Problem (4): An "enhanced" Voitage Regulator with a BJT "Current Booster". By adding to it a "power BJT" device as shown, the ordinary Zener Voltage regulator can now provide a load RL a regulated 12.3 V DC with significantly higher DC current of 0.3A! The energizing source CL consisting of 18V DC (with ripple 2 = 0.5 sin 240xt volts), is obtained by F.W. rectification and filtering of a portion of the 60-Hz AC Line voltage. A resulting ripple Vis produced across Re. INDUT RIPPLE 0.SK R Vin 18y + "DC SOURCE" 18 Q Ro . 0.3A + R12.3+2 LOAD OUTPUT RIPPLE Dz O /2 = ? Q B = 49 Von) = 6.7V Velsax)=0,2V VA = 60V V = 25 mV (a) By what factor is the DC "output current capability of this Voitage regulator being increased through the use of the "current-booster" BJT ? Hint without Q. Is would play the role of "output DC current". Right? (b) Determine the required zener diode voltage V, DC Voltage VCE, and DC currents IB 4 Iz. (Now you know the "Operating point" of Q!) (C) Draw the small-signal equivalent circuit. Derive a SYMBOLIC expression for the resurting ripple voleage acrass RL Then, for the given vi calculate Vo, and hence the "" CUTPUT RIPPLE 100 x(v (P-t-P)/12.3). By comparing with the "% INPUT RIPPLE" 100 x (2 (p-t-p)/18), determine their ratio -- OUTOUT RIPPLE NPUT RIPPLE INDUT RIPPLE 18 V 0.SK R Iz d "DC SOURCE" + VEE Q 18 0.3A R12.3+V Ro. + 33 "LOAD" OUTPUT RIPPLE