2 The electrons in a piece of n‐type semiconductor take 10 ps to cross from one end to another end when a potential of 1 V is applied across it. Find the length of the semiconductor bar. Solution L 10 ps V V L 2 L 10 ps nV 10 ps 1350 1 But V n E and E L 1.16 104 cm 1.16 m current of 0.05 μA flows through an n‐type silicon bar of length 0.2 μm and cross section area of 0.01 μm 0.01 μm when a voltage of 1 V is applied across it. Find the doping level at room temperature. 3 A Solution I 0.05 A L 0.2 m A 0.01 m 0.01 m 1 1012 cm 2 I tot I drift q nn p p AE n 2 1.08 10 p i n n 2 n 1350cm / (V.s) ----(1) 10 2 p 480 cm 2 /(V.s) V 1V 5 104 V / cm L 0.2 m Substituting these values in Eq. (1) we get E (1.08 1010 ) 2 480 12 4 0.05 106 A 1.6 1019 n 1350 10 5 10 n n 4.63 1015 cm 3 . 4. Repeat Problem 2.3 for Ge. Assume μn=3900cm2/(V·s) and μp =1900cm2/(V·s). In order to obtain the doping level proceed as follows. We use the formula 1 q ( n n p p ) For n-type semiconductor n>>p, and hence 1 nqn Given:I 0.05 A l 0.2 m A (0.01 m)2 V 1V n 3900cm 2 / (V s) p 1900cm 2 / (V s) Resistance of the Ge bar; R V I 1V 0.05 A 20MΩ Therefore resistivity, R A l 20MΩ (0.01 m)2 0.2 m 1cm Again, or, n 1 e n 1 nen Substituting the values, we get 1 1cm 1.6 10 3900cm 2 / (V s) 1 6.24 1016 cm3 1.6 1015 cm3 ND n 19 Since at room temperature all the impurities are ionized and hence carrier concentration n is equal to the doping density. A Si semiconductor cube with side equal to 1 μm is doped with 4 1017 cm–3 phosphorous impurities. Calculate the drift current when a voltage of 5 V is applied across it. 9 Solution n 4 1017 cm 3 ni2 (1.08 1010 ) 2 291.6 n 4 1017 n 1350 cm 2 /(V.s) and p 480 cm 2 /(V.s) p E V 5V 5 104 V /cm L 1m A 1 m 1 m 108 cm 2 Since nn p , we can write I tot qnn AE 0.0432 A One side of pn junction is doped with pentavalent impurities which gives a net doping of 5 1018 cm–3. Find the doping concentration to be added to the other side to get a built‐in potential of 0.6 V at room temperature. 10 Solution V0 VT ln N A ND ni2 N D 5 1018 cm 3 , VT 26 mV, ni 1.08 1010 cm 3 , V0 0.6 V N 5 1018 0.6 0.026 ln A N A 2.45 1011 cm 3 10 2 (1.08 10 ) The built‐in potential of an equally doped pn junction is 0.65 V at room temperature. When the doping level in n‐side is doubled, keeping the doping of the p‐side unchanged, calculate the new built‐in potential and the doping level in P and N region. 11 Solution V0 VT ln N A ND ni2 N A N D , VT 26 mV, ni 1.08 1010 cm 3 , V0 0.65 V NA NA 0.65 0.026 ln N A 2.89 1015 cm 3 10 2 (1.08 10 ) 2 2.89 1015 2.89 1015 V0 0.026 ln 0.668 V (1.08 1010 ) 2 A silicon pn junction diode having ND = 1015 cm–3 and NA= 1017 cm–3 gives a total depletion capacitance of 0.41 pF/m2. Determine the voltage applied across the diode. 12 Solution V0 VT ln C j0 Cj NA ND 1015 1017 0.026 ln 0.7144 V (1.08 1010 ) 2 ni2 si q N A N D 2 N A ND 1 11.7 8.854 1014 1015 1017 1.072 108 F/cm 2 2 (1015 1017 ) 0.7144 V0 C j0 1 VR V0 C 2 1.072 108 j0 1 0.7144 4.167 V VR V0 8 C j 0.41 10 Two identical pn junction diodes are connected in series. Calculate the current flowing through each diode when a forward bias of 1 V is applied across the series combination. Assume the reverse saturation current is 1.44 10–17 A for each diode. 14 Solution .5 VVF 0.0026 17 T I D I S e 1 1.44 10 e 1 3.23 10 9 A Consider a pn junction in forward bias. Initially a current of 1 mA flows through it, and the current increases by 10 times when the forward voltage is increased by 1.5 times. Determine the initial bias applied and reverse saturation current. 16 Solution VF I D I S e VT 1 mA I S e VF VT 10 mA I S e 1.5VF VT VT ln(100) 0.239 V 0.5 From Eq. (1), we get I S 107 A VF (1) (2)