Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- b) For correct operation as an ac amplifier a npn BJT must be correctly DC biased to establish an appropriate Q-point, i.e a negative voltage VBC between the base and collector terminals and a positive voltage VBE between the base and emitter terminals. Figure 3.1 shows a circuit that can achieve this requirement. Explain the role of each resistive element in this circuit. Vin RB1 Cin HH iin RB2 + Rc VB RE + Vcc Vc Cout VE ww 11 Figure 3.1 th Vout Ce İoutarrow_forwardHand Calculation: Since you now have the values of R1 and R2 resistors, obtain numeric value of voltage gain Av=Vo/Vi, and numerical values of input and output resistances Ri and Ro. Note that you need to DC analysis first. Do a hand calculation and calculate the DC collector current or IC. This is needed in order to calculate the small signal parameters gm and rπ and in turn to obtain the Av , Ri and Ro Hence, first perform a DC analysis (use the circuit in Figure 1 and take all capacitors open) and assume your transistor has a β=140.arrow_forward2. A single transistor amplifier is shown in the circuit to the right. The input is a 100 kHz sine wave from a very low impedance source, with a 1 mV peak-to-peak amplitude. hfe, the forward current gain of the transistor, is 300. The base reverse leakage current is 1 nA. The ideality factor for the base-emitter diode is 2, so nkT= 50 mV. a. What are the voltages relative to ground and the currents flowing into, into, and out from the collector, the base, and the emitter, respectively? b. What is the impedance of the 1 µF capacitors at 100 kHz and what effect will this have on the gain of the amplifier? c. What is the gain of the amplifier and how does it depend on the hfe of the transistor? Luff Vin maits Ik { +V₁ = +15V املا Vo Sik Y 1 Fb V₂ I m Vout MF 2K = 250 hfő - -15V --Y₂arrow_forward
- Which of the following is the basic feature of the emitter-connected amplifier circuit? I. It has voltage gain II. It has current gain III. There is a 180 degree phase difference between the input signal and the output signalarrow_forward........ (Figure-1) R. RB= 380kN,Rc= 1kN B = 100, VBB = Vcc=12V RB ww Vec CC ......... I, V CE СЕ V ВЕ BB Q-1-b) Describe briefly the input / output characteristics and application of Common Emitter BJT Configurationarrow_forward
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