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Solving (2) for Vb 0 e and substituting the result into (1) yield VTh gm VS sC 1 RS gm RS sC GE g 3

Deactivating (shorting) VS also shorts E to B 0 . Consequently, Vb 0 e 0, the dependent current source is open-circuited, and ZTh 1=sC . Now, the Thevenin equivalent and voltage division lead to VL RC kRL V RC kRL ZTh Th 4

(d) Dx (sec 1 x 3 ) =

Substitution of (3) into (4) and rearrangement give the desired voltage-gain ratio: Av VL gm RC kRL VS sC RC kRL 1 sC RS RS gm g GE 1 5

(b) Since (5) involves the upper frequency range, it describes the ampli er as a low-pass (midfrequency) lter with break frequencies at !1 1 C RC kRL and !2 RS gm g GE 1 C RS 6

Let y = sec 1 x. Then, sec y = x Dx (sec y) = Dx (x) dy =1 [by the chain rule] (sec y tan y) dx dy 1 1 = = dx sec y tan y x tan y But the identity 1 + tan2 y = sec2 y gives 1 + tan2 y = x 2 or tan2 y = x 2 1 or tan y = x 2 1

The $28,000 will become the estimated realizable value of the ac counts receivable at that date. The bad debt expense will appear as an op erating expense in the income statement.

(a) Apply the results of Section 8.6 to the small-signal equivalent circuit of Fig. 8-9(a) to determine the Miller capacitance. (b) Using the Miller capacitance, draw the associated equivalent circuit and from it nd an expression for the high-frequency voltage-gain ratio.

(a) First, the gain KF must be found with capacitor C and load resistor RL removed. VL gm Vb 0 e RC the desired gain is KF VL gm RC Vb 0 e YF 1 gm RC C s 1 Since

By de nition of sec 1 x, the angle y lies in the rst or third quadrant, where the tangent is positive. Hence, tan y = + x 2 1 and 1 dy = dx x x2 1

since comparison of Figs. 8-9(a) and 8-12(a) shows that C forms a feedback path analogous to YF . (b) The output shunt capacitance, as suggested by (8.48), must also be determined. Since hre 0 underlies the hybrid- model, the reverse voltage-gain ratio KR 0, hence: Yo Y2 1 KR YF % Y2 YF Y2 sC 3

Comparison of Fig. 8-9(a) with Fig. 8-12(b) and the use of (1) to (3) lead to the equivalent circuit of Fig. 8-25. Let Ceq CM C 1 gm RC C C Then, by voltage division, Vb 0 e and by Ohm s law, VL RC kRL g V 0 s RC kRL C 1 m b e 5 r = sr Ceq 1 r = r r V x V rx r = sr Ceq 1 s s rx kr Ceq 1 s 4

37.8 Prove the following formulas for antiderivatives: (a) (b) (c) x dx 1 = tan 1 + C a a a2 + x 2 dx a2 x2 = sin 1 x +C a [a > 0] [a > 0]

Substitution of (4) into (5) and rearrangement yield the desired voltage-gain ratio: Av s VL gm RC kRL r = rx r s RC kRL C 1 s rx kr Ceq 1 Vs

There are two generally accepted methods of calculating the amount of uncollectible accounts. One method is to use a at percentage of the net sales for the year. The other method takes into consideration the ages of the individual accounts at the end of the scal year. Percentage of Sales Method Under the percentage of sales method, a xed percentage of the total sales on account is taken. For example, if charge sales were $200,000 and ex perience has shown that approximately 1 percent of such sales will be come uncollectible at a future date, the adjusting entry for the bad debt account would be: Bad Debt Expense Allowance for Bad Debt 2,000

+ LL _

dx f (x)

Fig. 8-25