Can anyone help me to find the relation for sensitivity of wheatstone bridge using transfer function or any other methods. And why is a wheatstone bridge more sensitive when all resistors have equal value? I think what matters is the ratio of resistances and not their individual values.


2 Answers 2


Start with the Wheatstone bridge equation from wikipedia which is the subtraction of two voltage dividers.

$$ V_G = V_s( \frac{R_x}{R_3+R_x} - \frac{R_2}{R_1+R_2}) $$

Sensitivity is the derivative with respect to \$R_x\$. You'll notice immediately that the voltage divider that doesn't have the element of interest doesn't impact the sensitivity at all. The only function of the second voltage divider is to create a set point to compare the first voltage divider to. If we take the derivative with respect to \$ R_x\$ we get: $$ \frac{R_3}{(R_3+R_x)^2} $$ Note that I've chosen to ignore \$ V_s\$ as it is a scaling factor here. It should be noted that as your \$ V_s \$ increases, your sensitivity does increase as well. If you were to plot this you would see that there isn't really a normal value for the component under test that gives a maximum sensitivity because there is no inflection point in the graph.
Wolfram Alpha Plot of Sensitivity where \$R_3 =1 \$ enter image description here

This actually shows that the maximum sensitivity is where \$R_x\$ trends towards 0. This assumes you don't have negative value components.

If we plot the sensitivity equation with \$ R_3\$ as the variable and \$ R_x \$ as a constant, we'll find there is a maximum right at our constant value we used for \$R_x\$:
Wolfram Alpha Plot of Sensitivity where \$R_x=1\$ enter image description here

This is likely where the notion that you should keep \$R_3\$ as close to the same value as \$R_x\$ comes from. The other voltage divider again just needs to match closely with the test component's voltage divider to keep the voltage difference of 0 between them.

From these two plots we can conclude that if we want to design a Wheatstone bridge with the maximum sensitivity, we should have as small of component values as possible while still keeping all of the components at approximately the same value.

  • \$\begingroup\$ Is this sufficient to prove that the sensitivity of a Wheatstone bridge is maximum when balanced? \$\endgroup\$
    – Aditya P
    Commented Apr 24, 2019 at 11:32
  • \$\begingroup\$ @Aditya I dunno. I'm an engineer, not a mathematician. It's proof enough for me, but probably not to a mathematician. \$\endgroup\$
    – horta
    Commented Apr 25, 2019 at 1:03

In my opinion your question is a little bit too generic, because you don't specify the actual configuration of your Wheastone's Bridge (WB), i.e. how many resistors are variable and which ones. For the case of one resistor look at the answer from horta. I'd like to add here some short information about other uses with sensors. I think that WB is more suitable when you have one push-pull resistors/sensors or a pair of push-pull sensors. Suppose the left arm of you WB is composed of \$R_1\$ and \$R_2\$ and let you right arm be composed of \$R_3\$ and \$R_4\$. Let your bridge be supplied between \$V_{cc}\$ and GND. The output of you WB is $$\tag{1} V_O=V_{cc}\Big (\frac{R_2}{R_1+R_2}-\frac{R_4}{R_3+R_4}\Big) $$ Now suppose \$R_1\$ and \$R_2\$ are two linear sensors in push-pull configuration, i.e. \$R_1=R_0(1+x)\$ and \$R_2=R_0(1-x)\$. Here \$x\$ is the sensitivity of the sensor. Now take \$R_3=R_4=R_{dummy}\$. If you substitute in (1) you get: $$\tag{2} V_O=V_{cc}\frac{x}{2} $$ So that the sensitivity of your WB with respect to \$x\$ is \$S=V_{cc}/2\$ Alternatively, if you are in the situation for which \$R_1=R_4=R_0(1+x)\$ and \$R_2=R_3=R_0(1-x)\$ then you get $$\tag{3} V_O=V_{cc}x $$ and your sensitivity is \$S=V_{cc}\$. In this case you double the sensitivity with respect to (2).


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