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11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show your work and record the values in Table 1. a. Frequency due to the input capacitor. The source resistance in this case is in series with Rin(tot) fel 2π(600+ fez 1 27 (Rs+(R₂||R₂||(Bac(r'e+RE₁))))C₁ fc3 = 1 68000 2000 94.6(1.38+3 b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. 1 2m (Rallr'e+RE+ 1 2n(R₂ +RL) C₂ (100-10-6) = 479.36 R₁||R₂||RS)C₂ Bac c. Frequency due to the output capacitor. Equation 3 = 4.57

11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show your work and record the values in Table 1. a. Frequency due to the input capacitor. The source resistance in this case is in series with Rin(tot) fel 2π(600+ fez 1 27 (Rs+(R₂||R₂||(Bac(r'e+RE₁))))C₁ fc3 = 1 68000 2000 94.6(1.38+3 b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. 1 2m (Rallr'e+RE+ 1 2n(R₂ +RL) C₂ (100-10-6) = 479.36 R₁||R₂||RS)C₂ Bac c. Frequency due to the output capacitor. Equation 3 = 4.57

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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Can you help me with these equations based on the given circuit?

11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show your work and record the values in Table 1. a. Frequency due to the input capacitor. The source resistance in this case is in series with Rin(tot) fel 2T (600+ fez = 2n(Rs+(R₂||R₂|| (Pac(r'e+RE1))))C₁ fc3 = 1 1 1 68000 22000 94.6(1.38+33). b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. 1 2π(R₂ +R₂) C₂ 2n(RE₂r'e+RE+R₂R₂ KS)C= 1 (100-10-6) Bac = 479.36 c. Frequency due to the output capacitor. Equation 3 = 4.57 1
Transcribed Image Text:11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show your work and record the values in Table 1. a. Frequency due to the input capacitor. The source resistance in this case is in series with Rin(tot) fel 2T (600+ fez = 2n(Rs+(R₂||R₂|| (Pac(r'e+RE1))))C₁ fc3 = 1 1 1 68000 22000 94.6(1.38+33). b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. 1 2π(R₂ +R₂) C₂ 2n(RE₂r'e+RE+R₂R₂ KS)C= 1 (100-10-6) Bac = 479.36 c. Frequency due to the output capacitor. Equation 3 = 4.57 1
Vs 10mVpk 5kHz 0° 2. R7 m 6000 Rs C1 DE 100nF R1 68kQ R2 Σ22kΩ VCC 15.0V R3 3.9KQ Rc Q1 2N3904 R4 330 RE1 R5 >1.5kΩ RE2 +₁ C3 DE 100uF C2 100uF R6 35.6kΩ RL H1₁
Transcribed Image Text:Vs 10mVpk 5kHz 0° 2. R7 m 6000 Rs C1 DE 100nF R1 68kQ R2 Σ22kΩ VCC 15.0V R3 3.9KQ Rc Q1 2N3904 R4 330 RE1 R5 >1.5kΩ RE2 +₁ C3 DE 100uF C2 100uF R6 35.6kΩ RL H1₁
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