Question

Need help with 5-9 please

1. Connect four diodes for proper use of full wave rectifier. Show Connection for 60Hz, 20 Vrms AC input. 2. Determine waveshape of rectified output. Determine PIV on the diodes. 3. Find the peak to peak output ripple if filter capacitor C=200uf, and Ro-50002 4. Determine Vpc. Can we use Vz=20v, Izk=5mA Zener regulator, explain? 5. Determine series resistor Rs for proper regulation if Pz=5W and show the Circuit 6. Determine, IL max 7. For variable Rl at what value, maximum power would be dissipated in Zener? Pz? 8. For what value of Ri minimum power would be dissipated in Zener? Pz-? 9. For Ru=? 12, dissipated power in zener would be Pz=2W

Answer 1

Solution! - 0 A Regulated consists of three blocks power supply ② Rectifier Filter ③ Regulator Kebie It is an electronic circuit which converts pure Ac signal into pulsating De signal (or) which converts Bidirectional signal Into unidirectional signal properties of pulsating DC! mm x mmnxm © It varies periodicelly (2) It is Unidirectimal ice It lies only on one side of the time axis ☺ Non-zero average @V) Dc value 6 It contain's Harmonics filter! Filter it is an It is an electronic circuit used to eliminate the undesired Ac components present in the pulsating De output of a Rectifier practically the output of a filter is a fluctuating DC regnats. An electronic circuit which maintains the DC output Voltage of a paver supply stable con constant irrespective of Scanned with CamScanner

in AC Supply and the variations in the load fluctuations current. classificetim of Rectifier's Basically the Rectifiers are classified into two typee ☺ Low Voltage Rectifiers (formed using Diodes) 6 High Voltage Rectifiers (formed using SCR) Low Voltage Rectifiers! These are Basically classified into two types ☺ Halb wake Rectifier ② Full wave Rectifier y centre tapped Rectifier u Bridge Rectifier 2x m Bridge Rectifier Bridge Rectifier uses ordingry step down Transformer, 4 Normal diodes. ใ ” levi DqA 2. Vi : pure AC Voltage with Reduced Rms value Vi = Um sina pure AC (200 Vrms, 50 Hz N: N₂ (N,>N2) Scanned with CamScanner

Operatim! Vicinput) Cased)! osatu Vi is positive Node a is positive with respect to Nodeb . D & D are Forward Biaget = short circuit Dz & D4 are Reverse Biased ~ open circuit fb Vn- ocus H current How occurs in the loop Consisting of Di, RL, D3 and secondary winding - Volop i casecii)! <2<211 Vi is negative Modeb is positive with respect to Node a D,&d are Reverse Biased Topen circuit Dz & Dq are forward Bisexy short circuit current flow occurs in the loop containing Pa, kh , Dq and Secondary winding peak inverse Voltage of each diode (PIX) = Vn prv = 1 Wiode Imax in Reverse Bics 814 = Um Note! piror Each diode von = 20 » Vm = 2002 = 28.28 = pir = 28.75 Scanned with CamScanner

Given that - Capacitor c= 200 UF load resistor k = son filter is a capacita filter For a full wave Rectifier with Capacitor filter peak to peak ripple Yoltage Vr = loc 2 foc VDC . R Ipc = vera Jpc = 2,50 (ou Voc = 2VM (Bridge Rectifier) VDC = 2X28.28 = 18.012 y Joc = 18.012 Cc0.0360 og 36.02mA 50 peak to peak ripple Voltage = 20c 2. fo.c 36.02 X 50 2x 30 x 200 x vote Note For Bridge Rectifier rutput frequency is [vw=3.001) Half of the input frequency) output voltage CDC voltage) Voc = 21m 2 T of Bridge Rectifier - 2X28.28 - 18.012V - 3.14 Scanned with CamScanner

Given the zener voltage Vz = 20v Izmin = 5mA De output voltage Voc = Von – VE 2 m - Io e la = 28.28 - 3.001 - = 28.28 – 1.500 = 26.77 zener Regulator J Ilmax TERL=500m 86.77 Izmin = 5mA Iimax = level = 40mA Conditim Is Izmin & Ilmax rener Regulator Is = 40+5 =45mA zener Berives Regulator seres Resistance Rs = 5 Is = 26. 77-20 -rom = 150.444 Scanned with CamScanner

Is > Izmin & Ilmax Ilmax = Is Izmin = 45-5 = 40mA. Is = Izmay + Il min Given Pzz5 Vz. Izmax = 5 Izmax = 5=250mA. Is = 25+ Ilmin Ilmin = 45-25 = 20mA | Remax = 1 - 2 oma = oken] fz = Vz. Izmax =5X25 = 125mw = 0.125w through the minimum value RL maximum current flows load. e Ilmax = 40mA - Rlmin = 20 = 500m minimum pawer Pz= Vz. Iz = 5*5 = 25mw 40 Scanned with CamScanner

paver dissippates in the zener diode Pe=2w Ya. Iz max = 20 Izmax = 3 = 0.1A comA Is = Izmax t Iumin Il min = Ismax-Is 1a45 55mA | Remax = Ro ma = 363.6302 Note! A zener diode com be used as Voltage Regulator it it is operated in Breakdown Region Voltage Regulator is a circuit which maintain's the de output volteye on a power supply constant irrespective of fluctuations in Ac Supply Voltage and Variations in load current. Scanned with CamScanner