Question

Problem 3: Assuming that the diodes in the circuits are ideal, find the values of the labeled voltages and currents. + 3 V + 3V 35kn 10 kn JD, D D , D2 Joy Joy > 10 kn 35 kn - 3V - 3V +10 V 10 kn 11102 D. Dv DZI' D sk 10 kn -10V -10V

Answer 1

Assume both diodes are conducting.. 0.3MA 7+30 3 10Kn to novo . J E Skr 0.6mm By inspecting. -3V V-0 o = 0.6mA. carrent through oz will be 3 - V Tok lok 3 -0 Tok TOK a. : 0.3mA. curunt in lokn 1 . leads to negative count in Di. x+3 we can observe that the current through Hu SK2 resistor will exceed that drawn through the loka resistor, which leads to negative co It follows that the above assumption is wrong. And dI is off and D2 is on. I = 0 [:D1=OFF] JBL Ix Ix = 10+5 - 15 con o 2 v=-11 § 5 kr V = 5K12 - 3 = 5K *0.4m-3 = 2-3 --IV I=0, VO-IV { loka Imamo

a) In (b), the two resistory are interchanged. we can see that the test current through okr rusiston is exceed that the current through 12k2, leaving sufficient currunt to keep Di conducting so assuming Oi and D2 au conducting. +3V Tv-ov (due to greund] - ห๕ ร K- tito2 orta 0-3 MÁ 0.3movsor 3 loka *3V currunt through 5ks = :0.6mA. through currunt lokr o -(-3) lok SOSMA KCL at node a. 0. GMA = I to 3mA => 1 = 0. 3mA c) Assuming both diodes are conducting. HOV 18 = 0 → V-0. 10KS - + ELOIDZ V - carrent through Dz is Ioze E IMA. & 5kr Nodal at B 0-(-10) - JOV I tlm= 5k. I - 2m -im assumption is -ima. true. nice I = IMA, above I IMAV OV

2) Assume both diodes or conducting +TOV Skor VBov -> VOV & 10k2 -lov current through Dz is 102= 3k Nodal eet B : I + IDR = 0-(-10) Tok I MA -2mA -Ima Iz-IMA which is not possible, ows assumption is wrong. Again assume Di->OFF , D2 →ON. + +IOV I current through on 5k2 10-(-40) IDZ 5K+IDK = 20 = 1.33mA. I { loks I=0. -Iov Nodal at B VB +10 TOK + VG -10 5x = 0 VB + (a + 2(m - 20 - - ZUB=10 VB = 3.33r. =) V: Up: 3.337. In0A, VB = V = 3.33V|

+6V ++6V 5 OK >IV > Pot 20K Szokr - zokog -" Apply voltage division VH 6 x 20k 20K+10K rule. 6 x 20 30 10x20 R2 6.66 kn Rth = lok ll 20K : 20 Thevenin's equivalent circuit : For LMV ideal diode Ron = 6.66 koma . Rithu I no ET 320 Кл 6.66km & zokh 1 Vth Run+20K 0.15 ma. ou celok 120 = 0.15 m 1 = - 6.66K+20K 0.15 m - 3V V = 20K * L = 20K * I=0.15 MA, V-3V b) q+6V +5V Apply vollage division oule to find the okr this a tha . 6x IOKR Vthe = 10k +10K 3V 3 JOKI Ry , Tok ll lok . 10k x 10K TOK + 10K = 5kr

Apply voltage division rule to find the VHh2 = 5 x IOK Tok + lok = 2.50 Ritha lok X lok = lokil 10K = - lok +IOK 5K2 Rinia Riaz = 5K2 mukhi 'th 2 = 2.5 V - Vthe = zv Diode Diis (cathed) terminal ueverse biased since negative is connected to high voltage. Ren=5K2 Ito Riha= 5km Lubuh - V + Į th=3v Ì Utha = 2.50. I = 0 va than the = 2.5-3 = -0.5v I=0, V = -0.5V