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Determining BJT's voltages

  1. Jun 18, 2014 #1
    Hi everyone,


    I'm new to transistors analyzing, and I need a way to correctly identify the voltages.

    Let's take this circuit for example:

    20140618_112601.jpg

    Assuming that the Emitter is the common element (The one with arrow which is connected to the ground), then VBE would be:

    22222.jpg

    My questions are:

    1- If I want to know the common element of the transistor, shall I look for the element that is directly connected to the ground?
    2- For this circuit, I've VBE - VCE + VCB =0, On which basis I can get the these voltages?
     
  2. jcsd
  3. Jun 18, 2014 #2

    Drakkith

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    Staff: Mentor

    Is the "12" at the top right the voltage applied to the circuit? Or is that something else?
     
  4. Jun 18, 2014 #3

    phyzguy

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    You analyze it like this:

    (1) The BE junction is forward biased, so Vbe is about 0.7 V.
    (2) The voltage across the base resistor is then 12.0 - 0.7 = 11.3 V.
    (3) The current through the base resistor Ib = 11.3 / 376 K = .03 mA = 30 uA.
    (4) The colector current Ic is beta times this. Assuming beta = 100, Ic = 3 mA.
    (5) Vc = 12 - (3mA * 2 k) = 6.0 V.

    Does this make sense?
     
  5. Jun 18, 2014 #4
    Drakkith:
    Yes, it's applied on the circuit.

    phyzguy:
    That makes sense. Now I want to determine Vcb, which is Vc - Vb. Now I've Vc, is Vb the same as Vbe (0.7)? And what's the operating point of the transistor?
     
  6. Jun 18, 2014 #5

    phyzguy

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    Since you now know the voltages on all three terminals (Ve, Vb, and Vc) calculating the differences is easy. Vbe = Vb - Ve, Vce = Vc-Ve, etc. What do you think the operating point is? What's the definition of operating point? Are the BE and CE junctions forward biased or reverse biased?
     
  7. Jun 18, 2014 #6
    Actually, that what I'm asking for, the definition of the operating point. As far as I know, the operating point should be one of four regions: 1- Active (forward biased diode and reversed biased one) 2- Cut off (two reversed biased diodes) - 3- Saturation (two forward biased diodes) 4- Inverted (inverse of active).

    I know that BE resamples forward biased diode, but I'm not sure about CE
     
  8. Jun 18, 2014 #7

    NascentOxygen

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    Staff: Mentor

    In your second pic, those red + and ─ signs are wrong way 'round.
     
  9. Jun 18, 2014 #8
    NascentOxygen

    Thanks :)
     
  10. Jun 18, 2014 #9

    phyzguy

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    Well, what is Vc? What is Vb? Is the CB junction therefore forward biased or reverse biased? Given that it is an NPN transistor, you should be able to figure this out.
     
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