# Maximising sensitivity in a voltage divider

1. Nov 21, 2008

### iainfs

Hi folks,

1. The problem statement, all variables and given/known data

We have a voltage divider (more specifically, a resistive divider), with $$V_i$$ volts input and resistors $$R_1$$ and $$R_2$$ in series. The output voltage, $$V_o$$, is measured over $$R_2$$. $$R_1$$ is a fixed ohmic resistor; $$R_2$$ is a potentiometer. I'm looking for an algebraic proof that, for a constant $$V_i$$ and given range of resistances on $$R_2$$, the sensitivity (range of $$V_o$$) is maximised when the middle range of $$R_2 = R_1$$.

I have read this in a textbook but have yet to be satisfied by a proof!

2. Relevant equations

Well, I suppose we have $$V = IR$$ and $$V_o = \frac{V_i \cdot R_2}{R_1+R_2}$$.

3. The attempt at a solution

I've guess that this requires calculus, which I should be OK with as long as we don't get into anything too complicated. I'm not entirely sure how to approach this, but I'll give it a shot anyway!

As far as I can see, I want to maximise the rate of change of $$V_o$$ with respect to $$R_2$$.

$$\frac{dR_2}{dV_o}\;({\frac{V_i \cdot R_2}{R_1+R_2}})$$

I can't differentiate that because I don't know how. The constant seems to be all wrapped up with the variable. Some help here would be appreciated.

I would then go on to maximise this by finding $$\frac{d^2V_o}{d{R_2}^2}\;({\frac{V_i \cdot R_2}{R_1+R_2}}) = 0$$; hopefully solving the problem.

As you can see, I think I have a viable method, but I'm not able to follow it through. Any help would be greatly appreciated!

Many thanks,

2. Nov 21, 2008

### Staff: Mentor

I think you need to define what you mean by the "sensitivity (range of)" means. You get the max range when the R2 potentiometer is much larger than R1, so that the output voltage ranges from almost Vin to zero.

You would get some power transfer benifits if you size R2 = R1, but so far I don't see that entering into the question...

3. Nov 24, 2008

### iainfs

Yes, that would make sense. I think I have been misled as to the benefits of R1 = R2. For pure sensitivity (volts/ohm), a very large pot is best.

What, then, are the benefits of R1 = R2?

Many thanks.

4. Nov 24, 2008

### Staff: Mentor

Calculate the power transferred to R2 in terms of the ratio of R2:R1. Do you see anything useful?

5. Nov 29, 2008

### iainfs

I'm sorry, I'm struggling with that one. Do you mean power as would be given by P = I^2 x R?

Another possible point of confusion I have noticed is that a common type of potential divider has R1 + R2 = constant = the total resistance on the potentiometer. Mine uses a fixed resistor for R1; R2 is a variable resistor.

6. Dec 1, 2008

### iainfs

I had another attempt. Still no joy unfortunately.

$$I = \frac{V_i}{R_1+R_2}$$

$$I^2 = \frac{V_i^2}{(R_1+R_2)^2}$$

$$P_{R_2} = I^2R = \frac{V_i^2 \cdot R_2}{(R_1+R_2)^2}$$

Alternatively, using P = VI:
$$I = \frac{V_i}{R_1+R_2}$$

$$V_o = \frac{V_i \cdot R_2}{R_1+R_2}$$

$$P = V_oI = \frac{V_i^2 \cdot R_2}{(R_1+R_2)^2}$$

which gives the same. I can't seem to get it in terms of a ratio R1:R2. Assistance would be appreciated because it's driving me nuts!

7. Dec 1, 2008

### turin

This is a strange problem, and I wonder when you would ever encounter it in practice. Anyway ...

Let's call the range in in pot resistance as $\Delta{}R_2$. And, let's call the middle of this range simply $R_2$. So, the range in the output voltage, as a function of the middle of the pot range, is

$$\Delta{}V_o\left(R_2\right) = V_o^{hi}-V_o^{lo} = V_i\frac{R_2+\frac{1}{2}\Delta{}R_2}{R_1+R_2+\frac{1}{2}\Delta{}R_2} - V_i\frac{R_2-\frac{1}{2}\Delta{}R_2}{R_1+R_2-\frac{1}{2}\Delta{}R_2} = V_i\frac{R_1\Delta{}R_2}{\left(R_1+R_2\right)^2-\frac{1}{4}\Delta{}R_2^2}$$

We are looking for the maximum $\Delta{}V_o$ w.r.t. $R_2$. This looks like a resonance bump with a resonance at $R_2=-R_1$. I will check my work again.

UPDATE: Ah. It is not in the shape of a resonance bump, at least not w.r.t. $R_2$.

I conclude that $\Delta{}V_o$ is maximized w.r.t. $R_2$, for given $\Delta{}R_2$, $R_1$, and $V_i$, when $R_2=\frac{1}{2}\Delta{}R_2$.

Last edited: Dec 1, 2008