Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Solution Manual Electric Circuit Analysis and Design ch 15, Exercises of Electronic Circuits Analysis

This solution manual was provided by Maria Geven at Assam University for Electrical Circuital Analysis course. It includes: Low, Frequency, Gain, Hysteresis, Width, Output, Switches, Duty, Cycle, Recovery, Time

Typology: Exercises

2011/2012

Uploaded on 07/06/2012

shazli
shazli 🇵🇰

5

(3)

16 documents

1 / 28

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
pf3
pf4
pf5
pf9
pfd
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c

Partial preview of the text

Download Solution Manual Electric Circuit Analysis and Design ch 15 and more Exercises Electronic Circuits Analysis in PDF only on Docsity!

Electronic Circuit Analysis and Design, 2! edition Solutions Manual Chapter 15 Exercise Solutions E15.1 L dpole |] = = 6.0 dB fas * aE 1 (2) "RC yit as RO = aap = he = 159 10 1 2xfsaa — 2e(10*) 4-pote |] = eee = 7.24 BB La C =0.01 uF > 2= 159k je 10 Then E1s.4 Cy, = 0.03546 uF t Cz = 0.01392 uF Reg = Fcc Cy = 0.002024 uF a 7 3 * or fol = gm = Te a2 10 1 1 If C = 10 pF = fc = 20 kHz jTl= = 4(£ 1+{% E1S.$ faa Ww CQ 30 [T| = 0.124 or [7] = 18.1 dB Low-frequency gdias T= — 7 = — 3 _ fol 100 % 10°} (5 x 10777) E152 fuss = Fee Da(12 x 10") =— = foan = 6.63 kHe faa = Teng > RC = © ase =— = “6 E156 RE = Seip 7 1B 10 \ Let C= 0.001 wF =i aF => £=2i8k0 fo= Re Then Ry = 294k Ro = 3.44 Ba = 1.22 9 Re = 8,31 kD IT| = 0.01 = L L 1 6 RkO= = = 6.13 « 10 tehv3 2e(18 x 103),78 Let @ = 0.001 pF = 1 nF Thea R= 6.13 KO so Ry = BR = 49 kD B1S.7 fos 1 _ 1 °F ar VERG | 2xv/A(10*)(100 x 10-"7) = fp 265 ki Ra = 298 = 29(10°) => Ry = 290 kQ £15.8 —_ -—. f= apne 7°" ih l = = eee OE 0.02 pF C= ae * Se Ry = 2K, = 2(10) > Rp = 20 ER