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The amplifier circuit that is given in form-3, is B = 100, C o, VT = 25.845mV, VBE = VEB = 0.7V; calculate the values of re1,

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DC Analysis: sources be comes open cicust, Capacitor and AC voltage becomes shorted to ground VBE ON = 0.7V (Ve), - (VBEI = 0VB2 = 10 - Ioz (lok) VB2 = 10- 0.5 mA (lok) > lo - 0.5 (9.995V 101 101 Vea = V82 - ONTV VE2 = 9.295 V VE, = VES = 9.295 V Icq* Ac Analysis shorted to ground. slapen sources De voltage → Capacitor shortedo to go cucut Current Source open o=o0 Ť Outputsmall signal equivalent Tok. vo Ucz) B2=0 Biba ra VOD w VED iba JOE 9іті VE) ibi TOK Note + VIN= Vsignal Bibi - V sign vin ucV - Vip віь, — віb2 + lus Сь, у = -(-) tok + этх Та. Ток +22, ve Mй [++P) + ve ( ) iba = - log 1 та Tok+ул 193 - 0 (0 [1+E) =

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