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.
Electronic – Sensitivity of wheatstone bridge
wheatstone-bridge
Related Solutions
The whole point is to match 2 different resistor values to each other, correct?
In that case, get 2 high quality fairly closely matched resistors R1, R3.
Then irregardless of the actual values of R2 and Rx, as long as they are closely matched the voltage divider circuits R1/R2 and R3/Rx will produce a near-zero voltage Vg. The smaller the voltage, the closer the match (assuming very good R1/R3 matching). You can now use R2/Rx in your circuits and they will be pretty well matched to each other.
There is another issue with resistor matching, though: temperature coefficients. These are often specified as max values and there's no telling if two given resistors will drift at the same rate or even in the same direction.
OK, problem solved. The bridge is connected like this. Only one resistance in the load cells is variable, the other is fixed.
Why the confusion above? I was measuring resistance of a load cell which came from a different scale. The cells looked pretty similar, therefore I thought they were the same. But they were not! Eureka!
simulate this circuit – Schematic created using CircuitLab
Best Answer
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 \$
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\$
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.