Recall Lecture 7 Rectification transforming AC signal into

Recall Lecture 7 • Rectification – transforming AC signal into a signal with one polarity – Half wave rectifier • Full Wave Rectifier – Center tapped – Bridge

Rectifier Parameters Relationship between the number of turns of a step-down transformer and the input/output voltages The peak inverse voltage (PIV) of the diode is the peak value of the voltage that a diode can withstand when it is reversed biased Duty Cycle: The fraction of the wave cycle over which the diode is conducting.

Duty Cycle: The fraction of the wave cycle over which the diode is conducting.

EXAMPLE 1 – Half Wave Rectifier Determine the currents and voltages of the half-wave rectifier circuit. Consider the half-wave rectifier circuit shown in Figure. Assume VB = 6 V, R = 120 Ω , V = 0. 6 V and vs(t) = 18. 6 sin t. Determine the peak diode current, maximum reverse-bias diode voltage, the fraction of the wave cycle over which the diode is conducting. -VR + VB + 18. 6 = 0 VR = 24. 6 V - VR + + A simple half-wave battery charger circuit

This node must be at least 6. 6 V 6 V

The peak inverse voltage (PIV) of the diode is the peak value of the voltage that a diode can withstand when it is reversed biased

Type of Rectifier PIV Half Wave Peak value of the input secondary voltage, vs (peak) Full Wave : Center-Tapped 2 vs (peak)- V Full Wave: Bridge vs (peak) - V

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 centre-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. 6 - 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 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 • Hence for half wave rectifier Vr = ( VMTp) / RC Vr = VM / ( f RC) l For full wave rectifier Vr = ( VM 0. 5 Tp) / RC Vr = VM / ( 2 f RC)

MULTIPLE DIODE CIRCUITS

Example: 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 VI , ID 1 and ID 2. Assume that all diodes are forward-biased. End of Chapter 3