Instruction/s: 1. Draw, Illustrate and label your schematic diagram before solving the problem.4.) Given a Voltage -divider biased transistor circuit .Assume Beta DC is 100, ,Voltage at Base-emitter junction is 0.7v ,voltage at common collector is +15v, resistor RB1 is 10k ohms, resistor RB2 is 6k ohms, collector resistor is 1.5k ohms ,emitter resistor is 560 ohms. Determine base current ,collector current and voltage collector-emitter junction.These might help as a guide in answering the problem.. Voltage Dividervoltagedividerbiasingconfiguration is the most widely usedtransistor biasing method. The emitterThisVccRTh = RB = Rg1||RB2RB2Imdiode of the transistor is forward-Vrh = VB = VccReRcRB1 + RB2Vrh – VBEbiased by the voltage value developedacross resistor RB2. Also, voltagedivider network biasing makes theBoc oVoVINRthVoIB =RTh + (B + 1)RĘIc = BIBVCe = Vcc - Ic(Rc + RE)VBE Vtransistorcircuit independentofRwa RERchanges in beta as the biasing voltagesset at the transistors base, emitter, andcollector terminals are not dependenton external circuit values.GNDGND 3.Voltage DividerDetermine RTh, Vrh, IB,Ic, and Vce-Step 1: Solve for RTh and VTh:Step 2: Solve for Ig and Ic.RTh = RB1||RB239kN × 3.9kNVTh - VBERTh + (B + 1)RE2v – 0.7vO+22vIR =39kN + 3.9kNRrh = 3.55kNRB2IR =3.55kN + (100 + 1)1.5kN39kn10kNVTh = VccIB = 8.38µARB1 + RB23.9kNIc = BlB100(8.38µA)Ic = 0.84MAB = 100VoVIN O-= 22v%3D(39kN + 3.9kN,VTh = 2v3.9k1.5kNStep 3: Solve for VcE.VCEVcc – Ic(Rc + RẸ)GND=Vce= 22v – 0.84MA(10kN+ 1.5kN)VCE= 22v – 9.66v= 12.34v

Brief description : In the above given question they have mentioned a BJT. For the given voltage…

Answered: Instruction/s: 1. Draw, Illustrate and…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Instruction/s:
1. Draw, Illustrate and label your schematic diagram before solving the problem.

4.) Given a Voltage -divider biased transistor circuit .Assume Beta DC is 100, ,Voltage at Base-emitter junction is 0.7v ,voltage at common collector is +15v, resistor RB1 is 10k ohms, resistor RB2 is 6k ohms, collector resistor is 1.5k ohms ,emitter resistor is 560 ohms. Determine base current ,collector current and voltage collector-emitter junction.

These might help as a guide in answering the problem..

Voltage Divider
voltage
divider
biasing
configuration is the most widely used
transistor biasing method. The emitter
This
Vcc
RTh = RB = Rg1||RB2
RB2
Im
diode of the transistor is forward-
Vrh = VB = Vcc
Re
Rc
RB1 + RB2
Vrh – VBE
biased by the voltage value developed
across resistor RB2. Also, voltage
divider network biasing makes the
Boc oVo
VIN
Rth
Vo
IB =
RTh + (B + 1)RĘ
Ic = BIB
VCe = Vcc - Ic(Rc + RE)
VBE V
transistor
circuit independent
of
Rwa RE
R
changes in beta as the biasing voltages
set at the transistors base, emitter, and
collector terminals are not dependent
on external circuit values.
GND
GND

3.Voltage Divider
Determine RTh, Vrh, IB,
Ic, and Vce-
Step 1: Solve for RTh and VTh:
Step 2: Solve for Ig and Ic.
RTh = RB1||RB2
39kN × 3.9kN
VTh - VBE
RTh + (B + 1)RE
2v – 0.7v
O+22v
IR =
39kN + 3.9kN
Rrh = 3.55kN
RB2
IR =
3.55kN + (100 + 1)1.5kN
39kn
10kN
VTh = Vcc
IB = 8.38µA
RB1 + RB2
3.9kN
Ic = BlB
100(8.38µA)
Ic = 0.84MA
B = 100
Vo
VIN O-
= 22v
%3D
(39kN + 3.9kN,
VTh = 2v
3.9k
1.5kN
Step 3: Solve for VcE.
VCE
Vcc – Ic(Rc + RẸ)
GND=
Vce
= 22v – 0.84MA(10kN
+ 1.5kN)
VCE
= 22v – 9.66v
= 12.34v

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