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A transistor has a Pp(max) = 5 W at 25°C. The derating factor is 10 mW/°C. What is the Pp(max) at 70°C? Related Problem %3D

A transistor has a Pp(max) = 5 W at 25°C. The derating factor is 10 mW/°C. What is the Pp(max) at 70°C? Related Problem %3D

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Solve the related problems in this examples.please
EXAMPLE 4-3 Sketch an ideal family of collector curves for the circuit in Figure 4-11 for Ig = 5 µA to 25 µA in 5 µA increments. Assume Bpc = 100 and that VCE does not exceed breakdown. FIGURE 4-11 Re Poc = 100 Vcc Solution Using the relationship lc = BoclB, values of Ic are calculated and tabulated in Table 4-1. The resulting curves are plotted in Figure 4-12. TABLE 4-1 5 µA 10 μΑ 15 μΑ 0.5 mA 1.0 mA 1.5 mA 20 μΑ 2.0 mA 25 μΑ 2.5 mA Ic (mA) 2.5- I = 25 µA 2.0- h= 20 µA 15- L = 15 µA 1.0- 4 = 10 µA 0.5- h = 5 µA VCE o 0,7 V A FIGURE 4-12 Related Problem Where would the curve for Ig =0 appear on the graph in Figure 4-12, neglecting col- lector leakage current?
Transcribed Image Text:EXAMPLE 4-3 Sketch an ideal family of collector curves for the circuit in Figure 4-11 for Ig = 5 µA to 25 µA in 5 µA increments. Assume Bpc = 100 and that VCE does not exceed breakdown. FIGURE 4-11 Re Poc = 100 Vcc Solution Using the relationship lc = BoclB, values of Ic are calculated and tabulated in Table 4-1. The resulting curves are plotted in Figure 4-12. TABLE 4-1 5 µA 10 μΑ 15 μΑ 0.5 mA 1.0 mA 1.5 mA 20 μΑ 2.0 mA 25 μΑ 2.5 mA Ic (mA) 2.5- I = 25 µA 2.0- h= 20 µA 15- L = 15 µA 1.0- 4 = 10 µA 0.5- h = 5 µA VCE o 0,7 V A FIGURE 4-12 Related Problem Where would the curve for Ig =0 appear on the graph in Figure 4-12, neglecting col- lector leakage current?
A certain transistor has a PD(max) of 1 W at 25°C. The derating factor is 5 mW/°C. What is the Ppmax) at a temperature of 70°C? EXAMPLE 4-7 Solution The change (reduction) in Pp(max) is APD(max) = (5 mW/°C)(70°C - 25°C) = (5 mW°C)(45°C) = 225 mW Therefore, the PD(max) at 70°C is 1W - 225 mW = 775 mW Related Problem A transistor has a Pp(max) = 5 W at 25°C. The derating factor is 10 mW/°C. What is the PD(max) at 70°C?
Transcribed Image Text:A certain transistor has a PD(max) of 1 W at 25°C. The derating factor is 5 mW/°C. What is the Ppmax) at a temperature of 70°C? EXAMPLE 4-7 Solution The change (reduction) in Pp(max) is APD(max) = (5 mW/°C)(70°C - 25°C) = (5 mW°C)(45°C) = 225 mW Therefore, the PD(max) at 70°C is 1W - 225 mW = 775 mW Related Problem A transistor has a Pp(max) = 5 W at 25°C. The derating factor is 10 mW/°C. What is the PD(max) at 70°C?
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