DIODE THERMOMETER Objective: Design a simple electronic thermometer using the temperature characteristics of a diode. Specifications: The temperature range is to be 0 to 100 ◦F. Design Approach: We will use the forward-bias diode temperature characteristics as shown in Figure 1.20. If the diode

Question

DIODE THERMOMETER Objective:

Design a simple electronic thermometer using the temperature characteristics of a diode.

Specifications: The temperature range is to be 0 to 100 ◦F. Design Approach: We will use the forward-bias diode temperature characteristics as shown in Figure 1.20. If the diode current is held constant, the variation in diode voltage is a function of temperature.

Choices: Assume that a silicon pn junction diode with a reverse-saturation current of [latex]I_{S} = 10^{−13} A[/latex] at T = 300 K is available.

Accepted Answer

Neglecting the (−1) term in the diode I–V relation, we have

[latex]I_{D} = I_{S} e^{V_{D}/V_{T}} ∝ n_{i}^{2} e^{V_{D}/V_{T}} ∝ e^{ −E_{g} /kT} . e^{V_{D}/V_{T}}[/latex]

The reverse-saturation current [latex]I_{S}[/latex] is proportional to [latex]n_{i}^{2}[/latex] and in turn [latex]n_{i}^{2}[/latex] is proportional to the exponential function involving the bandgap energy [latex]E_{g}[/latex] and temperature.
Taking the ratio of the diode current at two temperature values and using the definition of thermal voltage, we [latex]have^{3}[/latex]

[latex]\frac{I_{D1}}{I_{D2}} = \frac{ e^{ −E_{g} /kT_{1}} . e^{V_{D1}/V_{kT_{1}}}}{ e^{ −E_{g} /kT_{2}} . e^{V_{D2}/V_{kT_{2}}}}[/latex] (1.35)

where [latex]V_{D1}[/latex] and [latex]V_{D2}[/latex] are the diode voltages at temperatures [latex]T_{1}[/latex] and [latex]T_{2}[/latex] respectively. If the diode current is held constant at the different temperatures, Equation (1.35) can be written as

[latex]e^{e V_{D2}/kT_{2}} = e^{ −E_{g} /kT_{1}} e^{+E_{g} /kT_{2}} e^{e V_{D1}/k{T_{1}}}[/latex] (1.36)

Taking the natural logarithm of both sides, we obtain

[latex]\frac{e V_{D2}}{kT_{2}} = \frac{- E_{g}}{kT_{1}} + \frac{E_{g}}{kT_{2}} + \frac{e V_{D1}}{kT_{1}} [/latex] (1.37)

or

[latex]V_{D2} = \frac{- E_{g}}{e} \left(\frac{T_{2}}{T_{1}} ight) + \frac{E_{g}}{e} + V_{D1} \left(\frac{T_{2}}{T_{1}} ight)[/latex] (1.38)

For silicon, the bandgap energy is [latex]E_{g} /e = 1.12 V[/latex] Then, assuming the bandgap energy does not vary over the temperature range, we have
[latex]V_{D2} = 1.12 \left(1 - \frac{T_{2}}{T_{1}} ight) + V_{D1} \left(\frac{T_{2}}{T_{1}} ight)[/latex] (1.39)

Consider the circuit shown in Figure 1.47. Assume that the diode has a reversesaturation current of [latex]I_{S} = 10^{-13} A[/latex] at T = 300 K. From the circuit, we can write

[latex]I_{D} = \frac{15 - V_{D}}{R} = I_{S} e^{V_{D}/V_{T}}[/latex]

or

[latex]\frac{15 - V_{D}}{15 imes 10^{3}} = 10^{-13} e^{V_{D}/0.026}[/latex]

By trial and error, we find

[latex]V_{D} = 0.5976 V[/latex]

and

[latex]I_{D} = \frac{15 - 0.5976}{15 imes 10^{3}} ⇒ 0.960 mA[/latex]

In Equation (1.39), we can set [latex]T_{1} = 300 K[/latex] and let [latex]T_{2} ≡ T[/latex] be a variable temperature. We find

[latex]V_{D} = 1.12 - 0.522 \left(\frac{T}{300} ight)[/latex] (1.40)

so the diode voltage is a linear function of temperature. If the temperature range is to be from 0 to 100 ◦F, for example, the corresponding change in kelvins is from 255.2 to 310.8. The diode voltage versus temperature is plotted in Figure 1.48.
A simple circuit that can be used was shown in Figure 1.47. With a power supply voltage of 15 V, a change in diode voltage of approximately 0.1 V over the temperature range produces only an approximately 0.67 percent change in diode current. Thus the preceding analysis is valid.

Comment: This design example shows that a diode connected in a simple circuit can be used as a sensing element in an electronic thermometer. We assumed a diode reverse-saturation current of [latex]I_{S} = 10^{-13} A[/latex] at T = 300 K(80 ◦F). The actual reversesaturation current of a particular diode may be different. This difference simply means that the diode voltage versus temperature curve shown in Figure 1.48 would slide up or down to match the actual diode voltage at room temperature.

[latex]^3[/latex] Note that e in, for example, [latex]e^{−Eg/kT}[/latex] represents the exponential function whereas e in the exponent, for example, [latex]eV_{D1}/kT_1[/latex] is the magnitude of the electronic charge. The context in which e is used should make the meaning clear.

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