Advance Electronics - Regular

8/14/2019 Advance Electronics - Regular

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SemiconductorDiodes

Introduction

A diode is an electrical device allowing current to move

through it in one direction with far greater ease than in the

other. The most common type of diode in modern circuit designis the semiconductor diode, although other diode technologies

exist. Semiconductor diodes are symbolized in schematic

diagrams as such:

When placed in a simple battery-lamp circuit, the diode will

either allow or prevent current through the lamp, depending on

the polarity of the applied voltage:

When the polarity of the battery is such that electrons are

allowed to flow through the diode, the diode is said to be

forward-biased. Conversely, when the battery is "backward"

and the diode blocks current, the diode is said to be reverse-

biased. A diode may be thought of as a kind of switch: "closed"

when forward-biased and "open" when reverse-biased.

C H A P T E R

+ + + +

Meter check of a diode

Diode Ratings

Diode Approximations

Series/Parallel DiodesConfiguration with DC Cir

AND/OR Gates

Clippers

Clampers

Voltage Multipliers

Zener Diodes

Special Purpose Diode+ + + +

1111


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Chapter 1 Semiconductor Diodes 2

Building skills for success

Oddly enough, the direction of the diode symbol's "arrowhead" points against the direction of

electron flow. This is because the diode symbol was invented by engineers, who

predominantly use conventional flow notation in their schematics, showing current as a flow

of charge from the positive (+) side of the voltage source to the negative (-). This

convention holds true for all semiconductor symbols possessing "arrowheads:" the arrow

points in the permitted direction of conventional flow, and against the permitted direction ofelectron flow.

Diode behavior is analogous to the behavior of a hydraulic device called a check valve. A

check valve allows fluid flow through it in one direction only:

Check valves are essentially pressure-operated devices: they open and allow flow if the pressure

across them is of the correct "polarity" to open the gate (in the analogy shown, greater fluid pressure

on the right than on the left). If the pressure is of the opposite "polarity," the pressure difference

across the check valve will close and hold the gate so that no flow occurs.

Like check valves, diodes are essentially "pressure-" operated (voltage-operated) devices. The

essential difference between forward-bias and reverse-bias is the polarity of the voltage dropped

across the diode. Let's take a closer look at the simple battery-diode-lamp circuit shown earlier, this

time investigating voltage drops across the various components:


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When the diode is forward-biased and conducting current, there is a small voltage dropped across it,

leaving most of the battery voltage dropped across the lamp. When the battery's polarity is reversed

and the diode becomes reverse-biased, it drops allof the battery's voltage and leaves none for the

lamp. If we consider the diode to be a sort of self-actuating switch (closed in the forward-bias mode

and open in the reverse-bias mode), this behavior makes sense. The most substantial difference here

is that the diode drops a lot more voltage when conducting than the average mechanical switch (0.7

volts versus tens of millivolts).

This forward-bias voltage drop exhibited by the diode is due to the action of the depletion region

formed by the P-N junction under the influence of an applied voltage. When there is no voltage applied

across a semiconductor diode, a thin depletion region exists around the region of the P-N junction,

preventing current through it. The depletion region is for the most part devoid of available charge

carriers and so acts as an insulator:


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If a reverse-biasing voltage is applied across the P-N junction, this depletion region expands, further

resisting any current through it:

Conversely, if a forward-biasing voltage is applied across the P-N junction, the depletion region will

collapse and become thinner, so that the diode becomes less resistive to current through it. In order for

a sustained current to go through the diode, though, the depletion region must be fully collapsed by the

applied voltage. This takes a certain minimum voltage to accomplish, called the forward voltage:

For silicon diodes, the typical forward voltage is 0.7 volts, nominal. For germanium diodes, the forward

voltage is only 0.3 volts. The chemical constituency of the P-N junction comprising the diode accounts

for its nominal forward voltage figure, which is why silicon and germanium diodes have such different

forward voltages. Forward voltage drop remains approximately equal for a wide range of diode

currents, meaning that diode voltage drop not like that of a resistor or even a normal (closed) switch.

For most purposes of circuit analysis, it may be assumed that the voltage drop across a conducting

diode remains constant at the nominal figure and is not related to the amount of current going through

it.


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In actuality, things are more complex than this. There is an equation describing the exact

current through a diode, given the voltage dropped across the junction, the temperature of

the junction , and several physical constants. It is commonly known as the diode equation:

The equation kT/q describes the voltage produced within the P-N junction due to the actionof temperature, and is called the thermal voltage, or Vt of the junction. At room temperature,

this is about 26 millivolts. Knowing this, and assuming a "nonideality" coefficient of 1, we

may simplify the diode equation and re-write it as such:

You need not be familiar with the "diode equation" in order to analyze simple diode circuits.

Just understand that the voltage dropped across a current-conducting diode does change

with the amount of current going through it, but that this change is fairly small over a wide

range of currents. This is why many textbooks simply say the voltage drop across a

conducting, semiconductor diode remains constant at 0.7 volts for silicon and 0.3 volts for

germanium. However, some circuits intentionally make use of the P-N junction's inherent

exponential current/voltage relationship and thus can only be understood in the context of

this equation. Also, since temperature is a factor in the diode equation, a forward-biased P-

N junction may also be used as a temperature-sensing device, and thus can only be

understood if one has a conceptual grasp on this mathematical relationship.

A reverse-biased diode prevents current from going through it, due to the expanded

depletion region. In actuality, a very small amount of current can and does go through a

reverse-biased diode, called the leakage current, but it can be ignored for most purposes.

The ability of a diode to withstand reverse-bias voltages is limited, like it is for any

insulating substance or device. If the applied reverse-bias voltage becomes too great, the

diode will experience a condition known as breakdown, which is usually destructive. A


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diode's maximum reverse-bias voltage rating is known as the Peak Inverse Voltage, or PIV,

and may be obtained from the manufacturer. Like forward voltage, the PIV rating of a diode

varies with temperature, except that PIV increases with increased temperature and

decreasesas the diode becomes cooler -- exactly opposite that of forward voltage.

Typically, the PIV rating of a generic "rectifier" diode is at least 50 volts at room

temperature. Diodes with PIV ratings in the many thousands of volts are available for modest

prices.

REVIEW:REVIEW:REVIEW:REVIEW:

A diode is an electrical component acting as a one-way valve for current.

When voltage is applied across a diode in such a way that the diode allows current, thediode is said to be forward-biased.

When voltage is applied across a diode in such a way that the diode prohibits current, the

diode is said to be reverse-biased.

The voltage dropped across a conducting, forward-biased diode is called the forward

voltage. Forward voltage for a diode varies only slightly for changes in forward current

and temperature, and is fixed principally by the chemical composition of the P-N

junction.

Silicon diodes have a forward voltage of approximately 0.7 volts.

Germanium diodes have a forward voltage of approximately 0.3 volts.

The maximum reverse-bias voltage that a diode can withstand without "breaking down"

is called the Peak Inverse Voltage, or PIV rating.


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Meter check of a diode

Being able to determine the polarity (cathode versus anode) and basic functionality of a diode is a veryimportant skill for the electronics hobbyist or technician to have. Since we know that a diode is

essentially nothing more than a one-way valve for electricity, it makes sense we should be able to

verify its one-way nature using a DC (battery-powered) ohmmeter. Connected one way across the

diode, the meter should show a very low resistance. Connected the other way across the diode, it

should show a very high resistance ("OL" on some digital meter models):

Of course, in order to determine which end of the diode is the cathode and which is the anode, you

must know with certainty which test lead of the meter is positive (+) and which is negative (-) when

set to the "resistance" or "" function. With most digital multimeters I've seen, the red lead becomes

positive and the black lead negative when set to measure resistance, in accordance with standard

electronics color-code convention. However, this is not guaranteed for all meters. Many analog

multimeters, for example, actually make their black leads positive (+) and their red leads negative (-)

when switched to the "resistance" function, because it is easier to manufacture it that way!

One problem with using an ohmmeter to check a diode is that the readings obtained only have

qualitative value, not quantitative. In other words, an ohmmeter only tells you which way the diode

conducts; the low-value resistance indication obtained while conducting is useless. If an ohmmeter

shows a value of "1.73 ohms" while forward-biasing a diode, that figure of 1.73 doesn't represent

any real-world quantity useful to us as technicians or circuit designers. It neither represents the

forward voltage drop nor any "bulk" resistance in the semiconductor material of the diode itself, but

rather is a figure dependent upon both quantities and will vary substantially with the particular

ohmmeter used to take the reading.


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For this reason, some digital multimeter manufacturers equip their meters with a special "diode check"

function which displays the actual forward voltage drop of the diode in volts, rather than a "resistance"

figure in ohms. These meters work by forcing a small current through the diode and measuring the

voltage dropped between the two test leads:

The forward voltage reading obtained with such a meter will typically be less than the "normal" drop of

0.7 volts for silicon and 0.3 volts for germanium, because the current provided by the meter is of trivial

proportions. If a multimeter with diode-check function isn't available, or you would like to measure a

diode's forward voltage drop at some non-trivial current, the following circuit may be constructed

using nothing but a battery, resistor, and a normal voltmeter:

Connecting the diode backwards to this testing circuit will simply result in the voltmeter indicating the

full voltage of the battery.

If this circuit were designed so as to provide a constant or nearly constant current through the diode

despite changes in forward voltage drop, it could be used as the basis of a temperature-measurement

instrument, the voltage measured across the diode being inversely proportional to diode junction

temperature. Of course, diode current should be kept to a minimum to avoid self-heating (the diode

dissipating substantial amounts of heat energy), which would interfere with temperature measurement.


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Beware that some digital multimeters equipped with a "diode check" function may output a very low

test voltage (less than 0.3 volts) when set to the regular "resistance" () function: too low to fully

collapse the depletion region of a PN junction. The philosophy here is that the "diode check" function is

to be used for testing semiconductor devices, and the "resistance" function for anything else. By using

a very low test voltage to measure resistance, it is easier for a technician to measure the resistance of

non-semiconductor components connected to semiconductor components, since the semiconductor

component junctions will not become forward-biased with such low voltages.

Consider the example of a resistor and diode connected in parallel, soldered in place on a printed

circuit board (PCB). Normally, one would have to unsolder the resistor from the circuit (disconnect it

from all other components) before being able to measure its resistance, otherwise any parallel-

connected components would affect the reading obtained. However, using a multimeter that outputs a

very low test voltage to the probes in the "resistance" function mode, the diode's PN junction will not

have enough voltage impressed across it to become forward-biased, and as such will pass negligible

current. Consequently, the meter "sees" the diode as an open (no continuity), and only registers the

resistor's resistance:

If such an ohmmeter were used to test a diode, it would indicate a very high resistance (many mega-

ohms) even if connected to the diode in the "correct" (forward-biased) direction:

Reverse voltage strength of a diode is not as easily tested, because exceeding a normal diode's PIV

usually results in destruction of the diode. There are special types of diodes, though, which aredesigned to "break down" in reverse-bias mode without damage (called Zener diodes), and they are

best tested with the same type of voltage source / resistor / voltmeter circuit, provided that the voltage

source is of high enough value to force the diode into its breakdown region. More on this subject in a

later section of this chapter.


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REVIEW:REVIEW:REVIEW:REVIEW:

An ohmmeter may be used to qualitatively check diode function. There should be low

resistance measured one way and very high resistance measured the other way. When using anohmmeter for this purpose, be sure you know which test lead is positive and which is negative!

The actual polarity may not follow the colors of the leads as you might expect, depending on

the particular design of meter.

Some multimeters provide a "diode check" function that displays the actual forward voltage of

the diode when it's conducting current. Such meters typically indicate a slightly lower forward

voltage than what is "nominal" for a diode, due to the very small amount of current used during

the check.

Diode ratings

In addition to forward voltage drop (Vf) and peak inverse voltage (PIV), there are many other ratings of

diodes important to circuit design and component selection. Semiconductor manufacturers provide

detailed specifications on their products -- diodes included -- in publications known as datasheets.

Datasheets for a wide variety of semiconductor components may be found in reference books and on

the internet. I personally prefer the internet as a source of component specifications because all the

data obtained from manufacturer websites are up-to-date.

A typical diode datasheet will contain figures for the following parameters:

Maximum repetitive reverse voltage = VRRM, the maximum amount of voltage the diode can withstand in

reverse-bias mode, in repeated pulses. Ideally, this figure would be infinite.

Maximum DC reverse voltage = VR or VDC, the maximum amount of voltage the diode can withstand in

reverse-bias mode on a continual basis. Ideally, this figure would be infinite.

Maximum forward voltage = VF, usually specified at the diode's rated forward current. Ideally, this

figure would be zero: the diode providing no opposition whatsoever to forward current. In reality, the

forward voltage is described by the "diode equation."

Maximum (average) forward current = IF(AV), the maximum average amount of current the diode is able

to conduct in forward bias mode. This is fundamentally a thermal limitation: how much heat can the PN

junction handle, given that dissipation power is equal to current (I) multiplied by voltage (V or E) and

forward voltage is dependent upon both current and junction temperature. Ideally, this figure would be

infinite.

Maximum (peak or surge) forward current = IFSM or if(surge), the maximum peak amount of current the

diode is able to conduct in forward bias mode. Again, this rating is limited by the diode junction's

thermal capacity, and is usually much higher than the average current rating due to thermal inertia (the

fact that it takes a finite amount of time for the diode to reach maximum temperature for a given

current). Ideally, this figure would be infinite.

Maximum total dissipation = PD, the amount of power (in watts) allowable for the diode to dissipate,

given the dissipation (P=IE) of diode current multiplied by diode voltage drop, and also the dissipation

(P=I2R) of diode current squared multiplied by bulk resistance. Fundamentally limited by the diode's

thermal capacity (ability to tolerate high temperatures).


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Operating junction temperature = TJ, the maximum allowable temperature for the diode's PN junction,

usually given in degrees Celsius (oC). Heat is the "Achilles' heel" of semiconductor devices: they must

be kept cool to function properly and give long service life.

Storage temperature range = TSTG, the range of allowable temperatures for storing a diode

(unpowered). Sometimes given in conjunction with operating junction temperature (TJ), because themaximum storage temperature and the maximum operating temperature ratings are often identical. If

anything, though, maximum storage temperature rating will be greater than the maximum operating

temperature rating.

Thermal resistance = R(), the temperature difference between junction and outside air (R() JA) or

between junction and leads (R()JL) for a given power dissipation. Expressed in units of degrees Celsius

per watt (oC/W). Ideally, this figure would be zero, meaning that the diode package was a perfect

thermal conductor and radiator, able to transfer all heat energy from the junction to the outside air (or

to the leads) with no difference in temperature across the thickness of the diode package. A high

thermal resistance means that the diode will build up excessive temperature at the junction (where it's

critical) despite best efforts at cooling the outside of the diode, and thus will limit its maximum power

dissipation.

Maximum reverse current = IR, the amount of current through the diode in reverse-biasoperation, with

the maximum rated inverse voltage applied (VDC). Sometimes referred to as leakage current. Ideally,

this figure would be zero, as a perfect diode would block all current when reverse-biased. In reality, it

is very small compared to the maximum forward current.

Typical junction capacitance = CJ, the typical amount of capacitance intrinsic to the junction, due to the

depletion region acting as a dielectric separating the anode and cathode connections. This is usually a

very small figure, measured in the range of picofarads (pF).

Reverse recovery time = trr, the amount of time it takes for a diode to "turn off" when the voltage

across it alternates from forward-bias to reverse-bias polarity. Ideally, this figure would be zero: the

diode halting conduction immediately upon polarity reversal. For a typical rectifier diode, reverse

recovery time is in the range of tens of microseconds; for a "fast switching" diode, it may only be a fewnanoseconds.

Most of these parameters vary with temperature or other operating conditions, and so a

single figure fails to fully describe any given rating. Therefore, manufacturers provide

graphs of component ratings plotted against other variables (such as temperature), so that

the circuit designer has a better idea of what the device is capable of.

The construction, characteristics, and models of semiconductor diodes were introduced in

Chapter 1. The primary goal of this chapter is to develop a working knowledge of the diode

in a variety of configurations using models appropriate for the area of application. By

chapters end, the fundamental behavior pattern of diodes in DC and in AC networks should

be clearly understood. The concepts learned in this chapter will have significant carryover inthe chapters to follow. For instance, diodes are frequently employed in the description of the

basic construction of transistors and in the analysis of transistor networks in the DC and AC

domains.

The concept of this chapter will reveal an interesting and very positive side of the study of a

field such as electronic devices and systems - once the basic behavior of a device is

understood, its function and response in an infinite variety of configurations can be


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determined. The range of applications is endless, yet the characteristics and models remain

the same. The analysis will proceed from one that employs the actual diode characteristics

to one that utilizes the approximate models almost exclusively. It is important that the role

and response of various elements of an electronic system be understood without continually

having to resort to lengthy mathematical procedures. This is usually accomplished through

the approximation process, which can develop into an art itself. Although the results obtainedusing a series of approximations, keep in mind that the characteristics obtained from a

specification sheet may in them be slightly different form the device in actual use.

In other words, the characteristics of a 1N4001 semiconductor diode may vary from one

element to the next in the same lot. The variation may be slight, but it will often be sufficient

to validate the approximations employed in the analysis. Also consider the other elements of

the network: Is the resistor labeled 100ohms exactly 100 ohms? Is the applied voltage

exactly 10 V or perhaps 10.08V? All these tolerances contribute to the general belief that a

response determined through an appropriate set of approximations can often be as

accurate as one that employs the full characteristics. In this book the emphasis is toward

developing a working knowledge of a device through the use of appropriate approximations,

thereby avoiding an unnecessary level of mathematical complexity. Sufficient detail willnormally be provided, however, to permit a detailed mathematical analysis if desired

2.2 Load Line Analysis2.2 Load Line Analysis2.2 Load Line Analysis2.2 Load Line Analysis

The applied load will normally have an impact on the region of operation of a device. I the

analysis is performed in a graphical manner, a line can be drawn on the characteristics of the

device that represents the applied load. The intersection of the load line with the

characteristics will determine the point of operation of the system. Such an analysis is, for

obvious reasons, called load-line analysis. Note through the majority of the diode networks

analyzed in this chapter do not employ the load line in this chapter do not employ the loadline analysis approach, the technique is one used quiet frequently in subsequent chapter, and

this introduction offers the simplest application of the method. It also permits a validation of

the approximate technique described throughout the remainder of the chapter.

Figure 2.1 Series Diode Configuration (a) circuit, (b) characteristics.


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Consider the network of Figure 2.1a employing a diode having the characteristics of figure

2.1b. Note in figure2.1a that the pressure established by the battery is to establish a

current through the series circuit in the clockwise direction. The fact that this current and

the defined direction of conduction of the diode are a match reveals that the diode is in the

on state and conduction has been established. The resulting polarity across the diode will

be as shown and the first quadrant (VD and ID positive) of figure 2.1b will be in the region ofinterest the forward bias region.

Applying Kirchoffs voltage law to the series circuit of figure 2.1a will result in

E VD VR = 0Or E = VD + IDR equation 2.1

The two variables of equation 2.1, VD and ID are the same as the diode axis variables of

figure 2.1b. This similarity permits a plotting of equation 2.1 on the same characteristics of

figure 2.1b.

The intersections of the load line on the characteristics can easily be determined in one

simply employs the fact that anywhere on the horizontal axis ID = 0A and anywhere on the

vertical axis VD = 0V.

If we set VD = 0V in equation 2.1 and solve for ID, we have the magnitude of ID in the vertical

axis. Therefore, with VD = 0V, equation 2.1 becomes

E = VD + IDR

= 0V + IDR

and ID = E / R; provided that VD = 0V equation 2.2

as shown in figure 2.2. If we set ID = 0A in equation 2.1 and solve for VD, we have the

magnitude of VD on the horizontal axis. Therefore, with ID = 0A. Equation 2.1 becomes

E = VD + IDR

= VD + (0A) Rand VD = E; provided that ID = 0A equation 2.3


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Figure 2.2 Drawing the load line and finding the point of operation.

as shown in figure 2.2. A straight line drawn the two points will define the load line as

depicted in figure 2.2 change the level of R (the load) and the intersection on the vertical

axis will change. The result will be a change in the slope of the load line and different point

of intersection between the load line and the device characteristics.We now have a load line defined by the network and a characteristic curve defined by the

device. The point of intersection between the two is the point of operation for this circuit. By

simply drawing a line down to the horizontal axis the diode voltage VD can be determined,

whereas a horizontal line from the point of intersection to the vertical axis will provide the

level of ID. The current ID is actually the current through the entire series configuration of

figure 2.1a the point of operation is usually called the quiescent point (abbreviated as the

Q-pt) to reflect its still, unmoving qualities as defined by a DC network.

The solution obtained at the intersection of the two curves is the same that would be

obtained by a simultaneous mathematical solution of equations 2.1 and 1.4. Since the curve

for a diode has a nonlinear characteristic the mathematics involved would require the use ofnonlinear techniques that are beyond the needs and scope of this book. The load-line

analysis described above provides a solution with a minimum of effort and a pictorial

description of why the levels of solution for VD and ID were obtained. The next two

examples will demonstrate the techniques introduced above and reveal the relative ease with

which the load line can drawn using equations 2.2 and 2.3.

EExxaammppllee 22..11

For the series diode configuration of figure 2.3a employing the diode characteristics of figure

2.3b determine:

a. VDQ and IDQb. VR


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Figure 2.3 a. Circuit, b. Characteristics

SSoolluuttiioonn

a. applying equation 2.2ID = E / R

= 10V / 1Kohm

= 10mA

Applying equation 2.3

VD = E

= 10 V

The resulting load line appears in figure 2.4 the intersection between the load line and the

characteristics curve defines the Q point as

VDQ = 0.78V

b.VR = IRR

= E VD

= 10V 0.78

= 9.22 Vthe difference in results is due to the accuracy with which the graph can be read. Ideally, the

results obtained wither way should be the same.


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Figure 2.4 Solution to example 2.1


Repeat the analysis of example 2.1 with R = 2Kohm

SSoolluuttiioonn

a. applying equation 2.2ID = E / R

= 10V / 1Kohm

= 10mA

applying equation 2.3

VD = E

= 10 VThe resulting load line appears in figure 2.5. Note the reduced slope and levels of diode

current for increasing loads. The resulting Q-point is defined by

VDQ = .07V

IDQ = 4.6mA

b.VR = IRR

= IDQR

= (4.6mA)(2Kohm)

= 9.2V

with VR = E = VD


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= 10V 0.7V

= 9.3 V

The difference in levels is again due to the accuracy with which the graph can be read.

Certainly, however, the results provide an expected magnitude for the voltage VR.

Figure 2.5 Solution to Example 2.2

As noted in the examples above, the load line is determined solely by the applied network

while the characteristics are defined by the chosen device. If we turn to our approximate

model for the diode and do not disturb the network, the load line will be exactly the same as

obtained in the examples above. In fact, the next two examples repeat the analysis of

examples 2.1 and 2.2 using the approximate model to permit a comparison of the results.


Repeat example 2.1 using the approximate equivalent model for the semiconductor diode.

SSoolluuttiioonn

The load line is redrawn as shown in figure 2.6 with the same intersections as defined in

example 2.1 the characteristics of the approximate equivalent circuits for the diode have also

been sketched on the same graph. The resulting Q-point:

VDQ = 0.7VIDQ = 9.25mA


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Figure 2.6 Solution to example 2.1 using the approximate model.

The results obtained in example 2.3 are quite interesting. The level if IDQ are exactly the

same as obtained in example 2.1 using a characteristic curve that is greater deal easier to

draw than that appearing in figure 2.4. The level of VD versus 0.78V from example 2.1 is of

different magnitude to the hundredths place, but they are certainly in the same neighborhood

if we compare their magnitudes to the magnitudes of the other voltages of the network.


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Repeat example 2.2 using the approximate model for the semiconductor diode.

SSoolluuttiioonn

The load line is redrawn as shown in figure 2.7 with the same intersection define in example

2.2. The characteristics of the approximate equivalent circuit for the diode have also been

sketched on the same graph. The resulting Q-point:

VDQ = 0.7V

IDQ = 4.6mA

Figure 2.7 Solution to example 2.2 using the diode approximation model.

In example 2.4 the results a obtained for both VDQ and IDQ are the same as those obtained

using the full characteristics in example 2.2. these examples above have demonstrated that

the current and voltage levels obtained using the approximate model have been very close to

those obtained using the full characteristics. It suggests, as will be applied in the sections to

follow, that the use of appropriate approximations can result in solutions that are very close

to actual response with a reduced level of concern about properly reproducing the

characteristics and choosing a large enough scale. In the next example we go to a step

further and substitute the idea model. The results will reveal the conditions that must be

satisfied to apply the idea equivalent properly.


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Repeat example 2.1 using the ideal diode model.

SSoolluuttiioonn

As shown in figure 2.8 the load line continues to be the same, but the ideal characteristics

now intersects the load line on the vertical axis. The Q-point is therefore defines by

VDQ = 0V

IDQ = 10mA

Figure 2.8 Solution to example 2.1 using the ideal diode model.

The results are sufficiently different from the solutions of example 2.1 to cause someconcern about their accuracy. Certainly, they do provide some indication of the level of

voltage and current to be expected relative to the other voltage levels of the network, but

the additional effort of simply including 0.7V offset suggests that the approach of example

2.3 is more appropriate.

Use of the ideal diode model therefore should be reserved for those occasions when the

roles of a diode is more important than the voltage levels that differ by tenths of a volt and in

those situations where the applied voltages are considerably larger than the threshold

voltage VT. In the next few sections the approximate model will be employed exclusively

since the voltage levels obtained will be sensitive to variations that approach VT. In later

sections the ideal model will be employed more frequently since the applied voltages will

frequently be quite a bit larger than VT and the authors want to ensure that the role of the

diode is correctly and clearly understood.


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2.3 Diode Approximations

In section 2.2 we revealed that the results obtained using the approximate piecewise linearequivalent model were quite close, if not equal to the response obtained using the full

characteristics. In fact, if one considers all the variations possible due to tolerances,

temperature, and so on, one could certainly consider one solution to be as accurate as the

other. Since the use of the approximate model normally results in a reduced expenditure of

time and effort to obtain the desired results, it is the approach that will be employed in this

book unless otherwise specified. Recall the following:

The primary purpose of this book is to develop a general knowledge of thebehavior, capabilities, and possible areas of application of a device in a mannerthat will minimize the need for extensive mathematical developments.

The complete piecewise- linear equivalent model in chapter 1 was not employed in the loadline analysis because Rav is typically much less than the other series elements of the

network. If Rav should be close in magnitude to the other series elements of the network, the

complete equivalent model can be applied in much the same manner as described in section

2.2.

In preparation for the analysis to follow, table 2.1 was developed to review the important

characteristics, models and conditions of application for the approximate and ideal diode

models. Although the silicon diode is used exclusively due to its temperature characteristics,

the germanium diode is still employed and is therefore included in table 2.1. As with the

silicon diode a germanium diode is approximated by an open circuit equivalent for voltages

less than VT. It will enter the on state when VD VT = 0.3V.

TTaabbllee 22..11 AApppprrooxxiimmaattee aannddIIddeeaall SSeemmiiccoonndduuccttoorr DDiiooddee MMooddeellss


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Keep in mind that the 0.7V and 0.3V in the equivalent circuits are not independent sources of

energy but are there simply to remind us that there is a price to pay to turn on a diode. An

isolated diode on a laboratory table will not indicate 0.7V or 0.3V if a voltmeter is placed

across its terminals. The supplies specify the voltage drop across each when the device is

on and specify that the diode voltage must be at least the indicated level before conduction

can be established.

In the next few sections we demonstrate the impact of the models of table 2.1 on the

analysis of diode configurations. For those situations where the approximate equivalent

circuits will be employed, the diode symbol will appear as shown in figure 2.9a for the siliconand germanium diodes. If conditions are such that the ideal diode can be employed, the diode

symbol will appear as shown in figure 2.9b.

Figure 2.9 (a) Approximate model notation; (b) Ideal diode notation


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2.4 Series Diode Configurations with DC circuits

In this section the approximate model is utilized to investigate a number of series diodeconfigurations with DC inputs. The content will establish a foundation in diode analysis that

will carry over into the sections and chapters to follow. The procedure described can, in fact

be applied to network with any number of diodes in a variety of configurations.

For each configuration the state of each diode must first be determined. Which diodes are

on and which are off? Once determined, the appropriate equivalent as defined in section

2.3 can be substituted and the remaining parameters of the network determined.

In general, a diode is in the on state if the current established by the applied sources is

such that its direction matches that of the arrow in the diode symbol, and VD 0.7V for

silicon and VD 0.3V for germanium diode.

For each configuration, mentally replace the diodes with resistive elements and note the

resulting current direction is a match with the arrow in the diode symbol, conduction

through the diode will occur and the devices is in the on state. The description above is, of

course, contingent on the supply having a voltage greater than the Turn-On voltage VT of

each diode.

If the diode is in the on state, one can either place a 0.7V drop across the element, or the

network can be redrawn with the VT equivalent circuit as defined in table 2.1. In each time

the preference will probably simply be to include the 0.7V drop across each diode and draw

a line through each diode in the off: or open state. Initially, however, the substitution

method will be utilized to ensure that the proper voltages and current levels are determined.

Figure 2.10 Series Diode configuration

Figure 2.11 Determining The state of thediode of figure 2.10

The series circuit of figure 2.10 described in some detail in section 2.2 will be used to

demonstrate the approach described in the paragraphs above. The state of the diode is first

determined by mentally replacing the diode with a resistive element as shown in figure 2.11.

The resulting direction of I is a match with the arrow in the diode symbol, and since E >V T

the diode is in the on state. The network is then redrawn as shown in figure 2.12 with the

appropriate equivalent model for the forward-biased silicon diode. Note for future reference


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that the polarity of VD is the same as would result if in fact the diode were a resistive

element. The resulting voltage and current levels are the following.

VD = VT equation 2.4

VR= E VT equation 2.5

ID = IR= VR/ R equation 2.6

Figure 2.12 Substituting the equivalent model for the on diode of figure 2.10.

In figure 2.13, the diode of figure 2.10 has been reversed. Mentally replacing the diode with

a resistive element as shown in figure 2.14 will reveal that the resulting current direction

does not match the arrow in the diode symbol. The diode is in off state, resulting in the

equivalent circuit of figure 2.15. Due to the open circuit the diode current is 0A and the

voltage across the resistor R is the following

VR = IRR = IDR = (0A)R = 0V

Figure 2.13 Reversing thediode of figure 2.10 Figure 2.14 Determining

the diode of figure 2.13

Figure 2.13 Substitutingthe equivalent model forthe off diode of figure

2.13.

The fact that VR = 0V will establish E volts across the open circuit as defined by Kirchoffsvoltage law. Always keep in mind that under any circumstances DC, AC, instantaneous

values, pulses, and so on Kirchoffs voltage law must be satisfied.


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For the series diode configuration of figure 2.16 determine VD,VR and ID.

Figure 2.16 Circuit for example 2.6

SSoolluuttiioonn

Since the applied voltage establishes a current in the clockwise direction to match the arrow

of the symbol and the diode is in the on state,

VD = 0.7 V

VR = E VD = 8V 0.7V = 7.3V

ID = IR = VR / R

= 7.3V / 2.2 Kohms

= 3.32mA


Repeat example 2.6 with the diode reversed.

SSoolluuttiioonn

Figure 2.17. Determining the unknown quantities for Example 2.7

Removing the diode, we find that the direction of I is opposite to the arrow in the diode

symbol and the diode equivalent is the open circuit no matter which model is employed. The

result is the network of figure 2.17, where ID = 0A due to the open circuit. Since VR = IRR,

VR = (0)R = 0V. Applying Kirchoffs voltage law around the closed loop yields


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E VD VR = 0

And VD = E VR = E 0 = E = 8V

In particular, note in Example 2.7 the high voltage across the diode even though it is in off

state. The current is zero, but the voltage is significant. For review purposes keep thefollowing in mind for the analysis to follow:

1. An open circuit can have any voltage across its terminals, but the current is always0A.

2. A short circuit has a 0V drop across its terminals, but the current is limited by thesurrounding network.

In the next example, the notation of figure 2.18 will be employed for the applied voltage. It is

a common industry notation and one with which the reader should become very familiar.

Figure 2.18 Source notation


For the series diode configuration of figure 2.19, determine VD, VR and ID.

Figure 2.19 Series diode circuit for example 2.8

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Although the pressure establishes a current with the same direction as the arrow symbol

the level of applied voltage is sufficient to turn the silicon diode on. The point of operation

on the characteristics is shown in figure 2.20, establishing the open-circuit equivalent as the

appropriate approximation. The resulting voltage and current levels are therefore the

following:

ID = 0 A


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VR = IRR = IDR = (0A) 1.2k = 0V

And VD = E = 0.5V

Figure 2.20 Operating point with E = 0.5V


Determine VO and ID for the series circuit of figure 2.21


SSoolluuttiioonn

An attack similar to that applied in example 21.6 will reveal that the resulting current has the

same direction as the arrowheads of the symbols of both diodes, and the network of figure

2.22 results because E = 12V > (0.7V + 0.3V) = 1V. Note the redrawn supply of 12V and the

polarity of VO across the 5.6Kohm resistor. The resulting voltage

VO = E VT1 VT2 = 12V 0.7V 0.3V = 11V

And ID = IR = VR / R = VO / R = 11V / 5.6k = 1.96 mA


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Figure 2.22 Determining the unknown for example 2.9


Determine ID, VD2, and VO for the circuit of figure 2.23


SSoolluuttiioonn

Removing the diodes and determining the direction of the resulting current I will result in thecircuit of figure 2.24. There us a match in current direction for the silicon diode but not for

the germanium diode. The combination of a short circuit in series with an open circuit always

results in an open circuit and ID = 0A, as shown in figure 2.25.

Figure 2.24 Determing the state of the didoeof figure 2.23

Figure 2.25 Substituting the equivalent statefor the open diode

The question remains as to what to substitute for the silicon diode. For the analysis to follow

in this succeeding chapters, simply recall for the actual practical diode that when ID = 0A, VD

= 0V (and vise versa), as described for the no-bias situation in chapter 1. The conditions

described by ID = 0A and VD = 0V are indicated in figure 2.26.


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Figure 2.26 Determining the unknown quantities for the circuit of Example 2.10

Applying Kirchoffs voltage law in a clockwise direction gives us

E VD1 VD2 VO = 0

And VD2 = E VD1 VO = 12V 0 0 = 12V

With VO = 0V


Figure 2.27 Circuit for Example 2.11

SSoolluuttiioonn

Figure 2.28 Determining the state of thediode for the network of figure 2.27

Figure 2.29 Determining the unknownquantities for the network of figure 2.27.

The sources are drawn and the current direction indicated as shown in figure 2.28. The

diode is in the on state and the notation appearing in figure 2.29 is included to indicate this

state. Note that the on state is noted simply by the additional VD = 0.7V on the figure. This

eliminates the need top redraw the network and avoids any confusion that may result form

the appearance of another source. As indicated in the introduction to this section, this is


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probably the path and notation that one will take when a level of confidence has been

established in the analysis of diode configurations. In time the entire analysis will be

performed simply by referring to the original network. Recall that a reverse biased diode can

simply be indicated by a line through the device.

The resulting current through the circuit is,

I = (E1 + E2 - VD) / (R1 + R2)

= (10V + 5V 0.7V) / (4.7K + 2.2K) = 14.3V / 6.9K

= 2.07mA

And the voltages are

V1 = IR2 = (2.07mA)(4.7K) = 9.73V

V2 = IR2 = (2.07mA)(2.2K ) = 4.55V

Applying Kirchoffs voltage law to the output section in the clockwise direction will result in

-E2 + V2 VO = 0

and VO = V2 E2 = 4.55 V 5V = -0.45 V

The minus sign indicates that VO has a polarity opposite to that appearing in figure 2.27.

2.5 Parallel and Series-Parallel Configurations

The methods applied in Section 2.4 can be extended to the analysis of parallel and seriesparallel configurations. For each area of application; simply match the sequential series of

steps applied to series diode configurations.


Determine VO, I1 , ID1 and ID2 for the parallel diode configuration of figure 2.30.

Figure 2.30 Network for Example 2.12

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For the applied voltage the pressure of the source is to establish a current through each

diode in the same direction as shown in figure 2.31. Since the resulting current direction

matches that of the arrow in each diode symbol and the applied voltage is greater than 0.7V,

both diodes are in the on state. The voltage across parallel elements is always the same

and

VO = 0.7V

Figure 2.31 Determining the unknown for the network of example 2.12

The current

I1 = VR / R = (E - VD) / R = (10V 0.7V) / 0.33K = 28.18 mA

Assuming diodes of similar characteristics, we have

ID1 = ID2 = I1 / 2 = 28.18 mA / 2 = 14.09 mA

Example 2.12 demonstrated one reason for placing diodes in parallel. If the current rating of

the diodes of figure 2.30 is only 20mA, a current of 28.18mA would damage the device if itappeared alone in figure 2.30. By placing two in parallel, the current is limited to a safe value

of 14.09mA with the same terminal voltage.


Determine the current I for the network of figure 2.32

Figure 2.32 Network for example 2.13

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Redrawing the network as shown in figure 2.33 reveals that the resulting current direction is

such as to turn on diode D1 and turn off diode D2. The resulting current I is then

I = (E1 E2 - VD) / R = (20 V 4V 0.7 V) / 2.2K = 6.95 mA

Figure 2.33 Determining the unknown quantities for the network of example 2.13.


Determine the voltage VO for the network of figure 2.34.

Figure 2.34 Network for example 2.14

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Initially, it would appear that the applied voltages will then both diodes on. However, if

both were on, the 0.7V drop across the silicon diode would not match the 0.3V across the

germanium diode as required by the fact that the voltage across parallel elements must be

the same. The resulting action can be explained simply by realizing that when the supply is


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turned on it will increase from 0V to 12V over a period of time although probably

measurable in milliseconds. At the instant during the rise that 0.3V is established across the

germanium diode it will turn on and maintain a level of 0.3V. The silicon diode will never

have the opportunity to capture its required 0.7V and therefore remains in its open-circuit

state as shown in figure 2.35. The result:

VO = 12V 0.3V = 11.7V

Figure 2.35 Determining VOfor the network of figure 2.34.


Determine the currents I1, I2 and ID2 for the network of figure 2.36.

Figure 2.36 Network for example 2.15.

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The applied voltage (pressure) is such as to turn both diodes on, as noted by the resulting

current directions in the network of figure 2.37. Note the use of the abbreviated notation for

on diodes and that the solution is obtained through an application of techniques applied to

DC series-parallel networks.


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I1 = VT2 / R1 = 0.7V / 3.3K = 0.212mA

Figure 2.37 Determining the unknown quantities for example 2.15.

Applying Kirchoffs voltage law around the indicated loop in the clockwise direction yields

-V2 + E VT1 VT2 = 0

and V2 = E VT1 VT2 = 20 V 0.7V 0.7V = 18.6 V

with I2 = V2 / R2 = 18.6 V / 5.6 K = 3.32 mA

At the bottom node (a)

ID2 + I1 = I2

ID2 = I2 I1 = 3.32mA 0.212 mA = 3.108 mA

2.6 AND/OR GATES

The tools of analysis are now at our disposal and the opportunity to investigate a computer

configuration is one that will demonstrate the range of applications of this relatively simple

device. Our analysis will be limited to determining the voltage levels and will not include a

detailed discussion of Boolean algebra or positive and negative logic.

The network to be analyzed in Example 2.16 is an OR gate for positive logic. That is, the

10V level of figure 2.38 is assigned a 1 for Boolean algebra while the 0V input is assigned

a 0. An OR gate is such that the output voltage level will be a 1 if either or both inputs is a

1. The output is a 0 if both inputs are at the 0 level.

The analysis of AND/OR gates is made measurably easier by using the approximate

equivalent for a diode rather than the ideal because we can stipulate that the voltage across

the diode must be 0.7V positive for the silicon diode (0.3V for Ge) to switch to the ON state.

In general, the best approach is simply to establish a gut feeling for the state of the diodes

by noting the direction and the pressure established by the applied potentials. The analysis

will then be to verify or negate your initial assumptions.


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Figure 2.38 Positve logic 0


Determine the VO for the network of figure 2.38.

Figure 2.38 Positve OR gate Figure 2.39 Redrawn network of figure 2.38

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First note that there is only one applied potential: 10V at terminal 1. Terminal 2 with 0V

input is essentially at ground potential, as shown in the redrawn network of figure 2.39.

Figure 2.39 suggests that D1, is probably in the on state due to the applied 10V while D2

with its positive side at 0V is probably off. Assuming these states will result in the

configuration of figure 2.40.


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Figure 2.40 Assumed diode states for figure 2.38

The next step is simply to check that there is no contradiction to our assumptions. That is,

note that the polarity across D1 is such as to turn it on and the polarity across D2 is such as

to turn it off. For D1 the on state establishes VO at VO = E VD = 10V 0.7V = 9.3V. With

9.3V at the cathode ( - ) side of D2 and 0V at the anode (+) side, D2 is definitely in the off

state. The current direction and the resulting continuous path for conduction further confirm

our assumption that D1 is conducting. Our assumptions seem confirmed by the resulting

voltages and current, and our initial analysis can be assumed to be correct. The output

voltage level is not 10V as defined for an input of 1, but the 9.3V is sufficiently large to be

considered a 1 level. The output is therefore at a 1 level with only one input, which suggests

that the gate is an OR gate. An analysis of the same network with two 10V inputs will result

in both diodes being in the on state and an output of 9.3V. A 0V input at both inputs will not

provide the 0.7V required to turn the diodes on and the output will be a 0 due to the 0V

output level. For the network of figure 2.40 the current level is determined by

I = (E - VD) / R = (10V 0.7V) / 1K = 9.3mA


Determine the output level for the positive logic AND gate of figure 2.41.

SSoolluuttiioonn

Figure 2.41 Positive logic AND gate.


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Note in this case that an independent source appears in the grounded leg of the network. For

reasons soon to become obvious it is chosen at the same level as the input logic level.

The network is redrawn in figure 2.42 with our initial assumptions regarding the state of the

diodes. With 10V at the cathode side of D1 it is assumed that D1 is in the off state eventhough there is a 10V source connected to the anode if D1 through the resistor. However,

recall that we mentioned in the introduction to this section that the use of the approximate

model will be an aid to the analysis. For D1, where will the 0.7V come from if the input and

source voltages are at the same level and creating opposing pressures? D2 is assumed to

be in the on state due to the low voltage at the cathode side and the availability of the 10V

source through the 1Kohm resistor.

For the network of figure 2.42 the voltage at VO is 0.7V, due to the forward biased diode D2.

With 0.7V at the anode of D1 and 10V at the cathode, D1 is definitely in the off state. The

current I will have the direction indicated in figure 2.42 and a magnitude equal to

I = (E - VD) / R = (10 V 0.7V) / 1K = 9.3 mA

Figure 2.42 Substituting the assumed states for the diodes of figure 2.41.

The state of the diodes is therefore confirmed and our earlier analysis was correct. Although

not 0V as earlier defined for the 0 level, the output voltage is sufficiently small to be

considered a 0 level. For the AND gate, therefore, a single input will result in a 0-level

output. The remaining states of the diodes for the possibilities of two inputs and no inputs

will be examined in the problems at the end of the chapter.

2.7 Sinusoidal Inputs; Half-Wave Rectification

Rectifier circuits


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Now we come to the most popular application of the diode: rectification. Simply defined, rectification is

the conversion of alternating current (AC) to direct current (DC). This almost always involves the use

of some device that only allows one-way flow of electrons. As we have seen, this is exactly what a

semiconductor diode does. The simplest type of rectifier circuit is the half-waverectifier, so called

because it only allows one half of an AC waveform to pass through to the load:

For most power applications, half-wave rectification is insufficient for the task. The harmonic content

of the rectifier's output waveform is very large and consequently difficult to filter. Furthermore, AC

power source only works to supply power to the load once every half-cycle, meaning that much of its

capacity is unused. Half-wave rectification is, however, a very simple way to reduce power to a

resistive load. Some two-position lamp dimmer switches apply full AC power to the lamp filament for

"full" brightness and then half-wave rectify it for a lesser light output:

In the "Dim" switch position, the incandescent lamp receives approximately one-half the power it would

normally receive operating on full-wave AC. Because the half-wave rectified power pulses far more

rapidly than the filament has time to heat up and cool down, the lamp does not blink. Instead, its

filament merely operates at a lesser temperature than normal, providing less light output. This principle

of "pulsing" power rapidly to a slow-responding load device in order to control the electrical power

sent to it is very common in the world of industrial electronics. Since the controlling device (the diode,

in this case) is either fully conducting or fully nonconducting at any given time, it dissipates little heat

energy while controlling load power, making this method of power control very energy-efficient. This

circuit is perhaps the crudest possible method of pulsing power to a load, but it suffices as a proof-of-

concept application.

If we need to rectify AC power so as to obtain the full use of bothhalf-cycles of the sine wave, a

different rectifier circuit configuration must be used. Such a circuit is called a full-waverectifier. One

type of full-wave rectifier, called the center-tap design, uses a transformer with a center-tapped

secondary winding and two diodes, like this:


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This circuit's operation is easily understood one half-cycle at a time. Consider the first half-cycle,

when the source voltage polarity is positive (+) on top and negative (-) on bottom. At this time, only

the top diode is conducting; the bottom diode is blocking current, and the load "sees" the first half of

the sine wave, positive on top and negative on bottom. Only the top half of the transformer's secondary

winding carries current during this half-cycle:

During the next half-cycle, the AC polarity reverses. Now, the other diode and the other half of the

transformer's secondary winding carry current while the portions of the circuit formerly carrying

current during the last half-cycle sit idle. The load still "sees" half of a sine wave, of the same polarity

as before: positive on top and negative on bottom:

One disadvantage of this full-wave rectifier design is the necessity of a transformer with a center-

tapped secondary winding. If the circuit in question is one of high power, the size and expense of a

suitable transformer is significant. Consequently, the center-tap rectifier design is seen only in low-

power applications.

Another, more popular full-wave rectifier design exists, and it is built around a four-diode bridge

configuration. For obvious reasons, this design is called a full-wave bridge:


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Current directions in the full-wave bridge rectifier circuit are as follows for each half-cycle of the AC

waveform:

Remembering the proper layout of diodes in a full-wave bridge rectifier circuit can often be frustrating

to the new student of electronics. I've found that an alternative representation of this circuit is easier

both to remember and to comprehend. It's the exact same circuit, except all diodes are drawn in a

horizontal attitude, all "pointing" the same direction:


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The diode analysis will now be expanded to include time varying functions such as the

sinusoidal waveform and the square-wave. There is no question that the degree of difficulty

will increase, but once a few fundamental maneuvers are understood, the analysis will be

fairly direct and follow a common thread.

The simplest of networks to examine with a time-varying signal in figure 2.43. For the

moment we will use the ideal model (note the absence if the Si or Ge label denote ideal

diode) to ensure that the approach is not conclude by additional mathematical complexity.

Figure 2.43 Half Wave rectifier

Over one full cycle, defined by the period T of figure 2.43, the average value (the algebraic

sum of the areas above and below the axis) is zero. The circuit of figure 2.43, called a half-

wave rectifier, will generate a waveform VO that will have an average value of particular use

in AC-to-DC conversion process. When employed in the rectification process a diode is

typically much higher than that of diodes employed in other applications, such as computers

and communication systems.

During the interval t = 0 to t/2 in figure 2.43 the polarity of the applied voltage Vi is such as

to establish pressure in the direction indicated and turn on the diode with the polarity

appearing above the diode. Substituting the short circuit equivalence for the ideal diode will

result in the equivalent circuit of figure 2.44, where it is fairly obvious that the output signal

is an exact replica of the applied signal. The two terminals defining the output voltage are

connected directly to the applied signal via the short-circuit equivalence of the diode.


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Figure 2.44 Conduction Region (0 T/2)

For the period t/2 T, the polarity of the input Vi is as shown in figure 2.45 and the

resulting polarity across the ideal diode produces an off state with an open-circuit

equivalent. The result is the absence of a path for charge to flow and VO = IR = (0)R for

the period t/2 t. the input vi and the output VO were sketched together in figure 2.46 for

comparison purposes. The output signal VO now has a net positive area above the axis over

a full period and an average value determined by

VDC = (0.318)VM equation 2.7

Figure 2.45 Non-conduction Region (T/2 T)

Figure 2.46 Half Wave rectifies signal

The process of removing one-half the input signal to establish a DC level is aptly called

half-wave rectification.


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The effect of using a silicon diode with VT = 0.7V is demonstrated in figure 2.47 for the

forward bias region. The applied signal must now be at least 0.7V before the diode can turn

on. For levels of Vi less than 0.7V the diode is still in an open-circuit state and VO = 0V as

shown in the same figure. When conducting, the difference between VO and Vi is a fixed level

of VT

= 0.7V and VO

= Vi V

T, as shown in the figure. The net effect is a reduction in area

above the axis, which naturally reduces the resulting DC voltage level. For situations where

Vm >> VT, equations 2.8 can be applied to determine the average value with a relatively high

level of accuracy.

VDC = 0.318 (VM - VT) equation 2.8

Figure 2.47 Effect on VT on Half wave rectified signal.

In fact, if Vm is sufficiently greater than VT, equation 2.7 is often applied as a first

approximation for Vdc.


a. Sketch the output Vo and determine the DC level of the output for the network offigure 2.48.

b. Repeat part (a) if the ideal diode is replaced by a silicon diode.c. Repeat parts (a) and (b) if Vm is increased to 200V and compare solutions using

equations 2.7 and 2.8.

Figure 2.48 Network for example 2.18.

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a. In this situation the diode will conduct during the negative part if the input as shownin figure 2.49, and VO will appear as shown in the same figure. For the full period, the

DC level is

VDC = -0.318 VM = -0.318 (20V) = -6.36V

The negative sign indicates that the polarity of the input is opposite to the defined polarity of

figure 2.48.

Figure 2.49 Resulting Vo for the circuit of example 2.18.

a. Using a silicon diode, the output has the appearance of figure 2.50 andVdc = -0.318 (VM= 0.7V) = -0.318 (19.3V) = -6.14 V

b. Equation 2.7: VDC = -0.318 VM = -0.318 (200V) = -63.6 VEquation 2.8: VDC = -0.318 (VM - VT) = -0.318(200V 0.7V)

= -(0.318)(199.3 V ) = -63.38 V

The resulting drop in DC level is 0.22V or about 3.5%.which is a difference that can certainly

be ignored for most applications. For part C the offset and drop in amplitude due to VT would

not be discernible on a typical oscilloscope if the full pattern is displayed.

figure 2.50 Effect of VT on output of figure 2.49.


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PIVPIVPIVPIV

The peak inverse voltage (PIV) or peak-reverse voltage rating of the diode of primary

importance in the design if rectification systems. Recall that it is the voltage rating that must

not be exceeded in the reverse-bias region or the diode will enter the Zener avalanche

region. The required PIV rating for the half wave rectifier can be determined from figure

2.51, which displays the reverse biased diode of figure 2.43 with, maximum applied voltage.

Applying Kirchoffs voltage law, it is fairly obvious that the PIV rating of the diode must not

equal or exceed the peak value of the applied voltage, therefore:

PIV rating VM; for the half wave rectifier equation 2.9

Figure 2 Determining the required PIV rating for the half wave rectifier.

2.8 Full wave rectification

Bridge NetworkBridge NetworkBridge NetworkBridge Network

The DC level obtained from a sinusoidal input can be improved 100% using a procedure

called full-wave rectification. The most familiar network for performing such a function

appears in figure 2.52 with its four diodes in a bridge configuration. During the period t = 0

to t/2 the polarity of the input is as shown in figure 2.53. The resulting polarities acroos the

ideal diodes are also shown in figure 2.53 to reveal that D2 and D3 are conducting while D1

and D$ are in the off state. The net result is the configuration of figure 2.54, with its

indicated current and polarity across R. Since the diodes are ideal the load voltage Vo = Vi

as shown in the same figure.


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Figure 2.52 Full wave bridge rectifier.

Figure 2.53 Network of figure 2.52 for the period 0 T/2 of the input voltage Vi.

Figure 2.54 Conduction path for the positive region of Vi.

For the negative region of the input the conducting diodes are D1 and D4, resulting in the

configuration if figure 2.55. The important result is that the polarity across the load resistor

R is the same as in figure 2.53, establishing a second positive pulse, as shown in figure 2.55.

Over one full cycle the input and output voltages will appear as shown in figure 2.56.

Figure 2.55 Conduction path for the negative region of VT.

Figure 2.56 Input and output waveforms for a full wave rectifier.


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Since the area above the axis for one full cycle is now twice that obtained for a half-wave

system, the DC level has also been doubled and

VDC = 2(equation 2.7) = 2(0.318VM)

VDC = 0.636VM; full wave rectifier equation 2.10

If silicon rather than ideal diodes are employed as shown in figure 2.57, an application of

kirchoffs voltage law around the conduction path would result in

vi VT vo VT = 0

vo = vi 2VT

The peak value of the output voltage VO is therefore

Vomax = VM 2VT

For situations where V >> 2VT, equations 2.11 can be applied for the average value with a

relatively high level of accuracy.

VDC = 0.636 (VM 2VT) equation 2.11

Figure 2.57 Determining VOMAX for silicon diodes in the bridge configuration.

Then again, if Vm is sufficiently greater than 2VT, then equations 2.10 is often applied as a

first approximation for VDC .

PIVPIVPIVPIV

The required PIV of each diode (ideal) can be determined from figure 2.58 obtained at the

peak of the positive region of the input signal. For the indicated loop the maximum voltage

across R is Vm anf the PIV rating is defined by

PIV VM; full wave rectifier equation 2.12


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CenterCenterCenterCenter----Tapped TransformerTapped TransformerTapped TransformerTapped Transformer

A second popular full-wave rectifier appears in figure 2.59 with only two diodes but

requiring a center-tapped transformer to establish the input signal across each section of the

secondary of the transformer. During the positive portion of Viapplied to the primary of the

transformer, the network will appear as shown in figure 2.60. D1 assumes the short-circuit

equivalent and D2 the open-circuit equivalent, as determined by the secondary voltages and

the resulting current directions. The output voltage appears as shown if figure 2.60.

Figure 2.59 Center Tapped transformer full wave rectifier.

Figure 2.60 Network for the positive region of vi.

During the negative portion of the input the network appears as shown in figure 2.61,

reversing the roles of the diodes but maintaining the same polarity for the voltage across the

load resistor R. The net effect is the same output as that appearing in figure 2.56 with the

same DC levels.

Figure 2.61 Network conditions for the negative region of vi.


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PIVPIVPIVPIV

Figure 2.62 Determining the PIV level for the diodes of the CT transformer full wave rectifier

The network of figure 2.62 will help us determine the net PIV for each diode for this full-

wave rectifier. Inserting the maximum voltage for the secondary voltage and Vm as

established by the adjoining loop will result in

PIV = Vsecondary + VR

= Vm + Vm

PIV 2VM; CT transformer, full wave rectifier equation 2.13


Determine the output waveform for the network of figure 2.63 and calculate the output DC

level and the required PIV of each diode.

Figure 2.63 Bridge network for example 2.19

SSoolluuttiioonn

The network will appear as shown in figure 2.64 for the positive region of the input voltage.

Redrawing the network will result in the configuration of figure 2.65, where VO = Vi or

VOmax = Vimax = (10V) = 5V, as shown in figure 2.65. For the negative part of the input

the roles of the diode will be interchanged and VO will appear as shown in figure 2.66.


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Figure 2.67 Series clipper

Figure 2.68 Series clipper with a dc supply

There is no general procedure for analyzing networks such as the type in figure 2.68, but

there are a few mental thoughts to keep in mind as you work toward a solution.

1. Make a mental sketch of the response of the network based on the direction of thediode and the applied voltage levels.

For the networks of figure 2.68, the direction of the diode suggests that the signal must be

positive to turn it on. The DC supply further requires that the voltage Vi be greater than V

volts to turn the diode on. The negative region of the input signal is pressuring the diode

into the off state, supported further by the DC supply. In general, therefore, we can be

quite sure that the diode is an open circuit (off state) for the negative region of the input

signal.

2. Determine the applied voltage (transition voltage) that will cause a change in thestate for the diode.

For the ideal diode the transition between states will occur at the point on the characteristics

where Vd = 0V and Id = 0 A. Applying the condition Id = 0 at vd = 0 to the network of figure

2.68 will result in the configuration of figure 2.69, where it is recognized that the level of v i

that will cause a transition in state is

vi = V equation 2.14


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Figure 2.69 Determining transition level for the circuit of figure 2.68

For an input voltage greater than V volts the diode is in the short-circuit state, while for

input voltages less than V volts it is in the open circuit or off state.

3. Be continually aware of the defined terminals and polarity of Vo.When the diode is in the short-circuit state, such as shown in figure 2.70, the output voltage

vo can be determined by applying Kirchoffs voltage law in the clockwise direction:

vi V vo = 0 (CW direction)

vo = vi V equation 2.15

Figure 2.70 Determining vo

4. It can be helpful to sketch the input signal above the output and determine the outputat instantaneous values of the input.

It is then possible that the output voltage can be sketched from the resulting data points of vo

as determined in figure 2.71. Keep in mind that an instantaneous value v i the input can be

treated as a DC supply of that value and the corresponding DC value (the instantaneous

value) of the output determined. for instance, at v i = Vm for the network of figure 2.68, the

network to be analyzed appears in figure 2.72. For Vm > V the diode is in the short circuit

state and Vo = Vm V, as shown in figure 2.71.


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Figure 2.71 Determining levels of vo

At vi = V the diodes change state and Vi = -Vm, vo = 0V, and the complete curve for vo can

be sketched as shown in figure 2.73.

Figure 2.72 Determining vo when vi = Vm

Figure 2.73 Sketching vo


Determine the output waveform for the network of figure 2.74.

Figure 2.74 Series clipper for example 2.20

SSoolluuttiioonn

Past experience suggests that the diode will be in the on state for the positive region of Vi

especially when we note the aiding effect of V = 5V. The network will then appear as


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shown in figure 2.75 and Vo = Vi + 5V. Substituting id = 0 at vd = 0 for the transition levels,

we obtain the network of figure 2.76 and vi = -5V.

figure 2.75 vo with diode in the on state Figure 2.76 Determining the transition levelfor the clipper of figure 2.74

For vi more negative than -5V the diode will then enter its open-circuit state, while for

voltages more positive than -5V the diode is in the short circuit state. The input and output

voltages appear in figure 2.77.

Figure 2.77 Sketching vo for example 2.20

The analysis of slipper networks with square-wave inputs is actually easier to analyze than

with sinusoidal inputs because only two levels have to be considered. In other words, the

network can be analyzed as if it had two DC level inputs with the resulting output v o plotted

in the proper time frame.


Repeat Example 2.20 for the square wave input of figure 2.78.

Figure 2.78 Applied signal for Example 2.21


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Figure 2.82 Response to a parallel clipper


Determine Vo for the network of figure 2.83.

Figure 2.83 Example 2.22

SSoolluuttiioonn

The polarity if the DC supply and the direction of the diode strongly suggest that the diode

will be in the on state for the negative region of the input signal. For the region the

network will appear as shown in figure 2.84, where the defined terminals of vo require that

vo = V = 4V.

Figure 2.84 vo for the negative region of vi


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The transition voltage can first be determined by applying the condition id = 0A at vi = VD =

0.7V and obtaining the network of figure 2.88. Applying Kirchoffs voltage law around the

output loop in the clockwise direction, we find that

vi + VT V = 0

And vi = V VT = 4V 0.7V = 3.3 V

Figure 2.88 Determining transition level for the network of figure 2.83

For input voltages greater than 3.3V, the diode will be an open circuit and vo = vi. For input

voltages of less than 3.3V, the diode will be in the on state and the network of figure 2.89

results, where

Vo = 4V 0.7V = 3.3V

Figure 2.89 Determining vo for the diode of figure 2.83 in the on state

The resulting output waveform appears in figure 2.90. Note that the only effect of VT was to

drop the transition level to 3.3V to 4V.


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Figure 2.90 Sketching vo for example 2.23

There is no question that including the effects of VT will complicate the analysis somewhat,

but once the analysis is understood with the ideal diode, the procedure, including the effects

of VT, will not be that difficult.

SummarySummarySummarySummary

A variety of series and parallel clippers with the resulting output for the sinusoidal input areprovided in figure 2.91. In particular, note the response of the last configuration, with its

ability to clip off a positive and a negative section as determined by the magnitude of the DC

supplies.


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2.11 Clampers

The clamping network is one that will clamp a signal to a different DC level. The network

must have a capacitor, a diode, and a resistive element, but t can also employ an independent

DC supply to introduce an additional shift. The magnitude of R and C must be chosen such

that the time constant t = RCt = RCt = RCt = RCis large enough to ensure that the voltage across the capacitor

does not discharge significantly during the interval the diode is nonconducting. Throughout

the analysis we will assume that for all practical purposes the capacitor will fully charge or

discharge in five time constants.

The network of figure 2.92 will clamp the input signal to zero level (for ideal diodes). The

resistor R can be the load resistor or a parallel combination of the load resistor and a

resistor designed to provide the desired level of R.

Figure 2.92 Clamper

Figure 2.93 Diode on and the capacitor charging to V volts

During the interval 0 to t/2 the network will appear as shown in figure 2.93, with the diode in

the on state effectively shorting out the effect of the resistor R. The resulting RCRCRCRCtime

constant is so small (R determined by the inherent resistance of the network) that he

capacitor will charge to V volts quickly. During this interval the output voltage is directly

across the short circuit and vo = 0V.

When the input switches to the V state, the network will appear as shown in figure 2.94,

with the open circuit equivalent for the diode determined by the applied signal and stored

voltage across the capacitor both pressuring current through the diode from cathode to

anode. Now that R is back in the network the time constant determined by the RC product is

sufficiently large to establish a discharge period 5T much greater than the period t/2 t and


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it can be assumed on an approximate basis that the capacitor holds onto all its charge and

therefore voltage (since V = Q/C) during this period.

Figure 2.94 Determining vo with the diode off

Since vo is in parallel with the diode and resistor, it can also be drawn in the alternative

position shown in figure 2.94. Applying Kirchoffs voltage law around the input loop will

result in

-V V vo = 0

and vo = -2V

The negative sign resulting from the fact that the polarity of 2V is opposite to the polarity

defined for vo. The resulting output waveform appears in figure 2.95 with the input signal.

The output signal is clamped to 0V for the interval 0 to t/2 but maintains the same total

swing (2V) as the input.

Figure 2.95 Sketching vo for the network of figure 2.92

For the clamping netwoFor the clamping netwoFor the clamping netwoFor the clamping networrrrk:k:k:k:

The total swing of the output signal is equal to the total swi

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