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

3. 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.

4. 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.

5. 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.

6. 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.

7. 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

8. 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 = VD
120 – 0.7 = V
Vr = – 100 = 19.3 V
19.3 = / (2*60*2500*C)
C = 20.6 F

9. 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

10. 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

11. Clipper Circuit

12. 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

13. 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

14. 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’

15. 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

16. 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

17. 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

18. 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