SaM - Half Bridge • Calculation for the Maximum Sensitivity of a Resistive Bridge

Remeber:

IMPORTANTE We have seen the resistive bridge, let’s see how the equations change when we consider a brige with 2 resistive sensors, instead of one:

• We put the two sensors in “opposite sensing direction”, so if we were to describe them we could say:$\begin{array}{l}{R}_1=R_{10}(1+x_1)\\ R_2=R_{20}(1-x_2)\end{array}$
• Then we define $R_3$ and $R_4$ such that the bridge is balanced, so we define the $K$-ratio:$K=\frac{R_4}{R_{10}}=\frac{R_3}{R_{20}}$
• Finally if we calculate $V_0$ we will obtain:$V_0=V\left(\frac{1+x_1}{1+x_1+K}-\frac{1-x_2}{1-x_2+K}\right)$If we consider $(1+K)\gg x_1$ and $(1+K)\gg x_2$, we can simplify the equation, resulting in:$V_0=V\cdot \frac{x_1+x_2}{1+K}$Finally if $x_1=-x_2$, so they both measure the same change:$V_0=V\cdot \frac{2x}{1+K}$

IMPORTANTE It is called a half-bridge because if we perform the same calculation as before for “Optimizing the Relative Sensitivity of the Bridge”, so:

1. Calculate:$S(x)=\frac{1}{V}\frac{d{V}_{TH}}{dx}$
2. Calculate $S(x=0)$, than calculate:$\frac{dS(x=0)}{dK}$
3. Find $K$ such that:$\frac{dS(x=0)}{dK}=0$
4. Find $K$ such that $S(x=0)$ is maximied, and we will obtain that for $K=1$ we will have $S_{MAX}$ and the final result will be:$S{|}_{x=0{}_{MAX}}=\frac{1}{2}$That’s why it is called “half” brige.

The simple resistive bridge or (”$\frac{1}{4}$ brige”) is called like that because if we perfom the same calculations we will obtain:$S{|}_{x=0{}_{MAX}}=\frac{1}{4}$.

What are the gains of using a half bridge instead of a $\frac{1}{4}$ brige?

• ==If $(1+K)\gg x_1$ and $(1+K)\gg x_2$ the half bridge behaves linearly==.
• It has higher sensitivity.NOT_SURE_ABOUT_THIS (higher relative sensitivity perhaps, the sensitivity of the “quarter bridge” is not linear, so at specific points it might be higher that that of the “half bridge”)
• However it requires two sensors, so requires more space, and has a higher cost.

---

Memory Card

---

Define our two sensors:

Consider it a balanced bridge, so:

As usual, I consider the output $V_o$ and consider to work with no load (or an infinite load given by an ideal differential amplifier, infinite load ⇒ open ciruit) so I have the output which is equal to the open circuit differential voltage:

So we calculate $V_{TH}$ and we obtain:

• Notice how the two sensor oppose each other: $x_1-x_2$

This is easily fixed by inverting one of the two:

An half-bridge has some advantages with respect to the $\frac{1}{4}$ brige:

It is called a half-bridge because if we perform the same calculation as before for “Optimizing the Relative Sensitivity of the Bridge”, so:

1. Calculate:$S(x)=\frac{1}{V}\frac{d{V}_{TH}}{dx}$
2. Calculate $S(x=0)$, than calculate:$\frac{dS(x=0)}{dK}$
3. Find $K$ such that:$\frac{dS(x=0)}{dK}=0$So $K$ such that $S(x=0)$ is maximied.

And we will obtain that for $K=1$ we will have $S_{MAX}$, such that:$S{|}_{x=0{}_{MAX}}=\frac{1}{2}$ The simple resistive bridge or (”$\frac{1}{4}$ brige”) is called like that because if we perfom the same calculations we will obtain:$S{|}_{x=0{}_{MAX}}=\frac{1}{4}$.