Next Report - Application circuits for BJTs

Discussion in 'Homework Help' started by PsySc0rpi0n, Apr 23, 2014.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hi...

    I need to start the theoretical part of my next report about the application circuits for BJTs.

    I have the attached schematics, which is (or pretend to be) a NAND port, to simulate, but I have some questions.

    The teacher asks to plot Vγ=f(Vx) when 1V<= Vx<=5V. He also says that we should consider the following voltages as the limit voltages to the logic values:

    4V <= Level 1 <= 5V and 0V <= Level 0 <= 1V

    My questions are:

    What should I place in A and B of the schematics?
    What is and where should I measure it in the schematics?
    How do I control the Vx voltage between 1V and 5V???
     
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  2. ericgibbs

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    Jan 29, 2010
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    hi,
    For starters, look up 'bv' in LTS, Behavioral Voltage Source..

    Vy =F(Vx) of the 'bv' source.

    Also a hint, create some method of controlling Inputs A and B on your Sim.
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    You mean search in LTSpice help for Behavioral Voltage Source?????

    About controlling the inputs A and B I was trying to limit their values to 0V~1V and 4V~5V with some directive associated to 2 voltage sources, one for each input! But I don't know how to do it!
     
    Last edited: Apr 23, 2014
  4. ericgibbs

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    Hi,
    When you have your circuit displayed, press F2, look thru the list of components.

    You should see 'bv' , place that on your circuit, you then can define the behaviour of that 'bv' by using a standard voltage source.. Vy= f(Vx)

    Where Vy is the 'bv' source and Vx is the standard voltage source or if you know the values for the Vx values for the f(vx), write them as the conditions within the Vy = conditions.

    Make an attempt yourself and post your sim.
     
  5. ericgibbs

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    Jan 29, 2010
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    hi Psy,
    Unless I am missing something, reading thru your posted question,Post #1, it does not make any sense.?
    Its either incorrect or incomplete.

    Would you please recheck with your tutor.?
    E
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Well, I need to restart this simulation... I misread the teacher question.

    Ok, starting over again.

    The teacher is asking the same as before but only for the right side of the circuit, meaning that there are no diodes... Circuit is attached!
     
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  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Humm, ok... I'll try that when I have the need!
     
  8. ericgibbs

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    hi,
    Thats the same circuit you posted originally, all diodes connected etc.??:confused:

    Its important that you get the questions and circuits correct, otherwise I am wasting my time in trying to figure out what you are asking.

    Before what.? I have no point of reference.:rolleyes:
     
    Last edited: Apr 23, 2014
  9. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    EricGibbs, my fault again.

    I thought I have saved the correct .asc file but it turned out to be the same as in the beginning... I'm really sorry...

    I'm going to post the question and the circuit from the beginning again. Hope I get it right this time!

    Teacher question:

    Simulate the sub-circuit from the right (from X point to the right side) and plot the characteristic curve of Vγ=F(Vx) when Vx goes from 0V up to 5V.
    Also consider the following voltages as the limit voltages to the logic values:

    4V <= Level 1 <= 5V and 0V <= Level 0 <= 1V

    My questions are:

    How do I setup V1 so that it goes up from 0V up to 5V? Should it be with a DC sweep or with a directive like ".step param value list 0 5"?????

    What should I put in X-axis and Y-axis to get something like the image attached?
     
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  10. Jony130

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    Try this LTspice file. And my result look like this
     
  11. ericgibbs

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    OK,
    Consider your V1 source, Right click it, then use the PWL option.

    set t=0 Vx=0, t=1 Vx=5.

    Xaxis = v(x)

    Use .tran = 1

    Try that and see what the sim plot shows.
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hum, ok...

    It looks good now!


    I got this and it is similar to what we have in our notes. I have just changed the X-axis to the Vx.
     
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  13. ericgibbs

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    You should try changing the hFE [BF] and the 100K and see the effect on the plot.
     
  14. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Actually I changed the hFE but not yet the Rb...

    Changing hFE to lower values makes the transition, from level 1 to level 0, occur later or at higher Vx voltages.
     
  15. PsySc0rpi0n

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    Mar 4, 2014
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    Ok, next question is to calculate the max value for Vx to keep the circuit (Vc) at level 1 attending that 4V <= Level 1 <= 5V!

    What I have done was using the equation of the "output net" (I can't remember how it says in English) Vcc = RcIc + Vce and calculated Ic.

    5 = 2.2Ic + 4 <=> Ic = 455μA

    Then

    Ic = β.Ib <=> Ib = 1.77μA

    Then I have used the "input net" equation to calculate Vx

    Vx = RbIb + Vbe --> Is this correct?

    Vx = 100*1.77 + 0.7 <=> Vx = 877mV

    What you think of my calculations???
     
    Last edited: Apr 24, 2014
  16. ericgibbs

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    Hi,
    I have just typed in a reply and its just vanished.!!
    E
    Will retype
     
  17. ericgibbs

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    Hi,
    Given that Vc limits are 4V to 5V, when using a 5Vsupply, I would choose for Vc 4.5V.

    So you have Vc= 4.5V , you are give Rc as 2K2, calc the Ic value.

    The Beta value is 257, so calc Ib.

    I would not use a Vbe of 0.7Vbe for such a low Ic value, more like 0.6V

    Your equations look OK to me,

    To check your answers why not use LTSpice, change the V1 source to a current source of 1.77uA.
    E
     
  18. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I need to clarify some points of your reply...

    When you choose 4.5V is to have some margin, 0.5V in this case, to ensure the Level 1 state? I have chosen the limit which is 4V.

    Then you say Rc = 2k2? What means the number "2" after the K?

    A friend of mine was saying that Vx is Vbe! He is right?

    So, if I consider 0.6V for Vbe and 4.5V for the Level 1 state limit I have:

    5 = 2.2Ic + 4.5 <=> Ic = 227uA

    Ib = Ic/β <=> Ib = 227/256.7 <=> Ib = 884nA

    Vx = RbIb + Vbe <=> Vx = 100k * 884nA + 0.6 <=> Vx = 688mV

    But I can't make the simulation... Anything comes up in the plot window but vertical line at 688mV at X axis and from 19mV up to 27mV at Y axis! Probably I'm not doing this simulation he right way!
     
    Last edited: Apr 24, 2014
  19. ericgibbs

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    The 'K' in the 2K2 replaces the decimal point '.', I will post a web link explaining.

    Is your friends Rb value = 0 ohms,:rolleyes:

    You have a Ib current of 1.77uA thru a 100K resistor, how could this happen if Vx = Vbe.??

    EDIT:

    A link explaining the resistor M,K,R codes.
    http://www.google.co.uk/url?sa=t&rc...zB3rOhO3ZC9Q3BA&bvm=bv.65397613,d.ZGU&cad=rja
     
    Last edited: Apr 24, 2014
  20. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ah ok I got the K meaning... Don't need the link explaining!
    Thanks anyway!

    What am I doing wrong in my simulation?
     
    Last edited: Apr 24, 2014
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