The simplest circuit that can convert alternating current to direct current is the half-wave rectifier, shown in Fig. 4-3. Line voltage from an ac power outlet is applied to the primary winding of the transformer. Usually, the power plug has a third prong to ground the equipment. Because of the turns ratio, the peak voltage across the secondary winding is
V N N V p2 p 2 1 1
Recall the dot convention used with transformers. The dotted ends of a transformer have the same polarity of voltage at any instant in time. When the upper end of the primary winding is positive, the upper end of the secondary winding is also positive. When the upper end of the primary winding is negative, the upper end of the secondary winding is also negative.
Here is how the circuit works. On the positive half cycle of primary voltage, the secondary winding has a positive half sine wave across it. This means the diode is forward-biased. However, on the
negative half cycle of primary voltage, the secondary winding has a negative half sine wave. Therefore, the diode is reverse-biased. If you use the ideal-diode approximation for an initial analysis, you will realize that the positive half cycle appears across the load resistor, but not the negative half cycle.
For instance, Fig. 4-4 shows a transformer with a turns ratio of 5:1. The peak primary voltage is
Vp1 1 2 0 0 7 0 7 1 7 0 V V .
Figure 4-3 Figure 4-4 Half-wave Rectifier 5:1 Turns Ratio
The peak secondary voltage is
Vp2
5 1 7 0 V
3 4 V
With the ideal-diode approximation, the load voltage has a peak value of 34 V.
Figure 4-5 shows the load voltage. This type of waveform is called half-wave signal because the negative half cycles have been clipped off or removed. Since the load voltage has only a positive half cycle, the load current is unidirectional, meaning that it flows only in one direction. Therefore, the load current is a pulsating direct current. It starts at zero at the beginning of the cycle, then increases to a
maximum value at the positive peak, then decreases to zero where it sits for the entire negative half cycle.
Figure 4-5 Half-wave Signal
Period
The frequency of the half-wave signal is still equal to the line frequency, which is 60 Hz. (In Europe, line frequency is 50 Hz.) Recall that the period, T, equals the reciprocal of the frequency. Therefore, the half-wave signal has a period of
T f
1 1
6 0 H z
0 .0 1 6 7 s 1 6 .7 m s
This is the amount of time between the beginning of a positive half cycle and the start of the next positive half cycle. This is what your
would measure if you looked at a half-wave signal with an oscilloscope.
DC or Average Value
If you connect a dc voltmeter across the load resistor of Figure 4-5, it will indicate a dc voltage of Vp/ , which may be written as
Vdc = 0.318Vp (4-5)
where Vpis the peak value of the half-wave signal across the load
resistor. For instance, if the peak voltage is 34 V, the dc voltmeter will read
Vdc= 0.318(34 V) = 10.8 V
This dc voltage is sometimes called the “average” value of the half- wave signal because the voltmeter reads the average voltage over one complete cycle. The needle of the voltmeter cannot follow the rapid variations of the half-wave signal, so the needle settles down on the average value, which is 31.8 percent of the peak value. (The 31.8 percent can be proved with calculus.)
Approximations
Because the secondary voltage is much greater than the knee voltage, using the second approximation will improve the analysis only slightly. If we use the second approximation, the half-wave signal has a peak of 33.3 V. Furthermore, since the bulk resistance of a 1N4001 is only 0.23 compared to a load resistance of 1 k , there is no increase in accuracy when using the third
approximation. In conclusion, either the ideal diode or the second approximation is adequate in analyzing this circuit.
Example 4–4
In Europe, a half-wave rectifier has an input voltage of 240 V rms with a frequency of 50 Hz. If the step-down transformer has a turns ratio of 8:1, what is the load voltage?
Solution
You can divide 240 V by 0.707 to get the answer. Here is an alternative way to get the peak voltage. Since the rms voltage is twice as large as previous examples, the peak voltage is twice as large as before:
Vpl = 2(170 V) = 340 V
Because of the 8:1 step down, the secondary voltage has a peak value of
Vp2
8 3 4 0 V
4 2 .5 V
Ignoring the diode drop means that the load voltage is a half-wave signal with a peak value of 42.5 V.
The period of the rectified output voltage is slightly longer:
T 1
5 0 H z
0 .0 2 s 2 0 m s
This is what you would measure with an oscilloscope.