Electronics 2nd Edition Hambley Pdf Files

• DaladierTypewritten textLIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRORESUELTOS Y EXPLICADOSDE FORMA CLARA VISITANOS PARADESARGALOS GRATIS. • Solutions Manual Errata for Electronics, 2nd ed. Hambley Problem 1.17 In line two, change 3.135 W to 3.125 W. Problem 1.29 In line one, inside the first integral, delete the exponent 2 on i1. In line four, change 8 2/20 to ( )8 2/202. In line five, change Iiavg to I1avg. Problem 1.49 Toward the end of the solution, change when Rs changes from 1 M to 10 k to when RL changes from 1 M to 10 k.

Edition solutions manual [pdf] electrical engineering hambley 2nd edition solutions. Download free pdf files,ebooks and documents - electronics hambley.

Problem 1.50 Change when Rs changes from 0 to 100 to when RL changes from 0 to 100. Problem 1.62 In line two of Part (b), change ( ) Lmm RGG 2121 + to ( ) Lmm RGG 21.

Make the same change in line two of Part (c). Problem 2.12 In Part (d), change )99( 1Cj to )101( 1Cj. In the last paragraph, change 99-pF to 101-pF. Problem 2.14 After the figure, change vo = 8vin to vo = 8vin and change the gain from 8 to 8. Problem 2.16 At the end of the solution, after For AOL = 105, change Av = 9.998 to Av = 9.9989 and change R2/R1 = 10 to R2/R1 = 10. • Problem 2.33 The problem statement should have specified Wspace = 10 m instead of 5 m. Problem 2.38 In the figure, change the upper 10-k resistor (connecting the inverting input to the output of the first op amp) to 15 k.

Problem 2.43 In Part (b), Equation (4), change R1 to R2. In Part (c), in the first equation after the figure, change vi to vi.

Problem 2.53 In the first line, change f0CL to fBCL. Problem 2.73 Before the figure, add the sentence: The PSpice simulation is stored in the file named P2_73.

Problem 2.75 Delete the sentence stating that the plot of vo(t) is on the next page. Problem 3.10 In line three (an equation), change iD/R to vD/R. Problem 3.53 In the sentence beginning with The dynamic resistance, change nVT/ICQ to nVT/IDQ. Problem 3.56 In Part (a), change nVT/ICQ to nVT/IDQ. Problem 3.57 The solution uses rd for the diode resistance rather than rz as specified in the problem statement.

• Problem 3.58 In Part (c), line two (an equation), change the minus sign inside the parentheses to a plus sign. Problem 3.70 In line one, change electon to atom. Problem 3.90 In Part (c), line one, change the denominator of the fraction in parentheses from IR to IR. Problem 3.92 At the end of the solution, add: Larger capacitance produces less output voltage ripple and higher peak diode current. Problem 4.10 In line five of the solution (an equation), change 10 0.6585 to 0.6585 10.

Problem 4.25 In the equation for Is (line seven of the solution), each of the two denominators should end with ) 1 instead of 1). Problem 4.34 In the line for part (d) with = 100, we should have I = 9.53 mA (instead of 10 mA) and V = 9.53 V (rather than 10 V). Problem 4.45 Change Avo = RL/r to Avo = RC/r. Problem 4.50 At the end of step one, add: Set all other independent signal sources to zero. Problem 4.54 Next to the figure, change V EQ to VBEQ.

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• Problem 4.60 In the first line after the figure, second equation, change IBEQ to IBQ. Problem 4.65 In the first line after the figure, insert an equals sign after IB. Problem 5.3 Calculation of the drain currents was omitted. The drain currents are: (a) iD = K(vGS Vto) 2 = (W/L)(KP/2)(vGS Vto) 2 = 2.25 mA (b) iD = K[2(vGS Vto)vDS (vDS)2] = (W/L)(KP/2)[2(vGS Vto)vDS (vDS)2] = 2 mA (c) iD = 0 Problem 5.7 In the last sentence, change K = 25 to K = 25 A/V2.

Problems 5.23 The last line of part (a) should read: VDSQ = 20 2IDQ = 12 V. Problem 5.25 Change the second equation from RSIDSQ = 6 V to RSIDQ 6 V. Problem 5.46 Change greater than zero to greater than unity.

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Problem 5.65 In the third-to-last sentence, change K(vGS5 Vto) to K(vGS5 Vto)2. Problem 5.74 In the sentence after the opening equation, change saturation to triode region. In part (c) before the table, insert Using the value of C given in part (d) of the problem, we have: • Problem 6.16 At the beginning of the solution, insert The following solution is for an inverter operating at 400 MHz. At the end of the solution, add For an inverter operating at 400 Hz, Pdynamic = 3.6 10 10 W. Problem 6.23 In the third line, change IOL = 1 mA to IOL = 1 mA. Problems 6.24 In the first line, change Pdynamic = If to Pdynamic = Kf. Problem 6.25 In the equation for Energy, change (42 12) to (52 02) and change 150 pJ to 250 pJ.

In the equation for Pdynamic, change 150 to 250 and change 3.75 mW to 6.25 mW. Problem 6.32 In the circuit diagram, the device should be an enhancement MOSFET rather than a depletion MOSFET. Problem 6.36 Change the middle of the fourth line to read VIH = 2.04 V, VIL = 1.08 V. Problem 6.51 At the end of the first paragraph, just before the figure, insert the following: [Note: The solution assumes (W/L)p = 1.