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. MULTIPLE DIODE CIRCUITS

12. Problem-Solving Technique: Multiple Diode Circuits
Assume the state of the diode.
If assumed on, VD = Vg
If assumed off, ID = 0.
2. Analyze the ‘linear’ circuit with assumed diode states.
3. Evaluate the resulting state of each diode.
4. If any initial assumptions are proven incorrect, make new assumption and return to Step 2.

13. Example 1
The figure shows a multiple diode circuit. If each diode cut-in voltage is Vγ= 0.7 V, determine iD1, iD2 and vO
+5 V
- 5 V 0V
= 5 k
= 10 k
Let’s assume both diodes are on
KVL at L1:
5 iD (iD2+ iD1) – 5 – 5 = 0
5 iD (iD2+ iD1) = (1)
KVL at L2:
(iD2+ iD1) – 5 = 0
10 (iD2+ iD1) = 4.3 replace in (1)
5 iD = 9.3
iD2 = 1 mA
Hence,
Using 10 (iD2+ iD1) = 4.3
so iD1 = 0.43 – 1 = mA
a negative diode current means that the diode is actually OFF.
Therefore, we need to calculate again, but now we know that diode D1 is actually off which means that iD1 = 0

14. D1 is off and D2 is on
+5 V
KVL at L1:
5 iD (iD2) – 5 – 5 = 0
5 iD (iD2) = 9.3
iD2 = 0.62 mA
= 10 k
KVL at L3:
vo – iD2 = 0
vo = 5 – 3.1 = 1.9 V
= 5 k
- 5 V

15. Example 2
The figure shows a multiple diode circuit. If each diode cut-in voltage is Vγ= 0.7 V, determine iD1, iD2 and vO
+5 V
- 5 V
4 V
= 5 k
= 10 k
Let’s assume both diodes are on
KVL at L1:
5 iD (iD2+ iD1) – 5 – 5 = 0
5 iD (iD2+ iD1) = (1)
KVL at L2:
(iD2+ iD1) – 5 -4 = 0
10 (iD2+ iD1) = 8.3 replace in (1)
5 iD = 9.3
iD2 = 0.2 mA
Hence,
10 (iD2+ iD1) = 8.3
so iD1 = 0.83 – 0.2 = 0.63 mA
Both currents have positive values which means that our assumption is correct
So calculate vo :
KVL at L3:
vo – iD2 = 0
vo = 5 – 1 = 4 V

16. Example 3
The figure shows a multiple diode circuit. If each diode cut-in voltage is Vγ= 0.7 V, determine ID1, ID2 and VO for R2 = 1.1 k.
Answers:
ID1 = 1 mA
ID2 = 6 mA
VO = 2.3 V