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1- Common Emitter Fixed Bias Amplifier Please -Use Equation Operators in Highlighted  Thank you

1- Common Emitter Fixed Bias Amplifier Please -Use Equation Operators in Highlighted  Thank you

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
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1- Common Emitter Fixed Bias Amplifier

Please -Use Equation Operators in Highlighted 

Thank you

 

For the network of Fig. 8.6: (a) Determine re. (b) Find Z; (with r = ∞ ). (c) Calculate Zo (with ro= ∞ N). (d) Determine A, (with ro= ∞ 2). (e) Find A, (with r = ∞ ). (f) Repeat parts (c) through (e) including r, = 50 kn in all calculations and compare results. Solution (a) DC analysis: (b) Bre IB = IE V₂c r₂ = (d) A₂ = -- (c) Zo Rc = 3 k Rc (e) Since RB A₁ = Vcc – VBE RB (B + 1)IB = (101) (24.04 μA) = 2.428 MA 26 mV IE (100) (10.71 2) = 1.071 k Z₁ = RB||Bre= 470 kn||1.071 kn = 1.069 k re A₁ = -- F 10 μF A₁ = B = 100 = As a check: 470 ΚΩ = 26 mV 2.428 MA BRBO (ro + Rc)(RB + Bre) A₁ = 3 ΚΩ 10.71 Ω 10ẞr (470 kn> 10.71 kn) 12 V 3 ΚΩ -Av HE 10 μF 12 V - 0.7 V 470 ΚΩ B = 100 r = 50 ΚΩ (f) _Z = r||Rc = 50 kΩ||3 ΚΩ = 2.83 kΩ vs. 3 ΚΩ roRc 2.83 ΚΩ re 10.71 Ω = -280.11 = = 10.71 Ω V₂ Zo = 24.04 μα = -264.24 vs. 280.11 Figure 8.6 Example 8.1. Z₁ -(-264.24)(1.069 km2) 3 ΚΩ Av Rc which differs slightly only due to the accuracy carried through the calculations. (100)(470 ΚΩ)(50 ΚΩ) (50 ΚΩ + 3 ΚΩ)(470 ΚΩ + 1.071 ΚΩ) = 94.13 vs. 100 8.2 Common-Emitter Fixed-Bias Configuration = 94.16 EXAMPLE 8.1 341
Transcribed Image Text:For the network of Fig. 8.6: (a) Determine re. (b) Find Z; (with r = ∞ ). (c) Calculate Zo (with ro= ∞ N). (d) Determine A, (with ro= ∞ 2). (e) Find A, (with r = ∞ ). (f) Repeat parts (c) through (e) including r, = 50 kn in all calculations and compare results. Solution (a) DC analysis: (b) Bre IB = IE V₂c r₂ = (d) A₂ = -- (c) Zo Rc = 3 k Rc (e) Since RB A₁ = Vcc – VBE RB (B + 1)IB = (101) (24.04 μA) = 2.428 MA 26 mV IE (100) (10.71 2) = 1.071 k Z₁ = RB||Bre= 470 kn||1.071 kn = 1.069 k re A₁ = -- F 10 μF A₁ = B = 100 = As a check: 470 ΚΩ = 26 mV 2.428 MA BRBO (ro + Rc)(RB + Bre) A₁ = 3 ΚΩ 10.71 Ω 10ẞr (470 kn> 10.71 kn) 12 V 3 ΚΩ -Av HE 10 μF 12 V - 0.7 V 470 ΚΩ B = 100 r = 50 ΚΩ (f) _Z = r||Rc = 50 kΩ||3 ΚΩ = 2.83 kΩ vs. 3 ΚΩ roRc 2.83 ΚΩ re 10.71 Ω = -280.11 = = 10.71 Ω V₂ Zo = 24.04 μα = -264.24 vs. 280.11 Figure 8.6 Example 8.1. Z₁ -(-264.24)(1.069 km2) 3 ΚΩ Av Rc which differs slightly only due to the accuracy carried through the calculations. (100)(470 ΚΩ)(50 ΚΩ) (50 ΚΩ + 3 ΚΩ)(470 ΚΩ + 1.071 ΚΩ) = 94.13 vs. 100 8.2 Common-Emitter Fixed-Bias Configuration = 94.16 EXAMPLE 8.1 341
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Why answer has been changed, before it was with Equation Operators.

now I see it like this in screenshot.

Step 2 Part (b) - Zi=RB||Brezi-470x103||1.071×103Zi=470×103×1.071×103470×103+1.071×103Zi=503.37×103471.071 Zi=1.06856×103 ≈ 1.069 Zi-1.069 x103 Zi=1.069 k Step 3 Part (f) - Zo=ro||RCZo=50x103 ||3x103Z0-50×103×3×10350×103+3×103Z0-150×10353Z0- 2.83x103 Zo= 2.83 knandAv=-ro||RCreAv=-2.83×10310.71Av=-264.239-264.24Av=-264.24
Transcribed Image Text:Step 2 Part (b) - Zi=RB||Brezi-470x103||1.071×103Zi=470×103×1.071×103470×103+1.071×103Zi=503.37×103471.071 Zi=1.06856×103 ≈ 1.069 Zi-1.069 x103 Zi=1.069 k Step 3 Part (f) - Zo=ro||RCZo=50x103 ||3x103Z0-50×103×3×10350×103+3×103Z0-150×10353Z0- 2.83x103 Zo= 2.83 knandAv=-ro||RCreAv=-2.83×10310.71Av=-264.239-264.24Av=-264.24
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