EECE 310 Exam, Spring 2008

Test 1 (Oct 8, 2007) Determine the current I and the voltage across each diode, assuming identical, ideal Si pn junction diodes having IS = 10 nA,  = 2, with VT = 26 mV. Solution: The upper diode will be reverse biased and the lower diode forward biased. The forward-biased diode will have a voltage of at most 1 V, so that the reverse voltage across the other diode is at least 4 V. At this reverse voltage the reverse current through the diode is practically IS. Hence, I = IS = 10 nA. For the forward-biased diode, . Since I = IS, the 1 cannot be neglected compared to the exponential. Substituting I = IS, gives: V. The voltage across the reverse-biased diode is V. To solve the problem ‘exactly’, we have for the forward-biased diode: , where iD1 = I. For the reverse-biased diode, , where iD2 = –I and vD2 = –(5 – vD1). Substituting: , or . Taking the log of both sides: . This is a transcendental equation in vD1 which cannot be solved in closed form. But even if vD1 is as large as 1 V, . Neglecting the exponential gives as assumed above. Substituting in the equation for D1: , so that iD1 = IS. +5 V I +5 V I +5 V I + – D1 D2 iD1 vD1 + – iD2 vD2