2. Using the IV plots shown in Fig. 3 (and found in the reintroduction to PSpice) design a BJTbiasing circuit that results in the following parameters: VCE = 2 Vand ig = 40 μA. Wealso require the power supply to be fixed at 5 Volts (this is where the load line interceptsthe iB =ic = 0 line). You may use the circuit shown in Example 1. Note that all resistorvalues in Example 1 must be recalculated. Your solution for the base to ground and baseto collector resistors may not be unique. 16mA14mAB = 80 UA12mA10mA8mA6mAI₁ = 70 uAB = 60 UA1=50 UAI₁ = 40 UAB=30 UA4mA-2mA-OmAIB = 20 UAIB = 10 UA=0uA-2OV1 2 3 4 5 6 7 8 9Figure 3. The IV characteristics ("Curve Trace") of a 2N3904 NPN BJT. The plot shows theoutput from a PSpice simulation of the BJT.10V

Answered: Image /qna-images/answer/f5af4b5c-dbbc-4914-b4c8-7405ad42e187.jpg

Answered: 2. Using the IV plots shown in Fig. 3…

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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RB= 430 kilo ohm
Discussion 1. Calculate the theoretical DC voltages and currents for the transistor bias circuit and compare them with the practically measured values. Table-1: Measured Quantities for the DC Bias Circuit Parameter B VB VE Vc Icọ VCEQ VBEQ Te Value 65.63 1.20 0.58 7.32 1418mA 6.74 0,62 18.3356 ECTR Vcc +12V R5 R3 3.3kQ всз37 R2 R4 4.7kQ 4700 Figure 5: The DC Bias Circuit of the Common Emitter Amplifier
(a) What is the output current IO in the circuitshown if −VEE = −10 V andR = 20 ohm? Assume that the MOSFET is saturated.(b) What is the minimum voltage VDDneeded to saturate the MOSFET if VTN = 2.5 Vand K'n = 0.25 A/V2. (c) What must be the powerdissipation ratings of resistor R and the FET.
choose the correct answer 1-The output voltage of an emitter follower isa-In phase with Vin . b-Much greater than Vin . c-180° out of phase.d-Generally much less than Vin .2-The ac emitter resistance of an emitter followera-Equal the dc emitter resistance .b-Is larger than the load resistance .c-Is β times smaller than the load resistance. d-Is usually less than the load resistance.3-A common-base amplifier can be used when a-Matching low to high impedance.b-A voltage gain without a current gain is required. c-A high- frequency amplifier is needed.d-All of the above.
7. Bipolar Junction Transistor (BJT) is a solid-state current-controlled device which can be used to electronically switch a circuit. You can think of it as your normal fan or light switch, but instead of you turning it on manually it can be controlled electronically. a) Explain why an ordinary junction transistor is called bipolar? b) Explain why silicon type transistors are more often used than germanium type? Given a transistor with B =100, determine c) the region of operation if: i. IB= 50µA and /c= 3mA IB= 50µA and VCE= 5V VBE= -2V and VCE= -1V ii. ii.
(a) Determine IC and VCE for the circuit (a) given below.(b) Change β to 120 (50% increase), and determine the new values of IC and VCE for the circuit of Figure (a)(c) Determine the magnitude of the percent change in IC and VCE using the following equations:(d) Determine IC and VCE for the network of Figure (b).(e) Change β to 180 and determine the new value of IC and VCE for the network of Figure (b)(f) Determine the magnitude of the percent change in IC and VCE using the above equations. (g) In each of the above, the magnitude of β was increased 50%. Compare the percent change in IC and VCE for each configuration, and comment on which seems to be less sensitive to changes in β.
9.→ The figure below shows the drain characteristics for a 2N4416 JFET transistor as displayed on a curve tracer. Determine Vp for the characteristics of the figure using IDSs and ID at some value of VGs. That is, simply substitute into Shockley's equation and solve for Vp. Compare the result to the measured value of -3 V from the characteristics. ↓ DSS=9mA DSS = 4.5 mA 2 (VGS = -0.9 V) 1 mA div 22 1 V div VGS = 0V Vertical Sens. 1 mA per div. VGS = -0.5 V Horizontal Sens. 1 V per div. -VGS = -1 V 500 mV per step. &m 2 m per div. VGS = -1.5 V VGS =-2 V Vp-3V

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