Question

The doping and geometric parameters of a P–N junction diode are:
- ND = 10^20 cm−3, Wcathode = 1 μm <<Lp,
- NA = 5 × 10^15 cm−3, Wanode = 10 μm <<Ln,
- Junction area AJ = 500 × 500 μm2.
Calculate the reverse biasing current IS, and then use this result to obtain the
forward voltage that corresponds to a current of 100 mA, if the semiconductor
material is:
(a) Si (μn = 1450 cm2/V · s, μp = 500 cm2/V·s, ni = 1.02 × 10^10 cm−3)
(b) SiC (μn = 380 cm2/V · s, μp = 70 cm2/V·s, ni = 1.6 × 10^−6 cm−3)
(c) Calculate the new junction area of a Si diode if the reverse biasing current doubles in value.

Answer 1

- Problem i Reverse biesing Saturation curent is, Again from Einstein's relacion, B = KT or, D = KTM a so equation becomes, Io = AKT( Map t - Wond an Na Na ni a) Un = 1450 cm/v.s , U = 0.05 mvis mi z 1:02X10 m = 0.145 mi/Vis A. - 5x10mm Nd = 1026 m3, Na = 5x10² me, we = 100m, Wa = 105 in To = 5 x16 € x 1.38 xo^°x300 Go tha) (1-02x0oje 2017 X10-29 (5x50*42 +29000X10-22 ) x 1.0404 x 1092 - 2017 X1.0404 x 29005 xro 29 -22 +32 I, 0.62x1013 A 5x10x1.38 x163x3001 -23 0.038 0.007 b) 5 - 501016 / = 20.7x10? (3.8x102 + 14 eru %4022) X2:57 20.7x2157 ) 1403.8 > 16-19-12- To 0.74 x 10146 A V/KT I - Te o r, 100X100 = 0.62x10-13 a over the aen ( 1970.62) - 28 1.38x10 23 x 300 x 28 = 0.72 v (An or. Ve 1.38 10 23 300 X2 1.671619 1045 b) e en 104 M, V 32169 v (quin 1 . As we can see from relation , 1 Io a A So, gf Do' = 2 10 , AS2A - 20 1:48x10 "A