1. Negative Half Wave Rectifiers
Fig. shows a half wave rectifier with diode direction is reversed. In this circuit the diode will conduct on the negative half cycle of the input, and
VL = V2.

2. Negative Half Wave Rectifiers
As a result positive half cycle of the input is eliminate. The operating principle of the negative half wave rectifier is same as the positive half wave rectifiers. The only difference is the polarity of the output will be reversed.

3. GENERAL RULES
When the diode points towards the load (RL), the output from the rectifier will be positive.
When the diode points toward the transformer, the output from the rectifier will be negative.
The points made so far are summarized in the next slide

4. O O

5. CALCULATING LOAD VOLTAGE AND CURRENT VALUES.
The load voltage in half wave rectifier can be found as
VL(pk) = V2(pk) – VF
V2(pk) is the secondary voltage of the transformer, found as
V2(pk) = (N2/N1) V1

6. CALCULATING LOAD VOLTAGE and CURRENT VALUES.
where
N2/N1 =(The turn ratio of transformer secondary turns to primary turn)
V1(pk) = (The peak transformer primary voltage)
Above equation assumes that the transformer is given as a peak value.

7. CALCULATING LOAD VOLTAGE and CURRENT VALUES.
More often than not, source voltages are given as rms values. When this is the case, the source voltage is converted into peak value as follows:
Vpk = Vrms/0.707

8. EXAMPLE
Determine the peak load voltage for the circuit shown in the fig.

9. SOLUTION
First, the ac input to the transformer is converted to a peak value, as follows
V1(pk) = V1rms/0.707
= 120/0.707

10. SOLUTION
= 169.7Vpk
Now, the voltage values in the secondary circuit are found as
V2(pk) = N2/N1(V1pk)

11. SOLUTION = 1/5(169.7) =33.94Vpk And VLpk = V2(pk) – VF =33.94 – 0.7

12. Determine the peak load voltage for the circuit shown in fig.
EXAMPLE
Determine the peak load voltage for the circuit shown in fig.

13. The transformer is shown to have
SOLUTION
The transformer is shown to have
a 25Vac rating .This value of V2 is converted into peak form as
V2(pk) = V2rms/0.707

14. SOLUTION = 25Vrms/0.707 =35.36Vpk Now the value of Vpk is found as
VL(pk) = V2(pk) – 0.7
= – 0.7
= 34.66Vpk

15. CALCULATING LOAD CURRENT
Once the peak voltage is determined, the peak load current is found as
IL(pk) = VL(pk)/RL

16. Example
What is the peak load current for the circuit shown in fig.

17. SOLUTION
The input voltage is given an rms value. This value is converted to a peak value as follows:
V1(pk) = V1(rms)/0.707
= 200Vac/0.707
= Vpk

18. SOLUTION
Now, the load voltage and current are found, after fining peak voltage, as
V2(pk) = N2/N1 V1(pk)
= (1/5)(282.9 Vpk)
= 56.6Vpk

19. SOLUTION
Finally, the load voltage and current and current values are found as:
VL(pk) = V2(pk) – VF
=56.6 – 0.7
= 55.9Vpk

20. SOLUTION and the current will be IL(pk) = VL(PK)/RL = 55.9 Vpk/10k
= 5.59mApk

21. AVERAGE LOAD VOLTAGE AND CURRENT
Since rectifiers are used to convert ac to dc, Vave is a very important value. For a half wave rectifier, Vave is found as
Vave = Vpk/π
Another form of this equation is
Vave = 0.318(Vpk) (half wave rectified)

22. AVERAGE LOAD VOLTAGE AND CURRENT
Where = 1/π . Either of these equations can be used to determine the dc equivalent load voltage for a half wave rectifier . Let’s take some examples to illustrate this

23. EXAMPLE
Determine the value of Vave for the circuit shown in fig.

24. SOLUTION V1(PK) = V1(rms)/0.707 = 75Vac/0.707 = 106.1 Vpk
V2(pk) = N2/N1 V1(pk)
=1/2(106.1)
= 53.04Vpk

25. SOLUTION VL(pk) = 53.04 – 0.7 = 52.34Vpk Vave =Vpk/π = 52.34 /π
= Vdc

26. CALCULATING AVERAGE CURRENT
THE VALUE OF Iave can be calculated in one of two ways
We can determine the value of Vave and then use Ohm’s law as follows
Iave = Vave/RL
(2) We can convert Ipk to average from using same basic equations, that

27. CALCULATING AVERAGE CURRENT
we used to convert Vpk to Vave. The current form of these equations are
Iave = Ipk/π
And
Iave = 0.318(Ipk)

28. EXAMPLE
Determine the dc load current for the rectifier shown in fig.

29. SOLUTION
The transformer has 24 Vac rating. Thus, the peak secondary voltage is found as
V2(pk) = 24Vac/0.707
= 33.9 Vpk
The peak load voltage is now found as

30. SOLUTION VL(pk) = V2(pk) – 0.7 = 33.2 Vpk IL(pk)= VL(pk)/RL
= 1.66mApk

31. SOLUTION
Iave = Ipk/ π
= 1.66m/π
= micro A