ reaches an extreme value.

Notice that

$$\lim_{R_3 \to \infty} V_\text{in}\frac{R_2}{R_1\left(\frac{R_2}{R_3}+1\right) + R_2} = V_\text{in}\frac{R_2}{R_1 + R_2}$$

(i.e. the first case) since the \$R_2/R_3\$ term goes to 0 as \$R_3\$ increases to \$\infty\$.

Notice also that $$\lim_{R_3 \to 0} V_\text{in}\frac{R_2}{R_1\left(\frac{R_2}{R_3}+1\right) + R_2} = 0$$

since the \$R_2/R_3\$ term becomes infinite, resulting in an infinite denominator that causes the fraction to go to 0. This makes sense, as \$R_3 = 0\$ is shorting \$V_\text{out}\$.

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  • \$\begingroup\$ You've done a yeoman's job showing traditional algebra and a few good implications to consider as validating that the result does the right thing at the two extremes. But how is it that \$R_3\$ modifies only the denominator in the original equation and, just then, still only by an added term of \$\frac{R_1\,R_2}{R_3}\$? Gaining some intuition there might be useful. Why should the term be the product of the two initial resistors divided by the third? What's going on? It's the kind of question I feel one needs to get into practice asking themselves. Answers can lead to improved vision. \$\endgroup\$ Apr 10, 2023 at 18:39
  • \$\begingroup\$ I'm not complaining about your answer. I'm giving it +1 as the more complete of the two so far. And it is good. Don't get me wrong. I'm just thinking it is also good to isolate a term and explain it's addition into the denominator. What is the added term describing? What is its meaning? \$\endgroup\$ Apr 10, 2023 at 18:41
  • \$\begingroup\$ @periblepsis I've edited my answer to provide some additional commentary which hopefully addresses your suggestions. \$\endgroup\$
    – Null
    Apr 10, 2023 at 18:51
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    \$\begingroup\$ @periblepsis There is no need to have a dangling question on a Q&A site. I encourage you to post it as a question. If you already know a good answer then you can answer your own question, otherwise the rest of the community can provide answer(s). \$\endgroup\$
    – Null
    Apr 10, 2023 at 19:35
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    \$\begingroup\$ @periblepsis: You're making a mountain out of a molehill. Both Null and Andyaka answered the question completely and succinctly. This is distracting to both answers and the ability of the OP to have clarity \$\endgroup\$
    – RussellH
    Apr 10, 2023 at 20:15
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I do not understand how to get to the second equation

$$\color{red}{\boxed{\text{This is the only question raised by the OP when I made my answer}}}$$

With \$R_3\$ in parallel with \$R_2\$ your top formula becomes this: -

$$\dfrac{V_{OUT}}{V_{IN}}\hspace{0.2cm} = \hspace{0.2cm}\dfrac{R_2||R_3}{R_1 +R_2||R_3}\hspace{0.2cm} = \hspace{0.2cm}\dfrac{\frac{R_2\cdot R_3}{R_2+R_3}}{R_1 +\frac{R_2\cdot R_3}{R_2+R_3}}\hspace{0.2cm} =\hspace{0.2cm}\dfrac{R_2\cdot R_3}{R_1\cdot R_2+R_1\cdot R_3+R2\cdot R_3} \hspace{0.2cm}$$ $$=\hspace{0.2cm}\dfrac{R_2}{R_1\cdot\frac{R_2}{R3}+R_1+R_2}$$

Can you take it from here?

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    \$\begingroup\$ I don't understand the purpose of your edit. The only edit to the question has been to add a paragraph break -- the OP hasn't raised any other question as you are implying. \$\endgroup\$
    – Null
    Apr 10, 2023 at 19:42
  • \$\begingroup\$ I don't see any problem here @Null \$\endgroup\$
    – Andy aka
    Apr 10, 2023 at 19:46
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    \$\begingroup\$ sigh You never do. You are implying the OP has edited the question and at least partially invalidated your answer. Since the OP has not done so your edit is pointless and distracting commentary. \$\endgroup\$
    – Null
    Apr 10, 2023 at 19:55
  • \$\begingroup\$ No, you are inferring that; I'm not implying that. \$\endgroup\$
    – Andy aka
    Apr 10, 2023 at 20:07
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    \$\begingroup\$ Thank you @Andyaka, I upvoted your answer. And yes, it is clear. I was stuck approaching this computation in a wrong way, and could not wrap my head around from here. Thank you! \$\endgroup\$
    – Jetboy
    Apr 12, 2023 at 21:02

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