# Experimental Investigation of Controlled and Uncontrolled Rectifiers for Low-Power Wind Turbines
## 4 Single-Phase Half-Wave Uncontrolled Rectifier
The phase angle \(\theta\) is expressed explicitly as:  
\[\theta=\pi-\tan^{-1}(\omega RC) \tag{7}\]The capacitive filter is charged to 90% of its maximum value according to the charging time constant, which is the product of load (\(R\) ) times capacitance (\(C\) )  
\[\tau=RC \tag{8}\]  
Observing the relationship between the discharge time, time period, and charging time constant is interesting. As shown in Figure 6a, as the discharge start time decreases, the time constant \(\tau\) increases, showing an inverse relationship [18]. Similarly, in Figure 6b, the output voltage is directly proportional to the time constant. Since the resistive load is constant, the output voltage can be increased or decreased by increasing or decreasing the capacitance value [26]. This paper reports the effect of the time constant on the discharge start time and the output voltage. The above theoretical analysis is validated by developing a single-phase half-wave hardware prototype. The input AC 208 \(V_{RMS}/60\) Hz supply is stepped down by a transformer having a turn ratio of 7.43. The semiconductor diode 1N007 device rated at 1 A and having a PIV of 1000V is used. A ceramic resistance of 25 \(\Omega\) is used as a load. Figure 7a shows the recorded waveforms on the oscilloscope, where channel 1 shows the stepped-down AC sinusoidal voltage measuring 23.45 \(V_{RMS}\). Channel 2 records the peak-to-peak and average current flowing into the load. Since the current is measured through differential probes, the current values are indicated in mV on the oscilloscope trace of channels 2 and 3. In addition, the signals on channels 2 and 3 are amplified by gain 2. Therefore, the calibrated current flow into the load can be calculated from Equation (9). \[I_{actual}=\frac{I_{Measured}*2}{10*10^{-3}} \tag{9}\]  
The average current measured in Figure 7a (channel 2) is 0.53 A. While Figure 7a channel 3 shows the voltage waveform across the resistive load without a capacitor filter. The half-wave rectifier passes the positive AC cycle to the load and blocks the negative half-cycle, making the voltage RMS value 16.04 \(V_{RMS}\). Figure 6: (a) Time constant vs. discharge start time; (b) time constant vs. output voltage. Figure 7: Experimental results of half–wave rectification: (a) without output capacitor (b) with output capacitor (c) the voltage ripples and DC components  
Effect of Capacitive Filtering  
Capacitors can improve the output ripple performance and can also be used at the input side to smooth out the harmonic and glitches in the source AC supply. A 10 mF capacitor filter was connected in parallel to the load to improve the ripple performance of the half-wave rectifier circuit. At the same time, the same capacitor was also added to the input AC parallel to the transformer-winding, making the sinusoidal input become clamped at a certain voltage. The clamping happens because the diode is forward-biased, and the charging voltage becomes higher than the AC supply voltage [20]. Therefore, the capacitor is charged to a particular value which holds this charge, resulting in a constant clamped input voltage in a positive AC cycle. As a result, the output voltage is smoothed into a constant DC at 24.40 V (Channel 3, Figure 7b) instead of a pulsating DC. The following Table 1 shows the effect of capacitor addition on various parameters of the half-wave rectifier. In Figure 7c, channel 3 shows the DC component of the output voltage. By increasing the capacitance, the output voltage stops pulsating. Thus, the peak-to-peak ripple spans only 440 mV, which is only 2% of the output voltage.