A p - n junction diode has the characteristic i = i₀ ! (e^ eV/kT - 1 ) where i₀ = 20 A . The diode is operated at T = 300 K . If a forward bias voltage of 300 mV is initially applied across it, at what voltage does the current double?

Step-by-step solution

The diode current is given by i = i₀ (e^ eV/kT - 1 ) . At T = 300 K , the thermal voltage is kT e = 1.38 10⁻²³ 300 1.6 10⁻¹⁹ 25.87 mV . For V₁ = 300 mV , the term e^ eV₁/kT = e^ 300/25.87 e^ 11.6 1 . Thus, we can approximate the current as i i₀ e^ eV/kT . Let the new voltage be V₂ where the current becomes 2i₁ . 2i₁ = i₀ e^ eV₂/kT Substituting i₁ i₀ e^ eV₁/kT : 2 i₀ e^ eV₁/kT = i₀ e^ eV₂/kT e^ e(V₂ - V₁)/kT = 2 Taking natural logarithm on both sides: e(V₂ - V₁) kT = 2 V₂ - V₁ = kT e 2 V₂ - V₁ = 25.87 0.693 17.9 mV