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Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter21: Resistive-capacitive Series Circuits
Section: Chapter Questions
Problem 6RQ: A 15-F AC capacitor is connected in series with a 50 resistor. The capacitor has a voltage rating...
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Ques: AC input given: a) b) 210KN ERB C₁ 11 joelf Vcc= +12V For DC analysis- AC supply is grounded and capacitors are open X₁ = √ C 1 [Fox DC, W=0] WC X₁ = x circuit becomes- bc=2.2k52 C₂ It- 10.up JB 840k2 솥 Re quec=+121 IB B + 1.Je 3/R₁₁=2₁2K1 + VBE E AC' ont put UCE JE Applying KUL in input loop- -12 +240 IB + VBE = O UBE = 0,7V 12-0.7 MA 240 IB = 0.047 mMA Applying KUL in output loop- === (1+p) — B VCE = 12 - 2₁22c Je = BIB Je = so X 0,047 हु-2354mA IE= SIX 0.047 JES 2.401 MA pz 50 ; ; VCE = 12—2.2x2.354 |UCES = 6.82 V DC operating characterstics are (VCE₁1 ICS) (UCERI 9 ) 2 (6.821,2.354MA) circuited.
Transcribed Image Text:Ques: AC input given: a) b) 210KN ERB C₁ 11 joelf Vcc= +12V For DC analysis- AC supply is grounded and capacitors are open X₁ = √ C 1 [Fox DC, W=0] WC X₁ = x circuit becomes- bc=2.2k52 C₂ It- 10.up JB 840k2 솥 Re quec=+121 IB B + 1.Je 3/R₁₁=2₁2K1 + VBE E AC' ont put UCE JE Applying KUL in input loop- -12 +240 IB + VBE = O UBE = 0,7V 12-0.7 MA 240 IB = 0.047 mMA Applying KUL in output loop- === (1+p) — B VCE = 12 - 2₁22c Je = BIB Je = so X 0,047 हु-2354mA IE= SIX 0.047 JES 2.401 MA pz 50 ; ; VCE = 12—2.2x2.354 |UCES = 6.82 V DC operating characterstics are (VCE₁1 ICS) (UCERI 9 ) 2 (6.821,2.354MA) circuited.
d) 00 (1 V BE = 0.7 iii) VB = VE = 0,7 VB = 0,7 1 |VBC= Vc = 12-2.27c. Vc = 12-22 x 2,354 | V₁ = 6.88121 Vecz Vgvc VE=0 - 0.7-6.8212 6.1212 V Now Method to find out region of operation of BJT- active region and calculate Assume BJY to be active JeQ, & VCE. If VCE 20.2V ΟὝ VCE 2 -ve value } Saturation regions Saturation region, because CB junction becomes RB when VCE-=-ve value & EB function also FB when we assume вут to be in active region. If VCE 2 VCC xatoff region. was
Transcribed Image Text:d) 00 (1 V BE = 0.7 iii) VB = VE = 0,7 VB = 0,7 1 |VBC= Vc = 12-2.27c. Vc = 12-22 x 2,354 | V₁ = 6.88121 Vecz Vgvc VE=0 - 0.7-6.8212 6.1212 V Now Method to find out region of operation of BJT- active region and calculate Assume BJY to be active JeQ, & VCE. If VCE 20.2V ΟὝ VCE 2 -ve value } Saturation regions Saturation region, because CB junction becomes RB when VCE-=-ve value & EB function also FB when we assume вут to be in active region. If VCE 2 VCC xatoff region. was
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