Question

Problem 14) : An "enhanced" Voituge Regulator' with a BTT "Current Booster". By crocling to it a "power BJT" device as shown, the ordinary Zener ! Voltage regulator can now provide a lood RL a regulated 12.3v DC with a significantly higher DC current 67 0,3A ! The anergizing Source consisting of 18V DC (with mpple Y = 0.5 sin 2407€ votes), is obtained by F.w. rectification and filtering of a portion of the 60-HZ AC Line voltage.. A "V" is produced across R. resulting ripple INDUT RIPPLE . Dz Oumor B-99 Is it 12.3+”. Valon) = 0.7v Veelset) 0,2V VA = 60V. DC SOURCE" V = 25 mV Roi RIPPLE VE? (a) By what factor is the oc Youtput current capability of this Vitage regulator being increased through the use of the current-booster" BJT ? Hint without Q. Ig would play the role of "butput DC current". Right? Determine the required zener diode voltage V, DC Voltage VCE, and DC currents IB 4 I (Now you know the operating point" of Q!) 11 .' Draw the small-signal equivalent circuit. Derive a SYMBOLIC expression for the resulting ripple voltage across RL: Then for the given Vi calculate V, and hence. The CUTPUT RI 100 * Vo(P-t-P)/12-3). By comparing with the "% INPUT RIPPLE" 4 100 *( 2;(P-t-p)/18 ), determine their ratio % OUTPUT RIPPLE . . . INPUT RIPPLE (d). To evaluate how close to an 'IDEAL DC Voltage source the above Voltage Regulator comes, determine its incremental (small-signat ) Thevening 'output resistance" Roa (Veest lies seen by the load R . Note: Obtain first a SYMBOLIC expression for Ro. Then calculate value. (e) ij Determine the EFFICIENCY of the circuit by Computing the sc "input power" supplied by the 18v source and the sc "output power being delivered to the load Rei % Efficiency = 100* DC Output power .. ... DC input Power in The "less-than-hundred-percent." Eficiency you found in i) is due to power loss occuning in (check all that applies): . OR=0.5K32 resistor Zener(Dz) BSTIQ)

Answer 1

Solution:

Answer i Giren a Power BJT device. ALCE - 03A INPUT A RIPPLE IB sa 4123to. 184 DZ Dc Source. DZ Vzro (22? ß = 49 LIBE(On) -0.76 CE Sab) -0.26 MA 2604 (9=25my

(a) By what factor is the De "output Current Capability" of the coltage regulator being increased through the use of "Coloent -boosteo" BIT? let De current flowing thoughe load is TE: , TE = (B+ IB - 50 IB. .. Current Capability increases by a factor 5o. (1) Determine the required Zener diode voltage 12, De voltage UCE, and DC Currents IBLI2 is: from the above given circuit the Dc Characteristics are:

LCE=184 -12.3 - 5.76. 12 = 12.3+UBE V2 = 12.3.to.7=136. TE = 0.34 =CB+1) IB FB FOR 50.6 mt. Ibt [2 = 18 12 18 13 Sis Flomat .. I2 =4mt. S. 3B = 0.6 ma, 12a 4ma, NCE -5.74, 12-134. (c) Dow the Small Signal cavitalent ext is: replace from the Q with above cairalent given figure ac model,

Go-to tale dere Som slang talker. Gy= b = 1.39m = 1912 = 1.2m GL = 1213 - 19 GL=244m o 5m 5m f1.7m +2440412 240.8.140 upr - To determine ratio is T ole ripple s408.146 To ilp ripple 18 12.39 577.2769 tolp ripple < 597.2747 Toilp ripple

(d) Determine coltage Source the abole same ckt is taken win is grounded. fit 2 et = 44 (z tamtho] Rip = L = ab tgimthe Rim - Lamtiatom 783.ama 1.7m+1215m - -=83.2ma (C) (8) Determine Efficiency of the ckt: %. Efficiency = loox Y Efficiency = look DC output Power De input power

DC input power = 18 x 3in De output power = 18 [lomto.3] ud XIE 512.3 X0.3 12.3X0.3 7 Efficiency = -xloo 18 1oth to:3] :. l.Effievenly 766.137 ( The "less than a hundred-percent" Efficiency you find in is due to power loss occurring in DR -0.5km o Zerex Dz gesta) AR.IR "2X12 "CE XI we can observe that stal Power is same. i. Power delivered a power absorbed e-ple Power = ole power tloss.