As we indicated in Section 3-5, a diode is said to operate under large-signal conditions when the current and voltage changes it undergoes extend over a substantial portion of its characteristic curve, including portions where there is a significant change in slope. In every practical large-signal application, the diode is operated both in the region where it is well forward biased (above the knee) and into the region where it is either reverse biased or biased near zero volts. We have seen that such large excursions wiII change the resistance of the diode from very small to very large values.
When the resistance of a diode changes from a very small to a very large value. it behaves very much like a switch. An ideal (perfect) switch has zero resistance when closed and infinite resistance when open. Similarly, an ideal diode for large signal applications is one whose resistance changes between these same two extrim ~s. When analyzing such circuits, it is often helpful to think of the diode as a voltage-controlled switch: a forward-biasing voltage closes it, and a zero or reverse biasing voltage opens it. Depending on the magnitudes of other voltages in the circuit, the 0.3- or 0.7- Y drop across the diode when it is forward biased mayor may not be negligible. Figure 3-16 shows the idealized characteristic curve for a
silicon diode (a) when the O.7-V drop is neglected and (b) when it is not. In case (a), the characteristic curve is the same as that of a perfect switch
Rectifiers
One of the most common uses of a diode in large-signal operation is in a recttjier circuit. A rectifier is a device that permits current to flow through it in one direction only. It is easy to see how a diode performs this function when we think of it as a voltage-controlled switch. When the anode voltage is positive with respect to the cathode. i.e .. when the diode is forward biased, the “switch is closed” and current flows through it from anode to cathode. If the anode becomes negative with respect to the cathode, the “switch is open” and no current flows. Of course, a real diode is not perfect. so there is in fact some very small reverse current that flows when it is reverse biased. Also, as we know, there is a nonzero voltage drop across the diode when it is forward biased (0.3 or 0.7 V), a drop that would not exist if it were a perfect switch.
Consider the rectifier circuit shown in Figure 3-17. We see in the figure that an ac voltage source is connected across a diode and a resistor R, the latter designed to limit current flow when the diode is forward biased. Notice that no de source is present in the circuit. Therefore, during each positive half-cycle of the ac source voltage c(t). the diode is forward biased and current flows through it in the direction shown. During each negative half-cycle of e(f) the diode is reverse biased and no current flows. The wave forms of e(l) and i(f) arc sketched in the figure. We see that. (f) is a series of positive current pulses separated by intervals of zero current. Also sketched is the waveform of the voltage VII(t) that is developed across R as a
result of the current pulses that flow through it. Note that the net effect of this circuit is the conversion of an ac voltage into a (pulsating) de voltage, a fundamental SkI’ in the construction of a dc power supply..
If the diode in the circuit of Figure 3-17 is turned around, so that the anode will to the resistor and the cathode (0 ((le gerter~{ar, (hen the dioot wjJJ be forward biased during the negative half-cycles of the sine wave. The current would then consist of a sequence of pulses representing current flow in a counterclockwise, 0, negative, direction around the circuit
Example 3-5
Assume that the silicon diode in the circuit of Figure 3-18 has a characteristic like that shown in Figure 3-16(b). Find the peak values of the current i(t) and the voltage VR(/) across the resistor when
1. eft) = 20 sin tot. and
2. e(f) = 1.5 sin cot, In each case, sketch the waveforms for eir), i(/). and VI<(I)
Solution
1. When e(t) = 20 sin wt, the peak positive voltage generated is 20 V_ At the instant e(t) = 20 V, the voltage across the resistor is 20 – 0.7 = 19.3 V, and the current is i = 193/(1.5 kO) = 12.87 mA. Figure 3-19 shows the resulting wave forms. Note that because of the characteristic assumed in Figure 3-16(b), the diode does not begin conducting until c(t) reaches +0.7 V and ceases conducting when
e(I) drops below 0.7 V. The time interval between the point where e(l) = 0 V and e(t) = 0.7 V is very short in comparison to the half-cycle of conduction time. From a practical standpoint, we could have assumed the characteristic in figure 3-lo(a), i.c., neglected the O.7-V drop. and the resulting wave forms would have differed little from those shown
2. When e(/) = 1.5 sin wt, the peak positive voltage generated is 1.5 V. At that instant, VII(t) = 1.5 – 0.7 = 0.8 V and i(t) = (0.8 V)/(1.5 kfl) = 0.533 mA. The waveforms are shown in Figure 3-20. Note once again that the diode does not conduct until e(t) =:; 0.7 V. However, in this case, the time interval between e(t) = o V and e(t) = 0.7 V is a significant portion of the conducting cycle. Consequently, current flows in the circuit for significantly less time than one-half cycle of the ac waveform. In this case, it clearly would not be appropriate to use Figure 3-16(a) as an approximation for the characteristic curve of the diode.
Elementary DC Power Supplies
As already mentioned, an important application of diodes is in the construction of dc power supplies. It is instructive at this time to consider how diode rectification and waveform tiltcriag, the first two operations performed by every power supply, arc used to create an elementary de power source. (If desired, this entire discussion can be deferred to a more detailed theoretical analysis in Chapter 16.)
The single diode in Figure 3-17 is called a half-wave rectifier, because the waveforms it produces (i(t) and VII(t» each represent half a sine wave. These hal l-sinc waves arc a form of pulsating de and by themselves arc of little practical use. (They can, however. be used for charging batteries, an application in which a steady dc current is 1I0t required.) Most practical electronic circuits require a de voltage source that produces and maintains a constant voltage. For that reason, the pulsating half-sine waves must be converted to a steady de level. This conversion is accomplished by./i/t(‘fillg the waveforms. Filtering is a process in which selected frequency components of a complex waveform are rejected (filtered out) so that they do not appear in the output of the device (the filter) performing the filtering operation. The pulsating half-sine waves (like all periodic waveforms) can be regarded .IS wave forms that have both a de component and ac components. Our purpose in filtering these waveforms for a de power supply is to reject all the ac
components.
The simplest kind 01 filter that will perform the filtering task we have just described is a capacitor. Recall that a capacitor has reactance inversely proportional to frequency. X; = 1/2lT/C. Thus, if we connect a capacitor directly across the out!~~:i vI a half-wave rectifier, the ac components will “see” a low-impedance path to ground and will not therefore, appear in the output. Figure 3-21 shows a filter capacitor C connected in this way. In this circuit the capacitor charges to the peak value of the rectified waveform, V/,II’ so the output is the dc voltage VpR• Note that V/,II = EI, – Vv, where E/, is the peak value of the sinusoidal input and VD is the de voltage drop across the diode (0.7 V for silicon).
In practice, a power supply must provide de current to whatever load it is designed to serve, and this load current causes the capacitor to discharge and its voltage to drop. The capacitor discharges during the intervals of time between input pulses. Each time a new input pulse occurs, the capacitor recharges. Consequently, the capacitor vol tage rises and falls in synchronism with the occurrence of the input pulses. These ideas are illustrated in Figure 3-22. The output waveform is said to have a ripple voltage superimposed on its de level.
The sinusoidal input e in Figure 3-22 is 120 V rms and has frequency 60 Hz. The load resistance is 2 k!l and the filter capacitance is 100 JLF. Assuming light loading and neglecting the voltage drop across the diode,
1. find the dc value of the load voltage;
2. find the peak-to-peak value of the ripple voltage
Solution
1. The peak value of the sinusoidal input voltage is £/. = Vz (120 V rms) = 169.7 V. Since the voltage drop across the diode can be neglected, V/’R = £p =169.7 V. From equation 3-13,
2. From equation 3-14,
full-wave rectifier effectively inverts the negative half-pulses of a sine wave to produce an output that is a sequence of positive half-pulses with no intervals between them. Figure 3-23 shows a widely used full-wave rectifier constructed from four diodes and called a full-wave diode bridge. Also shown is the full-wave rectified output. (::~2~:pr.::!;16-·6 and 16-7 for a detailed explanation of how the bridge operates.) Note that on each half-cycle of input, current flows through two diodes so the peak value of the rectified output is VPR = E; – 2Vn or Ep – 1.4 V for silicon
As in the half-wave rectifier, the full-wave rectified waveform can be filtered by connecting a capacitor in parallel with load RL• The advantage of the full-wave rectifier is that the capacitor does not discharge so far between input pulses, because a new charging pulse occurs every half-cycle instead of every fun cycle. Consequently, the magnitude of the output ripple voltage is smaller. This fact is illustrated in Figure 3-24.
Equations 3-13 and 3-14 for VJc and VI’P are valid for both half-wave and full wave rectifiers. Note that f, in those equations is the frequency of the rectified
Although the elementary power supplies we have described can be used in applications where the presence of some ripple voltage is acceptable, where the exact value of the output voltage is not critical, and where the load does not change appreciably, more sophisticated power supplies have more elaborate filters and special circuitry (voltage regulators) that maintain a constant output voltage under a variety of operating conditions. These refinements are discussed in detail in Chapter 16.
DIODE SWITCHING CIRCUITS
In another very important large-signal application of diodes, the devices are switched rapidly back and forth between their high-resistance and low-resistance states. Digital logic circuits, the building blocks of digital computers, are a typical example
In these applications, the circuit voltages arc pulse-type waveforms, or square waves, that alternate between a “low” voltage, often 0 Y, and a “high” voltage, such as +5 Y. These essentially instantaneous changes in voltage between low and high cause the diode to switch between its “off” and “on” states. Figure 3-25 shows the voltage waveform that is developed across a resistor in series with a silicon
diode when a square wave that alternates between 0 Y and +5 Y is applied to the combination. When e(t) = +5 Y, the diode is forward biased, or “ON,” so current flows through the resistor and a voltage equal to 5 – 0.7 = 4.3 V is developed across it. When e'(r) = 0 V, the diode is in its high-resistance state, or “OFF,” and, since no current flows, the resistor voltage is zero. This operation is very much like rectifier action. However, we study digital logic circuits in just the extreme cases where the voltage is either low or high. In other words, we assume that every voltage in the circuit is at one of those two levels. Because the diode in effect performs the function of switching a high level into or out uf a circuit, these applications are often called switching circuits. Diode switching circuits typically contain two or more diodes, each of which is connected to an independent voltage source. Understanding the operation of a diode switching circuit depends, first, on determining which diodes, if any. art! forward biased and which, if any, are reverse biased. The key to this determination is remembering that a diode is forward biased only if its anode is positive with respect to its cathode. The important words here (the ones that usually give students the most trouble) arc “with respect to.” Stated another way, the anode voltage (with respect to ground) must be more positive than the cathode voltage (with respect to ground) in order for a diode to be forward biased. This is of course the same as saying that the cathode voltage must be more negative than the anode voltage. Conversely, in order for a diode to be reverse biased, the anode must be negative with respect to the cathode, or, equivalently, the cathode positive with respect to the anode. The following example should help clarify these ideas
Example 3-7
Determine which diodes are forward biased and which are reverse biased in each of the configurations shown in Figure 3-26. The schematic diagrams in each part of Figure 3-26 arc drawn using the standard convention of omitting the connection
line between one side of a voltage source and ground. In this convention, it is understood that the opposite side of each voltage source shown in the figure is connected to ground. If the reader is not comfortable with this convention, then he or she should begin the process now of becoming accustomed to it, for it is widely used in electronics. As an aid in understanding the explanations given below, redraw each circuit with all ground connections included. For example, Figure 3-27 shows the circuit that is equivalent to Figure’ 3-26(c).
Solution
1. In (a) the anode is grounded and is therefore at 0 V. The cathode side is positive by virtue of the +5-V source ~9.I}.n~~te~Uo.it through resistor.R. The cathode is therefore positive with respect to the anode; i.e., the anode is more negative than the cathode, so the diode is reverse biased: •
2. In (b) the anode side is more positive than the cathode side (+10 V> +5 V), so the diode is forward biased. Current flows from the lO-V source, through the diode, and into the 5-V source.
3. In (c) the anode side is more negative than the cathode side, so the diode is reverse biased. Note that (essentially) no current flows in the circuit, so there is no drop across resistor R. Therefore, the total reverse-biasing voltage across the diode is 15 V. (See also Figure 3-27, and note that the sources are series-aiding.)
4. In (d) the cathode side is more negative than the anode side (-12 V < -5 V), so the diode is forward biased. Current flows from the – 5-V source, through the diode, and into the -12-V source.’ .
5. In (e) the anode is grounded and is therefore at 0 V. The cathode side is more negative than the anode side (-10 V < 0 V), so the diode is forward biased. Current flows from ground, through the diode, and into the -10- V source.
Figure 3-28 shows a diode switching circuit typical of those used in digital logic applications. It consists of three diodes whose anodes are connected together and whose cathodes may be connected to independent voltage sources. The voltage levels connected to the cathodes are called inputs to the circuit, labeled A, B, and C, and the voltage developed at the point where the anodes are joined is called the output of the circuit. All voltages are referenced to the circuit’s common ground. The voltage source V is a fixed positive voltage called the supply voltage. The figure shows the conventional way of drawing this kind of circuit (a), and the complete equivalent circuit (b).
Let us assume that the inputs A, B, and C in Figure 3-28 can be either +5 V (high) or 0 V (low). Suppose further that the supply voltage is V = +10 V. If A, B, and C are all +5 V, then all three diodes are forward biased (+ 10 > +5) and are therefore conducting. Current flows from the IO-V source, through the resistor, and then divides through the three diodes. Suppose the dc “ON” resistance of each diode is 300 0 and the resistor R is 1 kO. Figure 3-29 shows the resulting equivalent circuits. The parallel combination of the three 5-V sources, each having a 300-0 series resistor, is equivalent to a single S-V source in series with a 300/3 = 100
resistor (Millman’s theorem). The circuit is’ therefore equivalent to that shown in Figure 3–29(b). The current I is found by dividing the total series resistance into the net voltage across it:
The output voltage is
When the de resistance of the diode is not known, an approximate solution can be obtained by assuming a 0.7-V_drop across each (silicon) diode. Under this assumption, the equivalent circuit appears as shown in Figure 3-30. Using this equivalent circuit, we can write Kirchhoff’s voltage law around the clockwise loop to obtain
Vo = 5 V + 0.7 V == 5.7 V. Suppose now that input A == 0 V and B = C = +5 V, as shown in Figure· 3-31(a). It is clear that the diode connected to input A is forward biased. If wetemporarily regard that “ON” diode as a perfect closed switch, then we see that the anode side of .111diodes will be connected through this closed switch to 0 V. Therefore, the other two diodes have +5 V on their cathodes and 0 V on their anodes, causing them to be reverse biased. In reality, the “ON” diode is not a perfect switch, so it has some small voltage drop across it (across the 300 fl) and the anodes are near 0 V rather than exactly 0 Y. The net effect is the same: one diode is forward biased and the other two are reverse biased.
Figure 3-31 (b) shows the equivalent circuit that results if we treat the reverse biased diodes as open switches. Notice that we now show the de “ON” resistance as 100 n rather than 300 n.This is an assumption, based on the fact that the diode now carries substantially more current than before. We have seen that the diode resistance decreases with increasing current. The output voltage can now be calculated using the voltage-divider rule
If we regard the drop across each diode as 0.7 Y, then V” = 0.7 Y. as shown in Figure 3-32(c). The diode circuit we have just analyzed is called a diode A 0 gate because the output is high if and only if inputs A and Band C arc all high.
We have seen that the first step in this kind of analysis is to determine which diodes arc forward biased and which are reverse biased. This determination is best accomplished by temporarily regarding each diode as a perfect, voltage-controlled switch. At this point one might legitimately question how we determined that the forward-biased diode shown in Figure 3-31 is the only one that is forward biased. After all, the other two diodes appear to have their anode sides more positive (10 Y) than their cathode sides (5 Y), and seem therefore to meet the criterion for forward bias. However, if this were the case, then these “ON” diodes would, as closed switches, connect +5 Y to the anodes, which in turn would forward bias the third diode (since its cathode is at 0 V). The third diode would then act as a closed switch connecting 0 V to the anodes. Here, then, is the dilemma: how can the anodes be simultaneously at 0 Y and at 5 Y? Obviously they cannot, and this contradiction proves that only the one di.ode shown i.n Figure. 3-3\ can be. forward biased. The fact that it is forward biased, and therefore connects 0 Y to the anodes, ensures that the other two diodes are reverse biased.
A rule that is useful for determining which diode is truly forward biased is to determine which one has the greatest forward-biasing potential measured from the supply voltage to its input voltage. For example, in Figure 3-31, the net voltage between the supply and input A is 10 Y – 0 Y = 10 Y, while the net voltage between the supply and inputs Band Cis + 10 Y – 5 Y = 5 V. Therefore, the first diode is forward biased and the other two diodes arc reverse biased.
Example 3-8
Determine which diodes are forward hiase-d and which are reverse biased in the circuits shown in Figure 3-33. Assuming a O.7-Y drop across each forward-biased diode, determine the output voltage.
Solution
1. In (a) diodes 01 and 03 have a net forward-biasing voltage between supply and input of 5 Y – 5 Y = 0 V. Diodes D2 and 04 have a net forward-biasing voltage
2. In (b) the net forward-biasing voltage between supply and input for each diode is
3. In (c) the net forward-biasing voltage between supply and input for each diode is
Notice that the diode positions are reversed with respect to those in (a) and (b), in the sense that the cathodes are joined together and connected through resistor R to a negative supply. Thus, the diode for which there is the greatest negative voltage between supply and input is the forward-biased diode. In this case, that diode is O2, 0\ is reverse biased, by virtue of the fact that its cathode is near +5 V and its anode is at -5 V. Figure 3-35 shows the equivalent circuit path between input and output. Writing Kirchhoff’s voltage law around the loop. we see that Vi’ = 5 V – 0.7 V = 4.3 V.
