Development of miniature heater assembly for compact battery powered thermo-luminescence reader

The resistive heating element having resistance of R_H (Ohm) generates a heat dH(t) (Calories) under current I_H (t) (Ampere) passing through it for time dt (s) can be derived in term of potential difference V(t) (Volt) developed across resistive element as

$$\begin{eqnarray}&&{\rm{dH}}\left({\rm{t}}\right)=\displaystyle \frac{V\,\,{{\rm{I}}}_{{\rm{H}}}({\rm{t}})}{{\rm{J}}}{\rm{dt}}{\rm{.}}\end{eqnarray} \tag{ 1 }$$

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dH}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{\rm{V}}\,{{\rm{I}}}_{{\rm{H}}}({\rm{t}})}{{\rm{J}}}.\end{eqnarray} \tag{ 2 }$$

Where, 'J' is mechanical equivalent of heat (J = 4.184 J cal⁻¹)

The associated potential difference can be expressed as

$$\begin{eqnarray}&&{\rm{V}}={{\rm{R}}}_{{\rm{H}}}.{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)\end{eqnarray} \tag{ 3 }$$

Incorporating equation (3) in (2), we get

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dH}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}({\rm{t}})}{{\rm{J}}}.\end{eqnarray} \tag{ 4 }$$

$$\begin{eqnarray}&&{\rm{dH}}\left({\rm{t}}\right)=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}({\rm{t}})}{{\rm{J}}}{\rm{dt}}{\rm{.}}\end{eqnarray} \tag{ 5 }$$

This relationship is known as Joule's first law or Joule–Lenz law [8, 9]. The most of the TL readouts have being carried out with various temperature profiles, apart from generally utilized iso-thermal (constant temperature) and linear heating [10, 11]. The desirable temperature verses time profiles have been achieved by feeding programmable analog voltage V (t) to the temperature control circuits. The most generalized form of voltage verses time profile used in such circuits are described as [12, 13]

$$\begin{eqnarray}&&{\rm{V}}\left({\rm{t}}\right)={{\rm{kt}}}^{{\rm{l}}}.\end{eqnarray} \tag{ 6a }$$

Where, k is constant having dimension of Volt s^−l, and l the dimensionless number termed as 'Time Base Power' (TBP), can have values 0 ≤ l ≤ ∞ [14]. This voltage profile will lead to a modulation of the heater DC current with use of MOSFET/transistor and under AC current supply, phase angle control by SCR (Silicon Control Rectifier) is used for this purpose. Therefore, for present heater element this DC current modulation can be expressed as

$$\begin{eqnarray}&&{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)=\xi {\rm{V}}\left({\rm{t}}\right)=\xi {{\rm{kt}}}^{{\rm{l}}}.\end{eqnarray} \tag{ 6b }$$

Where, ξ is proportionality constant with unit 'A/V' and takes constant value for most of designed heater circuit. In practice, the heating is carried out either by the linear increasing or constant temperature profiles with respect to time. These two cases can be expressed by assuming l = 1 or l = 0 in above equation (6b), respectively. Further the heating profiles can be given as

Linear

$$\begin{eqnarray}&&{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)=\xi {\rm{V}}\left({\rm{t}}\right)=\xi {\rm{kt}}{\rm{.}}\end{eqnarray} \tag{ 6c }$$

and constant heating linear

$$\begin{eqnarray}&&{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)=\xi {\rm{V}}\left({\rm{t}}\right)=\xi {{\rm{k}}}_{{\rm{o}}}.\end{eqnarray} \tag{ 6d }$$

Where, k_o: takes constant value with unit (V).

Here, we assumed that the resistive heating element is fully covered with heat insulation material, thus does not have any privilege of exchanging a thermal energy with the surrounding environment. Under above assumption equation (5), presents the amount of heat generated in resistive heating element due to current I_H(t) passing through it. This results in an increase of temperature dT. Now the amount of heat energy associated with the heating element can be expressed as

$$\begin{eqnarray}&&{\rm{dH}}={\rm{C}}\,{\rm{m}}\,{\rm{dT}}{\rm{.}}\end{eqnarray} \tag{ 7 }$$

Here, C is specific heat capacity of solid expressed in [kJ/kg/K] cal/g-K and 'm' is the mass of heating element expressed in gram [1 kJ kg⁻¹ = 0.2390 kcal kg⁻¹]. In present assuming all the heat energy generated in heating element contributes for increase in temperature. We can incorporate equation (6c) in equation (5) and equating to equation (7) as

$$\begin{eqnarray}&&{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)=\xi {\rm{V}}\left({\rm{t}}\right)=\xi {{\rm{kt}}}^{{\rm{l}}}.\end{eqnarray} \tag{ 8a }$$

$$\begin{eqnarray*}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}}{\mathrm{JC}\,m\,}\end{eqnarray*}$$

When 'C' is in SI unit the above equation can be further expressed as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}}{C\,m\,}.\end{eqnarray} \tag{ 8b }$$

Integrating equation (8b) from initial start temperature T_o to the final temperature T in time 't ' for maximum allowed current I_max, we get

$$\begin{eqnarray}&&\displaystyle {\int }_{{{\rm{T}}}_{{\rm{o}}}}^{{\rm{T}}}{\rm{dT}}({\rm{t}})=\displaystyle {\int }_{0}^{{\rm{t}}}\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\left(\xi {\rm{k}}\right){}^{2}{{\rm{t}}}^{2{\rm{l}}}}{{\rm{C}}\,m\,}{\rm{dt}}.\end{eqnarray} \tag{ 8c }$$

Here,

$$\begin{eqnarray}&&{\rm{t}}={\left[\displaystyle \frac{{{\rm{I}}}_{{\rm{\max }}}}{\xi {\rm{k}}}\right]}^{-{\rm{l}}}\end{eqnarray} \tag{ 8d }$$

$$\begin{eqnarray*}&&\therefore {\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {\rm{k}}\right)}^{2}}{{\rm{C}}\,{\rm{m}}}\displaystyle {\int }_{0}^{{\rm{t}}}{{\rm{t}}}^{2{\rm{l}}}\,{\rm{dt}}\end{eqnarray*}$$

$$\begin{eqnarray}&&\therefore {\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {\rm{k}}\right)}^{2}}{{\rm{C}}\,{\rm{m}}}\left[\displaystyle \frac{{{\rm{t}}}^{2{\rm{l}}+1}}{2{\rm{l}}+1}\right].\end{eqnarray} \tag{ 8e }$$

$$\begin{eqnarray}&&\therefore {\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}}{{\rm{C}}\,{\rm{m}}}\left[\displaystyle \frac{{\rm{t}}}{2{\rm{l}}+1}\right].\end{eqnarray} \tag{ 8f }$$

From equation (8a)

$$\begin{eqnarray}&&\therefore {\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}({\rm{t}})}{{\rm{C}}\,{\rm{m}}}\left[\displaystyle \frac{{\rm{t}}}{2{\rm{l}}+1}\right].\end{eqnarray} \tag{ 8g }$$

$$\begin{eqnarray}&&\therefore {\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{\rm{P}}({\rm{t}})}{{\rm{C}}\,{\rm{m}}}\left[\displaystyle \frac{{\rm{t}}}{2{\rm{l}}+1}\right].\end{eqnarray} \tag{ 9a }$$

Here, ${\rm{P}}\left({\rm{t}}\right)={{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}\left({\rm{t}}\right)$ is the electrical power (Watt) delivered to the resistive element. Further, re-arranging the terms in equation (9a), we get,

$$\begin{eqnarray}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\left[\displaystyle \frac{{\rm{P}}({\rm{t}})\,\,}{{\rm{C}}\,{\rm{m}}\,(2{\rm{l}}+1)\,}\right]{\rm{t}}{\rm{.}}\end{eqnarray} \tag{ 9b }$$

Let's define $\beta ({\rm{t}})=\tfrac{{\rm{P}}({\rm{t}})\,\,}{({\rm{C}}\,{\rm{m}})(2{\rm{l}}+1)}$ is Effective Heating (EH) rate expressed in K s⁻¹. Under this assumption equation (9b) becomes

$$\begin{eqnarray}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\beta \left({\rm{t}}\right){\rm{t}}{\rm{.}}\end{eqnarray} \tag{ 9c }$$

Here, the value of β(t) increases with increase in the power delivery to the resistive element and decreases with both, increase in mass as well as heat capacity of the heating element along with TBP (l). For the condition of constant current (I_o) passing though resistive element with respect to time, i.e. l = 0; the equation (6d), reduces to

$$\begin{eqnarray}&&{{\rm{I}}}_{{\rm{H}}}\left({\rm{t}}\right)={{\rm{I}}}_{{\rm{o}}}=\xi {{\rm{k}}}_{{\rm{o}}}.\end{eqnarray} \tag{ 10a }$$

Equating equations (9c) and (10a), we get constant heating rate and heating profile as

$$\begin{eqnarray}&&{\beta }_{{\rm{o}}}=\displaystyle \frac{{{\rm{P}}}_{{\rm{o}}}\,\,}{{\rm{C}}\,{\rm{m}}}.\end{eqnarray} \tag{ 10b }$$

$$\begin{eqnarray}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+{\beta }_{{\rm{o}}}{\rm{t}}{\rm{.}}\end{eqnarray} \tag{ 10c }$$

The equation (10c) gives the linear profile of temperature with constant heating rate ${\beta }_{{\rm{o}}}$ for given constant heater element current. The whole heater assembly is composite in nature and consists of different materials, having different heat capacities as shown in figure 1. Thus, it is more practical to consider the combined effect of all the different components contained inside the heater system. Therefore, lets re-define the EH rate as

$$\begin{eqnarray}&&{\beta }_{{\rm{o}}}=\left[\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\right]{{\rm{I}}}_{{\rm{H}}}^{2}\left({\rm{t}}\right).\end{eqnarray} \tag{ 10d }$$

Where, (Cm)_H is the effective combined heat capacity - mass product which is a weighted average for all the different heater components including TL phosphor material. In practice for such miniature heater a TL phosphor of considerable large mass will affects the performance of heating system. However, the temperature feedback mechanism will defiantly negate this effect up to a greater extent. Further simplifying the equation (10d) as

$$\begin{eqnarray}&&{\beta }_{{\rm{o}}}=\psi {{\rm{I}}}_{{\rm{H}}}^{2}({\rm{t}})\end{eqnarray} \tag{ 10e }$$

Where, $\psi =\left[\tfrac{{{\rm{R}}}_{{\rm{H}}}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\right]$ is a new parameter termed as 'Heating Rate to Current Coefficient' (HRCC) expressed in K/(s.A²). The HRCC can be measured experimentally by applying a constant current to the heating element under feedback mechanism deactivated. The figure 3 shows the temperature profiles with respect to time of designed TL heater strip under various constant heater currents I_H for above discussed condition. The data corresponding to the initial ∼5% of the temperature profiles (as shown in figure 3) is considerably good approximation for fitting equation (10d). This assumption is being supported by the fact that at a higher temperature the heat associated with heating assembly increases and leads to deviation of the temperature profile from equation (10d). The EH rate β_o for different constant heater current can be determined from linear fit to initial temperature profiles discussed above. The determination of slope of the line fitted to experimental data (figure 4) of heating rate 'β' versus heater current I²_H, gives the HRCC parameter value ∼8.39 (K/s.A²) with intercept at β – axis of ∼0.47. This offset of EH rate (β_o) is due to the variation in ambience temperature of the heater assembly. Considering the measured value of HRCC for 2 Ω resistance of the nichrome wire the value of (Cm)_H for this kanthal planchet can be estimated from the definition of ψ introduced in equation (10e) as 0.23 J K⁻¹. For present kanthal strip planchet (shown in figure 1, dimension 2.0 × 1.2 × 0.025 cm , density 7.1 g/cc) with assuming specific heat capacity(C) at 200°C as 0.56 J g⁻¹ K⁻¹, the calculated value of (Cm)_H turned out to be ∼0.22 J K⁻¹. In this analysis we have ignored the contribution to composite heat capacity from nichrome wire due to its negligible mass with compare to the kanthal planchet. However, the close agreement between measured and estimated values of (Cm)_H shows imperceptible effect of a composite structure on the heat capacity as compared to kanthal planchet. Therefore, the basic thermal load other than kanthal planchet can be ignored in further discussion. Let us consider a more generalized form of the heating profiles for different TBP (l) values as expressed in equation (8e)

$$\begin{eqnarray*}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {\rm{k}}\right)}^{2}}{{\rm{C}}\,{\rm{m}}}\left[\displaystyle \frac{{{\rm{t}}}^{2{\rm{l}}+1}}{2{\rm{l}}+1}\right]\end{eqnarray*}$$

Now introduce a new parameter 'ω' as a parameter having dimension of K/s^2l+1

$$\begin{eqnarray}&&\omega ={\rm{\Psi }}{\left(\xi {\rm{k}}\right)}^{2}\end{eqnarray} \tag{ 11a }$$

Thus equation (8e) become

$$\begin{eqnarray}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\omega \left[\displaystyle \frac{{{\rm{t}}}^{2{\rm{l}}+1}}{2{\rm{l}}+1}\right].\end{eqnarray} \tag{ 11b }$$

For a linear increase in heater current I_H(t) with respect to time i.e. TBP (l) = 1, we get

$$\begin{eqnarray}&&{\rm{T}}\left({\rm{t}}\right)={{\rm{T}}}_{{\rm{o}}}+\omega \left[\displaystyle \frac{{{\rm{t}}}^{3}}{3}\right].\end{eqnarray} \tag{ 11c }$$

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Therefore, the linear increase in heater current with respect to time results in non-linear increase in temperature as depicted from equation (11c). The numerically simulated plot of equation (11b) for the different temperature verses heater current profiles have been plotted in figure 5. It is evident from figure 5 that in order to obtain a linear heating profile one must use non-linear current profile (l < 1). More of this will be discussed in further sections of this paper.

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This is the most practical scenario, where the resistive heating element (heater assembly) is allowed to lose heat energy to the environment/surrounding by conduction, convection and radiation processes. Therefore, we can write the above process as

The sufficient thermal insulation has been incorporated heater assembly to minimize the conduction heat loss from kanthal strip to fixing components. As the complete heater assembly has been enclosed in a light tight drawer module no free air flow is possible around heater. Thus, the convection contribution to heat loss can be ignored here for a further discussion. The remaining conservation energy law can be represented as

$$\begin{eqnarray}&&{{\rm{H}}}_{{\rm{J}}}={{\rm{H}}}_{{\rm{T}}}+{{\rm{H}}}_{{\rm{B}}}+{{\rm{H}}}_{{\rm{C}}}.\end{eqnarray} \tag{ 12a }$$

The generalized from equation (4) can be derived in terms of equation (12a) by incorporating the conduction and radiative losses as

$$\begin{eqnarray}&&\begin{array}{l}\displaystyle \frac{{{\rm{dH}}}_{{\rm{J}}}\left({\rm{t}}\right)}{{\rm{dt}}}={{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}\left({\rm{t}}\right)={\left({\rm{C}}{\rm{m}}\right)}_{{\rm{H}}}\,\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}\\ \,+\,\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}+\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}\left({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\right)}{{\rm{L}}}.\end{array}\end{eqnarray} \tag{ 12b }$$

Where, ε : is an emissivity, A: the surface area of the kanthal strip (m²), σ: the Stefan-Boltzmann's stiffens constant (5.7 × 10⁻⁸ W m⁻² K⁻⁴), γ, is the thermal conductivity (W/m.K), Ac, the effective cross sectional area of the conduction and L, the length of kanthal stripe over which thermal gradient is observed (as shown in figure 1). With the above defined terms the equation (12b) can be reorganized as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{H}}}^{2}\left({\rm{t}}\right)}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}\left({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\right)}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 12c }$$

Incorporating equation (6c) in (12c), we get

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}\left({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\right)}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 12d }$$

Further incorporating equation (10e) in the above equation we get,

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\psi {\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}\left({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\right)}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 12e }$$

Let us introduce few terms namely Effective Heating (EH) here β as

$$\begin{eqnarray}&&\beta ({\rm{t}})=\psi {\left(\xi {{\rm{kt}}}^{{\rm{l}}}\right)}^{2}.\end{eqnarray} \tag{ 12f }$$

Similar to equations (11a) and (9c) with dimension (K s⁻¹) and having direct dependence on an electrical resistivity and current flowing though a heating element. On the other hand let us define η as

$$\begin{eqnarray}&&\eta =\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 12g }$$

Here, η having dimensions (K⁻³ s⁻¹) and it represents the heat loss constant, which depends on effective area A (As shown in figure 1) and the emissivity of heater kanthal strip. The term μ is defined as

$$\begin{eqnarray}&&\mu =\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\end{eqnarray} \tag{ 12h }$$

with dimensions (s⁻¹) and represents the conduction heat transfer from center of kanthal strip to fixing screws along and other metallic objects supporting the heating assembly( as shown in figure 1). Having defined the above parameters equation (12e) can be further expressed as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\beta ({\rm{t}})-\eta {{\rm{T}}}^{4}\left({\rm{t}}\right)-\mu \left[{\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\right].\end{eqnarray} \tag{ 12i }$$

The equation (12i) represents a generalized rate of change of temperature with the heater current for a condition of heater assembly losing thermal energy to the surrounding environment. This rate equation is a first order non-linear ordinary differential equation and can be solved numerically by using Euler's method.

2.2.3.1. Analysis under no temperature feedback from heater planchet assembly

Let us considering a case of constant heater current (with respect to time), under this condition l = 0, I_H = I_o and equation (12c) becomes

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{o}}}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}})}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 13a }$$

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\left[\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {{\rm{k}}}_{{\rm{o}}}\right)}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\right]-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}})}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 13b }$$

Let us introduce a parameters, ${\omega }_{{\rm{o}}}=\tfrac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi {{\rm{k}}}_{{\rm{o}}}\right)}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}$ (similar as in case of equation (11a)) as constant EH having dimension (K s⁻¹). Under above assumptions the equation (13b) can be written in form of equation (12i) as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}={\omega }_{{\rm{o}}}-\eta {{\rm{T}}}^{4}\left({\rm{t}}\right)-\mu \left[{\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\,\right].\end{eqnarray} \tag{ 13c }$$

The evaluations of various constant parameters appearing in the equation (13c) are being described below. The value of ω_o can be obtained by applying constant current to heating elements (as discussed in previous section 2.2.2) along with given values of $\tfrac{{{\rm{R}}}_{{\rm{H}}}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.$ The experimentally measured value of this parameter was found to be ∼ 8.39 K s⁻¹ A⁻², yielding the estimated value of (Cm)_H ∼ 0.23 J K⁻¹. The values of ω_o can be selected for different constant current from measured (HRCC) as given in equations (10d) and (10e). The value of $\mu =\tfrac{\lambda {{\rm{A}}}_{{\rm{c}}}}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}$ in present setup is being calculate by assuming the thermal conductivity (γ) = 14.3 W m⁻¹ K⁻¹ (average value for kanthal @350 °C), Ac ∼2.3 × 10⁻⁶ m² and L = 0.02 m (figure 1).

The value of μ obtained after incorporating the values of γ, A_c and L was found to be ∼7.15 × 10⁻³ (s⁻¹). The value of η was also being calculated as ∼3.51 × 10⁻¹¹(s⁻¹ K⁻³) by assuming ε = 0.6 and emission area A = 2.4 × 10⁻⁴ m² for kanthal planchet as shown in figure 1. The differential equation (13c) has been solved by using Euler's method after incorporating the evaluated values of ω_o, μ and η parameters . The solution of the equation (13c) has been plotted in figure 6(B), for various values of constant heater current I_H (i.e. I_o). There is considerable good agreement observe between the experimentally measured and numerical simulated temperature profiles as shown in figures 3 and 6(A), (B). This validates the theoretical model developed under the assumptions made in above discussion for heater assembly. Figure 7 shows the optimized numerical fit of the equation (13c) for an experimental data corresponding to I_o = 1.4 A. The numerical fit has incorporated the evaluated values of the various parameters, such as μ = 1.5 × 10⁻¹⁰ s⁻¹ K⁻³, η = 0.032 s⁻¹, HRCC = 8.39 K s⁻¹.A⁻² and ω_o = 24.5 K s⁻¹.

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The slight difference in the calculated and simulated temperature profiles happened due to over simplified assumptions made in the estimation of parameters associated with heating assembly. The comparison of the experimental and numerically simulated temperature profiles for TBP(l) = 1 have been depicted in figure 8 showing good agreement among them. It is noticed that the last two terms of the equation (13c) are responsible for introducing sub-linearity in generated heating rate at higher planchet temperatures. More specifically at relatively lower planchet temperature the 3rd conduction term will dominates than after the 2nd radiation term which takes over at even higher temperatures. The generalized value of TBP (l) (0 < l ≤ ∞) will results in complicated solutions of equation (13c). Numerically simulated solutions of the equation (13c) has been plotted in figure 9, i.e. the variation of heater planchet temperature with different values of TBP (l). It is conclusive that for the value of TBP (l) < 1 (here l = 0.3) results in linear temperature profile even though the heater current I_H is non-linear. Thus, the heater current profile TBP (l) < 1 has desirable impact of overcoming from a non-linearity of the temperature profiles. This suggested that the temperature feedback correction in the input heater current profile should be designed in such a way to get a heater current profile as TBP < 1. This will produce desirable input voltages to current relation for the generation of a linear temperature profile.

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2.2.3.2. Analysis under negative feedback from heater planchet assembly

Figure 10 shows schematic diagram of generated negative temperature feedback from thermocouple sensor. Under this assumption the equation (8a) can be further modified to incorporate the heater temperature feedback corrected current ${\rm{I}}{\left({\rm{t}}\right)}_{{\rm{Hfb}}}$ as

$$\begin{eqnarray}&&{\rm{I}}{\left({\rm{t}}\right)}_{{\rm{Hfb}}}=\xi \left\{{\rm{V}}\left({\rm{t}}\right)-{\rm{V}}{\left({\rm{T}}\right)}_{{\rm{fb}}}\right\}=\xi \left\{{{\rm{kt}}}^{{\rm{l}}}-\theta \,{\rm{T}}\left({\rm{t}}\right)\right\}.\end{eqnarray} \tag{ 14a }$$

Here,

$$\begin{eqnarray}&&{\rm{V}}{\left({\rm{T}}\right)}_{{\rm{fb}}}=\theta \,{\rm{T}}\left({\rm{t}}\right).\end{eqnarray} \tag{ 14b }$$

and θ: is defined as a fractional constant (with dimension of V/K). This parameter determines the dependency of a temperature feedback voltage at any instantaneous time. The value of θ is to be set by tuning the hardware resisters of the temperature feedback amplifier. Further, incorporating equation (14a) in equation (12d) we get

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}\,{{\rm{I}}}_{{\rm{Hfb}}}^{2}({\rm{t}},{\rm{T}})}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}-\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}})}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{eqnarray} \tag{ 14c }$$

$$\begin{eqnarray}&&\begin{array}{l}\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\left[\displaystyle \frac{{{\rm{R}}}_{{\rm{H}}}{\left(\xi \left\{{{\rm{kt}}}^{{\rm{l}}}-\theta {\rm{T}}\left({\rm{t}}\right)\right\}\right)}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\right]-\displaystyle \frac{\varepsilon \sigma \,{\rm{A}}\,{\rm{T}}{\left({\rm{t}}\right)}^{4}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}\\ \,-\,\displaystyle \frac{\gamma {{\rm{A}}}_{{\rm{c}}}({\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}})}{{\rm{L}}{({\rm{C}}{\rm{m}})}_{{\rm{H}}}}.\end{array}\end{eqnarray} \tag{ 14d }$$

Equation (14c) can be expressed in a form equation (13c) as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{dT}}\left({\rm{t}}\right)}{{\rm{dt}}}=\omega {({\rm{T}}({\rm{t}}))}_{{\rm{o}}}-\eta {\rm{T}}{({\rm{t}})}^{4}-\mu \left[{\rm{T}}\left({\rm{t}}\right)-{{\rm{T}}}_{{\rm{o}}}\,\right].\end{eqnarray} \tag{ 14e }$$

Where, $\omega {({\rm{T}}({\rm{t}}))}_{{\rm{o}}}=\tfrac{{{\rm{R}}}_{{\rm{H}}}{(\xi \{{{\rm{kt}}}^{{\rm{l}}}-\theta {\rm{T}}\left({\rm{t}}\right)\})}^{2}}{{({\rm{C}}{\rm{m}})}_{{\rm{H}}}};$ is a feedback corrected Real Time Temperature Dependent Dynamic (RTDD) heating rate having unit of (K s⁻¹). Other parameters are having same meaning as discussed above in section 2.2.3.1. The experimental linear temperature profile recorded for l = 1, with set value of $\xi \theta \sim 0.01\,{\rm{A}}\,{{\rm{K}}}^{-1}$ for the developed miniature heart planchet with temperature feedback has been shown as shown in figure 11(A). A similar linear heating profile has been numerically simulated by using equation (14e), as shown in figure 11(B), with the value of $\xi \theta$ ∼ 0.01 A K⁻¹, μ and η as determine from numerical fitting (discussed in previous section 2.2.3.1. The validity of the derived equation (14e) is evidenced from the observed significant agreement between the experiential and numerically simulated temperature profiles. Thus, the proposed theoretical model expressed by equation (14e) can be used for designing of such a miniature heating assembly with limits of the assumptions made in above discussion.

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The figures 12(A)–(D) depicts the experimental temperature profiles for the constant heater current under negative temperature feedback condition. It was fund that the PID controller tries to increase the transient current in heating element in order to achieve an appropriate step function of the temperature profile. However, to protect heating wire maximum current supply to the heater element was restricted to 1.6 A. This leads to saturation of current for higher temperature step function as seen from figures 12(A)–(C). But, for the lower temperature step function, we still get the iso-thermal temperature profile with maximum possible heating rate as shown in figure 12(D). It can observed from this figure that for set target temperature of 50 °C, the current profile reaches to the maximum in transient response, thereafter, settles down at an around 400 mA. This settling in current has becomes possible due to the temperature feedback effect, which was active in the temperature profile of figure 3 (case of no feedback).

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