Generally speaking, electronic devices can be considered to “operate” (be used) in one of two distinct kinds of applications: small signal and large signal. In small signal applications, the current and voltage changes in the device occur over a very limited range of its /- V characteristic curve. In other words, the quantities LlV and LlJ to which we have referred in earlier paragraphs are very small in comparison to the overall voltage and current ranges through which the device is capable of operating. In practice, small-signal operation can be considered to be that which occurs when the voltage and current changes are over a range of the /- V curve that is essentially linear. Note in Figure 3-2 that the diode’s characteristic curve is essentially linear for currents greater than about 5 mA, that is. in the region above the knee.
In contrast, large-signal operation occurs when the voltage and current changes in a device occur over substantially its entire /- V characteristic. Large-signal applications usually require voltage and current changes over a region of the J- V curve in which there is a significant change in its slope. For example, a circuit in which the voltage across a silicon diode varies from -5 V to +0.7 V would be considered a large-signal application. In this case the diode would change in nature from an extremely high resistance when reverse biased (negative voltage region) to a moderately high resistance when slightly forward biased, to a very small resistance when biased above the knee. In this section we will study equivalent circuits for a diode operated in smal -signal applications, and in the next section we will do the same for large-signal applications.
Consider the circuit shown in Figure 3-7(b). Nste that the circuit contains a de voltage source in series with an ac voltage source. The dc voltage source produces the constant voltage E volts, and the ac voltage source produces the sinusoidal wave e = A sin wt, where the peak value A is assumed to be small in comparison to E, and w is the frequency in radians per second. Therefore, the total voltage U(/) applied to the series combination of the resistor R and the diode is the sum of the dc and ac voltages: u(t) == E + A sin wt. The total voltage v(t) is called an ac voltage with a de level of E volts. It is sketched in Figure 3-7(a). Note that v(t) has maximum value E + A and minimum value E – A. We wish to determine the voltage across
Recall that a circuit containing two voltage sources can be analyzed using the principle of superposition. That is, we can determine the current in the circuit due to each source acting alone and then add these results to find the actual current when both sources are present. The superposition principle can be applied only when all the circuit components are linear and since we are assuming small-signal operation, we are justified in applying it to the present problem. We should first determine the de diode current in Figure 3-7(b) because this current is needed to calculate the ac resistance rn (equation 3-6). Eliminating (shorting) the ac voltage source in Figure 3-7(b), we obtain a circuit like that shown in Figure 3-4. and therefore, assuming a silicon diode, I = (E – O.7)/R. We can then find the ac resistance of the diode using equation 3-6: I’ll = 0.026/1. Eliminating the de voltage source, the ac equivalent circuit is then as shown in Figure 3-R Note that the ac equivalent circuit represents the diode by a single equivalent resistance equal to r() ohms. We see that the ac current in Figure 3-R can be calculated from a direct application of Ohm’s law:
The ac diode voltage U() can be calculated from U() = iro or by application of the voltage-divider rule:
Combining the results of our dc ‘and ac analyses, we find (he total current and voltage in the diode to be
and
Example 3-3
Assuming that the silicon diode in Figure 3-9 is biased above its knee and has a bulk resistance of 0.1 0. find the total current in and total voltage across the diode. Sketch the current versus time.
Solution.
Shorting the ac source we find the dc current to be
Figure 3-10 is a sketch of the total current. Note that the maximum current is 19.63 + 7.37 = 27 mA and the minimum current is 19.63 – 7.37 = 12.26 mA. It would he difficult to sketch uJ)(t) accurately because its de level (0.7 Y) is so much greater than its ac component. The ac variation is only ±1O mY because the diode’s ac resistance is so small
The Load Line
Small-signal diode analysis can be performed graphically using the diode’s /- V characteristic curve. Although this method is not frequently used in practice, it deserves to be studied because it provides insights to the combined static and dynamic (de and ac) behavior of the diode circuit. Consider the circuit shown in Figure 3-11. Here we show the de equivalent circuit that results when the ac source
in Figure 3-7 is shorted, except instead of treating the de diode voltage as a constant. we consider it to he the variable quantity V.
By Kirchhoff’s voltage law, E = IR + V. Solving for I, we obtain
In equation 3-10, we regard 1 and Vas uariables, whereas E and R are constants. For example, if equation 3-10 were applied to Figure 3-9, we would have
or
Compare equations 3-10 and 3-11 with the general form (equation 3-2) of a linear . relation between 1 and V. Recall also that the general equation. for the graph of a straight line plotted in an xY, coordinate system is
where m is the slope of the line and b is its y-intercept.
We see that the variable I in (3-10) corresponds to (he variable y in (3-12), while the variable V in (3-10) corresponds to the variable x in (3-12). Further comparison of equations 3-10 and 3-12 reveals that the slope III in (3-12) corresponds to -IIR in (3-10), while the y-intercept b in (3-12) corresponds (0 thc Z-intcrccpt ElR in (3-10). We conclude that the graph of equation 3-10 is a st raight line on /- V axes and has slope –1/ R and I-intercept EIR. This line is called the de load line. In the example (equation 3-11) we note that the load line has slope -3.7 X 10-.1and I-intercept 0.0222 A = 22.2 mA.
Figure 3-12 shows the graph of the load line corresponding to equation 3-11. Note that a straight line having a negative slope is always a line that goes downward for increasing values of- the horizontal variable (V in this case). Note also that the V-intercept (Vo) of the graph can be found by setting I equal to 0 in equation 3-10 and solving for V:
The significance of the de load line is that every possible combination of the current / a71d voltage V in the circuit of Figure 3-11 is a point that lies somewhere on the line, Given a particular diode, whose characteristic /- V curve we happen to have, our objective is to find the current-voltage combination that results when
that diode is inserted in the circuit. We can find that point by plotting the de load line on the same /- V axes that contain the diode’s characteristic curve. The intersection of the dc load line and the characteristic curve gives us the actu I diode CUI -nt r.nd voltage that result when the diode is used in the circuit. What \c have accomplished. in effect, is a graphical solution to the two simultaneous equations
and
Figure 3- U shows the solution obtained by plotting the load line for the circuit of Figure 3-lJ (i.e .. equation 3-11) on the same set of’ axes as a hypothetical /- V characteristic. In the figure. we see that the diode voltage is 0.66 V and that the corresponding current is 19.8 mA. This point of intersection. labeled Q. is called the quiescent point (Q-point). or the opcruting point of the diode. It represents the de current and voltage in the circuit when only the de source voltage E = 6 V is present in the circuit. that is, when the ac source voltage in Figure 3-9 is O. Recall that all of our analysis so far has been under the assumption that the ac SOl r e is shorted out. The quiescent point is sometimes called the bias point because it represents’ the voltage and current in the diode when it is forward biased by the dc source.
Shown next is the plot produced by a program run. {Portions of the complete printout produced by SPICE have been omitted to conserve space.) The peak value of the current is seen to be 30.32 mA, occurring at t = 1.25 ms. We can compare this result with that calculated using equation 3-8: Since the de current is (12 – 0.7 V)/470 0. = 24 mA, the approximate value of r o is 0.026/(24 mA) = 1.08 D.. Thus, by equation 3-8, the peak value of the current is
We see that there is very favorable agreement between the SPICE plot (30.32 mA) and equation 3-8 (30.37 mA).
