SaM - MEMORIZE (Resistive Bridge • Wheatstone)

List of things to memorize:

SaM - Resistive Bridge • Wheatstone

• SaM - Definition of a Resistive Bridge • Balanced Bridge • Thevenim Equivalent of the Resisitive Bridge
• SaM - Half Bridge • Calculation for the Maximum Sensitivity of a Resistive Bridge
• SaM - Full Bridge
• SaM - Design of a Resistive Bridge
• SaM ~ Resistive Bridge Example • Strain Gauge
• SaM ~ Half Bridge Example • Two Strain Gauges
• SaM - Wire Resistance in a Resistive Bridge (2 and 3 Wires Layout)
• SaM ~ Complete Sensor • Load Cell with a Resistive Bridge
• SaM ~ Complete Sensor • Accelerometer with a Resistive Bridge

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SaM - Definition of a Resistive Bridge • Balanced Bridge • Thevenim Equivalent of the Resisitive Bridge

• Circuit:
  
• Sensor $(R_1)$ formula:$R_1=R_{10}(1+x)$
• Output formula:$V_0=V\left(\frac{R_1}{R_1+R_4}-\frac{R_2}{R_2+R_3}\right)$
• K-ratio:$K=\frac{R_4}{R_{10}}=\frac{R_3}{R_2}$
• Output formula with K-ratio:$V_0=V\left(\frac{1+x}{1+x+K}-\frac{1}{1+K}\right)$
• Balanced bridge $(K=1)$ formula:$V_0=V\left(\frac{1+x}{2+x}-\frac{1}{2}\right)$
• Maximum Realative Sensisitivity $(\text{for}K=1)$: $S{|}_{x=0{}_{MAX}}=\frac{1}{4}$
• Real World Measures:
  • $x=\alpha T$, if we consider a RTD sensor, specifically for the PT100 sensor:
    • $x\in \left[-0.78÷3.3\right]$
    • $R_0=100\text{ohm}$
  • $x=G{\epsilon }_l$, if instead we consider a metal strain gauge
    • $x\in \left[0÷0.1\right]$
    • $R_0\in \left[50\text{ohm}÷700\text{ohm}\right]$

• Terminology:

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

• Circuit:
  
• Sensors:$\begin{array}{l}{R}_1=R_{10}(1+x_1)\\ R_2=R_{20}(1-x_2)\end{array}$
• K-ratio:$K=\frac{R_4}{R_{10}}=\frac{R_3}{R_{20}}$
• Output formula:$V_0=V\left(\frac{1+x_1}{1+x_1+K}-\frac{1-x_2}{1-x_2+K}\right)$
• For: $(1+K)\gg x_1$, $(1+K)\gg x_2$ and $x_1=-x_2$:$V_0=V\cdot \frac{2x}{1+K}$
• Maximum Realative Sensisitivity $(\text{for}K=1)$: $S{|}_{x=0{}_{MAX}}=\frac{1}{2}$
• Calculate the maximum relative sensitivity:
  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 you’ll find $K=1$.
  5. Calculate $S_{MAX}$:$S{|}_{x=0{}_{MAX}}=\frac{1}{2}$That’s why it is called “half” brige.

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SaM - Full Bridge

• Circuit:
  
• Sensors:$\begin{array}{l}{R}_1=R_3=R_0(1+x)\\ R_2=R_4=R_0(1-x)\end{array}$
• Output formula $(\text{for}K=1)$:$\frac{V_{TH}}{V}=x$
• Maximum Realative Sensisitivity $(\text{for}K=1)$:$S{|}_{x=0{}_{MAX}}=1$

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SaM - Design of a Resistive Bridge

• We take in example an half bridge.

1. Set the output range $V_{T{H}_{MAX}}$, from this find $x_{\max }$.
2. Check the power (self heating):$P={\left(\frac{V}{R_1+R_4}\right)}^2\cdot R_1$
3. Check accuracy or maximum error:$\Delta V_{TH}\approx \frac{d{V}_{TH}}{dx}\Delta x>\text{Noise level, IZE (Amplifier), drifts}$
4. Linearity of the bridge: so if $1+K\gg x$.
   The non-linearity error:$\text{NL}=\frac{x}{1+K}$So assert that:${\text{NL}}_{\text{accettable}}>\frac{x_{\max }}{1+K}$

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SaM ~ Resistive Bridge Example • Strain Gauge

• Example of the design process of a ”$\frac{1}{4}$ balanced resistive bridge” using a single strain gauge sensor.
• (1.) Find $x_{\max }$:
  $x_{\max }=G\cdot {\epsilon }_{\max }=4\cdot 50000\mu \epsilon$
  ⇒ $x_{\max }=0.2$
• (3.) Check accuracy or maximum error:
  $\Delta \epsilon =10\mu \epsilon$ ⇒ $\Delta x=4\cdot 10^{-5}$
  Then:${\Delta V_{TH}|}_{x=0}=\frac{V}{4}\Delta x$And for $V=10\text{V}$ we obatain: $\Delta V_{TH}=100\mu \text{V}$

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SaM ~ Half Bridge Example • Two Strain Gauges

• Structure:
  
• Sensors formula (with temperature offset):
  
• Output formula:$\frac{V_{TH}}{V}\approx \frac{2G{\epsilon }_l}{1+K}$

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SaM - Wire Resistance in a Resistive Bridge (2 and 3 Wires Layout)

• Circuit:
  
• Output formula:$V_{TH}=V\frac{R_0(1+G\epsilon )+2R_W}{2R_0+R_{0}G\epsilon +2R_W}-\frac{V}{2}$
• Offset and sensitivity:NOT_SURE_ABOUT_THIS (redo the sensitivity calculations)$\begin{array}{lc}\text{Offset}:& {V_{TH}|}_{\epsilon =0}=V\frac{R_0+2R_W}{2R_0+2R_W}\\ \text{Sensitivity}:& {\frac{d{V}_{TH}}{d\epsilon }|}_{\epsilon =0}=V\frac{R_{0}G}{2R_0+2R_W}\end{array}$The offset in theory could be corrected.
  The sensitivity is different from the nominal one (it depends on $R_W$).
• Using a current suplly - circuit:
  
• Formulas:$\begin{array}{l}{V}_{TH}=I_{DC}{R}_0(1+G\epsilon )+I_{DC}2R_W-I_{DC}{R}_0\\ \text{Offset}:{V_{TH}|}_{\epsilon =0}=I_{DC}2R_W-I_{DC}{R}_0\\ \text{Sensitivity}:{\frac{d{V}_{TH}}{d\epsilon }|}_{\epsilon =0}=I_{DC}{R}_{0}G\end{array}$
  The sensitivity is constant, this is a linear sensor (good).
  The offset still depends on $R_W$ (bad).
  ==This is a good solution for AC measurements==.
• 3 wires layout - circuit:
  
• Formulas:NOT_SURE_ABOUT_THIS (redo the sensitivity calculations)$\begin{array}{l}{V}_{TH}=V\frac{R_0(1+G\epsilon )+R_W}{2R_0+R_{0}G\epsilon +2R_W}-\frac{V}{2}\\ \text{Offset}:{V_{TH}|}_{\epsilon =0}=0\\ \text{Sensitivity}:{\frac{d{V}_{TH}}{d\epsilon }|}_{\epsilon =0}=V\frac{G{R}_0(R_W+R_0)}{{\left(2R_W+2R_0\right)}^2}\end{array}$We have removed the offset (good).
  But the sensitivity still depends on $R_W$ (bad).

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SaM ~ Complete Sensor • Load Cell with a Resistive Bridge

• Load cell with a full bridge of strain sensors:
  
  
• Sensors:$\begin{array}{l}{R}_1=R_0(1+G{\epsilon }_l)\\ R_2=R_0(1+G{\epsilon }_t)\\ R_3=R_0(1-G{\epsilon }_l)\\ R_4=R_0(1-G{\epsilon }_t)\end{array}$
• Output formula using strain:$V_{out}=V\frac{G(1+\nu )}{2}{\epsilon }_l$
• Output formula using force:$V_{out}=V\frac{G(1+\nu )}{2}\frac{E}{A}{F}_1$
• Real World Measures:
  • Usually $\nu \simeq 0.35$, specifically for platinum ${\nu }_{Pt}\simeq 0.38$.

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SaM ~ Complete Sensor • Accelerometer with a Resistive Bridge

• Strucutre:
  
• Sensor formulas:$\begin{array}{l}{R}_{1,3}=R_0\left(1-G\left(\frac{m_s}{K\cdot l}a\right)\right)\\ R_{2,4}=R_0\left(1+G\left(\frac{m_s}{K\cdot l}a\right)\right)\end{array}$Where:$\frac{m_s}{K\cdot l}a={\epsilon }_l$
• Considering the pre-deformation of the straing gauges:$\begin{array}{l}{R}_{1,3}=R_0\left(1-G\left(\frac{m_s}{K\cdot l}a\right)+G{\epsilon }_0\right)\\ R_{2,4}=R_0\left(1+G\left(\frac{m_s}{K\cdot l}a\right)+G{\epsilon }_0\right)\end{array}$

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