3-4
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 (RL 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 (RL). 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).