The results show that the midland gain is approximately 132.7 (at 10 kHz, about one decade above the lower cutoff frequency the actual gain may be slightly greater at higher frequencies) Thus the gain will be approximately 0.707(132.7) = 93.8 at the lower cutoff frequency The frequency at which the output is nearest 93.81.259 kHz (92.09) A more accurate estimate of the lower cutoff frequency can made by restricting the range and increasing the number of frequencies at which the output is calculated in the vicinity of cutoff Figure 10-30(b) shows the results of a PRINT statement when the AC statement is changed to UN 201KHZ 1.5KHZ producing output at 20 linearly spaced frequencies between 1 kHz and

1.5 kHz. We see that the lower cutoff frequency is between 1.289 kHz and 1.316 kHz. These results agree well with the value calculated in Example 10-12.The principal consideration in the design and analysis of amplifiers in the low-frequency region is the (Thevenin equivalent) resistance in series with each capacitor used in the circuit. We have seen that a break frequency occurs whenever the frequency becomes low enough to make the capacities reluctance equal to the equivalent resistance at the point of connection. Therefore. in any BJT amplifier the capacitor is most critical in determining the lower cutoff frequency is the one that “sees” the smallest resistance. In the case of the common-emitter amplifier. Cf: is that capacitor, In a common-base amplifier. the input resistance is quite small so the input coupling capacitor Ch is the most crucial. In the common-collector amplifier the output resistance is quite small, so the output coupling capacitor is crucial. Of course a large source resistance in the case of a CB amplifier or a large load resistance in the case of a CC amplifier will mitigate these circumstances. since each is in series with the affected capacitor. Figure 10-31 shows common-base and common-collector amplifiers and gives the equations for the break frequencies due to each coupling capacitor. These are derived in a straightforward manner by solving for the frequency at which the capacities repentance of each capacitor equals the Thevenin equivalent resistance in series with it.

A certain transistor has f3 == 1 =100 kU and r == 25 n. The transistor is used in each or the circuits shown in Figure 10-3\. Find the approximate lower cutoff frequency in each circuit, given the following component values: 1. (Common-base circuit)

Design Considerations

Capacitors in practical discrete circuits arc generally bulky and costly, or else they are unreliable if quality is sacrificed for cost. Therefore a principal objective in the design of discrete BfT amplifiers is selection of the smallest capacitors possible, ‘consistent with the low-frequency response desired. Equations lO-44 through 10-46 can be used to find minimum capacitor values for CE, Cll, and CC amplifiers in terms of break frequencies and circuit parameters. As ‘noted can each configuration has one capacitor whose value is the most critical: The one connected to the point where the equivalent resistance is smallest. This capacuor should be selector’ first to ensure that. the lower cutoff frequency is at least :IS low as the break frequency it produces. The other capacilor(s) can then be selected to break frequencies a I decade or so below clltoff, (With two identical break frequencies occurring 1 decade below the third, the gain is actually reduced by a factor of 0.7, rather than 0.707, at the third ‘frequency.)

A CE amplifier having R, = 330 kO. R2 = 47 kO, Rc = 3.3 kO. and RE = 1.8 kO is to have a 40wer cutoff frequency of 50 Hz. The amplifier drives a 10-kO load and is driven from a signal source whose resistance is 600 O. The transistor has {3= 90 and r; = 100 kO and is biased at. It: :: 0.5 mA. Find coupling and bypass capacitors necessary to meet the low-frequency response requirement. Assume that capacitors must be selected from a line having standard values of 1 ILF, 1.5 ILF, 2.2 JLF, 3.3 JLF. 4.7 ILF, 10 JLF, 50 ILF, 75 ILF, and 100 JLF To obtain a value for CEo we must first find the value of r.:To ensure that the lower cutoff frequency is no greater than 50 Hz, we choose the standard-value capacitor with the next higher capacitance, rather than the one closest to 54.36 p.F. Thus, we choose CE = 75 p.F. To find C, and C2, we lett.{C,) = t.{C2) = 50 HzllO = 5 Hz and use equations 10-44 to calculate C2 = 217(3.19 kD. + 10 kD.)5.Hz = 2.41 p.F Again choosing standard capacitors with the next highest values, we select C, 10 p.F and ~2 = 3.3 p.F. It is a SPICE programming exercise at the end of this chapter to find the actual reduction in gain at 50 Hz when these standard-value capacitors are used. Our

conservative choices result in an actual lower cutoff frequency of about 40 Hz stage