Question

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NOTE: In all problems in this test, we use the constant voltage drop model with Von = 0.6V for all diodes. 1. (20 points) Find the values of V and I in the circuits in Fig. 1. +5V w + biz I < Fig. 1

Answer 1

9 Sv . & li lok of - E Assume diode D, and Do both are in forward bias lok & AY RYY 0.67 1 -SV I- It I In 5-0.6 - 0.44 MA Tg = 0,6-(-5) 0,56 MAYO 10 I = I- Iz Ij = 6.44 -0.56 = -0.12 mA • I = -ve to means drode & is reverse bias. o and I tve to Means diode 2 is forward bias.

draw equvritent att & 5v Tot $5k on -5V I = 0 reverse bias of a Apply kvi 5- 10 I - 0.6-5 I +6 =0 - 15 Ist 9.4=0 I = 9.4/15 = 0.626 MA V=51-5 [v=- 1.866 I=0 vot

Answer 2

I = 0 A

V = -0.3 V

Detailed answer is given below.

- ussume both the diodes are forward let us bias. =) So, the equaralent ckt become ses follows, 3 litir =) Apply KVL, we get 5- lok (eter) -0.6=0 - 5- loke (inter) - 0.6-lokiz +5=0 lok . co 001 - From ① & ②, we get lokir= 5 il 0.5 mA =) from ② , we get est iz 4. 4 = 0.44 mA i ii- - 0.06 A ( which means D1 is reverse bius ) So, final ckt become as follows, 7 I = è, = oA =) 5-lok iz-0.6 - lokezt5 =0 I ix = 9.4 = 0.47 mA . tiz 13 lok & Also, { lok 5 - lokiz - 0.6-v ao v=64.4 - 10K (0.47 mA) Tv=-0.3 vl