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## Q. 7.6

To gain more insight into the operation of transistor amplifiers, we wish to consider the waveforms at various points in the circuit analyzed in the previous example. For this purpose assume that vi has a triangular waveform. First determine the maximum amplitude that vi is allowed to have. Then, with the amplitude of vi set to this value, give the waveforms of the total quantities iB(t), vBE(t), iC(t), and vC(t).

## Verified Solution

One constraint on signal amplitude is the small-signal approximation, which stipulates that vbe should not exceed about 10 mV. If we take the triangular waveform vbe to be 20 mV peak-to-peak and work backward, Eq. (7.81)

$v_{be} = v_{i}\frac{r_{\pi}}{r_{\pi} + R_{BB}}$

$= v_{i}\frac{1.09}{101.09} = 0.011 v_{i}$                                      (7.81)

can be used to determine the maximum possible peak of vi,

$\hat{v} _{i} = \frac{\hat{v} _{be}}{0.011} = \frac{10}{0.011} = 0.91 V$

To check whether the transistor remains in the active mode with vi having a peak value $\hat{v} _{i}$ = 0.91 V, we have to evaluate the collector voltage. The voltage at the collector will consist of a triangular wave vo superimposed on the dc value VC = 3.1 V. The peak voltage of the triangular waveform will be

$\hat{v} _{o} = \hat{v} _{i}$ × gain = 0.91 × 3.04 = 2.77 V

It follows that when the output swings negative, the collector voltage reaches a minimum of 3.1 − 2.77 = 0.33 V, which is lower than the base voltage by less than 0.4 V. Thus the transistor will remain in the active mode with vi having a peak value of 0.91 V. Nevertheless, to be on the safe side, we will use a somewhat lower value for $\hat{v} _{i}$ of approximately 0.8 V, as shown in Fig. 7.29(a), and complete the analysis of this problem utilizing the equivalent circuit in Fig. 7.28(d). The signal current in the base will be triangular, with a peak value $\hat{i} _{b}$ of

$\hat{i} _{b} = \frac{\hat{v} _{i}}{R_{BB} + r_{\pi }} = \frac{0.8}{100 + 1.09} = 0.008 mA$

This triangular-wave current will be superimposed on the quiescent base current IB, as shown in Fig. 7.29(b). The base–emitter voltage will consist of a triangular-wave component superimposed on the dc VBE that is approximately 0.7 V. The peak value of the triangular waveform will be

$\hat{v} _{be} = \hat{v} _{i} \frac{r_{\pi }}{r_{\pi } + R_{BB}} = 0.8 \frac{1.09}{100 + 1.09} = 8.6 mV$

The total vBE is sketched in Fig. 7.29(c).
The signal current in the collector will be triangular in waveform, with a peak value $\hat{i} _{c}$ given by

$\hat{i} _{c} = β\hat{i} _{b}$ = 100 × 0.008 = 0.8 mA

This current will be superimposed on the quiescent collector current IC (=2.3 mA), as shown in Fig. 7.29(d).
The signal voltage at the collector can be obtained by multiplying vi by the voltage gain; that is,

$\hat{v} _{o}$ = 3.04 × 0.8 = 2.43 V

Figure 7.29(e) shows a sketch of the total collector voltage vC versus time. Note the phase reversal between the input signal vi and the output signal vo.
Finally, we observe that each of the total quantities is the sum of a dc quantity (found from the dc circuit in Fig. 7.28b), and a signal quantity (found from the circuit in Fig. 7.28d).