Digital display thermometer

Take 15 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

The principle of a digital display thermometer is :

The temperature probe is a platinum resistor, whose resistance \(R_{Pt}\) varies with temperature according to the law :

\(R_{Pt}=R_{Pt,0}(1+\alpha t)\)

Where \(t\) is the temperature expressed in \(°C\), \(\alpha\) the coefficient of temperature of the resistivity (\(\alpha=4,00.10^{-3}\;C^{-1}\)), \(R_{Pt}\) the resistance in \(\Omega\) at the temperature \(t\) and \(R_{Pt,0}\) the resistance in \(\Omega\) at \(0°C\) (\(R_{Pt,0}=1,00.10^2\;\Omega\)).

This sensor is placed in one arm of a measuring bridge (Wheatstone bridge) fed by an ideal generator of direct current (\(I_0=1,00\;mA\)).

Resistances \(R_1\), \(R_2\), \(R'_3\) and \(R_4\) are assumed to be independent of temperature.

Question

Determining the voltage \(U_e\) as a function of \(I_0\), \(R_1\), \(R_2\), \(R_3=R'_3+R_{Pt}\) and \(R_4\).

Solution

The KVL applied to the loop \((I)\) defined in the figure above gives :

\({U_e} = - {R_1}{I_1} + {R_2}{I_2}\)

Yet, according to the rule of the current divider :

\({I_1} = \frac{{{R_2} + {R_4}}}{{{R_1} + {R_2} + {R_3} + {R_4}}}{I_0}\)

And :

\({I_2} = \frac{{{R_1} + {R_3}}}{{{R_1} + {R_2} + {R_3} + {R_4}}}{I_0}\)

Referring in expression of \(U_e\), we finally obtain :

\({U_e} = \frac{{{R_2}{R_3} - {R_1}{R_4}}}{{{R_1} + {R_2} + {R_3} + {R_4}}}{I_0}\)

Question

What value does it give to \(R'_3\) if the bridge is in equilibrium at \(t=0°C\), that is to say \(U_e=0\) at that temperature ?

Keep this value for the following exercises.

Solution

So that the bridge is in equilibrium at the temperature of \(0°C\), it is necessary that :

\({R_2}{R_3} = {R_1}{R_4}\)

So :

\(R{'_3} = \frac{{{R_1}{R_4}}}{{{R_2}}} - {R_{Pt,0}} = 900\;\Omega\)

Question

The temperature may vary between \(0°C\) and \(50°C\), determine \(U_e\) in terms of temperature \(t\).

Make the numerical application for \(t=0°C\), \(10°C\), \(20°C\), \(30°C\), \(40°C\) and \(50°C\).

Does \(U_e\) linearly vary with temperature \(t\) ?

Solution

At any temperature \(t\) (in \(°C\)), platinum resistance can be expressed as :

\({R_{Pt}} = {R_{Pt,0}} + \alpha {R_{Pt,0}}\,t\)

The voltage \(U_e\) can then be written :

\({U_e} = \frac{{{R_2}(R{'_3} + {R_{Pt,0}} + \alpha {R_{Pt,0}}\,t) - {R_1}{R_4}}}{{{R_1} + {R_2} + R{'_3} + {R_{Pt,0}} + \alpha {R_{Pt,0}}\,t + {R_4}}}{I_0}\)

Consequently, using the equilibrium condition of the bridge at \(0°C\) :

\({U_e} = \,\,\,\frac{{\alpha {R_{Pt,0}}\,t}}{{{R_1} + {R_2} + R{'_3} + {R_{Pt,0}} + {R_4} + \alpha {R_{Pt,0}}\,t}}{R_2}{I_0}\)

Numerically, we get :

\({U_e} = \frac{{{{4.10}^{ - 1}}\,t}}{{4\,000 + {{4.10}^{ - 1}}\,t}} = \frac{t}{{t + 10\,000}}\)

With \(t\) in \(°C\) and \(U_e\) in \(V\).

The voltage values \(U_e\) for the conditions suggested in the statement are listed in the table below :

Température \(t (°C)\)

\(0\)

\(10\)

\(20\)

\(30\)

\(40\)

\(50\)

Voltage \(U_e\) (V)

\(0\)

\(10^{-3}\)

\(2.10^{-3}\)

\(3.10^{-3}\)

\(4.10^{-3}\)

\(5.10^{-3}\)

It is found that in the range of considered temperatures, \(t<<10\;000\) and that, therefore, \(U_e \approx 10^{-4}\;t\) : the voltage \(U_e\) varies linearly with temperature.

Question

The signal \(U_e\) delivered is small, it is amplified. Which amplifier circuit do you use ? After amplification and shaping, the tension \(U_s\) is obtained :

\(t = 12,5({U_s} - 1)\)

With \(t\) in °C and \(U_s\) in \(V\).

Solution

A simple amplifier circuit can, for example, be a non-inverting amplifier circuit, made of an ideal operational amplifier and whose schema is recalled in the figure below, for which :

\(\frac{{U{'_e}}}{{{U_e}}} = 1 + \frac{{{R_2}}}{{{R_1}}}\)

As part of this exercise, the affine variation between \(t\) and \(U_s\) is :

\(t = 12,5({U_s} - 1)\)

With \(t\) in \(°C\) and \(U_s\) in \(V\).

Question

The voltage \(U_s\) is applied to an analog-digital converter (ADC) successive approximation - \(8\) bits - scale (\(0\) - \(5\) \(V\)).

The ADC can encode the analog voltage \(U_s\) into a number of \(8\) binary digits (\(8\) bits).

The transfer characteristic is given in the figure below.

  • Why do we use the base \(2\) not base \(10\) ? How many digital values ADC can it distinguish ?

  • What is the minimum variation \(\Delta U_s\) of \(U_s\) so that the numerical value \(N\) in base \(2\) is changed by one unit, that is to say the most significant bit ?

  • Deduce the minimum temperature variation \(\Delta t\) that can be enjoyed with this arrangement.

Solution

  • The base \(2\) is used here because it is, in general, used in electronic logic.

    The number \(N\) of \(8\) digits, which can be equal to \(0\) or \( 1\). Therefore, \(N\) may take \(2^8=256\) values.

  • The minimum voltage variation \(U_s\) that can be indicated will be given by :

    \(\Delta {U_s} = 4/256 = {1,56.10^{ - 2}}\;V = 15,6\,mV\)

    Which corresponds to a minimum change in temperature equal to :

    \(\Delta t = 12,5\;\Delta {U_s} \approx 0,20^\circ C\)