The breakdown in a reverse biased p–n junction diode is more likely to occur due to ______. a. large velocity of the minority charge carriers if the doping concentration is small. b. large velocity of the minority charge carriers if the doping concentration is large. c. strong electric field in a depletion region if the doping concentration is small. d. strong electric field in the depletion region if the doping concentration is large.

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a and d Explanation: Reverse biasing: Positive terminal of the battery is connected to the N-crystal and negative terminal of the battery is connected to P-crystal. (i) In reverse biasing width of the depletion layer increases (ii) In reverse biasing resistance offered R_("Reverse") = 105 Ω (iii) Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority carriers. (A very small reverse current may exist in the circuit due to the drifting of minority carriers across the junction) (iv) Break down voltage: Reverse voltage at which break down of semiconductor occurs. For Ge, it is 25 V and for Si, it is 35 V. So, we conclude that in reverse biasing, ionization takes place because the minority charge carriers will be accelerated due to reverse biasing and striking with atoms which in turn cause secondary electrons and thus more charge carriers. When doping concentration is large, there will be a large number of ions in the depletion region, which will give rise to a strong electric field. [Solution diagram: 329486_solution_1.png]

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