COMMON-BASE CHARACTERISTICS Electronics Help

In our introduction to the theory of transistor operation wc showed a bias circuit (Figure 4…4..) in which the base was treated as the ground, or “common” point of the circuit. In other words all voltages (collector-to-base and emitter-to-base) were referenced to the base. This bios arrangement results in what is called the base (CB) configuration for the transistor. It represents only one of three possible ways to arrange the external circuit to achieve a forward-biased base-to-emitter junction and & reverse-biased collector-to-base junction since anyone of the three terminals can be made the common point. We will study the other two configurations in later discussions.

The significance of having a common point in a transistor circuit is that it gives (1″ _ _ ‘\7– us a single re ere nee for both the input voltage to the transistor and the output V-Vll4>~ .voltage. In the CB configuration, the emitter-base voltage is regarded as the input voltage and the collector-base voltage is regarded as the output voltage, See Figure 4-9. For an NgN ![ansistor, VBE is positive and for a PNP transistor. VfB is positive. Similarly, VeB is a positive output voltage for an NPN and VBe is positive for a PNP. Emitter current is input current and collector current is output current.

In our analysis of the CB configuration, the “input” voltage will of course be the same as the cmiucr-basc bias voltage (Vn·), and the “output” voltage will be the same as the collector-base bias voltage (Vcd. In Chapter 5, we will adopt a more realistic viewpoint in which we will regard small (ae) variations in the emitterbase and collector-base voltages as the input and output. respectively. For the time being, we will concern ourselves only with the effects of changes jn Vu; and Vcc on the behavior of the transistor. Do not be confused by the fact that the “input” current in the NP circuit (Figure 4-9(a)) flows out of the emitter. Again, it will be small changes in the magnitude of l t: that we will ultimately regard as the “input.”

Our objective now is to learn how the input and output voltages and the input and output currents arc related to each other in a cn configuration. Toward that end, we will develop sets of characteristic curves called input characteristics and output characteristics. The.Input characteristics show t:….relation between input current and input voltage fgr different oalues of output voltage, and the aUfput charactefistics show the relation between output current ~md output voltage fur difterellt values of input current. As these statements suggest, there is in a transistor a certain feedback (the output voltage) that affects the input, and a certain “feedforward” (the input current) that affects the output.

Common Base Input Characteristics

Let us begin with a study of the CB input characteristics of a typical NPN transistor. Since the input is across the forward-biased base-to-emitter junction, we would expect a graph of input current (ltJ versus input voltage (Vild to resemble that of a forward-bin-co diode. That is indeed the case. However, the exact shape of this curve will. as we have already hinted, depend on the reverse-biasing output voltage. V( n- The reason Ior is endency is that the greater the value of Veil, the more readily minority carriers in t e are swept through the base-to-collector junction. (Remember that the reverse-biasing voltage enhances such current.) The increase in c.mtter-to-collector current resulting from an increase in Veil means that the (i,lput) emitter current will be greater for a given value of (input) base-toemitter voltage. Figure 4- \0 shows a typical set of input characteristics in which this feedback effect can be discerned. Figure 4-10 is our lirst example of a  very useful way to display transistor behavior graphically anu one that can provide rewarding insights if studied carefully. Although characteristic curves are seldom used in actual design or analysis problems, they convey a wealthof information and we will see many more of them in the future. Each set should he scrutinized and dwelled upon at length. Try to visualize how currents and/or voltages change when one quantity is held constant and the others arc varied. Note

 (Example -1-2)

1. Find the 0′ of the transistor neglecting
2. Repeat if I( 1.987 mA when Vec is replaced by a short circuit to ground.

Solution

1. In Figure 4-11, we see that VHF. = 0.7 V. From Figure 4-10, the vertical line corresponding to VIJE = 0.7 V intersect,s the VCIJ = 25 V curve at l« = 9.0 mA. Therefore, 0′ “‘” leI h = (8.94 mA)/(9.0 mA) = 0.9933. .
2. When Vcc is replaced by n short circuit, we have VCII = O. From Figure 4-10, ‘” = 2 mA at VU1 = 0 V and VIII’: = D.7 V. Therefore, (V “‘” 1(11r = (I.l)X7 ml\)1 (2.0 mA) ,= O.l)!).,).

common-Base Output Characteristics

consider now an experiment in which the collector (output) current is measured as V, /I (the output voltage) is adjusted for fixed settings of the emitter (input) current. Figure 4-12 shows a schematic diagram and a procedure that could be used to conduct such an experiment on an NPN transistor. Understand that Figure 4-12 docs not represent a practical circuit that could be used Ior any purpose other than investigating transistor characteristics. Practical transistor circuits contain resistors and have input and output voltages that arc different from the de bias voltages. However. ::i this point in our study of transistor theory we are interested in the transisi-i, itself. We are using characteristic curves to gain insights to how the \'(\I:~.l!,es and currents relate to each other in the device, rather than in the external circuit. Once we have gleaned nil the device information we can from studying characteristic curves, we will have a solid understanding of what a transistor . really is and can proceed to study practical circuits. When lc is plotted versus VCB for differe»: values of Ir., we obtain the family of curves shown in Figure 4-13: the outpu, characteristics for the CB configuration. A close examination of these curves will reveal some new facts about transistor behavior.

We note first in Figure 4-1J that each curve starts at Ie = 0 and rises rapidly for a small positive increase. in VCII. In other words, Ie increases rapidly just as V( /I begins to increase slightly beyond its initial negative value. Since each curverepresents a fixed value of h. this means that while Ie is increasing. the ratio l.l l , must also he increasing. But 1<1h equals 0′, so the implication is that the value of a for a transistor is not constant. Alpha starts at 0 and increases as VCII increases. The reason for this fact is that a very small portion of the emitter current is able

to enter the collector region until the reverse-biasing voltage Vell is allowed to reach a value large enough to propel all carriers across the junction. When Vca is nr gative, the junction is actually forward biased, and minority carrier flow is inhibited. The proportion of the carriers that are swept across the junction (ex) depends directly on the value of VCR until VCll no longer forward biases the junction. The portion ofthe plot wtiere Vcs is negative is called the saturation region of the transistor. By definition (no matter what the. transistor configuration), a transistor is saturated , when both its collector-to-base junction and emitter-to-base junction are forward biased.

Once V reaches a value large enough \0 ensure that a large portion of carriers enter the collector (close to 0 in the figure), we see that the curves· more or less level off. In other words, for a fixed emitter current, the collector current remains essential y constant for further increases in Veil’ Note that this essentially constant value of I,· is, for each curve, very nearly equal to the value of I” r1presented by the curve. In short, the ratio Ic/h:, or ex, is very close to 1 and is essentially constant. These observations correspond to what we had previously assumed about the nature of ex. and the region of the plot where this is the case is called the active region. In its active region. a transistor exhibits those “normal” properties (transistor action) that we have associated with a forward-biased emitter-base junction and a reverse biased collector-base junction. Apart from some special digital-circuit applications

a transistor is normally operated (used) in its active region. Note that we can detect a slight rise in the curves as they proceed to the right through the active region. Each curve of constant 1/.. approaches a horizontal line that intersects the fe-axis at a value equal to h, implying that It: approaches lc, and that a approaches I, for increasing Veu. This we attribute to the increased number of minority carriers swept into the collector, which increases the collector current, a’s the reverse-biasing value of Veu is increased.

There is one other region of the output characteristics that deserves comment. Note that the curve corresponding to h = 0 is very close-to the horizontal axis, i.c., to the Ic: = 0 line- When the emitter current is made 0 (by opening the external emitter circuit), no minority carriers arc injected into the base. Under those conditions, the only collector current that flows is the very small leakage current, lceo . as we have previously described (see Figure 4-8). With the scale used to plot the output characteristics in Figure 4-13, a horizontal line corresponding to Ie = low coincides with the Ie = 0 line, for all practical purposes. The region of the output characteristics lying below the h = 0 line is called the cutoff region, because the collector current is essentially 0 (cut off) there. A transistor is defined to he cut off when both the collcctor–base ami emitter-base junctions are reverse biased. Except for special digital circuits, a transistor is not normally operated in its cutoff region.

Example 4-3

A certain NPN transistor has the CB input characteristics shown in Figure 4-10 and the CB output characteristics shown in Figure 4-13.

I. Find its collector current when Veil == \0 V and VIII: == 0.7 V.
2. Repeat when Veil == 5 V and h == 5.5 mA.

Solution

1. From Figure 4-10, we find h == 4 mA at VHf = 0.7 V and VC/J = 10 V. In Figure 4-1}, the vertical line Veil == 10 V intersects the I,: == 4 mA curve at approximately I( == 3.R5 mA. The collector current under these conditions is therefore 3.85 mA.
2. The condition~ given require that we interpolate the output characteristics along the vertical line Veil = 5 V, between le = 5 mA and ft.’ = 6 mA. The value of Ie that is halfway between the IE = 5 mA and It: = 6 mA curves is approximately 5.4 mA. Note that high accuracy is not possible when using characteristic curves in this way. In most practical situations, we could simply assume that lc = It: = 5.5 mA without seriously affecting the accuracy of other computations.

Posted on November 18, 2015 in BIPOLAR JUNCTION TRANSISTORS

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