Review Half Wave Full Wave Rectifier Rectifier Parameters Center tapped Bridge Rectifier Parameters PIV Duty Cycle

Filters 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 During the 1st quarter positive cycle, diode is forward biased, and C charges up. VC = VO = VS - VD 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.

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

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

Hence for half wave rectifier 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 For full wave rectifier Vr = ( VM 0.5Tp) / RC

Example Consider a full wave center-tapped rectifier 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 = Vo peak = 120 - VD 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 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 = Vo peak = 50 – 2 VD 50 – 1.4 = 48.6 V Vr = 48.6 / (2*60*10x103*20.3x10-6) Vr = 2 V

The full-wave rectifier circuit is shown in the figure below The full-wave rectifier circuit is shown in the figure below. The output peak current of the circuit is 200 mA when the peak output voltage is 12 V. Assume that input supply is 120 V(rms), 60 Hz and diode cut-in voltage Vγ = 0.7 V. Find the required value of C for limiting the output ripple voltage, Vr = 0.25 V. Answer: C = 6.67 mF

Clipper Circuit

Clipper is used to eliminate portion of a signal that are above or below a specified level i.e the clip value Step #1: Determine the clip value Step #2: Set the conditions

Clip value = V’. To find V’, use KVL at L1 Clipper circuits, also called limiter circuits, are used to eliminate portion of a signal that are above or below a specified level – clip value. The purpose of the diode is that when it is turn on, it provides the clip value Clip value = V’. To find V’, use KVL at L1 The equation is : V’ – VB - V = 0  V’ = VB + V Then, set the conditions If Vi > V’, diode conducts, hence Vo = V’ If Vi < V’, diode off, open circuit, no current flow, Vo = Vi Vi V’ = VB + V L1

Clippers Other clipping circuits can be constructed by reversing the diode, the or the polarity of the voltage VB. V’ = VB - V conditions: Vi > V’  off, Vo = Vi Vi < V’  conducts, Vo = V’ V’ = - VB + V conditions: Vi > V’  conducts, Vo = V’ Vi < V’  off, Vo = Vi V’ = - VB - V conditions: Vi > V’  off, Vo = Vi Vi < V’  conducts, Vo = V’

Parallel Based Clippers Positive and negative clipping can be performed simultaneously by using a double limiter or a parallel-based clipper. The parallel-based clipper is designed with two diodes and two voltage sources oriented in opposite directions. This circuit is to allow clipping to occur during both cycles; negative and positive

Example 1 Consider the parallel clip circuit shown below. Assume the VZ1 = 6V and VZ2 = 4V and V = 0.7V. Given Vi = 10 sin t, sketch VO

CHAPTER 3 II = IZ + IL Equate them and assume IZmin = 0.1IZmax Range of supply voltage Voltage Regulator The load resistor sees a constant voltage regardless of the current VZ = VL Zener Diode CHAPTER 3

Vm is the peak value of the output voltage Half Wave V r = V m T P RC If the ripple is very small, we can approximate T’ = Tp where Tp is the period of the cycle VC = Vme – t / RC Full Wave V r = V m T P 2RC Vm is the peak value of the output voltage Capacitor Discharge V r = V m T ′ RC Clipper Ripple Voltage, Vr Vo = Vs - VD Filter Vo = Vs - 2VD PIV = 2Vspeak - VD Center-tapped Duty Cycle Bridge PIV = Vspeak - VD Full Wave Vo = Vs - VD CHAPTER 3 Rectifier Half Wave PIV = Vspeak