Module 6—Introduction to Electronic Emission, Tubes, and Power Supplies
i - ix
, 1-1 to 1-10
1-11 to 1-20
, 1-21 to 1-30
1-31 to 1-40
, 1-41 to 1-50
1-51 to 1-56
, 2-1 to 2-10
2-11 to 2-20
, 2-21 to 2-30
2-31 to 2-39
, 3-1 to 3-10
3-11 to 3-20
, 3-21 to 3-30
3-31 to 3-40
, 3-41 to 3-50
3-51 to 3-60
, AI-1 to AI-3, Index
Q13. What is the average
voltage (Eavg) output of a full-wave rectifier that has an output of 100 volts peak?
The Bridge Rectifier
When four diodes are connected as shown in figure 3-12, the circuit
is called a BRIDGE RECTIFIER
. The input to the circuit is applied to the diagonally opposite
corners of the network, and the output is taken from the remaining two corners.
Figure 3-12.—Bridge rectifier circuit.
During one half-cycle of the applied voltage, point A becomes positive with respect to point B by the
amount of voltage induced into the secondary of the transformer. During this time, the voltage between points A
and B may be considered to be impressed across V1, the load resistor RL
, and V3, in series. The voltage
applied across these tubes makes their plates more positive than their cathodes, and current flows from point B
through tube V1 in an upward direction across the load resistor, through tube V3, to point A. This path is
indicated by the solid arrows. The waveform is shown as numbers (1) and (2).
One half-cycle later, the
polarity across the secondary reverses, making the plates of V1 and V3 negative with respect to their cathodes. At
the same time, the plates of V2 and V4 become positive with respect to their cathodes, and current flows in the
direction indicated by the dashed arrows. The current through RL is always in the same direction. This current, in
flowing through RL, develops a voltage corresponding to that shown in waveform (5) of the figure. The bridge
rectifier is a full-wave rectifier since current flows through the load during both half cycles of the applied
One advantage of a bridge rectifier over a conventional full-wave rectifier is that
with a given transformer, the bridge rectifier produces a voltage output that is nearly twice that of the
conventional full-wave circuit. We can show this by assigning values to some of the components as shown in figure
3-13, views (A) and (B). Assume that the same transformer is used in both circuits. The peak voltage developed
between points X and Y is 1,000 volts in both circuits. In the conventional full-wave circuit,
view (A), the peak voltage from the center tap to either X or Y is 500 volts. Since only one diode can
conduct at any instant, the maximum voltage that can be rectified at any instant is 500 volts. Therefore, the
maximum voltage that appears across the load resistor is nearly, but never exceeds, 500 volts (because of the
small voltage drop across the tube). In the bridge rectifier of view (B), the maximum voltage that can be
rectified is the full secondary voltage, 1,000 volts. Therefore, the peak output voltage across the load resistor
is nearly 1,000 volts. Thus, with both circuits using the same transformer, the full-wave bridge circuit produces
a higher output voltage than the conventional full-wave rectifier.
Figure 3-13.—Comparison of conventional full-wave and bridge rectifiers:
A. Conventional full-wave circuit
A second advantage of the bridge rectifier is the low ratio of peak inverse voltage to average
output voltage. For this reason bridge rectifiers that use vacuum tubes are widely used in high-voltage power
If directly heated diodes are used in a bridge rectifier, three separate filament
transformers are required. This is due to the different potentials existing at the filaments of the diodes. The
filaments of V2 and V3 in figure 3-14 are at the same potential, but the filament of V1 is at a different
potential from either V2 or V4. The three filament transformers must be well insulated from each other, and from
ground, because of the high potentials to which they are subjected. The use of indirectly heated diodes would
solve the filament transformer problem, but the high potential difference between cathode and heater would be
likely to result in arcing.
Figure 3-14.—Bridge rectifier with filament transformers.
Q14. What is the main disadvantage of the conventional full-wave rectifier?
Q15. What main advantage does a bridge rectifier have over a conventional full-wave rectifier?
While the output of a rectifier is a pulsating dc, most electronic circuits
require a substantially pure dc for proper operation. This type of output may be provided by placing single or
multisection filter circuits between the output of the rectifier and the load.
There are four basic types
of filter circuits:
· Simple capacitor filter
· LC choke-input filter
capacitor-input filter (pi-type)
· RC capacitor-input filter (pi-type)
We will cover the
function of each of these filters in detail later in this chapter.
Filtering is done by using various
combinations of capacitors, inductors, and resistors. Inductors are used as series impedances to oppose the change
in flow of alternating (pulsating dc) current. Capacitors are used as shunt elements to bypass the alternating
components of the signal around the load (to ground). Resistors are used in place of inductors in low current
Let's briefly review the properties of a capacitor. First, a capacitor opposes any change
in voltage. The opposition to a change in voltage is called capacitive reactance (XC
) and is measured
in ohms. The capacitive reactance is determined by the frequency (f) of the applied voltage and capacitance (C) of
From the formula, you can see that if frequency or capacitance is increased, the XC will decrease. Since
filter capacitors are placed in parallel with the load, a low XC will provide better filtering than a high XC.
This is done by providing a better shunting effect of the ac around the load, as shown in figure
Figure 3-15.—Capacitor filter. Fast charge time
To obtain a steady dc output, the capacitor must charge almost instantaneously to the value of applied
voltage. Once charged, the capacitor must retain the charge as long as possible. The capacitor must have a short
charge time constant (view A) and a long discharge time constant (view B). This can be
done by keeping the
internal resistance of the power supply as small as possible (fast charge time) and the
resistance of the load
as large as possible (slow discharge time).
From your earlier studies in basic electricity, you may
remember that one capacitor time constant is
defined as the time it takes a capacitor to charge to 63.2
percent of the applied voltage or to discharge to
36.8 percent of its total charge. This can be expressed by
the following equation:
You should also recall that a capacitor is considered fully charged after five RC time constants.
Referring to figure 3-16, you should see that to obtain a steady dc output voltage, the capacitor should charge
rapidly and discharge as slowly as possible.
Figure 3-16.—RC time constant chart.
In filter circuits the capacitor is the common element to both the charge and discharge paths.
Therefore, to obtain the longest possible discharge time, you want the capacitor to be as large as possible.
Another way to look at this is: The capacitor acts as a short circuit around the load (as far as the ac component
is concerned), and since
the larger the value of the capacitor (C), the smaller the opposition (XC
) or resistance to
Now let's look at inductors and their application in filter circuits. Remember, AN INDUCTOR OPPOSES
ANY CHANGE IN CURRENT. In case you have forgotten, a change in current through an inductor produces a changing
electromagnetic field. The changing field, in turn, cuts the windings of the wire in the inductor and thereby
produces a counterelectromotive force (cemf). It is the cemf that opposes the change in circuit current.
Opposition to a change in current at a given frequency is called inductive reactance (XL
) and is
measured in ohms. The inductive reactance (XL
) of an inductor is determined by the applied frequency
and the inductance of the inductor. Mathematically,
XL = 2!fL
From the preceding formula, you know that if either frequency or inductance is increased, the XL
will increase. Since inductors are placed in series with the load (fig. 3-17), the larger the XL
larger the ac voltage developed across the inductor and the smaller the ac voltage developed across the load.
Figure 3-17.—Voltage drops in an inductive filter.
Now back to our circuit. As illustrated in figure 3-18, when the current starts to flow through the
coil, an expanding magnetic field builds up around the inductor. This magnetic field around the coil develops the
cemf that opposes the change in current. When the rectifier current decreases as shown in figure 3-19, the
magnetic field collapses and again cuts the turns (windings) of wire, thus inducing current into the coil. This
additional current adds to the rectifier current and attempts to keep it at its original level.
Figure 3-18.—Inductive filter (expanding field).
Figure 3-19.—Inductive filter (collapsing field).
Now that you have learned how the components in the filter circuits react to current flow from the
rectifier, let's discuss the different types of filter circuits in use today.
Q16. If you increase the value of the capacitor will the XC increase or decrease? Why?
The Capacitor Filter
The simple capacitor filter is the most basic type of power supply
filter. The use of this filter is very limited. It is sometimes used on extremely high-voltage, low-current power
supplies for cathode-ray and similar electron tubes that require very little load current from the supply. This
filter is also used in circuits where the power-supply ripple frequency is not critical and can be relatively
The simple capacitor filter shown in figure 3-20 consists of a single-filter element. This capacitor (C1) is
connected across the output of the rectifier in parallel with the load. The RC charge time of the filter capacitor
(C1) must be short and the RC discharge time must be long to eliminate ripple action when using this filter. In
other words, the capacitor must charge up fast with preferably no discharge at all. Better filtering also results
when the frequency is high; therefore, the full-wave rectifier output is easier to filter than the half-wave
rectifier because of its higher frequency.
Figure 3-20.—Full-wave rectifier with a capacitor filter.
To understand better the effect that filtering has on E avg
, compare the rectifier circuits
without filters in figure 3-21 to those with filters in figure 3-22. The output waveforms in figure 3-21 represent
the unfiltered outputs of the half-wave and full-wave rectifier circuits. Current pulses flow through the load
) each time a diode conducts. The dashed line indicates the average value of output
voltage. For the half-wave rectifier, Eavg
is less than half the peak output voltage (or approximately
0.318 of the peak output voltage). For the full-wave rectifier, Eavg
is approximately 0.637. This value
is still much less than the applied voltage. With no capacitor connected across the output of the rectifier
circuit, the waveform has a large pulsating component (ripple) compared with the average or dc component. Now
refer to figure 3-22. When a capacitor is connected across the output (in parallel with RL), the average value of
output voltage (Eavg
) is increased due to the filtering action of capacitor C1.
Figure 3-21.—Half-wave/full-wave rectifiers (without filters).
Figure 3-22.—Half-wave/full-wave rectifiers (with capacitor filters).
The value of the capacitor is fairly large (several microfarads); it thus presents a relatively low
reactance to the pulsating current and stores a substantial charge. The rate of charge for the capacitor is
limited only by the relatively low resistance of the conducting diode. The RC charge time of the circuit is,
therefore, relatively short. As a result, when the pulsating voltage is first applied to the circuit, the
capacitor charges rapidly and almost reaches the peak value of the rectified voltage within the first few cycles.
The capacitor attempts to charge to the peak value of the rectified voltage anytime a diode is conducting, and
tends to retain its charge when the rectifier output falls to zero. (The capacitor cannot discharge immediately).
The capacitor slowly discharges through the load resistance (RL
) during the time the rectifier is
The rate of discharge of the capacitor is determined by the value of capacitance and the
value of the load resistance. If the capacitance and load resistance values are large, the RC discharge time for
the circuit is relatively long.
From the waveforms shown in figure 3-22, you should see that the addition
of C1 to the circuit results in an increase in the average value of output voltage (Eavg
) and a
reduction in the amplitude of the ripple component (Er
) present across the load resistance.
Now, let's consider a complete cycle of operation using a half-wave rectifier, a capacitive filter (C1), and a
load resistor (RL
As shown in figure 3-23, C1 is assumed to be large enough to ensure a
small reactance to the pulsating rectified current. The resistance of RL
is assumed to be much greater than the reactance of C1 at the input frequency.
Figure 3-23.—Half-wave rectifier capacitor filter (positive input cycle).
When the circuit is energized, the diode conducts on the positive half cycle and current flows through
the circuit allowing C1 to charge. C1 will charge to approximately the peak value of the input voltage. The charge
is less than the peak value because of the voltage drop across diode V1. The charge on C1 is indicated by the
heavy, solid line on the waveform.
As illustrated in figure 3-24, the diode (V1) cannot conduct on the
negative half cycle because the plate of V1 is negative in respect to the cathode. During this interval, C1
discharges through load resistance RL. The discharge of C1 produces the downward slope indicated by the solid line
on the waveform in the figure.
Figure 3-24.—Half-wave rectifier capacitor filter (negative input cycle).
During the discharge period, in contrast to the abrupt fall of the applied ac voltage from peak value to
zero, the voltage across C1 (and thus across RL) gradually decreases until the time of the next half cycle of
rectifier operation. Keep in mind that for good filtering, the filter capacitor should
as possible and discharge
Since practical values of C1 and RL ensure a more or less gradual decrease of
the discharge voltage, a substantial charge remains on the capacitor at the time of the next half cycle of
operation. As a result, no current can flow through the diode until the rising ac input voltage at the plate of
the diode exceeds the voltage of the charge remaining on C1. The charge on C1 is the cathode potential of the
diode. When the potential on the plate exceeds the potential on the cathode (the charge on C1), the diode again
conducts, and C1 commences to charge to approximately the peak value of the applied voltage.
capacitor has charged to its peak value, it begins to discharge. Since the fall of the ac input voltage on the
plate is considerably more rapid than the decrease on the capacitor voltage, the cathode quickly becomes more
positive than the plate, and the diode ceases to conduct.
The operation of the simple capacitor filter using a full-wave rectifier is basically the same as the
operation we discussed for the half-wave rectifier. Notice in figure 3-25 that because one of the diodes is always
conducting on either alternation, the filter capacitor charges and discharges during each half cycle. (Note that
each diode conducts only for that portion of time when the peak secondary voltage is greater than the charge
across the capacitor.)
Figure 3-25.—Full-wave rectifier (with capacitor filter).
We stated before that a major advantage of full-wave and bridge rectifiers over half-wave
rectifiers is the ease of filtering their output voltages. You can now see the reason for this. The ripple
frequency is doubled; therefore, the time period the capacitor is allowed to discharge is cut in half. This means
that the capacitor discharges less. Thus, ripple amplitude is less, and a smoother output voltage occurs.
Another thing to keep in mind is that the ripple component (Er
) of the output voltage is an ac voltage
and the average output voltage (Eavg
) is the dc component of the output. Since the filter capacitor
offers a relatively low impedance to ac, the majority of the ac component flows through the filter capacitor. The
ac component is therefore bypassed (shunted) around the load resistance and the entire dc component (or Eavg
flows through the load resistance. To clarify this statement, let's take a look at the formula for XC
in a half-wave and full-wave rectifier. First, you must establish some values for the circuit.
Introduction to Matter, Energy, and Direct Current,
to Alternating Current and Transformers, Introduction to Circuit Protection,
Control, and Measurement
, Introduction to Electrical Conductors, Wiring Techniques,
and Schematic Reading
, Introduction to Generators and Motors
Introduction to Electronic Emission, Tubes, and Power Supplies,
Introduction to Solid-State Devices and Power Supplies
Introduction to Amplifiers, Introduction to
Wave-Generation and Wave-Shaping Circuits
, Introduction to Wave Propagation, Transmission
Lines, and Antennas
, Microwave Principles,
, Introduction to Number Systems and Logic Circuits, Introduction
to Microelectronics, Principles of Synchros, Servos, and Gyros
Introduction to Test Equipment
, Radar Principles,
The Technician's Handbook,
Master Glossary, Test Methods and Practices,
Introduction to Digital Computers,
Magnetic Recording, Introduction to Fiber Optics