Graphical Analysis of the Small Signal CE Amplifier Electronics Help

Our study of transistor amplifiers begins with an analysis of the common-emitter circuit, because that is the most widely used configuration. Figure 5-10 shows the CE bias circuit we studied in Chapter 4 modified by the inclusion of an ac signal source in s-erieswith the base. For the sake of completeness we include a coupling capacitor, but we will assume for the time being that it is large enough to have negligible effect on the ac signal. We will also temporarily postpone discussion of source and load resistance.

currents v ith lowercase letters to represent ac quantities. The small variations in the transistor input voltage, which in turn cause small Val :’It;0n<; in the base current. As the base current decreases and increases, the collector current does the same. Since the collector current is approximately /3 times as large as the base current, we achieve current gain between input and output. The transistor’s ability to produce gain can be attributed to the difference in impedance levels at input and output: A small current into a low-resistance” forward-biased junction controls current flow across u high-resistance, reverse-biased junction  It is instructive to view ac operation in terms of the variation in output voltage and output current that occurs along a load line plotted on a set of transistor output characteristics. As a specific example, we will perform a graphical analysis of the circuit shown in Figure 5-11:

To determine the total variation in base current caused by the signal source in Figure 5-11. we will use the input characteristic shown in Figure 5-12. Here, we neglect the feedback effect of Vc£ 011 the input characteristics and show only a single.average IB- VB£ characteristic for the transistor. Under the assumption that the base-emitter junction is forward biased by 0.65 V, the :to.03-V variation of Us causes VB£ to vary between 0.62 V and 0.68 V. We see in Figure 5-12 that the input voltage variation causes IB to vary sinusoidally between a minimum value of 20 p.A and a maximum value of 40 p.A. ‘ The load line for the circuit in Figure 5-11 intersects the Vc£-axis at Vcc = 18 V and the Ie-axis at (18 V)/(3 kG) = 6 mA. It is shown plotted on a set of output characteristics in Figure 5-13. The Q-point is located at the intersection of the load line with the base current curve corresponding to the base curre t that flows when Us = O. By definition, the Q-point locates l~,ede bias values of Vc£ and Ic, which are the output values when no ac signal is present. Since Figure 5-12 shows that Is = 30 p.A when VB£ = 0.65 V (us = 0), the Q-point is the intersection of the load line with the Is = 30 p.A curve. We see in Figure 5-13 that the transistor is biased .at Vel: = 9 V and Ie = 3 mA. Recall that the load line is a plot of all possible combinations.of Ic and Va. Therefore, as the base current alternates between its extremes of 20 p.A and 40 p.A. the values of Ic and Vc£ change along the load line between its intersections with the two curves Is = 20/LA and Is = 40 /LA. We see in Figure 5-13 that the collector current changes between Ie = 2-mA and lc = 4 mA as the, base current changes between 20 p.A and 40 /LA. Since the base c~rrent is changing sinusoidally, the collector current is doing likewise. The sinusoidal ic waveform is sketched in the figure.

We also see in Figure 5-13 that the valves of V(,£change between 6 V and 12 Vas the base current changes between 20 p,A and 40 p,A. Note carefully that Vn: decreases when IH and Ie are increasing, and vice versa. Therefore, the sinusoidal voltage Vet is 1800 out of phase with each of the sinusoidal currents ib and ie• Since ih is in phase with we conclude that Vet is.also 1800 out of phase with Vk’ In other words, the ac output voltage from a common-emitter amplifier is 1800 out of phase with the ac input voltage. This fact can also be deduced from the load line equation:

Since the term l R  in equation 5-15 subtracts from the constant Vcc, we observe that an increase in Ie causes a decrease in Vc£. Thus, as input voltage increases, Is increases, Ic increases, and Vc£ decreases. The common-emitter voltage amplifier is said to cause phase inversion, or to invert voltage. We can use the values we have obtained graphically to compute important characteristics of the amplifier. The current gain Ai is

The Effect of Q Point Location on AC Operation

Let us consider now how the location of the Q-point affects the ac operation of the ampr. Suppose the value of the base resistor RII in Figure 5-11 is changed from 576 k.o. to 3.47 MH. The quiescent value of the base current will then become

Figure 5-14 shows that the Q-point in this case is shifted down the load line to a point where it intersects the III = 5 p.,A curve. At the new Q-point, Ie = 0.5 mA and VeE = 16.5 V. When the base current increases 10 p.,A beyond the Q-point to 15 p.,A, it can be seen in Figure 5-14 that the collector current increases to 1.5 mA and Va decreases to 13.5 V. However, when the base current decreases 10 p.,A below ~he Q-point (to (5 p.,A) – (10 p.,A) = -5 p.,A), the transistor enters its cutoff region. C,learly lc cannot be l.ess than 0 and Va cannot be greater than Vee= 18 V

As shown in the figure, the output current prematurely becomes 0 in the sine wave cycle, and clipping results. At the same time, Va reaches its limit of 18 V and the output waveform shows positive clipping. With the Q-point in its new location, the output voltage change cannot exceed 18 V – 16.5 V = 1.5 V without positive clipping occurring. Thus, the maximum peak-to-peak output voltage is 2 x 1.5 = 3 Y, and we say that the amplifier has a maximum output swing of 3 Y. This reduced swing limits the usefulness of the amplifier and illustrates the importance of locating the bias point somewhere near the center of the load line.

If the Q-point is located too far up the load line, the output swing will be limited by the onset of saturation. This fact is illustrated in Figure 5-15(a). In this case, a substantial increase in base current beyond its quiescent value causes the transistor to saturate. The collector current cannot exceed its saturation value and Vc£ cannot be less than O. Consequently, the output voltage is a negatively clipped waveform, as shown in the figure. Even if the Q-point is located at the center of the. load line, positive and negative clipping can occur if the input signal is too large. Figure 5-15(b) shows what happens when the total change in base current is so great that the transistor is driven into saturation at one end and cut off at the other. We see that both positive and negative clipping occur due to the amplifier being overdriven

Linearity and Distortion

To be useful, an amplifier’s output waveform must be a faithful replica of the input waveform (or a phase-inverted replica of the input). That clearly is not the case when clipping occurs. Apart from clipping, the degree to which the output waveform has the same shape as the input depends upon the amplifier’s linearity. To be linear: any change in output voltage must be directly proportional to the change in input voltage that created it. For example, if 11V” = 1 V when 11ViII = 0.01 Y, then 11V” must equal 2 V when 11ViII = 0.02 V, and 11Vo must equal 0.5 Y when 11ViII = 0.005 V. The linearity of a transistor can be determined by examining the extent to which equal increments of base current correspond to equally spaced curves on the CE output characteristics, If we assume that input current is. directly proportional to input voltage (i.e., that the base-emitter junction is linear, in the sense discussed In Ci’l tcr 1), ther. changes in input voltage should C<lUSt; the output voltage to vary to a proportional extent along the load line. This will be the case only if the cur yes of constant base current are equally spaced. To demonstrate this fact, Figure .)-16 shows a S~( of CE output characteristics that have been intentionally distorted to exaggerate nonuniform spacing. Notice that the distance between the curves increases for larger values of base current. In the figure, the base current is assumed [0 vary sinusoidalJy 20 /-LAbelow and 20 /-LAabove its quiescent value of 50 /-LA. The variation from 50 /-LA to 70 /-LA causes a change in Ie from 4 mA to 8 mA, but the variation from 50 /-LA to 30 /-LAonly causes lc to change from 4 mA to 2 mA. Sirnnu: change in Vc£ is (12.5 V) – (2.5 V) = 10 V when IIJ increases

The active region of a transistor’s output characteristics is the region where the base current curves are generally found to have equal or nearly equal spacing. For this reason, the active region is often calied the linear region. Of course, the characteristics shown in Figure 5-16 are decidedly nonlinear. In this example, the nonlinearity is due to the fact that device parameters (such as f3) change significantly over the region of operation. Small-signal analysis of this de i’ we uld thus be rest rioted to a very small range of operation along he load line

Posted on November 18, 2015 in SMALL SIGNAL BJT AMPLIFIERS

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