1. CALCULATING CURRENTS
Once the peak and average load voltage values are known, it is easy to determine the values of IL(pk) and Iave with the help of Ohm’s law.

2. EXAMPLE
Determine the values IL(pk) and Iave for the circuit in last lecture.

3. SOLUTION
In the previous lecture, we calculated peak and average output voltages for the circuit. Using these value we can determine current as follows

4. SOLUTION IL(pk) = VL(pk)/RL = 20.5/5.1k =4.02 mApk And Iave = Vave/RL
= 2.56mA dc

5. EXAMPLE
Determine the values of VL(pk), Vave, IL(pk)and Iave for the circuit shown in the fig.

6. SOLUTION
Rated value of the transformer is given so the peak value can be calculated as
V2(pk) =24Vac/0.707
=33.94Vpk

7. SOLUTION Now VL(pk) = V2/2 – 0.7 =16.27Vpk And Vave = 2VL(pk)/ Π
=10.34Vdc

8. SOLUTION Now for currents IL(pk) = VL(pk)/RL =16.27/8k =2.03mApk And
Iave = Vave/RL
=10.34/8k
=1.29mA

9. NEGATIVE FULL WAVE RECTIFIER
The main differences between the positive and negative full wave rectifier are the direction that the diodes are pointing and the polarity of the output voltage.
The analysis of the negative full wave rectifier circuit is same as negative half wave rectifier.

10. NEGATIVE FULL WAVE RECTIFIER
If we reverse the direction of the diodes in the positive full wave rectifier, we will have a negative full wave rectifier as shown in the fig.

12. PEAK INVERSE VOLTAGE
When one of the diodes in a full wave rectifier is reverse biased, the voltage across it is approximately equal to V2.
The peak load voltage supplied by the full wave rectifier is equal to the one half of the secondary voltage V2. Therefore, the reverse voltage will be

13. PEAK INVERSE VOLTAGE twice the peak load voltage. By formula
PIV = 2VL(pk)
Since the peak load voltage is half the secondary voltage, we can also find value of PIV as
PIV = V2(pk)

14. PEAK INVERSE VOLTAGE
When calculating the PIV in practical diode circuit, the 0.7 drop will be also be taken into consideration. In the fig. when D1 is on and will have voltage drop of 0.7

15. PEAK INVERSE VOLTAGE
Since D1 is in series with D2 the PIV across D2 will be reduced by the voltage across D1. The equation
PIV = V2(pk) – 0.7
Gives a more accurate PIV voltage.

17. FULL WAVE BRIDGE RECTIFIER
The bridge rectifier is the most commonly used full wave rectifier circuit for several reasons
(1) It does not require the use of center-tapped transformer, and therefore can be coupled directly to the ac power line, if desired.

18. FULL WAVE BRIDGE RECTIFIER
(2) Using a transformer with the same secondary voltage produces a peak output voltage that is nearly double the voltage of the full wave center-tapped rectifier. This results in the higher dc voltage from the supply.

20. BASIC CIRCUIT OPERATION
The bridge rectifier shown in the fig. consists of four diodes and a resistor. The full wave rectifier produces its output by alternating the circuit conduction

21. BASIC CIRCUIT OPERATION
Between the two diodes . When one is on( conducting), the other if off( non conducting), and vice versa. The bridge rectifier works basically in the same way. The main difference is that the bridge rectifier alternates conduction between two diodes pairs.

23. BASIC CIRCUIT OPERATION
When D1 and D3 are on, D2 and D4 are off and vive versa. This circuit operation is illustrated in the fig.
The current direction will not change in the either condition.

24. CALCULATING the LOAD VOLTAGE and CURRENT VALUES
The full wave rectifier has an output voltage equal to the one half of the secondary voltage. The center-tapped transformer is essential for the full wave rectifier to work, but it cuts the output voltage to half.

25. CALCULATING the LOAD VOLTAGE and CURRENT VALUES
The bridge rectifier does not require the use of center-tapped transformer. Assuming the diodes to be ideal, the rectifier will have a peak output voltage of
VL(pk) = V2(pk) (ideal)

26. CALCULATING the LOAD VOLTAGE and CURRENT VALUES
When calculating the circuit output values, we will get more accurate results if we take the voltage drops across the two conducting diodes into account. To include these values,we will use of following equation
VL(pk) =V2(pk) – 1.4
The 1.4 volts represents the sum of diode voltage dropes.

27. EXAMPLE
Determine the dc load voltage and current values for the circuit shown in the fig.

28. SOLUTION
With the 12Vac rated transformer, the peak secondary voltage is found as
V2(pk) = 12/0.707
=16.97Vpk

29. SOLUTION The peak load voltage is now found as VL(pk) = V2 – 1.4
=15.57Vpk
The dc load voltage is found as

30. SOLUTION Vave = 2VL(pk)/Π =31.14/Π =9.91 Vdc
Finally the dc load current is found as
Iave = Vave/RL
= 9.91/12k
= 825.8microA