Concept explainers
A half−wave rectifier such as shown in Figure 2.2(a) has a
a.
The peak output voltage of a half-wave rectifier.
Answer to Problem 2.3P
Peak output voltage is 16.27 V
Explanation of Solution
Given information:
Load resistance
Input RMS voltage,
Input frequency,
Transformer turn ratio
Diode cut-in voltage
Calculation:
Since the turn ratio of the transformer is 10:1 the secondary voltage is
Then the peak secondary voltage,
The peak output voltage,
Conclusion:
The peak output voltage is
b.
The peak diode current of a half-wave rectifier
Answer to Problem 2.3P
Peak diode current is 8.14 mA.
Explanation of Solution
Given information:
Load resistance
Input RMS voltage,
Input frequency,
Transformer turn ratio
Diode cut-in voltage
Calculation:
Since the turn ratio of the transformer is 10:1 the secondary voltage is
Then the peak secondary voltage,
The peak output voltage,
The diode current is maximum when the output voltage is maximum, hence the peak diode current is,
Conclusion:
The diode maximum current is 8.14 mA.
c.
The fraction of a cycle where the output voltage always greater than zero.
Answer to Problem 2.3P
The fraction of a cycle (where output voltage is positive) is 48.7%
Explanation of Solution
Given information:
Load resistance
Input RMS voltage,
Input frequency,
Transformer turn ratio
Diode cut-in voltage
Calculation:
When diode is forward biased and started to conduct then there will be an output voltage which is greater than zero. This concept is used to find where diode starts conducting. Similarly, the time where the diode is going to reverse bias can be calculated. Then the diode conduction time can be found as a fraction from the output cycle.
Since the turn ratio of the transformer is 10:1 the secondary voltage is
Then the peak secondary voltage,
The output voltage can be written as,
When diode is forward biased and started to conduct,
Substituting the values
By symmetry it can be found where
Hence the diode conduction period (where
So, the fraction of a cycle such that
Fraction of a cycle =
Conclusion:
The fraction of cycle the diode is conducting is 48.7%.
d.
The average output voltage for the half-wave rectifier
Answer to Problem 2.3P
Average output voltage is5.06 V.
Explanation of Solution
Given information:
Load resistance
Input RMS voltage,
Input frequency,
Transformer turn ratio
Diode cut-in voltage
Calculation:
Since the turn ratio of the transformer is 10:1 the secondary voltage is
Then the peak secondary voltage,
The output voltage can be written as,
When diode is forward biased and started to conduct,
Substituting the values
By symmetry it will be found where
Hence the diode conduction period (where
The average output voltage can be determined by integrating the equation (1) over a one period,
During the period
Hence
Conclusion:
Average voltage is
d.
The average output current the half-wave rectifier.
Answer to Problem 2.3P
Average output current is2.53 mA
Explanation of Solution
Given information:
Load resistance
Input RMS voltage,
Input frequency,
Transformer turn ratio
Diode cut-in voltage
Calculation:
Since the turn ratio of the transformer is 10:1 the secondary voltage is
Then the peak secondary voltage,
Output voltage can be written as,
When the diode is forward biased and started to conduct,
Substituting the values
By symmetry,
Hence the diode conduction period (where
The average output voltage can be determined by integrating the equation (1) over a one period,
During the period
Hence
Then the average diode current is,
Conclusion:
Then the average diode current is
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