Question

For the given circuit; a) Determine V, IL, IR and Iz b) For R_ = 1k02, find V, IL, IR and Iz. C) For proper regulator operation, determine the minimum value of RL. R W 1 ke 2 + Iz IR RL 18 V PZmar V2 = 10 V = 150 mW 3kΩ VL

Answer 1

Question 1

IR R Iz I R2 lov Vzalov P = 15omu 3kn "Max (a) V, I, IR 9 Iz Find V by open zener diode R 188 } R₂ VL Apply vottoge division VL = R2 RtRq x 18 ID x18 13,5 1 +3 v> Vz so diode works in breakdown region. By parallel Connected to RL with Zener so VL = V₂ = lov

IR R V = 102 Iko IL Iz 118v 3kn I = VL 10=3.33mA 3k 3 IL = 3:33MA ! IR = 18-Vz 18-10 =&ma - IR=8mA by Kll, IR = Izt L Iz= IR - IL 8-3.33 Iz = 4.67mA こ (6) If RL=1 kn VL IL IR Iz IK2 18V { lkn VL By voltage division VL = I x 18 = 1+2 gv Y <ų (lov) So diode act as reverse bias region. so act as open ckt.

So | _=9V lkn IL IR Iz A 18V IR Apply KUL. 18=1IR + IIL I=0 so IR = It due to open circuit 18 = 2IR IR =gma IL=9mA ARS Iz= OMA V=9v (C) for Zener is work as Hegulator IR lkn 'Iz 7 IL Vs 1187 RL Rza Vasjable Vs=contt (18V) RL&Min) = VL IL (10) Ip (Fix) = I z (mith - IL (Max)

Zmax = Vzx Izmax Izmax = Pmax , 150 Vz 10 Iz (max) = 15mA But we want Iz (min) or knee current of Zener diode So Iz free OMA assume Leal conditions Izmin = OMA IL (max) = IR IR = 18-V2 18-10. 1 IR=8mA So ILE IR = 8 ma

Rmin= V2 V = V₂ Ilmax Rmin= to Rmin 1.25kn = An L

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