RECTIFIERS

From previous discussions, you know that rectification is the changing of an ac voltage to a pulsating dc voltage. Now let's discuss the process of rectification.

Since a diode vacuum tube will pass current in only one direction, it is ideally suited for converting alternating current to direct current. If an ac voltage is applied to a diode, the diode will conduct ONLY DURING THE POSITIVE ALTERNATION OF VOLTAGE when the plate of the diode is made positive with respect to the cathode.

Figure 3-4 shows a diode connected across the 120-volt ac line. During the positive alternation of the source voltage, the sine wave applied to the tube makes the plate positive with respect to the cathode. At this time the diode conducts and plate current flows from the negative supply lead, through the milliammeter, through the tube, and to the positive supply lead. This is indicated by the shaded area of the output waveform. This current exists during the entire period of time that the plate is positive with respect to the cathode (for the first 180 degrees of the input sine wave).

Figure 3-4. - Simple diode rectifier.

During the negative alternation of plate voltage (dotted polarity signs), the plate is driven negative and the tube cannot conduct. When conditions prevent the tube from conducting, the tube is said to be in CUTOFF. This is indicated by the dotted waveform. The tube will be in cutoff and no current will flow for the entire negative alternation.

For each 360-degree cycle of input voltage, the tube conducts for 180 degrees and is in cutoff for 180 degrees. The circuit current therefore has the appearance of a series of positive pulses, as shown by the shaded areas. Notice that although the current is in the form of pulses, the current always flows through the circuit in THE SAME DIRECTION. Current that flows in pulses in the same direction is called PULSATING DC. The diode has thus RECTIFIED the input voltage. Although the principle of rectification applies to all rectifier circuits, some rectifiers are more efficient than others. For this reason, we will explain the three rectifier circuits most commonly used in electronics today-the half-wave, full-wave, and bridge.

A Practical Half-Wave Rectifier

Figure 3-5 is a diagram of a complete half-wave rectifier circuit. For the diode to be used as a rectifier, it must be connected in series with a load device (R_L for this circuit), through which the direct current flows. Because Navy electronic equipment requires various input voltages, it is necessary to have a rectified voltage that is greater (or smaller in some cases) than the source voltage. The rectifier plate circuit is supplied power from a step-up (or step-down) transformer. Notice that the transformer has the two secondary windings mentioned earlier. The lower winding supplies high voltage to the plate and cathode of the diode, and the upper winding supplies a low ac voltage to the filaments of the diode. Notice also that the cathode of the diode is connected to the secondary winding of the transformer through the load resistor (R_L). Any current flowing through the tube also flows through the load resistor, causing a voltage to be developed across it. The magnitude of the voltage developed across the load resistor is directly proportional to the amount of current flowing through it (Ohm's law: E = IR).

Figure 3-5. - Half-wave rectifier circuit.

You will better understand the operation of the half-wave rectifier circuit if it is redrawn in the form of a simplified series circuit. As you can see in figure 3-6, the diode (V1) and load resistor (R_L) are in series with the secondary winding of the transformer. During the positive alternation of the input, as the voltage in the secondary winding increases, the current through diode (V1) and load resistor (R_L) increases. Since the diode tube and the load resistor form a series circuit, the same current flows through both the tube and the resistor. This current produces a voltage drop across the tube and the load resistor, which have polarities as shown. Since the plate resistance of the tube is only about 500 ohms and the resistance of the load resistor is 10,000 ohms, approximately 95 percent of the applied 425 volts is dropped across the load resistor (425 X .95 = 404 V) and 5 percent (425 X .05 = 21 V) across the tube.

Figure 3-6. - Simplified half-wave rectifier circuit and waveforms.

During the negative half of the alternation of input voltage, the tube cannot conduct and no current flows in the circuit. Since there is no current flow through R_L, the load voltage remains at zero volts throughout the negative alternation. During this time the entire negative alternation is felt across the tube. The reason for this is derived from Kirchhoff's law, which states:

E_L + E_b = E _a

The sum of the load voltage and diode voltage equals the applied voltage.

Since a half-wave rectifier conducts once for each full cycle of input voltage, the frequency of the pulses is the same as the frequency of the input sine wave. The output pulse frequency is called RIPPLE FREQUENCY. If the rectifier circuit is supplied power from a 60-hertz ac line voltage, 60 pulses of load current will occur each second. Therefore, THE RIPPLE FREQUENCY OF A HALF-WAVE RECTIFIER IS THE SAME AS THE LINE FREQUENCY.

If a series of current pulses like those obtained from a half-wave rectifier is applied to a load resistance, an average amount of power will be dissipated over a given period of time. This average dc power is determined by the amplitude of the pulses and the time delay between pulses. The higher the peak amplitude of the pulses or the less the time between pulses, the greater the average dc power supplied to the load. To determine average dc voltage (E_avg), it is necessary to know the average value of the pulses and the peak value of load voltage. This is illustrated in figure 3-7.

Figure 3-7. - Peak and average values for a half-wave rectifier.

Since current and voltage waveforms in a half-wave rectifier circuit are essentially half sine waves, we can develop a conversion factor. The formula for average value was discussed earlier in NEETS, module 2. By now you should know that the average value for a full sine wave is .637 times its peak or maximum value. Therefore, if you want the average value of a half-wave rectifier output, you should multiply half the value of .637 (.318) times the peak or maximum voltage, as expressed in the following equation:

E_avg (the average load voltage) = .318 X E_max

Where:

E_max = The peak value of the load voltage pulse

In most applications the drop across the rectifier tube is small compared to the load voltage, so we can assume E_max in our equation to be the same as the peak value of the input sine wave.

Since the load current has the same wave shape as the load voltage, we can modify the equation so that it applies to the load current. Thus,

I_avg (the average load current) = .318 X I_max

Where:

I_max = The peak load current

If a line is drawn through the rectified waveform at a point that is 0.318 of the distance from zero to maximum, the waveform will be divided so that area A is equal to area B (fig. 3-7). Therefore, current or voltage pulses with a value of .318 of the peak value have the same effect on the load as a steady voltage or current.

The half-wave rectifier uses the transformer during only one-half of the cycle. Therefore, for any given size transformer, less power is developed than if the transformer were used on both halves of the cycle. In other words, to obtain large amounts of power, the half-wave transformer must be relatively large in comparison to what it would have to be if both halves of the cycle were used. This disadvantage limits the use of the half-wave rectifier to applications that require a very small current drain. The half-wave rectifier is widely used for commercial ac and dc radio receivers and other applications where inexpensive voltage supplies will suffice. As you can see from your study on half-wave rectifiers, this type of circuit placed many limitations on electronic equipment. For this reason another type of rectifier circuit had to be developed. One of the factors that had to be considered was how to use the full output from the transformer to obtain the highest average voltage and current. Thus, the FULL-WAVE rectifier was developed.

Q.6 Does a rectifier tube conduct on the positive or negative alternation of the input signal?
Q.7 What term is used to describe the period when the diode is not conducting?
Q.8 Current that flows in pulses in the same direction is called _____.
Q.9 For a diode to act as a rectifier, should it be connected in series or parallel with the load?
Q.10 What is the Ripple frequency of a half-wave rectifier if the input frequency is 60 Hz?
Q.11 What is the equation for determining average voltage in a half-wave rectifier?