Determine the DC parameters for this BJT circuit. Μ +20 V R₁ 150 ΚΩ R₂ 20 ΚΩ Rc · 12 ΚΩ hFE = 200 RE • 2.2 ΚΩ Μ RL · 50 ΚΩ
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Vbe=0.7v
find Vb, Ve,Ie,Ib,Icq,Vce
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- Assume that the voltage drop across the resistor, ER, is 78 V; the voltage drop across the capacitor, EC, is 104 V; and the circuit has a total impedance, Z, of 20 . The frequency of the AC voltage is 60 Hz. Find the missing values. ET ER78V EC104V IT IR IC Z20 R XC VA P VARSC PF CInductive Circuits Fill in all the missing values. Refer to the following formulas: XL=2fLL=XL2ff=XL2L Inductance (H) Frequency (Hz) Inductive Reactance ( ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841Avresistor of 51.0 Q, an inductor of 22.0 pH and a capacitor of 150 pF are in parallel. The frequency is 1.00 MHz. Whatis the complex impedance, R_jX? A.51.0_j14.9. B.51.0_/149 C.46.2_j14.9. D.46.2_j14.9. ¥ Show Answer Answer: D
- Given an input ac-voltage, v = 120 sin120πt, if the SCR Vbr is 100V, What is the fire delay angle? a. 35 degrees b. 56 degrees c. 67 degrees d. 83 degreesIn the figure, R = 13.0 Ω, C = 6.26 μF, and L = 50.0 mH, and the ideal battery has emf = 33.0 V. The switch is kept in position a for a long time and then thrown to position b. What are the (a) frequency and (b) current amplitude of the resulting oscillations? Image is attached, thank youGet the following parameters: RC RE RB VCE VB
- 2) Find the real and imaginary part of the total equivalent circuit impedance at the signal frequency of 1 MHz (Unit: Ohm)Choose the correct answer 1. Common emitter configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 2. Common base configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 3. Function generators and AC source in Multisim are a) Similar in nature b) Function generators can generate more waveforms while AC source is a square wave c) AC source is just a 220V, 50Hz wave d) Function generate can generate more waveforms while AC source is a asine wave 4. Common collector configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 5. How can you control the output voltage of a common emitter amplifier BJT a) By RB b) By RE c) By VCC d) By coupling capacitorjust need fınal answer Since Vi = 20 mV, Vcc = 25 V, RB = 470 kΩ, RC = 3.2 kΩ, RE1 = 470 Ω, RE2 = 1.2 kΩ, RL = 42 kΩ and β = 110 in the circuit in the figure, the value of the output voltage (Vo) find it. NOTE-1: Capacitors are negligible at mid-band frequency. NOTE-2: The output impedance of the transistor (r0) will be neglected. a. -197,96 mV b. -222,71 mV c. -98,98 mV d. -148,47 mV e. -123,73 mV f. -173,22 mV g. -247,45 mV h. -74,24 mV
- Since Vi = 20 mV, Vcc = 25 V, RB = 470 kΩ, RC = 3.2 kΩ, RE1 = 470 Ω, RE2 = 1.2 kΩ, RL = 42 kΩ and β = 110 in the circuit in the figure, the value of the output voltage (Vo) find it. NOTE-1: Capacitors are negligible at mid-band frequency. NOTE-2: The output impedance of the transistor (r0) will be neglected. a. -197,96 mV b. -222,71 mV c. -98,98 mV d. -148,47 mV e. -123,73 mV f. -173,22 mV g. -247,45 mV h. -74,24 mV10. The higher the frequency, the easier it becomes to analyze electronics circuits.a. Not all the timeb. Truec. Falsed. In some cases only11. The __ ___ leads ensure low inductance and permit high currents.a. Short, fatb. Long, fatc. Long, thind. Short, thin12. Most microwave amplifiers have input and output impedances of _a. 75Ωb. 50Ωc. 300Ωd. 120Ω Note: Explanation is not needed just provide the correct answers for each number.The three forces, each of magnitude F, are applied to the crate. Determine F so that the three forces are equivalent to a single 3000-N Ans. F= 1141N
