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Ecet Week 2 Homework

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Chapter 3

Problems: 1, 2, 5, 6, 9, 13.

1. What is the exact value of IC for IE = 5.34 mA and IB = 47.5 μA?

Answer: IC = IE-IB = 5.34-0.0475 IC = 5.29mA

2. A certain transistor has an IC = 25 mA and an IB = 200 μA. Determine the βDC. Answer: BDC = IC/IB = 25/200 BDC = 0.125

5. Determine the IB, IC, and VC for the transistor circuit in Figure 3–66. Assume βDC = 75.

Answers: IB = (VBB-VB)/(RB) = (2.0-0.7)/(4.7) = 0.276mA

IC = βDC*IB = 75 * 0.276mA = 20.7mA

VC = VCC – IC*RC = 24V – (20.7mA) ( 0.43Ω) = 15.1V

6. Draw the dc load line for the transistor circuit in Figure 3–67.

Answer:

IC(Sat) = VCC/RC = 10V/1.0kΩ = 10mA

VCE = VCC = 10V

9. Repeat Problem 8 for βDC = 300. (Hint: The transistor is now saturated!) (8. For the base-biased npn transistor in Figure 3–68, assume βDC = 100. Find IC and VCE.)

Answers: IC = βDC(VRB/RB) = 300(9.3V/680kΩ) = 300(13.67x10^-6) = 4.103mA

VCE = VCC – (IC)(RC) = 10 – ( 4.108mA)(2.7kΩ) = 10 – 11.08 = -1.08V

13. For the voltage-divider biased circuit in Figure 3–71, determine

Answers:
R1 and R2 set up a voltage divider on the base.

VBE = VB = R2/(R1 +R2) *VCC = 6.8k/(6.8k+56k)* 15 VB = 1.62V

VBE +VE-VB= 0 VE = VB –VBE = 1.62- 0.7 = 0.92V

IE = VE/RE = 0.92/ 1.0k = 0.92mA IE =IC = 0.92mA

IC = VCC – VC/RC

VC = VCC – (IC)(RC) = 15 – (0.92x10^-3)(6.2x10^3) = 15 -5.73 = 9.27V

VCE = VC - VE = 9.27 -0.92 = 8.35V

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