Question

A depletion-layer detector has an electrical capacitance determined by the thickness of the insulating dielectric. Estimate the capacitance of a silicon detector with the following characteristics: area 1.5 cm², dielectric constant 10, depletion layer 40μm. What potential will be developed across the capacitance by the absorption of a 5.0 MeV alpha particle which produces one ion pair for each 3.5 eV dissipated?

Answer 1

If A is the area, d the thickness of depletion layer and K the dielectric constant then the capacitance is 

C = εAK/d = 8.8 × 10⁻¹² × 1.5 × 10⁻⁴ × 10/40 × 10⁻⁶ = 3.3 × 10⁻¹⁰ F 

The charge liberated 

q = 5 × 10⁶ × 1.6 × 10⁻¹⁹/3.5 = 2.286 × 10⁻¹³ Coulomb 

Potential developed 

V = q/C = 2.286 × 10⁻¹³/3.3 × 10⁻¹⁰ = 0.69 × 10⁻³ V 

= 0.69 mV