Question

Prove that the diode operating point (lo, Vd) in the circuit shown below is (3.2601 mA, 667.6242 mV) The diode model parameters are as follows: Is = 1.66 PA, n = 1.2; assume V1 = 26 mV. lo = A 560 A in ID V VD VD Vo= V 1910 6 m © PMCL Version 2.0.0

Full and clear solution is expected. Thank you!

Answer 1

+ Given circuit is 0.6 KM → + 2011 $0.91K2 + toid 0.918 to iroka . 1.44K . C 1 Vo Gmax O. AIKA 25.06V 7 di from circuit we can say that D: 20 = 5.46–vo - lutka And we know Roa dole volare) - given Do = 1.66810-th A n - equating or ?. 1.66X2012 [ Vo/1.2x20mv) l-arka Scanned with CamScanner

No 10.0312 => - 9. Có '* 2 5 - 8 - vo : 2. ( . » Lu ( - 6 - 9 ) : ln (3, 4 x 6") + '/a 02, a> du (s. 46-UD) = - 19.83+ o encab). Elscootencb) This algebraic equation can encea) = x] Solved using Newton. Raphson method for ln (5.46.") a by Taylor series tapansion. . for lucr. 46-40) a I way :- luc 5.46 0.1 ~ VD 15.46)) Incr.66) + en CI+ C-vos.cc]). Saylor sesies. In Ctr) = E E P In the form of Pal P aloccm) and ways a dirctly using Newton-Raphsoss method for en (5.46-vo). tem? . we get a value of 1.54 2n+2 = An- fean) i 1.54 = -19.83 + Vo 0.0312 Vo: 21.37 X 0.0312 Up = 0.666744 v) ~ 0.667 CS$canned with CamScanner

If The diode is open then 1.0 IN Va S.UGV > 0.6677 . asy? so diode is on. we can also say if tve diode, Then diode is on. I flows through S.QG -UD Io Lacs TUKN 5.46 -0.667 1:41км ,: 2 3.264VY A led -3.26MA Y Nos 0.66:v Scanned with CamScanner