A common emitter BJT circuit is designed as shown in Fig. 4(a). The output load line and defined Q-point of the circuit are shown in Fig. Determine the required values of Vcc, Rc and RB. .Assume that VBE (on)=0.7 V 12 e(ma) 10 las = 30 µA le= 20 uA le=10 pA 8. Opoint 6. ww VCE 4. VBE VaV) =10 V 10 15 20 25

A common emitter BJT circuit is designed as shown in Fig. 4(a). The output load line and defined Q-point of the circuit are shown in Fig. Determine the required values of Vcc, Rc and RB. .Assume that VBE (on)=0.7 V 12 e(ma) 10 las = 30 µA le= 20 uA le=10 pA 8. Opoint 6. ww VCE 4. VBE VaV) =10 V 10 15 20 25

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Q4. The BJT in the common-base (CB) amplifier in Figure 4 has +12 Figure 4 s kn B=100, VE = 0.7 V. in the active mode. The coupling capacitors are very large. The Ve output resistance r, of the transistor can be neglected. 20 k2 (a) For I = 1.2 mA, calculate the bias values of the collector voltage Vc and the emitter voltage VE ( 0.5 k2 VE (b) Calculate the small-signal parameters ( gm, ", ) and draw the small-signal equivalent cireuit of the amplifier. V sig (10 (c) Find the overall voltage gain G, defined as G, V. sig
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On a VCE - Ic graph: a) Draw a load line for a BJT transistor, and show the change of Q point with three different IB values. (IB1< IB2 < IB3) (Qı , Q2 , Q3). Explain the reason for the changes shortly. 4) b) Draw the load line and show the change of Q point with three different Rc (Rci< Rc2< Rc3) values (IB constant). Explain the reason for the changes shortly.
For the circuit given below, suppose that RB = 230k2, RE1 = 4002, RE2 = 02, Rc = 1002 and Vcc = +6V. Given that the BJT's have B=100, verify that the transistors are in active region by finding icı, VCE1, İc2 and VCE2. Active region: VBe = 0.7V Q+Vc RB Rc Q1 Q2 Re1 RE2
In common base configuration of BJT, if it is working in the active region then the collector current will be approximately equal to the emitter current. True False A Moving to another question will save this response. Activa Go to Se
Consider the circuit depicted in the figure shown below. This circuit is designed using a silicon bipolar junction transistor (BJT) with a forward current gain BF = 100, and two resistances RB = 1kN and Rc = 1kN. The supply voltage is Vcc = 10 volts. The currents Ig and Ic designate the base and collector currents, respectively, whereas the voltage VCẸ denotes the voltage between collector and emitter. Vcc = 10 volts Rc =1 k2 Rg = 1 k2 VCE GND Throughout this question, you will use VBE,on = 0.7 volt and VCE,sat = 0.2 volt. Recall that VBE.on is the threshold voltage of the base-emitter junction, while VCE.sat represents the voltage that is measured between collector and emitter when the BJT is saturated. a) What is the value of the base current In? Please enter the value expressed in milliamperes (mA) in the box below. For instance, if you think that IR = 0.1 A, enter 100 in the box. If you think that Ip = 10 mA, enter 10 in the box. mas-numbas-Iti.ncl.ac.uk Submit part Unanswered b)…
29 - Which of the following is true?A) In order for the BJT transistor to operate in the active region, the correct polarization must be between the Beyz-Collector.B) BJT transistors are 2 types of N-channel and P-channel.C) For the BJT transistor to operate in the active region, the reverse polarity between the Beyz-Emitter must be.D) For the BJT transistor to work in the cut-off area, the correct polarization between the Beyz-Emitter must be.E) For the BJT transistor to operate in the saturation region, the correct polarization must be between the Beyz-Collector.
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Find the emitter current and VCE in the following circuit constructed with an npn BJT having a ß of 100. Assume that VBE=0.7 V if the junction is forward biased and VCESAT=0.2 V.Ignore the early effect.
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Consider a common emitter amplifier with the following characteristics: Rc= 2200ohms, Vcq=14V, minimum collector voltage min node voltage at collector :Vc(min)= 3.2 volts Vcc = 24. The positive peak of the output voltage occurs when the input voltage is at its negative peak (enter "min") or positive peak(enter "max") ? Enter the maximum output voltage swings values max swing upward Volts, max swing downward Volts (answers to the tenths place) Find the max collector current: mA (ROUND to the TENTHS place) NO UNITS.