Chapter 3   DC Power Supply Circuits.

The power supply is connected to the 120 vAC power line by a plug. Many power supplies (those used in laboratory equipment) have a plug with a third wire which connects the metal chassis of the instrument to earth ground. The fuse and switch constitute the protection circuits. The fuse helps to protect the power supply against circuit component failures and mistaken connections. The switch permits the power supply to be turned on and off. The transformer steps the 120 volts of the AC line down to the value required by the devices being powered. The rectifier changes the AC from the transformer to pulsating DC. The filter reduces the magnitude of the pulsations and smoothes out the DC. The regulator reduces the pulsations to a very small value and also holds the output voltage constant regardless of the load current. The regulator also contains circuits which will shut down the power supply if the load current becomes too large or the temperature of the regulator becomes too high.

If the power supply is part of something else, such as an oscilloscope, an FM tuner, or an amplifier the output of the power supply is delivered to the internal circuits of the instrument and is not available to the operator. If the power supply is a laboratory bench power supply, its output is connected to binding posts on the supply's front panel for the purpose of powering experimental circuits. A laboratory bench power supply may also include a voltmeter and an ammeter to permit the operator to monitor the voltage and current which are supplied to the load.

Fuses are often placed in the primary circuit of the transformer. If a transformer develops an internal fault it can catch fire if the power is not shut off by an open fuse. There should always be a fuse in the primary of a power transformer. Many low-cost "wall transformers" do not have fuses in their primary circuits. These units present a constant fire hazard to any building in which they are used.

There are two types of fuses, fast-acting and time delay. Time delay fuses are often called slow-blow fuses. Power supplies which have very large filter capacitors and semiconductor rectifiers will draw a very large current for a very short time after being turned on. Have you ever noticed a flicker of the lights when you turn on a transistorized stereo receiver? You are seeing the effect of a large current pulse which is charging up the capacitors in the power supply. A fast-acting fuse would most likely blow out the moment the power switch was thrown. A slow-blow fuse is used in such a case. Slow-blow fuses are used only if there is a large current pulse at the time of turn-on.

When selecting the current rating of the fuse, measure the current drawn by the device under normal operation and multiply this figure by 2. A smaller fuse than this will tend to burn out when nothing is wrong.

Other protection devices are electro-mechanical circuit breakers. A circuit breaker may use a thermal sensor to detect the heating in a small wire or the current may be detected by allowing it to flow through a coil of wire and become an electromagnet. If the temperature of the wire gets too high or the electromagnet becomes too strong, the circuit breaker will "pop out" and the circuit will be opened. The advantage of a circuit breaker is that it can be reset instead of having to be replaced, as do fuses. The major disadvantage of circuit breakers is that they are more expensive, more complex, larger and heavier than fuses. As current gets smaller, the circuit breakers get larger and more complex. Circuit breakers are usually not used for currents below 1 ampere.

In a sense the on-off switch may be considered as a protection device. If the operator smells something burning, he can use the switch to turn the power off. In a few cases a circuit breaker may serve double duty as an on-off switch. Most circuit breakers are not designed for repeated on-off cycles and will quickly wear out if used as on-off switches.

Most instruments include a small light which glows when the power is on and goes out when the power switch is turned off or the fuse blows.

If you look in a catalog listing power transformers you will not find "turns ratio" mentioned even once. The specifications of power transformers are written in terms of primary and secondary voltages and maximum secondary current. You may read the following specification, "Primary, 117 vAC 50 - 60 Hz. Secondary, 24 vCT @ 1 A." The translation of this is "The transformer will operate on 117 volts ±10% at a line frequency of 50 or 60 Hz. If you apply 117 volts to the primary the voltage across the entire secondary will be 24 volts if the secondary current is 1 ampere. The maximum current which can safely be drawn from the entire secondary is 1 ampere." If this current is exceeded the transformer is in danger of burning out due to over heating.

Example 3.1

A transformer has the following specification: "Primary, 117 vAC 60 Hz. Secondary, 32 vCT @ 1.5 A." If the primary is connected to a source of 120 vAC, what voltage is measured from one side of the secondary to its center-tap?

Solution:

The ratio of secondary to primary voltage is a constant. 32 / 117 = V / 120. V = 32.82 v. The voltage measured from one side of the secondary to the center-tap is 1/2 that of the whole secondary or 16.41 v.

Electrical Isolation.

The argument might be made that one side of the 120 vAC line is connected to earth ground and no isolation is necessary. If a wiring error caused the leads to be reversed, the chassis of the instrument would be at a potential of 120 v with respect to earth ground. Such a wiring error did occur at the university I recently retired from. No one was injured but there was some soiled armor. It could be most unhealthy for anyone who might be unlucky enough to complete the circuit to ground.

In addition to safety considerations there are other reasons for not wanting a piece of equipment to be connected to the power line ground. The chief reason is noise. Noise is any electrical voltage or current in a place where it is not wanted. Noise voltages and currents are the weeds of electronics. In addition to bringing us the electric power we need, the power line brings us a lot of noise. One of the most effective noise removers is a transformer. The noise on a power line often has the same phase on both wires and, therefore, will not produce a magnetic field in the primary winding of a transformer. This is called common mode noise and will be discussed in more detail in the section on differential amplifiers.

In laboratories, recording studios, and radio stations, many different pieces of equipment are connected to work together. If each piece of equipment has its own earth ground and is connected to another piece of equipment, a phenomenon known as "ground-loop noise" appears. There are two possible solutions to this problem. One is to connect each piece of equipment to earth ground and make sure there is no electrical ground connection between individual pieces of equipment. The other is to connect only one piece of equipment to earth ground and make sure that there is only one ground wire from that piece to each of the other pieces of equipment in the setup. The isolation provided by the power transformer is absolutely vital to this "single point" grounding system.

Unfortunately, there is insufficient space here to cover the physics of semiconductors in detail. Whole books 1 have been devoted to this topic and there is a course in this department entitled "Solid State Physics" which interested students should enroll in. Meanwhile, we can only scratch the surface. Emphasis will be placed on visualization of the physical processes which are taking place.

1   Semiconductor Devices and Applications
    by R. A. Greiner McGraw-Hill 1961

A Semiconductor Is a Semiconductor Is a Semiconductor.

Silicon and germanium each have 4 valance electrons in an outer shell which has room for 8 electrons. These elements (and carbon as well) form very stable tetrahedral crystals with the shared valance electrons filling the outer electron shell of each atom. If every electron stayed in place within the crystal lattice the crystal would be a perfect insulator. The electrons do not stay home.

The width of the forbidden band is called the band-gap energy. The band-gap energy for silicon at room temperature (300 degrees K) is 1.106 electron-volts (ev) and for germanium it is 0.67 ev. At room temperature there is enough thermal energy in the crystal lattice to promote some electrons to the conduction band.

Holes and Electrons.

When an electron leaves its place in the crystal lattice it leaves behind a hole in the lattice. Strange as it may seem, the hole does not stay in place but moves as if it were a positively charged particle. It has energy and even momentum. A hole can move by a nearby valence electron jumping from its place in the crystal lattice to the hole without entering the conduction energy band. After the electron has jumped, it leaves a hole where it used to be. Thus, the hole can move through the crystal lattice.

Thermal energy in the crystal lattice creates hole- electron pairs which move through the crystal. If there is an electric field present, the electrons will move toward the positively charged region and the holes will move toward the negatively charged region. If there is no electric field present, the holes and electrons will move randomly.

If you visualize a crystal of silicon warming up from absolute zero, hole-electron pairs will begin to form. This process cannot continue indefinitely or the crystal would come apart because all of its valence bonds would be broken. When holes and conduction electrons (electrons not held in a valance bond) become plentiful enough, electrons will begin to fall into holes they happen to run into. This process is called recombination and goes on at the same rate as hole- electron pair formation. The equilibrium between pair formation and recombination causes silicon and germanium crystals to be semiconductors at room temperature.

In pure semiconductor material the concentration of holes is the same as the concentration of electrons in the conduction band. (The total number of electrons cannot change, what we mean when we talk about changing the concentration of electrons, is the number of electrons which are free for conduction.) In pure silicon or germanium

We may now define the intrinsic charge carrier concentration n_i as

Changing the Relative Concentration of Holes and Electrons.

If we add an element from group 5 in the periodic table the effect on a semiconductor is striking. The elements from group 5 have five electrons in the outer shell. Wherever one of these atoms appears in the crystal lattice, four of its outer electrons go into valence bonds and the fifth one becomes very loosely bound to its atom. A very small amount of energy (about 0.05 ev) is needed to promote this electron into the conduction band. Such an impurity is called a donor impurity because each atom donates an electron to the conduction energy band.

If we go back to pure silicon and add an element from group 3 in the periodic table the effect on a semiconductor is opposite to that for an element from group 5. The elements from group 3 have three electrons in the outer shell. Wherever one of these atoms appears in the crystal lattice, there are only three electrons in the outer shell to go into valence bonds. The unfilled valence bond is a hole. Once a wandering electron has fallen into this hole, a comparatively large amount of energy is required to promote that particular electron back into the conduction energy band. That electron and its three companions will remain in place and a hole will have been set free to wander through the crystal lattice. Such an impurity is called an acceptor impurity because each atom accepts an electron from the conduction energy band and causes a conduction hole to be created.

When a donor impurity loses its loosely-bound electron or an acceptor impurity captures a wandering electron (loses its loosely-bound hole), the impurity atom is said to be ionized. Ionization in a solid means the same thing it means in a liquid or gas, the loss of, or gain of an extra electron. When an impurity is added to a semiconductor, most of the impurity atoms are ionized most of the time, statistically speaking.

Because the impurity atoms are virtually 100 percent ionized, for every atom of donor impurity there is an additional electron in the conduction band. For every acceptor atom added, there is an additional conduction hole in the crystal. The meaning in this discussion is that donor impurity is added to one piece of semiconductor and acceptor impurity is added to another piece of semiconductor.

Recombinations occur when wandering holes and wandering electrons happen to collide. The probability of collisions goes up as the concentration of either holes or electrons goes up. When, for example, a donor impurity is added to a semiconductor, the increased concentration of electrons causes more frequent collisions and a higher rate of recombination. The recombination rate falls again when the hole concentration has been reduced to the point where a new equilibrium has been established. The end result is to decrease the concentration of holes in the crystal. It can be shown mathematically, in about ten pages, note 1, that

Some Terms to Be Remembered.

There are two types of charge carriers, holes and electrons. A majority carrier is the type of carrier which there are more of in a particular type (P or N) of semiconductor. A minority carrier is the type of carrier which there are fewer of in a particular type (P or N) of semiconductor.

Example 3.2

Identify each of the following as majority or minority carriers: (a) electrons in N type semiconductor, (b) holes in P type semiconductor, (c) electrons in P type semiconductor and (d) holes in N type semiconductor.

Solution:

The majority carriers have the same charge polarity as the type of semiconductor in which they are found. The one which is left is the minority carrier. Thus, (a) majority, (b) majority, (c) minority and (d) minority.

The P-N Junction

*   This is not how P-N junction diodes are manufactured.
    In fact, this could not even be done in the laboratory.
    This is merely a thought experiment which illustrates
    how the depletion region and contact potentials are
    formed in a P-N junction.

The region near the junction becomes depleted of charge carriers in the conduction band. As stated above, there are ions in the depletion region but these ions are held in the crystal lattice and are not in the conduction energy band. The depletion region constitutes an insulator and the regions on either side of the depletion region which are not depleted of charge are conducting regions. This concept is very important. Failure to understand the depletion region will result in failure to understand how P-N junctions conduct in only one direction.

Animation, which is not possible in a paper book could not be done in any medium when I was teaching this subject. We have sure come a long way baby.

The depletion region constitutes an insulator and the regions where there are charge carriers in the conduction energy band are conductors. This forms a capacitor. As the voltage changes the width of the depletion region changes, changing the amount of capacitance. Any P-N junction will show varying amounts of capacitance as the reverse voltage is changed. In rectifiers and small switching diodes this is only a few picofarads. Special diodes are available that have enhanced capacitance and are known as variable capacitance diodes or varicaps for short.

Figure 3.5b shows a P-N junction which is slightly forward biased. The first few tenths of a volt only serve to reduce the potential gradient across the junction and no conduction results. In Figure 3.5c the forward bias has been increased to the point where the contact potential has been overcome. The width of the depletion region has been reduced to zero, and current now flows in the forward direction. Notice that the positive terminal of the battery is connected to the P type semiconductor and the negative terminal of the battery is connected to the N type semiconductor. This condition is known as forward bias.

A Derivation.

Get in touch with your inner physicist.

For negative values of Vj (reverse bias) the limiting case is I_J = I_O. I_O is correctly called the reverse saturation current but is often called the reverse leakage current. For modern silicon P-N junction diodes the value of IO is of the order of 10^-12 amperes.

For positive values of Vj the current will rise exponentially. Diodes are never used in circuits where the circuit determines the voltage. Diodes are used in circuits where the circuit sets the current and the diode itself is allowed to set its own voltage drop. With current as the independent variable the voltage rises logarithmically and will easily stay within realistic bounds.

A Minute Amount of History.

Cold Cathodes.

Roentgen became curious about what would happen if the pressure was pumped down to as low as possible. He found that the arc stopped, leaving a faint greenish-blue glow. Measurements revealed that some current was flowing through the tube. The florescent material used to make watch dials glow in the dark was known in Roentgen's time and he had some in his laboratory. He happened to notice that this material glowed more brightly when the tube was in operation. He moved it closer to the tube and found that it glowed even brighter. Evidently the sample was in the form of a flat plate. Most likely it was coated on a glass plate like those used for photography in that time. When he interposed his hand between the florescent plate and the tube he probably got the shock, non electrical, of his life. He could see the outlines of the bones in his hand as shadows on the plate. That was the discovery of x-rays.

What was happening was that the strong electric field was pulling electrons out of the metal of the cathode, negative electrode, and they were being accelerated to a very high velocity and then impacting the anode. The incoming electrons had so much energy that they knocked other electrons out of the inner shells in the atoms of the anode metal. These electrons were replaced by electrons from the outer shells falling down to replace them. These large energy transitions caused the atoms to emit electro magnetic radiation of a very short wavelength, namely x-rays.

Hot Cathodes.

It didn't take long for Roentgen and friends to put this discovery to use. They surmised correctly that a heated filament could emit electrons better than a cold piece of metal and the hot cathode was born.

Bulbs or Tubes?

The Fleming Valve.

So What's a Diode?

Thermeonic Emission.

The atomic model proposed by Niels Bore is now considered hopelessly out of date. It is ridiculed by modern physicists but it is still a useful tool in understanding atoms for those who aren't going to get a PhD in the subject. The nucleus of the atom is positively charged and the negatively charged electrons spin around it a bit like the planets around the sun. It's a case of forces in balance like the solar system. The inner electrons are held very tightly and it takes a lot of energy to pull them out of orbit. The outer most are more loosely held and it doesn't take a lot of energy to pull them away from the nucleus.

At this point our solar system model starts to fall apart because an atom has things called shells. There is a maximum number of electrons that can exist in one shell. The innermost shell can hold up to 2 electrons, the second 8, the third 8, the fourth 18, the fifth 18 and the sixth 32. If the outermost shell is completely filled the atom is very unreactive, a kind of loner. These are the inert gases which refuse to form any sort of bonds with any other element. Carbon has its outer shell, which will hold 8 electrons, half filled. When several atoms of carbon get together they form very stable bonds with each other by sharing their outer electrons to make each atom feel as if its outer shell is completely filled. This is how diamonds are formed.

Many compounds are formed by sharing electrons in a partially filled outer shell. These are called covalent compounds. The bonds that hold the molecules together are known as covalent bonds. Some other compounds are formed by one, or sometimes two, electrons jumping from an atom of one element to an atom of another element. For example common table salt, sodium chloride, is formed when atoms of sodium give up the loan electron in their outer shells to an atom of chlorine which needs only one electron to fill its outer shell. The resulting positive and negative ions have a strong attraction for each other. These are known as ionic compounds.

If you could have a single atom of any element in isolation you would find that it takes a certain amount of energy to remove one electron from its outer shell. This energy varies considerably from element to element. Some need very large amounts of energy and others need much less. It always takes some energy to take an electron away from an atom in isolation.

When atoms of the same kind are clustered together, as is always the case, things change. In elements classed as metals the amount of energy required to remove an electron from its atom is zero. That's because the atoms are so close together that the electrons in the outer shells don't know to which atoms they belong. They all just wander around in there, moving randomly. If an electric field is applied externally the electrons will move in the direction the field wants them to and we now have an electric current in the metal. When there is no external field the movement is completely random.

All of these randomly moving electrons don't have the same velocity. There is a statistical value which depends on the temperature but some are faster, some are slower, and a very few are very fast. If one of these high speed electrons comes to the surface of the metal it can pass through the surface and leave the metal behind. If this happens many times the metal will develop a positive charge and subsequent would-be escapees will be attracted back to the surface.

When electrons escape from the surface of a metal it is called electron emission. If we set up a source of EMF and arrange things so the escaped electrons will be collected and the positive charge on the emitting metal is neutralized by the continuous replacement of electrons the process of electron emission will continue until the EMF source is removed. If the emitting metal is at room temperature the process is called cold emission. A very strong electric field is required to produce any significant current from cold emission. That means a high voltage applied between electrodes that are close together.

The piece of metal that is connected to the negative of the EMF and emits the electrons is called the cathode. The one that is connected to the positive and collects the electrons is called the anode. I am told that these names come from the Greek words for up and down but I couldn't begin to tell you which is which. It's all Greek to me.

Edison used to make experimental light bulbs with more than one filament to save work for his glass blowers. As related briefly above he found that conventional current would flow from a cold filament to a hot one but not the other way. This became known as the Edison effect. I don't think the existence of electrons had been recognized at this time. Conventional current flows opposite to electron current. This may well have been one of those serendipitous discoveries that seemed to occur often in early science.

As the temperature of the cathode is increased the electrons get to moving much faster than at room temperature. The likelihood of one having enough velocity to escape from the surface is tremendously increased. Hot emission does not require a strong electric field and will occur even in the absence of one. That can't go on very long because of the positive charge that develops on the cathode but it will build up a considerable space charge in an evacuated chamber. A space charge is a volume that has a lot of electrons floating around in it.

The carbonized filament that Edison used was not a very good emitter of electrons. The later tungsten filaments weren't either. Very early tubes used tungsten because that was all that was known. They glowed as bright as light bulbs. It was soon found that if a little thorium was alloyed with the tungsten the filament didn't need to run as hot and would still emit more electrons per square millimeter of filament surface. These are used to this day in transmitting tubes. They glow about the same color as the upper part of a candle flame.

The most efficient emitter is an oxide coated cathode. The discovery of this process made possible the heater-cathode type of tube. The electrical cathode is a small cylinder with the heater inside and electrically insulated from it. The heater is actually a filament but it is not the electrical cathode of the tube. Providing electrical insulation and thermal conductivity at the same time is not an easy thing to do. (The same battle is still being fought to provide cooling for transistors and microprocessors.) The end result is that the heater must be hotter than the cathode. If the cathode were thoriated, the heater would have to run so hot that it would not last long. In a heater-cathode tube the heater runs about the same temperature as a thoriated tungsten filament. The cathode runs a dull red but you can't see it in most tubes. The 6AS7G/6080WA has a gap in the plate and the grid wires can be seen. The cathode is visible between the grid wires and you need a fairly dark room to see its glow. Electrically separating the cathode from the heater had the effect of making circuits mor simple and meant that AC could be used to heat the tubes without inducing much hum in the signal path.

The hot cathode emits electrons in droves. Many more than are needed. They build up in the volume around the cathode and form a large pool of electrons just waiting for something positive to happen. This pool of electrons is called the space charge. When the anode (plate) is made positive some of the electrons are attracted to it. They impact on it, are absorbed into the metal and electrons flow out of the plate connection into the battery or what ever provided the positive voltage. The other end of the voltage source must be connected back to the cathode in some way.

If the polarity is reversed which makes the plate negative with respect to the cathode the electrons in the space charge are repelled away from the plate and no conduction takes place. The plate is cold and is made of a metal that is a very poor emitter of electrons. The voltages are not nearly as high as those used by Crooks so there is no cold emission from the plate. This makes the diode conduct current in only one direction.

Rectification.

The Space Charge.

Indirectly heated cathodes.

Rectification.

Detection

How I got hooked.

Linearity.

When a vacuum diode is used as a detector, where linearity is necessary, the current is set to a small enough value so the V term will not become significant compared to the input AC voltage. Below is a plot of this equation were A = 0.001 square meters and d = 0.00125 meters.

In power supply circuits the nonlinearity doesn't make a bit of difference. The only goal is to allow current to flow in one direction and a bit of nonlinearity doesn't matter.

This large disparity in forward voltage drop means that a power supply must be designed to use either tube or semiconductor diodes and you can't change later. There are still some of those so called "solid state tube replacements" floating around. DON'T USE THEM! Installing them in existing equipment can result in 100 volts or more of additional B+ and you can do serious damage to what ever you put them in. I have nothing against the use of semiconductor diodes as long as the circuit has been designed for them.

It is the function of a rectifier to change AC to DC. In single phase circuits the rectifier changes the AC to a pulsating DC and some additional circuits are needed to change the pulsating DC to a smooth DC. In this section we will look at a few rectifier circuits.

The schematic symbol for a P-N junction diode is a triangle with a line across its point as shown in Figure 3.7. The triangle is the P side and the line is the N side of the junction. The P side is called the anode and the N side is called the cathode. Conventional current will flow in at the anode and out at the cathode. Current will flow in the direction of the arrow. Current will not flow from cathode to anode. Electron current flows in the opposite direction as conventional current. When current is flowing in the forward direction, the anode is more positive than the cathode by about 0.7 volts. If the cathode is positive with respect to the anode, no measurable current will flow and the voltage can have any value up to the rated maximum PIV (peak inverse voltage) or PRV (peak reverse voltage) of the diode. Typical silicon diodes have PIV ratings in the range of 50 to 1000 volts. Vacuum diodes of the type found in consumer equipment may have PIV ratings from 350 to 1500 volts. Rectifier tubes designed for use in high power transmitters may have PIV ratings in the thousands of volts and maximum currents in Amps.

These drawings are being used because they already exist. When vacuum rectifiers are used the circuits aren't any different. Just substitute a vacuum diode for the semiconductor diode shown and provide power for the filament or heater.

On the first half-cycle when the top end of the secondary is positive with respect to the center-tap the bottom end is negative with respect to the center-tap. (Note that the load is returned to the center-tap, not to the bottom of the secondary.) Diode D₁ is forward biased and D₂ is reversed biased. D₁ conducts current which flows through the load and back to the center-tap. The first half-cycle appears across the load with the top end of the load resistor positive. On the second half-cycle the top of the secondary is negative with respect to the center-tap and the bottom is positive with respect to the center-tap. Diode D₁ is reversed biased and D₂ is forward biased. D₁ does not conduct but D₂ does. Remember that the bottom end of the transformer secondary is now positive and when D₂ conducts, current flows downward through the load resistor, making its top end positive. Current flows through the load on both halves of the input sine wave and both halves appear positive across the load. This process is called full-wave rectification and the circuit is called a full-wave center-tapped rectifier.

The peak output voltage V_P of this rectifier is

When the AC voltage goes positive on the first half-cycle, current flows through D₁, down through the load and back to the junction of D₃ and D₄. The current will not flow through D₃. That would be equivalent to a river flowing up a hill. The current flows through D₄ to a point of lower electrical potential. On the second half-cycle, current flows through D₂, down through the load and back to the junction of D₃ and D₄. The current will flow through D₃ back to the transformer secondary. Regardless of the direction of current in the transformer secondary winding the current always flows downward in the load resistor. The peak output voltage of this rectifier is given by.

The charge on C₁ is replenished 60 times per second as is the charge on C₂. The charges are replenished alternately instead of at the same time; therefore, the total charge is replenished 120 times per second, which qualifies the circuit as a full-wave rectifier.

Figure 3.10b is a variation of the voltage doubler. It has the advantage that one side of both the source and load can be grounded. It has the disadvantage that it is only a half wave rectifier. This circuit is often used in test equipment to obtain the peak to peak value of the incoming wave.

When the input goes negative D₁ conducts charging C₁ to the peak value of the input with the polarity shown. C₁ holds its charge and that voltage is added to the input voltage. When the input goes positive the charge on C₁ is added to the peak voltage from the source. This gives a voltage equal to twice the peak or the peak to peak voltage. The wave form at the junction of D₁ and C₁ is a sine wave which is entirely above the zero axis. D₂ rectifies this voltage and charges C₂ up to the peak to peak value of the input wave. The charge on C₂ is only replenished once for each input cycle or 60 times a second. The peak output voltage of both rectifiers is the same and is given by

Example 3.3

A full-wave center-tapped rectifier circuit (Figure 3.8) employs a transformer which is specified as 9 vCT @ 1 A. What is the peak voltage this rectifier will deliver with semiconductor rectifiers?

Solution:

The full-wave center-tapped circuit uses only half of the secondary at a time; therefore, the output will be the peak of half of the secondary.

V_P = (1.4 V_RMS / 2) - Rect Drop

V_P = 1.4 x 9 / 2 - 0.7 = 5.6 volts.

You could draw 1.4 A from this circuit because the two halves of the secondary are effectively connected in parallel but you are using the peak voltage.

Example 3.4

A full-wave center-tapped rectifier circuit (Figure 3.8) employs a transformer which is specified as 4.50 - 0 - 4.50 vAC @ 1.5 A. What is the peak voltage this rectifier will deliver?

Solution:

The specification means that the entire secondary voltage is 9 vAC; which is exactly the same as Example 3.3 above. The solution to this problem is exactly the same so there is no need to repeat it.

Example 3.5

A full-wave bridge rectifier (Figure 3.9) employs a transformer which has a 9 volt CT @ 1 A secondary. What is the peak output voltage of this rectifier circuit?

Solution:

The center-tap is not used. The peak output voltage is given by

V_P = 1.4 V_RMS - 2 x 0.7 = 1.4 x 9 - 1.4 = 11.2 volts.

Remember, these calculations are not all that precise. The available current is 0.7 A maximum.

Example 3.6

A voltage doubler rectifier (Figure 3.10) employs a transformer which has a 9 volt @ 1 A secondary. What is the peak output voltage of this rectifier circuit?

Solution:

The voltage doubler gives twice the peak voltage of the secondary

V_P = 2 x (1.4 V_RMS - Rect Drop) = 2 x (9 x 1.4 - 0.7) = 23.8 volts.

The original transformer is 9 V at 1 A which is 9 Watts. You can't ask for any more than 9 watts from the rectifier output.

I_Max = 9 W / 23.8 V = 0.37 A

This is the maximum current without damage to the transformer.

Maximum current.

The available current for the various connections is as follows.

The lower the resistance of the resistor (the greater the load current), the faster the capacitor will discharge and the lower will be the voltage when the next positive half-cycle comes along. A way to keep the voltage from falling so low between the times when the capacitor is charged is to make the capacitor bigger. Because a full-wave rectifier is being used the charging peaks occur more often than would be the case for a half-wave rectifier. Thus the capacitor will not have as much time to discharge and the voltage will not fall as low as it would with a half-wave circuit. The amount of ripple (variation) in the voltage across the capacitor is given by.

Example 3.7

A 25.2 vCT @ 1 A transformer is being used in a full-wave center-tapped power supply which is to deliver 150 mA to the heater of a 12AX7 (DC heater supply to reduce hum). (a) Calculate the correct capacitor to give a ripple of 1.2 volts or less, (b) calculate the DC voltage at the capacitor, (c) the value of the resistor required to apply 12.6 volts to the heater of the tube, and (d) the wattage of the resistor.

Solution:

(a) Using equation 3.21.1 for C gives

C = (0.15 A x 8 ms)/(1.2 V) = 1000 microfarads.

(b) The DC output voltage is

V_DC = 1.4 x 25.2 / 2 - 0.7 = 16.94 volts.

That's the peak voltage. The DC is the average voltage which is half of the ripple less than the peak voltage or 16.94 - 1.2 / 2 = 16.34 volts; let's call it 16.3 volts.

(c) The resistor is given by ohm's law as follows.

R = (16.3 - 12.6) / .15 amps = 24.7 ohms.

Use a 24 ohm resistor.

(d) The power dissipated by the resistor is given by

P = I squared R = .15 squared x 24 = .54 watts

Use a 1 watt resistor. Most designers would likely connect another 1000 microfarad capacitor from the other end of the 24 ohm resistor to common as shown in the figure below.

Example 3.8.1

A full-wave bridge rectifier is to be used with a transformer which is rated as 28 vCT @ 2 A. The capacitor is 2200 uf and the load current is 1.3 A. What is the ripple voltage?

Solution:

Using equation 3.19 we have,

Delta V = (I/C) Delta t = (1.3 A / 2200 rf) x 8 ms = 4.7 V.

As compared to physical reality, this is a pessimistic prediction.

Example 3.8.2

A full-wave bridge rectifier is to be used with a transformer which is rated as 280 vCT @ 250 mA. The capacitor is 22 mu f and the load current is 85 mA. What is the ripple voltage?

Solution:

Using equation 3.19 we have

Delta V = (I/C) Delta t = (85 mA / 22 mu f) x 8 ms = 30.9 V

As compared to physical reality, this is a pessimistic prediction.

Additional Filtering LC section.

Example 3.9

A filter like that of Figure 3.12.2 has two 22 microfarad capacitors and an 8 Henry filter choke. The ripple voltage across C₁ is 31 volts (Example 3.8.2). What is the ripple voltage across C₂?

Solution

Using Equation 3.25 gives.

V_C2 = (V_C1) / (4 pi squared f squared L C₂)

V_C2 = 31 vAC / (4 pi squared x 120 squared x 8 henrys x 22 mu f = 0.31 vAC.

Example 3.10

A DC power supply like that of Figure 3.13 uses a transformer with a high voltage winding having a rating of 325 - 0 - 325 volts at 100 mA. (This current rating is for a capacitor input filter so the factor of 1.4 has already been accounted for.) It uses a 5Y3 rectifier tube and an RC filter with two 40 mu f capacitors and a 1 k ohm resistor. The DC load taken off at C₁ is 80 mA and the load taken off at C₂ is 15 mA. What are (a) the ripple across C₁, (b) the ripple across C₂, (c) the DC voltage across C₁, and (d) the DC voltage across C₂?

Solution

(a) The ripple across C₁ is given by equation 3.19 which is.

Delta V = (I/C) Delta t

We must use the sum of the currents for I in this equation which is 95 mA.

Delta V = (95 mA/40 mu f) 8.3 ms = 19.7 volts.

Let's call it 20 volts.

(b) The ripple across C₂ is given by equation 3.28.

V_C2 = (V_C1) / (2 π f R C₂)

V_C2 = 20 / (2 π 120 1000 40 mu f) = 0.66 volts.

(c) Using equation 3.17 gives

V_DC = 1.4 V_RMS / 2 - Rect Drop = 1.4 x 650 / 2 - 125 = 330 volts

The graphs in the tube manual for the 5Y3 give an output voltage of 330 volts for a load of 95 mA and a voltage on each plate of 325 volts.

(d) The voltage across C₂ is given by equation 3.29 as

V_DC-C2 = V_DC-C1 - I_Load R = 330 V - 15 mA x 1000 Ω = 315 VDC

Preliminary, Classes of Voltage Regulators.

The Shunt Regulator.

Let us say that the regulator is operating normally. Then the voltage of the rectifier/filter increases. For the load voltage to remain constant there must be an increased voltage drop across the fixed resistor. Ohm's law tells us that the current through the resistor must increase. The regulator element, inside the red box, will change its resistance to draw more current and the output voltage remains constant.

Now suppose the load current decreases a little. This would cause a decreased drop across the fixed resistor but the regulator compensates by decreasing its resistance to take more current. If the load current goes all the way to zero the regulator will take it all.

If the load current increases the regulator current will decrease to keep the voltage across the output terminals constant. However if the load current increases to the point where the regulator current goes to zero that's as far is it can go. After all the regulator is a variable resistor so it can't generate additional voltage. If the load current keeps increasing after the regulator current has gone to zero the voltage will fall below the regulated value because the regulator is powerless to prevent it from falling below the set voltage. This is the chief disadvantage of a shunt regulator. The regulator will also lose control if the rectifier/filter voltage falls so low that the regulator current goes to zero.

A regulator of this type wastes a lot of power. The current through the regulator must be set above the maximum current that the load will ever take. If this happens for only a small percent of the time and operates at say 1/10 of that value most of the time the shunt regulator will always draw the highest current from the rectifier/filter. If the rectifier/filter were replaced by a battery, it would be discharged very quickly by the large amount of current demanded by the shunt regulator.

The Series Regulator.

Notice that there is no extra resistor in the circuit. Whatever current the load draws, that much current will flow through the regulator, and the same amount will be drawn from the rectifier/filter. The voltage from the rectifier/filter can change up or down and the series regulator element will adjust it's resistance to compensate. The load current can change up or down, even to zero, and the series pass element will adjust to compensate. Series regulators are more efficient than shunt regulators because in the absents of load current they draw very little current from the source. The amplifiers that make it work need a little current but it isn't very much. In general a series regulator can operate with its input and output voltages closer together than for a shunt regulator. That makes them more efficient and a better choice for regulating the voltage in battery operated equipment.

Shunt Zener Diode Regulator.

If the doping level (amount of impurity) is increased in a P-N junction the reverse breakdown * voltage will be decreased. When a P-N junction breaks down in the reverse direction, the voltage is held constant (regulated) even though the current may vary over a large range. The typical characteristic of a P-N junction is shown in Figure 3.13. The portion in the first quadrant is the same as any other P-N junction diode. The portion in the third quadrant is the reverse breakdown * characteristic. As the voltage increases in the negative direction, nothing happens until the voltage reaches the constant voltage breakdown point. When this point is reached, the current goes up very rapidly and the voltage will not increase any further. If an attempt is made to force the voltage higher the diode will overheat very quickly and burn out.

* The term "breakdown" does not mean a malfunction as it does
  to an auto mechanic. In semiconductor talk "breakdown"
  means the beginning of conduction of current. A diode
  which has broken down does not have to be replaced; it is
  just beginning to go to work.

If the doping has not been increased to the level where the reverse breakdown is below 6.2 volts, the mechanism is avalanche breakdown. When the electric field becomes intense enough to promote electrons to the conduction band they are accelerated by the field. These newly freed electrons collide with other valance electrons and promote them to the conduction band which in turn collide with other valence electrons promoting them to the conduction band et cetera, et cetera. Avalanche breakdown occurs if the breakdown voltage is greater than about 6.2 volts.

As you might expect the transition from Zener breakdown to avalanche breakdown is not a sudden one but a gradual one. That has interesting consequences because of the temperature coefficients of the two processes. Avalanche breakdown has a positive temperature coefficient and Zener breakdown has a negative temperature coefficient. At the voltage where the two effects play an equal role in reverse breakdown the temperature coefficient is exactly zero. This is at a voltage of 6.1 and some more decimal places.

It used to be literally true that diode manufacturers would make a batch of diodes and then test them to see what they had made. That is no longer completely true but when Zener diodes are being manufactured in the neighborhood of 6 volts, they are carefully tested. The ones which fall exactly on the magic zero temperature coefficient voltage are pulled out and sold for ten dollars a piece as reference diodes. The ones which don't pass this test are sold as 6.2 volt 5% units for seventy five cents or as 5.6 volt 10% units for forty cents. The manufacturing cost is a fraction of a cent per diode. The cost is in the testing. Even though the mechanism may be avalanche breakdown, the diodes are called Zener diodes.

Uses for the Zener Diode.

Zener diodes are also used as voltage regulators. They were formerly used to regulate voltage in both low and high power circuits. Units were available in power ratings from 1 to 50 watts. In modern electronics the Zener diode is used to provide voltage references to high power regulators. The Zener diode is a low power unit and a power transistor takes care of any heavy work. Power transistors cost much less than power Zener diodes. Zeners are now available as 0.4, 1, and 5 watt units.

A typical circuit for a Zener diode voltage regulator is shown in Figure 3.14. The input is an unregulated power supply consisting of a circuit such as Figure 3.11 without the resistor. The load current which is drawn from the output can range from a few nanoamps to several tens of milliamps.

How a voltage regulator works and what it does.

Changing load current is another difficulty. We have used resistors to indicate the load on a power supply. The real load on a power supply is a transistor circuit and/or a small motor or two. The current drawn by a transistor circuit may change depending on what the circuit is and what it is doing at the moment. The current drawn by a motor changes as its mechanical load changes. So a regulator may have a lot to do.

A regulator can't increase the voltage applied to it. All it can do is reduce it. It regulates the voltage by reducing it by the amount required to hold it at some preset value. The difference between the regulator input and the output is called the regulator drop. Each type of regulator has it's minimum regulator drop. If the input voltage falls so low as to make the regulator drop less than the minimum the regulator will lose control and the output voltage will fall.

For example, a 12 volt regulator IC has a minimum drop of 2 volts. That means that the input voltage must be 14 volts or higher. It regulates just fine for input voltages anywhere from 14 to 40 volts. But if the input voltage falls to 13 volts the output voltage will fall to 11 volts and it will no longer be regulated. It will just be 2 volts less than the input, ripple and all. If the ripple valley falls below the minimum required the output voltage will acquire ripple as shown in the graph below.

A measurement with a DC voltmeter is likely to indicate that the input voltage is high enough but the ripple valley is too low. Only an oscilloscope measurement will reveal this defect. I have taught the senior projects course in the electrical engineering department for several years and I have found this to be one of the most common mistakes in circuit design.

The Zener Diode Voltage Regulator.

The power can be increased to about 1/2 of the specified power by mounting the diode through a hole in a heat sink and coating it with heat sinking compound. The hole should be sized so the diode is a tight fit and the sink should be as thick as the diode is long. When the diode is operated in open air the maximum power is 1/4 of the specified power. Operation at higher power will cause the diode to overheat and fail.

The rules for design are as follows.

1. Calculate the Zener diode current for 25% of rated power.

2. Add any load current to this Zener current. This is the current through R_A.

3. Subtract the Zener voltage from the average expected input voltage.

4. Use results of 2 and 3 in Ohm's law to calculate the resistance of R_A.

5. Calculate the minimum input voltage for regulation to be maintained. This value must be below the ripple valley voltage of the power supply.

6. Calculate the power dissipation of the resistor based on the average ripple voltage. *

7. Calculate the maximum power dissipated by the zener diode based on minimum load current, often zero, and average input ripple.

* because the power is proportional to the square of the voltage, using the average voltage is not quite correct. However, for realistic values of ripple the error is so small as to be unimportant.

Example 3.10.

An unregulated power supply delivers a ripple peak of 21 volts and a ripple valley of 19 volts. The Zener diode regulator circuit is to deliver 12 volts at 2 mA and will use a 12 volt 400 milliwatt Zener diode. What is (a) the value of R_A, (b) the average power dissipated in R_A and (c) the minimum input voltage to maintain regulation?

Solution:

The ripple waveform may be approximated as a series of straight lines and the average is simply (V_P + V_V) / 2 or (21 + 19) / 2 = 20 v. 25% of 400 mW is 100 mW; therefore, I_Z = 100 mW / 12 v = 8.3 mA. (a) The value of R_A is

R_A = (V_IN - V_Z) / (I_Z + I_L)

R_A = (20 v - 12 v) / 8.3 mA + 2 mA) = 780 ohms.

(b) Power in R_A is P = (20 v - 12 v)² / 780 ohms = 0.082 W

(c) Let I_Z = 0.

V_IN = V_Z + I_L R_A = 12 v + 2 mA x 780 ohms = 13.6 volts.

This is much less than the ripple valley voltage of 19 volts, so the circuit will work.

The writer considers the memorization of the table of standard values to be a waste of time and effort. This is information which can always be looked up if and when it is needed. A table of standard values is given in the appendix for those who need to look them up.

Example 3.11.

An unregulated power supply delivers a ripple peak of 39 volts and a ripple valley of 25 volts. The Zener diode regulator circuit is to deliver 16 volts at 1 mA and will use a 16 volt 1 watt Zener diode. What is (a) the value of R_A, (b) the average power dissipated in R_A and (c) the minimum input voltage to maintain regulation?

Solution:

The average input voltage is (39 + 25) / 2 = 32 volts. The Zener diode current is I_Z = 0.25 x 1 W / 16 v = 15.6 mA. (a) R_A = (32 - 16) / (15.6 + 1) = 964 ohms. The next lowest standard value is 910 ohms and that is what we will use. (b) The power is (32v - 16v)² / 910 ohms = 0.28 watts. It is best to use a 1/2 watt resistor here. (c) The minimum input voltage is 16 v + 1 mA x 910 ohms = 16.91 volts. The circuit will work.

Zener diode voltage regulators are occasionally used to regulate the voltage for low current circuits when a design goal is to keep it simple. Because it is a shunt regulator it wastes too much power to be used in a battery operated device.

Zener diodes are most often used in conjunction with other circuits such as op amps and power transistors. We will return to these circuits later after we have studied these devices.

If You Need a High Power Zener.

Gas Discharge Regulator Tubes.

The workhorse set is the octal series, 0A3, 0B3, 0C3, and 0D3. The 0A3 is the 75 volt tube and the 0D3 is the 150 volt regulator. The current range is 5 to 40 mA for all tubes. There is a 7 pin miniature series 0A2, 0B2, and 0C2. Their voltages are 150, 105, and 75 respectively. Their maximum current is 30 mA.

The design procedure is almost the same as for the zener diode, with one exception. The glow discharge in the tubes require a higher voltage than the operating value to get the ionization started. This value is different for each tube and must be looked up in a tube manual. The initial voltage of the power supply must be above this starting value and the starting current of the load must be zero or very small. If the load is resistive, the voltage of the power supply must be large enough to place the starting voltage on the tube even with the load drawing current. The schematic symbol of a VR tube can be seen in a diagram at the end of this chapter.

These regulators regulate very well against changes in load current and changes in input voltage. Most of the ripple voltage which is present in the output of the unregulated power supply is regulated out by the IC VR. The IC contains a current sensor and the voltage will be shut down if the load tries to draw too much current. The IC also contains a temperature sensor and, if the IC gets too hot, the voltage will be shut down. When the power is turned off the IC will cool down and the power will be turned back on again. The observable results of an IC VR which is getting too hot is that the voltage will be turned off and on periodically.

One undesirable thing that these little wonders do rather well is to oscillate at a frequency of approximately 1 Megahertz. This oscillation can cause lots of trouble and headaches for research lab workers (graduate students). The way to prevent these oscillations is to connect a 0.1 uf capacitor on each side of the regulator, one from input to ground and the other from output to ground. These capacitors must be physically close to the regulator or they will not do any good. A regulator which does not have these capacitors will, not may, but WILL, oscillate. In addition to these capacitors, the unregulated power supply must have its filter capacitor and a capacitor in the neighborhood of 25 uf connected across the output of the voltage regulator. Figure 3.16 is the circuit of a typical 5 volt 1 amp power supply.

There is another series of regulators which are very useful in low power circuits. They are the LM78LXX and LM79LXX series. These regulators are supplied in a TO92 package and have a maximum current rating of 100 mA. These units usually do not need a heat sink. The small package makes it possible to construct a very small low-power power supply.

Adjustable Voltage Regulators.

Example 3.12.

If R₁ is 100 ohms and R₂ is 1100 ohms, what is the output voltage of the circuit?

Solution:

The voltage between the O and A terminals is 1.2 volts so the current in R₁ is,

I = 1.2 / 100 = 12.0 mA.

The current in R₂ is also 12 mA so the voltage across R₂ is,

V = 12.0 mA x 1100 ohms = 13.2 volts.

This is not the output voltage, it is the voltage between common and the A terminal. The voltage between the o terminal and common is the sum of the voltage between common and the A terminal and the voltage between the A and O terminals or,

13.2 + 1.2 = 14.4 volts.

A triple output 1.2 to 20, -1.2 to -20, and fixed 5 volt, power supply is shown in Figure 3.18. The diodes across the output are to prevent either supply from being pulled to the opposite sign by the other power supply. If this happens, the supply will latch at -0.7 volts and will not come up to its normal voltage.

Vacuum Tube Voltage Regulator.

This circuit will not go to zero voltage. One that will, needs a negative voltage supply and can get more complicated. This circuit is as simple as it can get and still be a true closed loop* regulator.

* A closed loop regulator is one in which the output voltage
  is sensed, compared to a reference voltage, and if the
  output voltage changes it is instantly corrected
  to the set voltage.

1. The unregulated portion of a power supply is delivering 18 volts at a current of 1 ampere. What size (current rating) fuse should be used in the primary of the transformer?

2. A power transformer has the ratings 34 - 0 - 34 v @ 3 amps. If this transformer is used with a full-wave center-tapped rectifier circuit, Figure 3.8, what is the peak output voltage VP? Take diode drops into account.

3. A power transformer has the ratings 36 vCT @ 2 amps. If this transformer is used with a full-wave center-tapped rectifier circuit, Figure 3.8, what is the peak output voltage VP? Take diode drops into account.

4. A power transformer has the ratings 21 vAC @ 5 amps. If this transformer is used with a full-wave bridge rectifier circuit, Figure 3.9, what is the peak output voltage VP? Take diode drops into account.

5. A full-wave voltage doubler circuit (Figure 3.10) is attached to a transformer whose secondary is rated at 120 vAC @ 10 amperes. (This is an isolation transformer.) What is the peak output of this circuit? Take diode drops into account.

6. A power transformer is rated as 40 - 0 - 40 vAC @ 1.2 amps. The transformer is equipped with a full-wave center-tapped rectifier using silicon diodes and a filter capacitor of 7,000 microfarads. What is the ripple valley voltage of this circuit if the load current is 0.8 ampere? Apply all corrections.

7. A power transformer is rated as 20 vAC @ 7 amps. The transformer is equipped with a full-wave bridge rectifier using silicon diodes and a filter capacitor of 20,000 microfarads. What is the ripple valley voltage of this circuit if the load current is 5 amperes? Apply all corrections.

8. Using the transformer and rectifier from problem 6, calculate the value of the filter capacitor to give a ripple valley voltage of 45 volts if the load current is 0.8 amps.

9. Using the transformer and rectifier from problem 7, calculate the value of the filter capacitor to give a ripple valley voltage of 17 volts if the current is 5 amps.

10. A 12 volt 300 mA transformer is being used with a bridge rectifier which is to deliver 10 mA to a load. An uninformed person uses a 20,000 uf capacitor in the power supply. What is (a) the peak output voltage and (b) the difference between the peak and valley voltages?

11. For the conditions of problem 10, what size capacitor should be used to give a ripple valley voltage of 14 volts?

12. A 6.2 volt 400 mW Zener diode is to be supplied from an unregulated power supply which has a peak output voltage of 26 volts and a ripple valley voltage of 22 volts. The load current on the regulator is approximately 1 picoampere. (a) Calculate the resistance of RA. (b) Select the proper value for RA from the table of standard values in the appendix. (c) Calculate the power dissipated in the resistor. (d) Calculate the minimum input voltage to maintain regulation. Restrict your answers to no more than 3 significant digits.

13. What is the highest tolerable peak output voltage from the rectifier for each of the following regulators: (a) LM7805, (b) LM7912, (c) LM7812, (d) LM7915, and (e) LM7824? Don't forget the signs.

14. What is the lowest tolerable ripple valley output Voltage, closest to zero, from the rectifier for each of the following regulators: (a) LM7805, (b) LM7912, (c) LM7812, (d) LM7915, and (e) LM7824? Don't forget the signs.

15. In the circuit of Figure 3.18, what is the value of R₂ to give an output voltage of 13.5 volts if R₁ is 120 ohms?

1. 3/10 Amp.

2. 46.9 volts.

3. 24.5 volts.

4. 28.0 volts.

5. 334.6 volts.

6. 54.4 volts.

7. 24.6 volts.

8. 621 microfarads.

9. 4170 microfarads.

10. (a) 15.4 volts. (b) 4.00 millivolts.

11. 57.1 microfarads.

12. (a) 1100 ohms, (b) 1100 ohms, (c) 0.288 watts, (d) 6.2 volts.

13. (a) +40, (b) -40, (c) +40, (d) -40, (e) +40 volts.

14. (a) +7, (b) -14, (c) +14, (d) -17, (e) +26 volts.

15. 1230 ohms.

Now, if you have really mastered this material you deserve some kind of award. Go have some ice cream.