Recall Lecture 8 Full Wave Rectifier Center tapped

Recall Lecture 8 • Full Wave Rectifier • Center tapped • Bridge • Rectifier Parameters • PIV • Duty Cycle

Example: Half Wave Rectifier Given a half wave rectifier with input primary voltage, Vp = 80 sin t and the transformer turns ratio, N 1/N 2 = 6. If the diode is ideal diode, (V = 0 V), determine the value of the peak inverse voltage. 1. Get the input of the secondary voltage: 80 / 6 = 13. 33 V 2. PIV for half-wave = Peak value of the input voltage = 13. 33 V

Example: Full Wave Rectifiers Calculate the transformer turns ratio and the PIV voltages for each type of the full wave rectifier a) center-tapped b) bridge Assume the input voltage of the transformer is 220 V (rms), 50 Hz from AC main line source. The desired peak output voltage is 9 volt; also assume diodes cut-in voltage = 0. 6 V.

Solution: For the center-tapped transformer circuit the peak voltage of the transformer secondary is required The peak output voltage = 9 V Output voltage, vo = vs - V Hence, vs = 9 + 0. 6 = 9. 6 V this is peak value! Must change to rms value Peak value = Vrms x 2 So, vs (rms) = 9. 6 / 2 = 6. 79 V The turns ratio of the primary to each secondary winding is The PIV of each diode: 2 vs (peak) - V = 2(9. 6) - 0. 6 = 19. 2 - 0. 6 = 18. 6 V

Solution: For the bridge transformer circuit the peak voltage of the transformer secondary is required The peak output voltage = 9 V Output voltage, vo= vs - 2 V Hence, vs = 9 + 1. 2 = 10. 2 V this is peak value! Must change to rms value Peak value = Vrms x 2 So, vs (rms) = 10. 2 / 2 = 7. 21 V The turns ratio of the primary to each secondary winding is The PIV of each diode: vs (peak) - V = 10. 2 - 0. 6 = 9. 6 V

Filters Ø A capacitor is added in parallel with the load resistor of a half-wave rectifier to form a simple filter circuit. At first there is no charge across the capacitor Ø During the 1 st quarter positive cycle, diode is forward biased, and C charges up. Ø VC = VO = VS - V. Ø As VS falls back towards zero, and into the negative cycle, the capacitor discharges through the resistor R. The diode is reversed biased ( turned off) Ø If the RC time constant is large, the voltage across the capacitor discharges exponentially.

Filters Ø During the next positive cycle of the input voltage, there is a point at which the input voltage is greater than the capacitor voltage, diode turns back on. Ø The diode remains on until the input reaches its peak value and the capacitor voltage is completely recharged.

Vp Vm Quarter cycle; capacitor charges up Capacitor discharges through R since diode becomes off VC = Vme – t / RC Input voltage is greater than the capacitor voltage; recharge before discharging again NOTE: Vm is the peak value of the capacitor voltage = VP - V Since the capacitor filters out a large portion of the sinusoidal signal, it is called a filter capacitor.

Ripple Voltage, and Diode Current ØVr = ripple voltage Tp ØVr = VM – VMe -T’/RC T’ where T’ = time of the capacitor to discharge to its lowest value Vr = VM ( 1 – e -T’/RC ) Expand the exponential in series, Vr = ( VMT’) / RC Figure: Half-wave rectifier with smoothing capacitor.

• If the ripple is very small, we can approximate T’ = Tp which is the period of the input signal • Hence for half wave rectifier Vr = ( VMTp) / RC l For full wave rectifier Vr = ( VM 0. 5 Tp) / RC

Example Consider a full wave center-tapped rectifier. The capacitor is connected in parallel to a resistor, R = 2. 5 k. The input voltage has a peak value of 120 V with a frequency of 60 Hz. The output voltage cannot be lower than 100 V. Assume the diode turn-on voltage, V = 0. 7 V. Calculate the value of the capacitor. VM = 120 – 0. 7 = 119. 3 V Vr = 119. 3 – 100 = 19. 3 V 19. 3 = 119. 3 / (2*60*2500*C) C = 20. 6 F

Example Consider a full wave bridge rectifier. The capacitor C = 20. 3 F is connected in parallel to a resistor, R = 10 k. The input voltage, vs = 50 sin (2 (60)t). Assume the diode turn-on voltage, V = 0. 7 V. Calculate the value of the ripple voltage. Frequency = 60 Hz VM = 50 – 1. 4 = 48. 6 V Vr = 48. 6 / (2*60*10 x 103*20. 3 x 10 -6) Vr = 2 V

MULTIPLE DIODE CIRCUITS

Example 1 Cut-in voltage of each diode in the circuit shown in Figure is 0. 65 V. If the input voltage VI = 5 V, determine the value of R 1 when the value of ID 2 = 2 ID 1. Also find the values of Vo, ID 1 and ID 2. Assume that all diodes are forwardbiased.

Example 2 The figure shows a multiple diode circuit. If each diode cut-in voltage is Vγ= 0. 7 V, determine ID 1, ID 2 and VO for R 2 = 1. 1 k.