1 answer

Fundamentals of Electronic Engineering Answer should be step by step and readable (clear handwriting) Given: •...

Question:

Fundamentals of Electronic Engineering

Given: • Power voltage: • Collector resistor: • Current amplification of BJT: • Source peak voltage: • Input/Output resistanc

Answer should be step by step and readable (clear handwriting)

Given: • Power voltage: • Collector resistor: • Current amplification of BJT: • Source peak voltage: • Input/Output resistance: • The corner frequency: Vcc = 12V; Rc = 4k22; B = 100; Vs = 10mV; Rvs, RL = 1 ks2; f-3dB = 100Hz. Vec=12V w • Rc www R1 HE Сс CC2 RL + R2 RE CE Find: ? Bias quiescent point: Load line, Q-point, VcEQ, Ico ? Emitter resistance and emitter capacitor: RE, CE ? Bias voltage divide circuit: R1, R2 ? Calculate coupling capacitors Cci, Cca (Capacitor values may be assumed, but assumption must be explained) ? Simulate calculated circuit in CircuitJS. What will be the voltage gain on the output?

Answers

Answer: Given that Vec 12U power Voltage collector resitor Re current amplifiation of BIT B = 100; source peak voltage V = lo26 mu VG Ic 17.33 1.5mA da pre - 1733 0 In general ve is 107 of the supply voltage VE = lot. vce = lot of 12 1.2 V VCE = Ve-VR, R, YU Bib Rc Rus (R, R) + = 1kn R13 110 RB X RB +r RB ART = Ika-RB + 12 T RB (7-1km) = 1knan I kn-ro - 8.36 + lkn dc cur1 V = VBE IERE VE + ve 07 +1.2 R₂ = 1.90 Re 19 = 0.15833 6.RO 12 =) R = 0.15933 R, +0 15333 $4, = 0-15933 R, >> - 5.315 = 5.3Capacitor Ce should abe of proximately short above Cuf- of frequency (1000 Hz ) REN looXCE RE 100 21fc. CE > RE 100 2T (CooleRis Roll = lkn Кsert ка) а 39 sen F+ Кs - 11 Р. 3 Чto- 11 1л zooka ва 4 Ikace, 32. Sayzoo xit gоо усе, 6 1 1274 + 1 - 2 х

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