Recall Lecture 7 Rectification transforming AC signal into

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 = 6V, R = 120 , 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 6.6V 6V 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 Half Wave Full Wave : CenterTapped Full Wave: Bridge PIV Peak value of the input secondary voltage, vs (peak) 2vs (peak)- V

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, N1/N2 = 6. If the diode is ideal diode, (V = 0V), 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 = 9V Output voltage, vo = vs - V Hence, vs = 9 + 0.6 = 9.6V 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: 2vs = 18.6 V

(peak) - V = 2(9.6) - 0.6 = 19.6 - 0.6 Solution: For the bridge transformer circuit the peak voltage of the transformer secondary is required The peak output voltage = 9V Output voltage, vo= vs - 2V this is peak value! Must change to rms value Hence, vs = 9 + 1.2 = 10.2 V 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 A capacitor is added in parallel with the load resistor of a halfwave rectifier to form a simple filter circuit. At first there is no

charge across the capacitor st During the 1 quarter positive cycle, diode is forward biased, and C charges up. VC = VO = VS - V. 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) If the RC time constant is large, the voltage across the capacitor discharges exponentially.

Filters 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. 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 Vr = ripple voltage Tp Vr = VM VMe T where T = time of the capacitor to discharge

to its lowest value Vr = V M ( 1 e -T/RC -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) For full wave rectifier Vr = ( VM 0.5Tp) / 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 R1 when the value of ID2 = 2ID1. Also find the values of VI , ID1 and ID2. Assume that all diodes are forward-biased. End of Chapter 3

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