# Archived Sensitivity in potential divider circuits.

1. Mar 13, 2009

### Sockpirate

1. The problem statement, all variables and given/known data
Why is sensitivity in a potential divider circuit greatest when the resistance of the fixed resistor is equal to the average resistance of the other resistor (in my case a LDR)?

This isn't a homework question, but I need to include a bit on why I chose the fixed resistor I did in my coursework - I know I chose it for the reason that it has the highest sensitivity, but I need to explain why this is.

2. Relevant equations

V2 = (R2 / (R1 + R2)) * V1

3. The attempt at a solution

...I don't know >.>; help much appreciated.

2. Feb 6, 2016

### Staff: Mentor

Suppose that we have a voltage divider with a fixed resistance $R_f$ and a sensor with an average resistance $R$ configured so that:

$\frac{V_{out}}{V_{in}} = \frac{R_f}{R_f +R}$

We'll call the sensitivity $S$ of the divider the magnitude of amount by which the voltage ratio changes with respect to small changes in R. So:

$S = \left| \frac{d}{dR}\left( \frac{R_f}{R_f +R} \right) \right| = \frac{R_f}{(R_f + R)^2}$

Now we'd like to maximize this sensitivity through a good choice of $R_f$. Differentiating again and setting equal to zero:

$\frac{dS}{dR_f} = \frac{(R_f - R)}{(R_f + R)^3} = 0$

We can see that it is zero when $R_f = R$, making the average sensor resistance value $R$ the best choice for $R_f$.

Note that I didn't check to prove that what was found was a maxima rather than a minima; Left as an exercise for the student

Last edited: Feb 6, 2016