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Proceed as in Problem 22.2. x 3 2xy + y3 = 8 Dx (x 3 2xy + y3 ) = Dx (8) 3x 2 2(xy + y) + 3y2 y = 0 3x 2 2xy 2y + 3y2 y = 0 Dx (3x 2 2xy 2y + 3y2 y ) = Dx (0) 6x 2(xy + y ) 2y + 3(y2 y + y (2yy )) = 0 6x 2xy 2y 2y + 3y2 y + 6y(y )2 = 0 Substitute 2 for x and 2 for y in (1), 12 4y + 4 + 12y = 0 Substitute 2 for x, 2 for y, and 1 for y in (2), 12 4y + 2 + 2 + 12y + 12 = 0 or 8y + 28 = 0 or y = 7 28 = 8 2 or 8y + 8 = 0 or y = 1 (2) (1)

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For the ampli er of Fig. 3-8, (a) nd the sensitivity factor and (b) show that the condition under which the sensitivity factor is reduced to zero is identical to the condition under which the emitter current bias is constant.

Rearranging gives ICQ VBB VBEQ VBB VBEQ RB 1 RB 1 RE RE @ICQ RB RE VBB VBEQ @ RB 1 RE 2 1

22.4 Let s = t 3 9t 2 + 24t describe the position s at time t of an object moving on a straight line. (a) Find the velocity and acceleration. (b) Determine when the velocity is positive and when it is negative. (c) Determine when the acceleration is positive and when it is negative. (d) Describe the motion of the object.

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UPC -A Barcode C# .NET Generation SDK. UPC -A, short for Universal Product Code version A, is a commonly used linear barcode , especially in America. It can only encode 10 characters, i.e., digit 0-9. In general, an UPC -A barcode will encode 11 data and 1 check digit.

Temperature variations can shift the quiescent point by a ecting leakage current and base-toemitter voltage. In the circuit of Fig. 5-1, VBB 6 V; RB 50 k ; RE 1 k ; RC 3 k , 75; VCC 15 V, and the transistor is a Si device. Initially, ICBO 0:5 A and VBEQ 0:7 V, but the temperature of the device increases by 208C. (a) Find the exact change in ICQ . (b) Predict the new value of ICQ using stability-factor analysis.

In order to transfer balances from an asset account to an expense account, an adjusting entry is re quired. Adjusting entries are used throughout the year to transfer these prepaid amounts to expense accounts as the asset is used.

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(a) Let the subscript 1 denote quantities at the original temperature T1 , and 2 denote quantities at T1 208C T2 . By (5.7), ICQ1 VBB VBEQ1 ICBO1 RB RE 6 0:7 0:5 10 6 51 103 3:1953 mA RB = RE 50 103 =75 1 103 ICBO2 ICBO1 2 T=10 0:5 10 6 220=10 2 A so Again by (5.7), ICQ2 Thus, (b) By (5.16) and (5.17), RB RE 50 1 30:6 RB = RE 50=75 1 75 0:6 10 3 SV RB RE 50 103 75 1 103 SI Then, according to (5.13), ICQ % SI ICBO SV VBEQ 30:6 1:5 10 6 0:6 10 3 0:04 0:0699 mA and ICQ2 ICQ1 ICQ 3:1953 0:0699 3:2652 mA VBB VBEQ2 ICBO2 RB RE 6 0:66 2 10 6 51 103 3:2652 mA RB = RE 50 103 =75 1 103 ICQ ICQ2 ICQ1 3:2652 3:1953 0:0699 mA VBEQ 2 10 3 T 2 10 3 20 0:04 V VBEQ2 VBEQ1 VBEQ 0:7 0:04 0:66 V

(a) v = ds = 3t 2 18t + 24 = 3(t 2 6t + 8) = 3(t 2)(t 4) dt dv = 6t 18 = 6(t 3) a= dt (b) v is positive when t 2 > 0 and t 4 > 0 or when t 2 < 0 and t 4 < 0; that is, when t > 2 and t > 4 or t < 2 and t < 4

Now, according to Section 5.2,

In Problem 5.11, assume that the given values of ICBO and VBEQ are valid at 258C (that is, that T1 258C . (a) Use stability-factor analysis to nd an expression for the change in collector current resulting from a change to any temperature T2 . (b) Use that expression to nd ICQ when T2 1258C. (c) What percentage of the change in ICQ is attributable to a change in leakage current

(a) Recalling that leakage current ICBO doubles for each 108C rise in temperature, we have ICBO ICBO jT2 ICBO jT1 ICBO j258C 2 T2 25 =10 1

which is equivalent to t > 4 or t < 2. v = 0 if and only if t = 2 or t = 4. Hence v < 0 when 2 < t < 4.

CHAP. 5]

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