Examcrazy Logo
HOME  SITEMAP CONTACT US LOGIN
HOME Engineering AIEEE GATE IES DRDO-SET BSNL-JTO
MBA in India CAT How to Prepare for Exams Technical Freshers Jobs
  Follow us|  twitter  Orkut  facebook
BJT Tutorials
   Bipolar Junction Transistor- NPN
   Bipolar Junction Transistor- PNP
   DC Analysis of BJT Circuits
   The BJT as a Signal Amplifier
   BJT Small-Signal Equivalent Circuit Models
   BJT Small-Signal Amplifier
   Graphical Analysis of a BJT Small-Signal Amplifier
   BJT Biasing - Current Mirror
   Common Emitter Amplifier
   Common Base Amplifier
   Common Collector (Emitter Follower) Amplifier
   BJT Internal Capacitances. High Frequency Circuit Model
Free Electronics Tutorials
   Diode Tutorials
   BJT Tutorials
   MOSFET Tutorials
   Electronics II Tutorials
   Applied Electromagnetics Tutorials
   Microwave Tutorials
GATE preparation tips
   GATE Books & How to prepare
   Objective Solving Tricks
   Other GATE links
   IES exam preparation
   All about DRDO-SET
More Engineering Links
   Directory of coaching Institutes
   Govt engg college rankings
   Private engg college rankings
   Admission notifications for Mtech/PhD
   All Engineering Colleges in India
DC Analysis of BJT Circuits

In this section we will consider a number of BJT circuits and perform the DC circuit analysis. For those circuits with an active mode BJT, we’ll assume that VBE = 0.7V (npn) or V EB = 0.7V (pnp). Example N12.1 (text example 5.4). Compute the node voltages and currents in the circuit below assuming ß = 100.

If the BJT is in the active mode, Vbe = 0.7V then

With Ic Iae = then

Consequently, using KVL

Vc=10-IeRc= 10-0.99*10-3>-4.7*103=5.3V Finally, using KCL IB+IC=IE,or IB=IE-IC=1-0.99=0.01mA
Now we’ll check to see if these values mean the BJT is in the active mode (as assumed).
• VCB = 5.3-4 =1.3 V. This is greater than zero, which means the CBJ is reversed biased.
• VBE =0.7 V. This is greater than zero, which means the EBJ is forward biased. Because the CBJ is reversed biased and the EBJ is forward biased, the BJT is operating in the active mode. Note that in the text, they show a technique for analyzing such circuits right on the circuit diagram in .

Example N12.2 (text example 5.5). Repeat the previous example but with VB =6 V. Assuming the BJT is operating in the active mode:

From the last calculation 2.57 C Vc = 2.57VÞ VCB = 3.43 V. Consequently, the BJT is not in the active mode because the CBJ is forward biased. A better assumption is the transistor is operating in the saturation mode. We’ll talk more about this later. For now, suffice it to say that in the saturation mode VCE|sat »0.2 V (see Section 5.3.4). Assuming this and reanalyzing the circuit:

Notice that

This ratio is often called forced ß . Observe that it’s not equal to 100, as this ratio would be if the transistor were operating in the active mode (see Section 5.3.4). Example N12.3 (text example 5.7). Compute the node voltages and currents in the circuit below assuming ß = 100. To begin, we’ll assume the pnp transistor is operating in the active mode.

Now check if the BJT is in the active mode:
• EBJ? Forward biased.
• CBJ? Reversed biased.
So the BJT is in the active mode, as originally assumed. Example N12.4 (text exercise D5.25). Determine the largest RC that can be used in the circuit below so that the BJT remains in the active mode. (This circuit is very similar to the one in the previous example.)

We’ll begin by assuming the BJT is operating in the active mode. In the active mode, the CBJ needs to be reversed biased. The lowest voltage across this junction for operation in the active mode is VCB=0 ÞVC=VB=0V
Therefore, by KVL
-10+R CIc=0
or

This value of RC and smaller is required for the BJT to operate in the active mode. Example N12.5 (text example 5.10). Determine the node voltages and currents in the circuit shown below. Assume the BJT is operating in the active mode with ß=100 .

First, we’ll use Thévenin’s theorem to simplify the base circuit

The Thévenin equivalent resistance and voltage are then

Using this Thévenin equivalent for the base circuit, the overall circuit is then

To find the emitter current, we’ll apply KVL over the loop shown giving 5=33.3*103.IB+0.7+3,000.IEThe quantity of interest is IB. With CB IC= Iß B=and IC =IaE for a BJT in the active mode, we find

Using this in the KVL equation
5-0.7=[33.3*103+3,000(ß+1)]Ib
With ß=100 then solving this equation we find

Next, by KCL

The node voltages are then

Lastly, let’s check if the BJT is operating in the active mode.
• BE B E VBE= VB- VE = 4.57 -3.87 =0.7 V This is 0.7 V originally assumed for a forward biased EBJ.
• VBC= VB -V C=4.57 -8.6=-4.03 V. This is less than zero, which means the CBJ is reversed biased. Therefore the BJT is operating in the active mode, as originally assumed.


Discuss about BJT here
   START NEW THREADS
Discussion Board for BJT
Discuss all your issues or difficulties on BJT
Thread / Thread Starter Last Post Replies Views
example of BJT small signal
I want do to this


Posted By :-
 socheat
Jul 1, 7:35:08 AM 0 8369
drdo previous papar
preparation for drdo


Posted By :-
 rinkumathur.1988
Oct 16, 2:00:04 PM 1 9971
FET
Its operation and manufacturing.


Posted By :-
 dharmendra608@yahoo.in
Apr 16, 6:54:20 PM 0 6455
diode
sssssssssssssssss


Posted By :-
 sunilkumarmishra88
Dec 2, 4:03:32 PM 0 6961

To start your new thread you must login here.
New user signup at ExamCrazy.com Exam Crazy
To reply/post a comment you need to login, Use your user name and password to login if you are already registered else register here

EXISTING USER LOGIN
(Members Login)
Username:
Password:
NEW USER REGISTERATION FORM
Login-Id
Email-ID
Password
Confirm-Password
Full-Name

  About us | Privacy Policy | Terms and Conditions | Contact us | Email: support@Examcrazy.com  
Copyright © 2014 Extreme Testing House, India. All rights reserved.  4285