1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find the state of Transistor

  1. Dec 27, 2017 #1
    1. The problem statement, all variables and given/known data
    Example1.jpg

    2. Relevant equations


    3. The attempt at a solution

    This is a solved example in my Physics textbook . I don't understand how this problem can be done without specifying the battery VBB in the Base Emitter loop .

    The author assumes that the 20 V battery is connected between both Base-Emitter as well as Collector Emitter ?

    Instead it should be VBE = VBB - IBRB ?

    VBB is the biasing battery in Base Emitter loop .

    Please help me understand this .
     
  2. jcsd
  3. Dec 27, 2017 #2

    cnh1995

    User Avatar
    Homework Helper
    Gold Member

    Here only one battery is used.
    What exactly is your confusion?
     
  4. Dec 27, 2017 #3
    How is 20V battery connected between Base and Emitter ?

    How does 20 V battery come in picture while writing KVL in the Base Emitter loop ?

    In fact where is the loop to write KVL ?
     
  5. Dec 27, 2017 #4

    cnh1995

    User Avatar
    Homework Helper
    Gold Member

    Like this:
    20171227_192409.jpg
     
  6. Dec 27, 2017 #5
    I am talking about Base-Emitter not Collector-Emitter .
     
  7. Dec 27, 2017 #6

    cnh1995

    User Avatar
    Homework Helper
    Gold Member

    How many paths from +ve terminal to -ve terminal do you see here? The battery can send current through BE junction. In fact, many of the practical transistor circuits use single power source.
     
  8. Dec 27, 2017 #7
    But not in the way the book seems to apply KVL :smile:

    Could you draw the loop which the book assumes so as to apply KVL in Base Emitter loop ?

    Please show me how is VBE = 20 - IBRB ?
     
  9. Dec 27, 2017 #8

    cnh1995

    User Avatar
    Homework Helper
    Gold Member

    Let's call the +ve and -ve terminals of the battery P and G respectively.
    Starting from P, you can see there's a path
    P→RB→Base→Emitter→G.

    Apply KVL to this path and see.
     
  10. Dec 27, 2017 #9
    You are right :smile:
     
  11. Dec 27, 2017 #10

    gneill

    User Avatar

    Staff: Mentor

    Very strange analysis given in yourbook for this example problem. Giving a circuit with all the part values including the transistor beta, then imposing a particular collector current is not right. Consider that a typical npn silicon transistor is unlikely to survive an 18.5 V base-emitter voltage! It's essentially a forward biased diode, and the resulting current would be huge.

    Instead, I would have started with an approximation for the base current, assuming forward bias:

    ##I_B = \frac{(20 - 0.7)\; V}{120\; kΩ} \approx 160 \; μA##

    Then found that if the transistor was operating in its linear region that this base current, via the transistor's β, would want to produce an ##I_C## that is closer to 32 mA rather than 2.5 mA . Clearly that can't happen because 32 mA flowing through the collector resistor would produce a potential drop of about 160 V, and there's only 20 V to work with. From this you should be able to draw your own conclusions about what state the transistor will be in, and calculate an approximate value for the actual collector current.
     
  12. Dec 28, 2017 #11
    Thanks gneill and cnh1995 .
     
  13. Dec 29, 2017 #12

    CWatters

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I agree with gneill, the circuit and the question are inconsistent with each other.

    If you just look at the circuit and ignore the problem statement you get a base current of about (20-0.7)/120*10^3=0.16mA. The transistor will try to achieve collector current of 200 times that or 32mA. However the collector resistor limits it to about (20-Vcesat)/5000 ~ 4mA. So according to the circuit the transistor is fully on/saturated.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...