Download Comparison of Microelectronics Technology: Diode #1 vs. Diode #2 - Prof. E. Fred Schubert and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! 1 ECSE-2210 Microelectronics Technology Class Activity 14 – Solution 1. Two silicon p+-n step junction diodes are physically identical except for the n-side doping and the lifetimes. Compare the operation of the two diodes by answering the questions below. Try to explain by using physical reasoning, rather than using formula (One or two sentences are enough). Diode #1 Diode #2 P (a) Which diode will exhibit the larger built-in voltage, Vbi (also called the contact potential)? Diode #2. From the doping concentrations we know that the EF – Ei in the n-side is larger for the second diode. Hence in equilibrium the bands have to bend more. As a result the Vbi is greater for the Diode #2. (b) Which diode will have a larger depletion layer width, W, for a given reverse bias? Diode #1. For a given reverse voltage, the E-vs.-x curve should occupy the same area since voltage is ∫ E dx. Since d E /dx (compare the values of ND !) is smaller for diode #1, the depletion region in diode #1 extends a longer distance away from the metallurgical junction. (c) Which diode will have a larger reverse saturation current, Io ? Diode #1. The reverse saturation current depends on the minority carrier concentrations pn0 and np0 present in either side of the diode (se equation below): Io = hole current + electron current = q A Dp /Lp × pn0 + q A Dn /Ln × np0 In a p+n junction you will always have pp >> nn . (Majority hole concentration is much larger than majority electron concentration) This results in (using the law of mass-action): np0 << pn0 You can therefore make the approximation: Io ≈ q A Dp /Lp × pn0 Since L = (D τ)1/2 we can also write Io = q A (Dp/τp)1/2 × pn0 In diode #1 the minority hole concentration pn0 is larger by a factor 100, and the lifetime τp is larger by a factor 2. So, Io will be larger by a factor of 100/√2 for diode #1. (d) If the diodes are forward biased, qualitatively plot the minority carrier profiles in the n-region of each diode (Your plot should show the differences between these two diodes). P+ N ND = 1014 cm-3 τP = 2µs P+ N ND = 1016 cm-3 τP = 1µs 2 τDL = As the lifetime of the minority carriers in the Diode 1 is larger than in Diode 2, the injected carriers will travel a larger distance before they can recombine in Diode1. Therefore, the profiles are as shown above. 2. An abrupt Si p-n junction has the following properties at 300 K. Assume a junction area A = 1cm2. P-side N-side NA= 2 × 1017cm-3 ND = 3 × 1015 ND = 1017 cm-3 NA = 2 × 1015 τn = 0.1 µs τp = 10 µs µP = 200 cm2/Vs µn = 1300 cm2/Vs µn = 700 cm2/Vs µP = 450 cm2/Vs a. Calculate the built-in voltage, Vbi. On the p-side NAeff = NA-ND= (2X1017 -1017)=1 x 1017cm-3 On the n-side NDeff = ND - NA = (3 × 1015 - 2 × 1015) = 1 x 1015 cm-3 Vbi = kT / q × ln [(NAeff)p-side × (NDeff)n-side / ni2] = 0.0259 V x = 0.715 V b. Calculate the reverse saturation hole current. First find Dp, Lp for n-side and Dn and Ln for p-side. Dp = 0.0259 V × 450 cm2/Vs = 11.6 cm2/s; Lp = (11.6 cm2/s × 10-5 s)1/2 = 0.0107 cm = 107 µm Dn = 0.0259x700=18.1 cm2/s; Ln = 18.1x0.1x 10-6=13.4 µm Io = hole current + electron current = q A Dp /Lp × pn0 + q A Dn /Ln × np0; pn0 = ni2 / (NDeff) = 105 cm-3 npo = ni2 / (NAeff) = 103 cm-3 q kTD = µ pn0 = 106 cm-3 pn0 = 104 cm-3 x x
