Final answer:
To double a diode's current, doubling the acceptor doping is the suggested option because it increases the concentration of holes in the p-type region, leading to a higher reverse saturation current and thus a higher forward bias current.
Step-by-step explanation:
Doubling the Current in a p⁺⁺n Diode To double the current in a p⁺⁺n diode, we have to consider the relationship between doping levels and the diode current equation. The current I in a forward biased p⁺⁺n diode is given by the Shockley diode equation:
I = I0(eeV/kT - 1), where I0 is the reverse saturation current, e is the charge of an electron, V is the voltage across the diode, k is the Boltzmann constant, and T is the temperature.
Increasing either the donor or acceptor doping concentrations increases the reverse saturation current I0, which in turn increases the forward current I when the diode is biased. Between the options given, we would choose (c) doubling the acceptor doping to increase the number of holes in the p-type region, which would lead to a higher reverse saturation current and hence a higher forward current thereby doubling the diode's current. This is because increasing the acceptor doping concentration directly impacts the p-side of the diode, which is critical for the carrier injection process in forward-bias.