Source: https://electricalassignments.com/graphical-characteristics-12941
Timestamp: 2019-04-23 16:29:29+00:00

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We are now ready to consider: more general cases in which operation is not necessarily in the active mode and the relationship ic = βiB does not neessarily apply. Instead, we shalrmake use of graphs that show the relationships of transistor variables to one another.
To begin, we recall that analysis of a transistor circuit requires calculation of four different voltages and currents. (There are three voltages and three currents associated with the transistor, but one current can be expressed in terms of the other two by means of Kirchhoff’s current law, and one voltage in terms of the others by Kirchhoff’s voltage law.) Presumably each of the variables could depend on all three of the others. This would make the subject complicated, and fortunately it is not the case. The usual variables chosen to describe operation are iB, VBE, iC and VCE; as we shall see, these depend on each other in fairly simple ways.
The currents and voltages in question are indicated in Fig. 11.28. Most often the high-power signal being controlled comes out of the right-hand port; thus we shall refer to it as the output port. The low-power control signal normally enters at the left, which we shall call the input port. We note that the emitter terminal is common to (meaning it is connected to) both the input and the output ports; hence this is referred to as the common-emitter connection. We are now ready to specify how iB and iC depend on the voltages.
We have already noted that in the active mode iE is approximately an exponential function of VBE’ as given in Eq. (11.7). A typical relationship between iB and VBE for an npn transistor is shown somewhat more generally in Fig. 11.29. The value of iB might be expected to depend not only on VBE but also on VCE, and from the figure we see this is the case; however, the dependence is not very strong. In fact, whenever VCE ≅ 0.3 V the graph of iB is nearly identical with the curve marked VCE ≥ 0.3 V in the figure. Note that this is the case in the active mode, since in that mode VCE ≥ 0.7 V. The right-hand curve in Fig. 11.29 is the exponential function (11.7). As VCE becomes less than 0.3 V we see that the iB – VBE curve does shift slightly, but this shift is not very important for circuits treated in this book. In fact, for our purposes Eq. (11.7) an” the right-hand curve of Fig. 11.29 are always good approximations.
The input characteristics of a typical npn silicon transistor in the common emitter configuration. If Vce ≥ 0.3 V, the curve on the right applies.
We now turn our attention to graphing the collector current ic. A typical dependence of ic on VCE for an npn transistor is shown in Fig. 11.30. Several curves are shown in this graph, corresponding to different values of the base current. It is necessary to use this complete set of curves because iC depends strongly on iB as well as on VCE. For example, if iB is known to be 20 μA and VCE is 6 V, we use the top curve in the figure and read off that ic = 2 mA.
The output characteristics of a typical npn transistor in the common-emitter configuration.
If iB is reduced to 15 μA and VCE is 6 V, we use the second curve from the top and find iB = 1.5 mA. If iB has some value between 15 and 20 – μA,one can interpolate a suitable curve between the 15-μA and 20-μA curves. We note that for VCE > 0.7 V, the curves are nearly horizontal. This means that when VCE > 0.7, iC is nearly independent of VCE and depends only on iB. This part of the graph. corresponds to the active mode of operation. (See Fig. 11.31.) We observe that in this part of the graph ic ≅ 100iB, independent of VCE’ in agreement with Eq. (11.5). Evidently Fig. 11.30 has been drawn for a transistor for which β ≅ 100.
The output characteristics of a typical npn transistor. The regions of operation are indicated.
The transistor in Fig. 11.32(a) has the output characteristics shown in Fig. 11.30. Let Vcc = 6 V. Sketch a graph of iC versus it. and thus find β for this transistor.
The value of iC corresponding to each iB is located by finding the intersection between a vertical line at VCE = 6 V and the curve corresponding to each value of iB.
(Note that i1 = iB and VcE = Vcc.) Points corresponding to iB = 0,5,10,15, and 20 μA are plotted and a line drawn through them. The slope of this line, which is the ratio iC/iB’ is equal to β In this case we see that β = 100.
At the left of Fig. 11.31 is the saturation region. This region corresponds to low values of VCE, so low, in fact, that the collector-base junction has lost its reverse bias. In this case the collector loses its ability to collect, and iC becomes less than ≅iB’ Any time VeE becomes less than 0.7 V, a transistor is likely to leave the active mode and become saturated. The effect is undesirable in amplifier circuits, but if figures usefully in digital circuits.
Now as we increase iB, VCE decreases, eventually approaching 0.7 V. At this point the collector junction has lost its reverse bias and the transistor enters saturation. Further increases of iB now do not result in proportional increases in ic because the relationship ic = βiB no longer applies. This prevents VCE from decreasing indefinitely, and in fact v CE usually approaches a minimum value in the neighborhood of 0.2 V. The minimum value reached by VCE in saturation is usually referred to as V CESAT.
To review, we note that in saturation (1) iB is greater than ic/3 and (2) VCE = V CESAT ≅ 0.2 V.
Find the collector current of the transistor in Fig. 11.35(a) for the following three values of i1: 0, 5, and 20 μA. Assume the transistor is the “typical” transistor whose characteristics are illustrated in Figs. 11.29 and 11.30.
To solve this problem we will use the graphical load-line technique. An imaginary break is made in the circuit, as shown in Fig. 11.35(b). The break in the circuit is purely imaginary, and in reality ic equals iR and Vc equals VCE’ We plot the curve of ic versus VCE and the curve of iR versus Vc on the same graph; the point of intersection of the curves is the desired solution. A set of curves of ic versus VCE for this transistor is already given in Fig. 11.30; we need only choose the appropriate curve from the known base current (iB = i1)’ The characteristic iR versus Vc is obtained from Kirchhoff’s voltage law: Vc = V2 – iRR. We now plot ic versus VCE and iR versus Vc on the same graph, as in Fig. 11.35(c), for the specified values of R, V2, and i1. For the case il = 5 μA, the point at which the conditions ic = iR and VCE = Vc are satisfied is the point ic = 0.5 mA, VCE = 2.5 V. The transistor is operating in the active region. For the case i1 = 20 μA, the solution is ic = 0.95 mA, VCE = 0.25 V, and the transistor is-in saturation. When it = 0, ic = 0 and VCE = 5 V, and the transistor is cut off.
If the entire set of curves of Fig. 11.30 is superimposed on the load line, as in Fig. 11.35(d), then ic may be quickly determined for any value of it. It is noteworthy that the transistor behavior is quite different in the different regions of operation. In the active region (that is, for VCE > 0.7 V) the collector current and collector- emitter voltage depend strongly on i1. In the saturation region the collector current becomes almost independent of i1> and the collector-emitter voltage is about 0.2 V. In the cutoff region, when i1 becomes very small, the collector current goes to zero and the collector-emitter voltage equals V2.

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