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An Exercise Question on BJT

  1. May 8, 2012 #1
    Hi, Everyone.

    I have an electronics I exams tomorrow, and I'm looking for help in solving the following exercise if you don't mind.


    The exercise tells that this circuit is to be fabricated with a transistor type whose β is in the range of 50-150. It asks to find the value of Rc, so that all fabricated circuits are guaranteed to work in active mode. It also asks to find the collector voltage range that the fabricated circuits may exhibit.

    The answer, as it's provided in the textbook, is Rc= 1.5KΩ and Vc= 0.3V to 4.8V.

    I have made a couple of attempts to solve it, but apparently I'm running out of time, so I need your help, experts!

    Thanks in advance.

    My attempt in solving the exercise
    Last edited: May 8, 2012
  2. jcsd
  3. May 8, 2012 #2


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    Gold Member
    2016 Award

    you need to show your own work and folks here will help you see where you are going wrong. We don't just feed you the answer. Read the forum rules.
  4. May 8, 2012 #3
    Sorry for not reading the forum rules. A picture of my pencil and paper attempt is in the link below:

    Last edited: May 8, 2012
  5. May 8, 2012 #4


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    How did your class define saturation mode and active mode for a BJT?
    The answer to the question will fall out of these definitions.
  6. May 8, 2012 #5
    Well, not the class, it's the textbook by Sedra&Smith. Talking about NPN BJT:

    Active mode: as long as VBC<= 0.4V, in other words, VCB>= - 0.4V.
    Saturation mode: as long as VBC>0.4V, in other words, VCB< - 0.4 V.

    In either cases, it's always assumed that VBE=0.7 V, just to simplify the calculations. Also, it's assumed that VCE=0.2 V in saturation mode.
    Last edited: May 8, 2012
  7. May 8, 2012 #6
    You have two pieces of information here that you need to juggle here. The first is the minimum collector voltage to stay within your definition of active mode: Vb is 0.7V and your definition of active mode requires that the drop from base to collector is less than 0.4V. That tells you [STRIKE]something about[/STRIKE] the minimum collector voltage. With that in mind, the second piece is the fact that the collector voltage is determined by how big the drop is over Rc. Use Ohm's law here for the worst case scenario (i.e. the biggest drop) to figure out the minimum value Rc must have.
    Last edited: May 8, 2012
  8. May 8, 2012 #7
    But to be able to use Ohm's law for the worst case scenario (the largest voltage drop across Rc), I must have the value of Ic. As shown in my pencil and paper attempt of solution (here), I used β=150, because it defines Ic in the worst case scenario since Ic=IB (β).

    Please check the solution that I made in the link, and tell me if it's correct or not that way. Also, how would I be able to get the maximum Vc value that the circuit may exhibit, which is specified as 4.8V in the solution? I tried to use the best case scenario, when β=50 (to get the minimum voltage drop across Rc which has become known as 1.5KΩ), and I did all the calculations to end up with a Vc value of 6.775 V.
    Last edited: May 8, 2012
  9. May 8, 2012 #8

    Let's go back to your first attempt. You were going in the right direction and the work you did was correct. You determined that the collector current was 2.15mA at a beta of 50. How can we use that collector current to find the voltage at the collector?
  10. May 9, 2012 #9
    I would be able to use the 2.15mA collector current to find the collector voltage, if I have the value of Rc. It's 10 - (2.15Rc), but Rc is a missing parameter, which has to be found, too.
    Last edited: May 9, 2012
  11. May 9, 2012 #10


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    Staff: Mentor

    You can find RC using Ohm's Law. RC carries a maximum of https://www.physicsforums.com/images/icons/icon5.gif [Broken] mA and has 10V at one end and how many volts at the other end of RC? https://www.physicsforums.com/images/icons/icon6.gif [Broken]
    Last edited by a moderator: May 6, 2017
  12. May 9, 2012 #11
    Well, it's ( (10 - Vc)/ 2.15mA). Now, Vc is missing. So?
  13. May 9, 2012 #12


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    Staff: Mentor

    I noticed I should not have used 2.15mA as the value of collector current here, so have edited my post.
  14. May 9, 2012 #13
    So, none of the attempts I made was going in the right direction. Is that it? Wouldn't be correct to take β=150 to get the highest possible value for Ic, and then assume that Vc=0.3V (the worst case scenario), and finally calculate Rc value?

    Rc= ((10-Vc) / β (IB)) --> ((10-0.3) / 150(0.043mA))= 1.5KΩ

    That was my second attempt, but is it correct this way? and what about the 4.8V for Vc (the highest value in the range of 0.3 to 4.8V)? how was it calculated?
  15. May 9, 2012 #14


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    Staff: Mentor

    That looks like the right idea.

    Now, keeping that as Rc, what is Vc when β is the lowest value expected?
  16. May 9, 2012 #15
    It's equal to ( 10 - ( β (IB) 1.5K)) --> ( 10 - ( 50 (0.043m) 1.5K)) = 6.775 V

    So, it's wrong, or maybe I did it wrong. What's the problem here?
  17. May 9, 2012 #16


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    Staff: Mentor

    That's right.
  18. May 9, 2012 #17
    This means that it's the textbook's value that's wrong. If this is the case, then thanks everybody for help.
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