Discovery of Joule Heating
First experimentally proven by James Prescott Joule in 1841, hence the name ‘Joule Heating’.
Joule passed a current through a wire of fixed length, then immersed it in a bath of water of fixed volume/mass. He then observed the temperature in the water varied with current, length of the wire, and time.
Thermal Energy Generation
$$ Q = I^2 R t, $$
where $Q [\text{joule}] = [\frac{\text{kg m}}{s^2}]$ amount of Energy produced (in the form of heat).
$I [\text{A}]$ is the Current passing through the wire.
$R [\Omega]$ is the Resistance of the wire.
$t [\text{s}]$ is the Time the current is passed through the wire.
Specific Heat of Capacity
$$ C = \frac{Q}{M \Delta T}, $$
where $C [\text{joule / kg / K}]$ is the Specific Heat of Capacity of the material.
$M [\text{kg}]$ is the Mass of the material.
$\Delta T [\text{K}]$ is the Temperature Change of the material.
Ideally, the temperature of the material rises continuously while we apply electrical current.
However, there are always heat losses!
- Radiation
- $\frac{dQ}{dt} = - \phi_{\text{rad}} \Delta T^4$.
- Convection
- $\frac{dQ}{dt} = - \phi_{\text{conv}} \Delta T$.
- Conduction
- $\frac{dQ}{dt} = - \phi_{\text{cond}} \Delta T = - A_{\text{total}} \kappa \Delta T$.
where $A_{\text{total}} [m^2]$ is the Total Area attaching with nearby bodies.
$\kappa [\text{W / m / K}]$ is the Thermal Conductivity of the material.
$\Delta T [\text{K}]$ is the Temperature Difference relative to the nearby material.
Joule Heating (cont.)
Heat losses increase with the level of the material temperature higher than its environment.
In the steady state, we can always find,
$$ \frac{dQ}{dt} = CM \frac{dT}{dt} = \text{Heat Generation} + \text{Heat Losses} = 0. $$
Which means,
$$ CM \frac{dT}{dt} = I^2 R - \phi_{\text{cond.}} \Delta T - \phi_{\text{conv.}} \Delta T - \phi_{\text{rad.}} \Delta T^4 = 0. $$
When $\Delta T$ is small enough such that radiation effect is negligible, we have,
$$ I^2 R = (\phi_{\text{cond.}} + \phi_{\text{conv.}}) \Delta T. $$
Therefore, the Joule Heating will generate an increment of the bulk material temperature. We only consider the conduction effect here, hence, $$ I^2 R = A_{\text{total}} \kappa \Delta T. $$
The resistance $R$ of a material can be calculated from its ‘resistivity’ $\xi$ and dimensions,
$$ R = \xi \frac{L}{A}, $$
where $\xi$ is the electrical resistivity of the material.
$L$ is the length of material.
$A$ is the cross-sectional area.
Therefore, we have the relation,
$$ \begin{align*} I^2 R & = A_{\text{total}} \kappa \Delta T \newline \frac{V^2}{R} & = A_{\text{total}} \kappa \Delta T \newline V^2 = \xi L (\frac{A_{\text{total}}}{A}) \kappa \Delta T. \end{align*} $$
Thermal Strain
If the temperature increases, generally a body expands, whereas if the temperature decreases, a solid body will contract.
If the body material is homogeneous and isotropic, it has been found that the thermal strain $\varepsilon_T$ caused by a change in temperature $\Delta T$ can be expressed as,
$$ \varepsilon_T = \alpha \Delta T, $$
where $\alpha$ is a property of the material, referred to as the coefficient of linear thermal expansion. The units of $\alpha$ measure strain per degree of temperature.
Therefore, we can convert electrical voltage to temperature, and to mechanical deformation/defection.
How do Thermal Actuators Work?
One component of MEMS is thermal actuators.
When a voltage is applied to the terminals, current flows through the device.
However, because of the different widths, the current density is unequal in the two arms. This leads to a different rate of Joule heating in the two arms, and thus to different amounts of thermal expansion.
The thin arm is often referred to as the hot arm, and the wide arm is often referred to as the cold arm.
So, we make use of differing arm thickness to provide movement.
But we have a problem, using the principles described, it is difficult to obtain a movement linear to the supply voltage.
- The wide arm
- Small resistance → Small heat generation.
- Large are → Heat loss due to convection.
- Temperature ~ environmental temperature.
- The anchors
- Physically connecting to the substrate (heat sink).
- Temperature ~ environmental temperature.
We approximate the length difference between the (extended) narrow arm and the wide arm during Joule heating as,
$$ V^2 \approx \xi L_1 \frac{A_{\text{total}}}{A} \kappa \Delta T. $$
Since the beam is contacting with other parts of the device on its both ends, we estimate $\frac{A_{\text{total}}}{A} \approx 2$, and the temperature rise becomes,
$$ \Delta T \approx \frac{V^2}{2 \xi L_1 \kappa}. $$
Thus, the length increment of the narrow arm is,
$$ \begin{align*} \Delta L & = L_1 \epsilon_T \newline & = \alpha L_1 \Delta T \newline & = \alpha \frac{V^2}{2 \xi \kappa}. \end{align*} $$
We further approximate the narrow arm is bent with the shape of an arc (this is a very rough approximation).
We neglect the geometric effects caused by the bending of the narrow beam at the end of the wide arm (again, this is a very rough approximation).
Thus, we can approximate the tip displacement $x$ as, $$ x \approx \Delta L \frac{L_1}{d} $$
Since we have applied a couple rough approximations, we should consider adding a correction factor for the expression of $x$, $$ \begin{align*} x & = \phi \Delta L \frac{L_1}{d} \newline & = \phi \alpha \frac{V^2 L_1}{2 \xi \kappa d}, \end{align*} $$
The correction factor has been determined by experiments, such that,
$$ x = \frac{\alpha v^2 L_1}{2 \xi \kappa d \sqrt{0.7707 + 0.3812(w_{\text{wide}}^2 / d^2)}}, $$
where $w_{\text{wide}}$ is the width of the wide arm.
Piezoresistivity
Consider a conductive block of length $L$, width $a$, and thickness $b$.
The resistance $R_0$ along the length is given by,
$$ R_0 = \frac{\xi L}{a b} = \frac{\xi L}{A}. $$
where $\xi$ is the resistivity of the material.
$A$ is the cross-sectional area.
Change of relative dimensions, as the resistance is related to length and cross-sectional area (local),
$$ R = R_0 + \frac{\partial R}{\partial L} \Delta L + \frac{\partial R}{\partial \xi} \Delta \xi + \frac{\partial R}{\partial a} \Delta a + \frac{\partial R}{\partial b} \Delta b. $$
A force applied lengthwise results in deformation in all three axes.
with $\sigma_x \neq 0$ and $\sigma_y = \sigma_z = 0$.
$$ \begin{align*} \Delta L & = \varepsilon L = \frac{\sigma_x L}{E} \newline -\Delta a & = \varepsilon v a = \frac{\sigma_x v a}{E} \newline -\Delta b & = \varepsilon v b = \frac{\sigma_x v b}{E}. \end{align*} $$
We also consider that $\Delta \xi \approx 0$.
This causes a change in resistance,
$$ R = [1 + (1 + 2v) \varepsilon] R_0. $$
Thus,
$$ \frac{\Delta R}{R_0} = (1 + 2v) \varepsilon = G \varepsilon. $$
Where $G = (1 + 2v)$ is the gauge factor.
$v$ is the Poisson’s ratio.
Applications at Macro Scale
Spot-weldable strain gauges are used with strain gauge sensors and a vibrating wire indicator or data logger to monitor strain in steel members.
Typical applications include:
- Monitoring structural members of buildings and bridges during and after construction.
- Determining load changes on ground anchors and other post-tensioned support systems.
- Monitoring load in strutting systems for deep excavations.
- Measuring strain in tunnel linings and supports.
- Monitoring areas of concentrated stress in pipelines.
- Monitoring distribution of load in pile tests.
Metal Strain Gauge
For metals, the resistivity is not changed significantly by the stres. The gauge factor is believed to be contributed by the change of dimensions. These may be made from thin wires or metal films that may be directly fabricated on top of micro structures.
Thin film strain gauges are typically fabricated on top of flexible plastic substrates and glued to surfaces.
Etched foil gauges consist of a conduction path etched onto metal clad plastic film. The strain gauges are designed to be glued, using very special procedures onto the component to be tested.
When the component stretches, the strain gauge will also stretch as will the etched conduction path.
Analysis of Accelerometer
Acceleration induced force $F = ma$.
The force induces stress at the fixed end of the cantilever beam.
The stress is detected by change in resistance.
Assumptions:
-
Assume entire resistance is concentrated at the anchor.
-
For moment of inertia at the end, ignore the thickness of the resistor.
-
Assume the stress on the resistor is the maximum value.
-
The proof mass is rigid. It does not bend because of the significant thickness and width.
Under a given $a$, the force has a magnitude $F = ma$.
The moment applied at the fixed end of beam is,
$$ M = F\left(l + \frac{L}{2}\right). $$
Therefore, the maximum strain, which is the strain experienced by the resistor, is, $$ \begin{align*} \varepsilon_{\text{max}} & = \frac{Mt}{2EI} \newline & = \frac{F\left(l + \frac{L}{2}\right)t}{\frac{Ewt^3}{6}} \newline & = \frac{6F\left(l + \frac{L}{2}\right)}{Ewt^3}. \end{align*} $$
The strain is applied in the longitudinal direction of the resistor.
Assuming the gauge factor is $G$, the change in resistance is,
$$ \frac{\Delta R}{R} = G \varepsilon_{\text{max}} = \frac{6GF\left(l + \frac{L}{2}\right)}{Ewt^3} = \left(\frac{6GF\left(l + \frac{L}{2}\right)}{Ew^2t^2}\right) a. $$