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I want to explain how the result becomes (735.1) Hz) and what are the steps and explain the reasons? Q6 The FET shown in Fig. 1.43 has gm = 3.4mS and ra =100 K. Find the approximate lower cutoff frequency. Ans: 735.1 Hz. 25V 2ΚΩ 1.5ΜΩ 0.02µF 0.02µF 20 ΚΩ 330kQ 820 ΩΣ OpF Fig. 1.43 Circuit for Q6. 40ΚΩ

I want to explain how the result becomes (735.1) Hz) and what are the steps and explain the reasons? Q6 The FET shown in Fig. 1.43 has gm = 3.4mS and ra =100 K. Find the approximate lower cutoff frequency. Ans: 735.1 Hz. 25V 2ΚΩ 1.5ΜΩ 0.02µF 0.02µF 20 ΚΩ 330kQ 820 ΩΣ OpF Fig. 1.43 Circuit for Q6. 40ΚΩ

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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I want to explain how the result becomes (735.1) Hz) and what are the steps and explain the reasons? Q6 The FET shown in Fig. 1.43 has gm = 3.4mS and ra =100 K. Find the approximate lower cutoff frequency. Ans: 735.1 Hz. 25V 2ΚΩ 1.5ΜΩ 0.02µF 0.02µF 20 ΚΩ 330kQ 820 ΩΣ OpF Fig. 1.43 Circuit for Q6. 40ΚΩ
Transcribed Image Text:I want to explain how the result becomes (735.1) Hz) and what are the steps and explain the reasons? Q6 The FET shown in Fig. 1.43 has gm = 3.4mS and ra =100 K. Find the approximate lower cutoff frequency. Ans: 735.1 Hz. 25V 2ΚΩ 1.5ΜΩ 0.02µF 0.02µF 20 ΚΩ 330kQ 820 ΩΣ OpF Fig. 1.43 Circuit for Q6. 40ΚΩ
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