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Then the use of (3.2) and (3.1) with negligible leakage current yields ICQ IBQ 80 30 10 6 2:4 mA and IEQ ICQ 2:4 10 3 2:43 mA 0:9876

Now KVL around the collector circuit leads to the minimum value of RC to ensure saturation: RC VCC VCEsat IEQ RE 15 0:2 2:43 1 5:154 k ICQ 2:4 10 3

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The Si transistor of Fig. 3-19 has 50 and negligible leakage current. Let VCC 18 V, VEE 4 V; RE 200 , and RC 4 k . (a) Find RB so that ICQ 2 mA. (b) Determine the value of VCEQ for VB of part (a).

12.14 A function f , de ned for all real numbers, is such that: (i) f (1) = 2; (ii) f (2) = 8; (iii) f (u + v) f (u) = kuv 2v 2 for all u and v, where k is some constant. Find f (x) for arbitrary x. 12.15 Let f (x) = 2x 2 + 3x for all x.

Fig. 3-19 (a) KVL around the base-emitter-ground loop gives VEE IBQ RB VBEQ IEQ RE Also, from (3.1) and (3.2), IEQ 1 ICQ 2 1

26,000 26,000

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Now, using (3.2) and (2) in (1) and solving for RB yields RB VEE VBEQ 50 4 0:7 1 RE 50 1 200 72:3 k ICQ 2 10 3

(a) Find the nonnegative value(s) of x for which the tangent line to the graph of f at (x, f (x)) is perpendicular to the tangent line to the graph at ( x, f ( x)). (b) Find the point of intersection of each pair of perpendicular lines found in part (a). 12.16 If the line 4x 9y = 0 is tangent in the rst quadrant to the graph of y = 1 x 3 + c, what is the value of c 3

(b) KVL around the collector-emitter-ground loop gives 1 VCEQ VCC VEE RC RE ICQ 50 1 200 2 10 3 13:59 V 18 4 4 103 50

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SaveAs(path); } // Now we try to read the barcode // Instantiate BarCodeReader object BarCodeReader reader = new BarCodeReader (path, BarCodeReadType.

The dc current source IS 10 A of Fig. 3-19 is connected from G to node B. The Si transistor has negligible leakage current and 50. If RB 75 k , RE 200 , and RC 4 k , nd the dc current-gain ratio ICQ =IS for (a) VCC 18 V and VEE 4 V, and (b) VCC 22 V and VEE 0 V.

1 12.17 For what nonnegative value of b is the line y = 12 x + b normal to the graph of y = x 3 + 1 3

CHAP. 3]

(a) A Thevenin equivalent for the network to the left of terminals B; G has VTh RB IS and RTh RB . With the Thevenin equivalent circuit in place, KVL around the base-emitter loop yields RB IS VEE IBQ RB VBEQ IEQ RE 1

Solved Problem 16.2 Adams, Bentley, and Carson have capital balances of $30,000, $25,000, and $20,000, respectively. Adams devotes 75% of his time; Bentley, 50% time; and Carson, 25% time. Determine their par ticipation in net income of $37,500 if income is divided (a) in the ratio of capital investments; (b) in the ratio of time worked. Solution: (a) Total capital is $75,000. Hence: Adams ($30,000 / $75,000) times $37,500 = $15,000 Bentley ($25,000 / $75,000) times $37,500 = $12,500 Carson ($20,000 / $75,000) times $37,500 = $10,000

12.18 Let f be differentiable (that is, f exists). De ne a function f # by the equation f # (x) = lim (a) Find f # (x) if f (x) = x 2 x. Hint: f (x + h) f (x h) h h 0

Using (3.2) and (2) of Problem 3.17 in (1), solving for ICQ , and then dividing by IS results in the desired ratio: ICQ RB IS VEE VBEQ 75 103 10 10 6 4 0:7 ! 237:67 3 RB 1 IS 6 75 10 50 1 200 RE IS 10 10 50 50 2

Note that the value of VCC must be large enough so that cuto does not occur, but otherwise it does not a ect the value of ICQ . (b) VEE 0 in (2) directly gives ICQ IS 75 103 10 10 6 0:7 10 10 6 75 103 50 1 200 50 50 ! 2:93

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