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

Remeber:

A resisitve bridge or “wheatstone” or ” $\frac{1}{4}$ bridge” has a structure like this:

$R_{1}$ is our sensor, and we will describe it with the formula: $R_{1} = R_{10} (1 + x)$ Where, based on the sensors we have seen so far $x$ can be seen as:

$x = α T$ , if we consider a RTD sensor.

$x = G ε_{l}$ , if instead we consider a metal strain gauge.

The output of a ” $\frac{1}{4}$ bridge” is like so: $V_{0} = V (\frac{R _{1}}{R _{1} + R _{4}} - \frac{R _{2}}{R _{2} + R _{3}})$ So a resistive bridge will have a DC offset equal to: $V (\frac{R _{2}}{R _{2} + R _{3}})$ .
And a changing part (given by our sensor) equal to: $V (\frac{R _{1}}{R _{1} + R _{4}})$

We can then define a particular resisitve bridge called a balanced bridge, that has the property that ==for $x = 0$ the outptut will be $V = 0$ ==. To achive this we just need to define the $K$ -ratio: $K = \frac{R _{4}}{R _{10}} = \frac{R _{3}}{R _{2}}$ Then the output will be: $V_{0} = V (\frac{1 + x}{1 + x + K} - \frac{1}{1 + K})$

IMPORTANTE The output $V$ is not grounded, this means the for the read out electronics for a resistive bridge we will need a differential amplifier. (In general we will need a read-out with at least two inputs).NOT_SURE_ABOUT_THIS We CANNOT use a one-input amplifeir.

We have also seen two variations of resistive bridges: half brige and full bridge.

What is the use of a balanced resistive bridge?

If we were to measure the resistance of a strain gauge the formula, as we have said would be: $R = R_{0} (1 + x)$ .
In a real world scenario:

The voltage $R_{0}$ assumes values in the order of $[50 \ohm \div 700 \ohm]$

And the variation of $x$ is around $[0 \div 0.1]$ , so the variation due the strain is really low.

While for RTD sensors, specifically we have seen the PT100 sensor:

$R_{0} = 100 \ohm$ .

$x \in [- 0.78 \div 3.3]$ , based on the range of temperature $[- 200° C \div 850° C]$ .

For startets, at rest the output will be:

For the normal passive sensor $R = R_{0}$ .

For the resistive brige: $V_{o u t} = 0 V$ .
⇒ The bridge corrects the DC offset of the passive sensor.

So the range of values $x$ can assume are quite limited, especially for strain gauges, to increase this range, we can use a resistive bridge with a “high” source voltage $V$ , so that if we consider the sensitivity, of the output value around $x = 0$ , with $K = 1$ , and $V = 10 V$ we will have:NOT_SURE_ABOUT_THIS

For the normal passive sensor $α = 1$ .

For the resistive brige: $α = \frac{1}{4} \cdot V$ , so $α = 2.5$ .

==If we use a balanced resisitve brige, we will also compensate for temperature, since changes in temperature will affect all the resistors in the bridge equally, maintaining the balance==.

However note that the resisitve brige, results in a non-linear sensor, if we plot how $\frac{V _{o u t}}{V}$ changes given $R_{1} = R_{10} (1 + x)$ , we will obtain somenthing like this:

Sources:

Youtube ’# Basic configurations #1 - Wheatstone bridge’ (best explenation)

Youtube ’# Why you need a Wheatstone Bridge to get accurate Strain Gauge Readings (simple fast explenation)

Youtube ’# Wheatstone bridge & its logic | Electric current | Physics | Khan Academy’ (calculations)

Memory Card

If we do a thevenim equivalent we find: $V_{T H} = V_{0} = V (\frac{R _{1}}{R _{1} + R _{4}} - \frac{R _{2}}{R _{2} + R _{3}})$ And: $R_{T H} = (R_{1} ∥ R_{4}) + (R_{2} ∥ R_{3})$ We can then divide $V_{T H}$ in DC offest ( $V_{C}$ ) that is all component that do not depend on $R_{1} (T)$ (our sensor), and actual sensor component, all those parts that depend on $R_{1}$ :