Semiconductor
Diodes
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 design is 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 Diodes Configuration with DC Circuits
AND/OR Gates Clippers Clampers
Voltage Multipliers Zener Diodes
Special Purpose Diodes + + + +
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 of electron 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:
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 all of 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:
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.
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 action of 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
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
decreases as 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.
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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, the diode 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.
Meter check of a diode
Being able to determine the polarity (cathode versus anode) and basic functionality of a diode is a very important 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 "Ω" 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 Ω 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.
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.
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" (Ω) 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 are designed 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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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 an ohmmeter 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).
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 the
maximum 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(Θ), the temperature difference between junction and outside air (R(Θ)JA) or
between junction and leads (R(Θ)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-bias operation, 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 few nanoseconds.
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 chapter’s 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 in the 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
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 obtained using 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 will normally be provided, however, to permit a detailed mathematical analysis if desired
2.2 Load Line Analysis
2.2 Load Line Analysis
2.2 Load Line Analysis
2.2 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 load line 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.
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 of
interest – the forward bias region.
Applying Kirchoff’s voltage law to the series circuit of figure 2.1a will result in
E – VD – VR = 0
Or
E = V
D+ I
DR
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 = V
D+ I
DR
= 0V + IDR
and
I
D= E / R; provided that V
D= 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 = V
D+ I
DR
= VD + (0A) R
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 it’s “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 of nonlinear 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.
E
E
x
x
a
a
m
m
p
p
l
l
e
e
2
2
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.
1
1
For the series diode configuration of figure 2.3a employing the diode characteristics of figure 2.3b determine:
a. V
DQand I
DQFigure 2.3 a. Circuit, b. Characteristics
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a. applying equation 2.2I
D= E / R
= 10V / 1Kohm
= 10mA
Applying equation 2.3V
D= 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
V
DQ= 0.78V
b.VR = IRR
= E – V
D= 10V – 0.78
= 9.22 V
the 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.
Figure 2.4 Solution to example 2.1
E
E
x
x
a
a
m
m
p
p
l
l
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e
2
2
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2
2
Repeat the analysis of example 2.1 with R = 2Kohm
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a. applying equation 2.2I
D= E / R
= 10V / 1Kohm
= 10mA
applying equation 2.3V
D= E
= 10 V
The 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
V
DQ= .07V
IDQ = 4.6mA
b.
VR = IRR
= IDQR
= (4.6mA)(2Kohm)
= 9.2V
with
V
R= E = V
D= 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.
E
E
x
x
a
a
m
m
p
p
l
l
e
e
2
2
.
.
3
3
Repeat example 2.1 using the approximate equivalent model for the semiconductor diode.
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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:
V
DQ= 0.7V
IDQ = 9.25mA
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.
E
E
x
x
a
a
m
m
p
p
l
l
e
e
2
2
.
.
4
4
Repeat example 2.2 using the approximate model for the semiconductor diode.
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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.
E
E
x
x
a
a
m
m
p
p
l
l
e
e
2
2
.
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5
5
Repeat example 2.1 using the ideal diode model.
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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
V
DQ= 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 some concern 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
2.3 Diode Approximations
In section 2.2 we revealed that the results obtained using the approximate piecewise linear equivalent 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 the behavior, capabilities, and possible areas of application of a device in a manner that will minimize the need for extensive mathematical developments.
The complete piecewise- linear equivalent model in chapter 1 was not employed in the load line 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.
T
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 silicon and germanium diodes. If conditions are such that the ideal diode can be employed, the diode symbol will appear as shown in figure 2.9b.
2.4 Series Diode Configurations with DC circuits
In this section the approximate model is utilized to investigate a number of series diode configurations 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 the diode 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 >VT
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
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.
V
D= V
Tequation 2.4
V
R= E – V
Tequation 2.5
I
D= I
R= V
R/ 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
V
R= I
RR = I
DR = (0A)R = 0V
Figure 2.13 Reversing the
diode of figure 2.10 Figure 2.14 Determining the diode of figure 2.13
Figure 2.13 Substituting the equivalent model for the “off” diode of figure
2.13.
The fact that VR = 0V will establish E volts across the open circuit as defined by Kirchoffs
voltage law. Always keep in mind that under any circumstances – DC, AC, instantaneous values, pulses, and so on – Kirchoff’s 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
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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
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Repeat example 2.6 with the diode reversed.
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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,
E – V
D– V
R= 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 the following in mind for the analysis to follow:
1. An open circuit can have any voltage across its terminals, but the current is always 0A.
2. A short circuit has a 0V drop across its terminals, but the current is limited by the surrounding 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
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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:
VR = IRR = IDR = (0A) 1.2kΏ = 0V
And
V
D= E = 0.5V
Figure 2.20 Operating point with E = 0.5V
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Determine VO and ID for the series circuit of figure 2.21
Figure 2.21 Circuit for example 2.9
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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
V
O= E – V
T1– V
T2= 12V – 0.7V – 0.3V = 11V
Figure 2.22 Determining the unknown for example 2.9
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Determine ID, VD2, and VO for the circuit of figure 2.23
Figure 2.23 Circuit for example 2.10
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Removing the diodes and determining the direction of the resulting current
I
will result in the circuit 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 didoe of figure 2.23
Figure 2.25 Substituting the equivalent state for 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.
Figure 2.26 Determining the unknown quantities for the circuit of Example 2.10
Applying Kirchoff’s voltage law in a clockwise direction gives us
E – V
D1– V
D2– V
O= 0
And
VD2 = E – VD1 – VO = 12V – 0 – 0 = 12V
With
V
O= 0V
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Figure 2.27 Circuit for Example 2.11
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Figure 2.28 Determining the state of the diode for the network of figure 2.27
Figure 2.29 Determining the unknown quantities 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
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 = (E
1+ E
2- V
D) / (R
1+ R
2)
= (10V + 5V – 0.7V) / (4.7K + 2.2K) = 14.3V / 6.9K
= 2.07mA
And the voltages are
V
1= IR
2= (2.07mA)(4.7K) = 9.73V
V2 = IR2 = (2.07mA)(2.2K ) = 4.55V
Applying Kirchoff’s voltage law to the output section in the clockwise direction will result in
-E
2+ V
2– V
O= 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 series parallel configurations. For each area of application; simply match the sequential series of steps applied to series diode configurations.
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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 it appeared 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.
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Determine the current
I
for the network of figure 2.32Figure 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 thenI = (E
1– E
2- V
D) / 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.
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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
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.
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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.
I1 = VT2 / R1 = 0.7V / 3.3K = 0.212mA
Figure 2.37 Determining the unknown quantities for example 2.15.
Applying Kirchoff’s 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)I
D2+ I
1= I
2ID2 = 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.
Figure 2.38 Positve logic 0
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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.
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 - V
D) / R = (10V – 0.7V) / 1K = 9.3mA
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Determine the output level for the positive logic AND gate of figure 2.41.
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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 even
though 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 toI = (E - V
D) / 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
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-wave rectifier, 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 both half-cycles of the sine wave, a different rectifier circuit configuration must be used. Such a circuit is called a full-wave rectifier. 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:
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:
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:
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 half-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.
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 forcomparison purposes. The output signal VO now has a net positive area above the axis over
a full period and an average value determined by
V
DC= (0.318)V
Mequation 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.
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 – VT, 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.
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a. Sketch the output Vo and determine the DC level of the output for the network of figure 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.
