NA cm-3 1E18 p-side Dn cm²/s p-side Op Rp S/cm Ω Ln cm On S/cm ND cm-3 6E15 n-side Rn Ω n-side Dp Lp cm²/s cm I(V)x R V Vo V Is A Emax(V) Js V/cm A/cm² Area cm² 1E-7 Area cm² 1E-7 I(V) A J(V) A/cm² + I(V) XR is the sum of the voltages dropped across each "neutral region". The ideal diode equation assumes that all of the applied voltage, V=0.6 volts, is dropped across the space charge region. Therefore, this value should be a small fraction of V=0.6 volts. Is it? 2. Given: a long base, silicon p*n abrupt junction diode, with No = 6E15 cm³ on the n-side of the junction, and NÃ= 1E18 cm³ on the p-side of the junction. The minority carrier lifetimes on both the n- and p-side, tp = Tn = 10 ns = 1E-8 seconds. Dimensions: junction area, A= 1E-7 cm²; distance from the junction to applied V contacts, W₁ = Wp = 25 μm = 2.5E-3 cm. Alert: this W is not the space charge width, W = Xn + Xp! Refer to the diagram on next page. NA p+ Wp Xp Xn n ND Wn Drawing is not to scale. a. With a forward bias, V = 0.6 volts, show that this diode is operating in low level injection (LLI), i.e. Ap(xn) = Pno [exp (qV/kT)-1] << nn = ND b. Using the given parameters, calculate values for each entry in the Tables below for this diode, using forward bias V=0.6 volts for those parameters that depend on bias. Note: D and L represent the minority carrier values on each side, Emax = the electric field at the metallurgical junction with V=0.6 volts, On = qu₂ND and op = qupNA are the conductivities of the n and p side neutral regions, respectively, which are governed by the majority carrier concentration and mobility in each region. R = Rp + R₁ = Wp/(A op) + Wn/(A σn) c. Write the expression for Jp(x) on the n-side of the junction, and determine its value at the contact, i.e. x=W₁ = 25 μm. Show that at the contact, Jn(x) = J - Jp(x) = J d. Assuming that the current density, J, at the contacts on either side of the junction, is entirely due to majority carrier drift current, calculate the actual electric field, E, inside the "neutral region" on each side, by applying J = = On En = Op Ep. Compare the resulting E values to Emax found in b, and comment on the validity of the "neutral region" assumption.

NA cm-3 1E18 p-side Dn cm²/s p-side Op Rp S/cm Ω Ln cm On S/cm ND cm-3 6E15 n-side Rn Ω n-side Dp Lp cm²/s cm I(V)x R V Vo V Is A Emax(V) Js V/cm A/cm² Area cm² 1E-7 Area cm² 1E-7 I(V) A J(V) A/cm² + I(V) XR is the sum of the voltages dropped across each "neutral region". The ideal diode equation assumes that all of the applied voltage, V=0.6 volts, is dropped across the space charge region. Therefore, this value should be a small fraction of V=0.6 volts. Is it? 2. Given: a long base, silicon p*n abrupt junction diode, with No = 6E15 cm³ on the n-side of the junction, and NÃ= 1E18 cm³ on the p-side of the junction. The minority carrier lifetimes on both the n- and p-side, tp = Tn = 10 ns = 1E-8 seconds. Dimensions: junction area, A= 1E-7 cm²; distance from the junction to applied V contacts, W₁ = Wp = 25 μm = 2.5E-3 cm. Alert: this W is not the space charge width, W = Xn + Xp! Refer to the diagram on next page. NA p+ Wp Xp Xn n ND Wn Drawing is not to scale. a. With a forward bias, V = 0.6 volts, show that this diode is operating in low level injection (LLI), i.e. Ap(xn) = Pno [exp (qV/kT)-1] << nn = ND b. Using the given parameters, calculate values for each entry in the Tables below for this diode, using forward bias V=0.6 volts for those parameters that depend on bias. Note: D and L represent the minority carrier values on each side, Emax = the electric field at the metallurgical junction with V=0.6 volts, On = qu₂ND and op = qupNA are the conductivities of the n and p side neutral regions, respectively, which are governed by the majority carrier concentration and mobility in each region. R = Rp + R₁ = Wp/(A op) + Wn/(A σn) c. Write the expression for Jp(x) on the n-side of the junction, and determine its value at the contact, i.e. x=W₁ = 25 μm. Show that at the contact, Jn(x) = J - Jp(x) = J d. Assuming that the current density, J, at the contacts on either side of the junction, is entirely due to majority carrier drift current, calculate the actual electric field, E, inside the "neutral region" on each side, by applying J = = On En = Op Ep. Compare the resulting E values to Emax found in b, and comment on the validity of the "neutral region" assumption.