Question

X=500, y=10

1 2 3 Soru 6 Vio 24 V Vo 1 ks ΧΩ y ΚΩ 6V ΤΟ ΚΩ Find: a) regulated output voltage (Vo) b) Power dissipation (PD) across the transistor (neglect currents other than the load current in your calculations) c) if the unregulated voltage increases 20% find the percent change in PD % x = last three digits of your student ID (Ex: 1234567, x = 567) % y = first two digits of your student ID (Ex: 1234567, y = 12) Yanitiniz: a BI D Aramak için buraya yazın

Answer 1

Gison X = 500, y Flo + NE 24 V lkn OX + I Na w Node ① OA Y. I- Tokne Iz gim Vine 24 7 6v so brecek down ano by Gemosor dinele is operated in pon can be replaced cons of ant withega figure innerting terminal is vz &z= 360 from the voltage ie hon Vz6v from vor fuell around concept [ Vt ve bu

and current into for ideal op. camp the terminals both are zero. He Ran easily observe that ut is directly appeared access tokis apply keet at hode 9 marked NOO in the figure h tar Iz II + 0 I 2 Vo-V yan loken Vo - V V-o loka vo 27 :7 vo Fex vt - 2X 6 vo = 12v So gunel voltage No 12v

b> medle (2) Vo Vin IE VE Vo Io OA Sykn Exn 1 Gv 'st foku Vo = 12 V GV then Ijs | 26 6 lokn kn 10kn I 0.6 ANA To a vo Хи 12 500 m to 24 MA NOD Kch at node apply IES It Io Oodmaat 24 mA 26.6 mA

took at figure the transistor in We Can. see that Vc: Vina 24V VE = Vo - 12 V TE u Ie Ibno So alleress transistor power dissipation Pdc = VCE X Ic = (ve-VE) Xile ( 29-12) V X 84.6 mA 12 x 24.6 mw 295. 2 mw.. Pde Pae 295. 2 mw ~ 0.295W e) if nereuse 20 voltage 200/0 unregulated 도 Vin C 100% 20%) 24 x 1280 = 1.2X24

Vin' 528.8 v Since vo is indepedent of change input voitenge No 12V IE = 24.6 mA Const) buf NCE : Vc. VE vin' No es.fr 12 Vce - of 6.8 NO power dissypatron VCE X IE 16.8 x 24.6 mw 413.28 "mw of change in po PD₂-Pol Por Xlen

11 ? .25 -8ܬ . 413 x-1. 295. 2- % increase in PO ܘ ܝ90 0