Question 7.17: A PYROELECTRIC RADIATION DETECTOR Consider the radiation det...
A PYROELECTRIC RADIATION DETECTOR Consider the radiation detector in Figure 7.49 but with a single element A. Suppose that the radiation is chopped so that the radiation is passed to the detector for a time Δt seconds every τ seconds, where Δt ≪ τ. If Δt is sufficiently small, then the temperature rise ΔT is small and hence the heat losses are negligible during Δt. Using the heat capacity to find the temperature change during Δt, relate the magnitude of the voltage ΔV to the incident radiation intensity I. What is your conclusion?
Consider a PZT-type pyroelectric material with a density of about 7 g cm^{−3} and a specific heat capacity of about 380 J K ^{-1} kg ^{-1} \text {. If } \Delta t=0.2 s and the minimum voltage that can be detected above the background noise is 1 mV, what is the minimum radiation intensity that can be measured?
Suppose that the radiation of intensity I is received during a time interval Δt and delivers an amount of energy ΔH to the pyroelectric detector. This energy ΔH, in the absence of any heat losses, increases the temperature by ΔT. If c is the specific heat capacity (heat capacity per unit mass) and \rho is the density,
\Delta H=(A L \rho) c \Delta T
where A is the surface area and L the thickness of the detector. The change in the polarization ΔP is
\Delta P=p \Delta T=\frac{p \Delta H}{A L \rho c}
The change in the surface charge ΔQ is
\Delta Q=A \Delta P=\frac{p \Delta H}{L \rho c}
This change in the surface charge gives a voltage change ΔV across the electrodes of the detector. If C=\varepsilon_{o} \varepsilon_{r} A / L is the capacitance of the pyroelectric crystal,
\Delta V=\frac{\Delta Q}{C}=\frac{p \Delta H}{L \rho c} \times \frac{L}{\varepsilon_{o} \varepsilon_{r} A}=\frac{p \Delta H}{A \rho c \varepsilon_{o} \varepsilon_{r}}
The absorbed energy (heat) ΔH during Δt depends on the intensity of incident radiation. Incident intensity I is the energy arriving per unit area per unit time. In time Δt, I delivers an energy ΔH = IA \Delta t. Substituting for ΔH in the expression for ΔV, we find
\Delta V=\frac{p I \Delta t}{\rho c \varepsilon_{r} \varepsilon_{o}}=\left(\frac{p}{\rho c \varepsilon_{r} \varepsilon_{o}}\right) I \Delta t [7.66] Pyroelectric detector output voltage
The parameters in the parentheses are material properties and reflect the “goodness” of the pyroelectric material for the application. We should emphasize that in deriving Equation 7.66 we did not consider any heat losses that will prevent the rise of the temperature indefinitely. If Δt is short, then the temperature change will be small and heat losses negligible.
For a PZT-type pyroelectric, we can take p=380 \times 10^{-6} C m ^{-2} K ^{-1}, \varepsilon_{r}=290, c=380 J K ^{-1} kg ^{-1} \text {, and } \rho=7 \times 10^{3} kg m ^{-3}, and then from Equation 7.66 with ΔV = 0.001 V and Δt = 0.2 s, we have
I=\left(\frac{p}{\rho c \varepsilon_{o} \varepsilon_{r}}\right)^{-1} \frac{\Delta V}{\Delta t}=\left(\frac{380 \times 10^{-6}}{(7000)(380)(290)\left(8.85 \times 10^{-12}\right)}\right)^{-1} \frac{0.001}{0.2}
=0.090 W m ^{-2} \text { or } 9 \mu W cm
We have assumed that all the incident radiation I is absorbed by the pyroelectric crystal. In practice, only a fraction η (called the emissivity of the surface), that is, ηI, will be absorbed instead of I. We also assumed that the output voltage ΔV is developed totally across the pyroelectric element capacitance; that is, the amplifier’s input impedance (parallel combination of its input capacitance and resistance) is negligible (i.e., infinite) compared with that of the pyroelectric crystal. As stated, we also neglected all heat losses from the pyroelectric crystal so that the absorbed radiation simply increases the crystal’s temperature. These simplifying assumptions lead to the maximum signal ΔV that can be generated from a given input radiation signal I as stated in Equation 7.66. It is left as an exercise to show that Equation 7.66 can also be easily derived by starting from Equation 7.63 for the pyroelectric current density J_{p}, and have J_{p} charge up the capacitance C=\varepsilon_{o} \varepsilon_{r} A / L of the crystal.