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Continuity Equation:
Explain continuity equation and discus the special case of continuity equation when concentration is independent of time and with zero electric field.
Or
Explain carrier of life time and continuity equation with required figured.
(b) If a P-type semiconductor bar has width and thickness of 3 mm each, the measured value of current and Hall voltage are 20µA and 100mV respectively. The Resistivity of the bar is 2 × 105?-cm and the applied magnetic field is 0.1 Wb/m2. Calculate the mobility of holes.
Sol. (a) The continuity equation states a condition of dynamic equilibrium for the density of mobile carriers in any elementary volume of the semiconductor. We know that no distributing the equilibrium concentrations of holes or electrons vary with time approaching the equilibrium value exponentially. In general, the carrier concentration in the body of a semiconductor is a function of both time and distance. The differential equation governing this functional relationship is called continuity equation. This equation is based on the fact that charge can neither be created nor destroyed. Consider an infinitesimal element of volume of area A and length dx as shown in fig. Let p be the average hole concentration within this volume. Considering the problem dimensional, let the hole current Ip is only a function of X. As shown in fig., the current entering the volume at x+ dx is Ip+time t while at the same Ip time, the current leaving the volume at x+ dx is Ip + dlp. Thus more current leaves the volume for positive value of dlp. Thus more current leaves the volume for a positive value of dlp. Hence within the volume, the decrease in the number of holes per second with in the volume is dlp/e, where e is the magnitude of charge so the decrease in hole concentration (holes per unit volume) per second due to current Ip is given by. The increase of holes per unit volume per second due to thermal generation g = p. while the decrease of holes per unit time per second due to recombination is P. hence the increase in holes per unit volume per second must be equal to the algebraic sum of the increase in hole concentration.
If a current of 20A flows for five minutes, find the quantity of electricity transferred. Quantity of electricity, Q = It = 20 x 5 x 60 = 6000C
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