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How BJT transistor works (with conventional current notation perspective)

  1. Nov 10, 2014 #1
    So far the explanation on BJT working principle always explained in electron flow perspective.
    I felt it's hard to relate with conventional current direction notation when the explanation explained in electron flow direction notation. For example in NPN transistor, the C-E current flow made me confused because there were depletion layer on collector-base area (especially if we substitute the transistor symbol with two diodes, there should be reverse bias on Collector-Base. But, current still manage to reach Emitter side Ie=Ic+Ib).
    Also, i feel off when it said transistor act as "amplifier" (amplifying base current). The highest current value in NPN transistor is on Emitter. Meanwhile, the common usage is to put Load Output on collector. Why collector? Why not emitter if the function is to amplify current which has the highest value for current?
    would it be easier to relate BJT transistors as a water faucet?
    Please enlighten me on this matter.
     
  2. jcsd
  3. Nov 10, 2014 #2
    Well I prefer a water analogy.
    The BJT work very similarly to the tap water valve.
    A water valve is always used as a device to control the flow of water. Similarly, always think of a bipolar transistor as a device used to control electric current flow by assistance of a base current. See this picture
    http://images.elektroda.net/94_1250754403.png
    If base current is flowing the BJT is ON and the collector current is BETA (β or Hfe) times larger than the base current (Ic = β*Ib).
    And Ic = β *Ib and Ie = Ib + Ic is a basic principle of a transistor "action".
    Of Course this "gain" current comes from power supply not from transistor them self. Transistor can only control the amount of collector current (Ic) that power supply supplies.
    cache.php.png

    35a.png

    The emitter current is always equal to Ib + Ic. But because the beta value is large we can ignore the base current and say that Ie = Ic.
    For example if β = 100 and Ib = 10μA we have Ic = 100 * 10μA = 1mA and Ie = Ib + Ic = 1.01mA
    So we don't do the big mistake/error if we ignore the Ib current and say that Ie = Ic = 1mA

    Sometime we use emitter as a output. But notice that the emitter voltage (output voltage) is always Vbe ≈ 0.6V lower the the base voltage (input voltage). See some examples:

    ef-png.50210.png

    As you can see the current gain is not everything, some times we need a voltage gain also. And in this case we must use collector as a output.
    Or if we want to use transistor as a switch it is better to use collector as a output because of a lower voltage drop across transistor.
     
    Last edited: Nov 10, 2014
  4. Nov 10, 2014 #3
    Hi Jony130, thank you for sharing your perspective and examples. It was very interesting :D
    This explanation is clear to me and the example pictures also great at explaining the function of Ib as controller for Ic.

    On this part, the example pictures showed that Ve keep getting bigger as Ib increasing, which i think that was a voltage gain on Emitter part. Wouldn't the voltage gain of Ve always bigger than output on Collector? Also, which voltage drop is lower? i still don't get this part
     
  5. Nov 10, 2014 #4
    But how can we say about the gain if emitter voltage is always Vbe lower then the base voltage ?
    In first fig Vb = 2V and Ve = 1.4V; in the middle Vb = 6V and Ve = 5.4V, and last fig Vb = 10.5V and Ve = 9.9V
    Why this voltage is lower? Well because base-emitter junction is nothing more than a ordinary diode. And Vbe is a forward voltage drop.
    Also notice that in this circuit the emitter voltage follow the base voltage minus Vbe voltage.
    And this is why we called this circuit emitter follower or voltage follower
    Please read this
    http://electronics.stackexchange.co...-voltage-on-base-limit-the-voltage-on-emitter

    Transistor without any resistor is nothing more then a base current controlled collector current source Ic = β * Ib .
    35a.png
    As you can see we have a "current gain", but with no voltage gain.
    What we can do about that ? Let as try to add Rc resistor and see what we get.
    Now we give Rc resistor very important task. His job will be to covert Ic current into voltage.
    Rc will act just like a current to voltage converter thanks to Ohms law Vrc = Ic*Rc

    35b.png
    Finally we have a voltage gain. Our amplified voltage is of-course taken between collector and emitter (Vce).
    Also notice that we can change voltage gain by changing Rc resistor value.
    The larger the Rc resistor value the larger the voltage gain. Why you may ask? Well because now we need smaller change in Ic current to get the same change in Vce. For example for Rc = 1k we need change in Ic current from 0A to 10mA to change Vce voltage from 10V to 0V. But now if we increase Rc resistor value from 1k to 2kΩ we need ΔIc = 5mA to get the same change in Vce.
    So we need smaller change in input voltage to get the same output voltage. So this means that our amplifier has larger voltage gain. I hope you understand this and the important role which lies on Rc resistor.
     
    Last edited: Nov 10, 2014
  6. Nov 11, 2014 #5
    Oh! sorry. I didn't see clearly on the picture that Ve=Vb-Vbe. My bad, sorry. Thank you for pointing this out :D

    Thank you very much for your explanation + pictures. They are very clear and easy to follow, especially the pictures :D. Now i prefer to look BJT transistors as electric current faucet rather than amplifier. So far the conclusion i manage to make are:
    1. BJT Transistors controls how much current drawn from Vcc.
    2. The Vce is the so-called transistor's output which looked like how much voltage drawn from Vcc (Vc=Vcc-Vrc).
    3. Emitter is only current residue (Ie=Ib+Ic) therefore, using Collector as output would be ideal since it is the original value of current (Ic) drawn from Vcc. I call it residue because Ie already mixed with Ib. Plus, Emitter's Voltage doesn't come from Vcc.
    And i agree with your main concept about water tap analogy. It is very much easier to relate BJT transistors that way :D
    Do i get it right? please correct me if i'm wrong
     
    Last edited: Nov 11, 2014
  7. Nov 11, 2014 #6
    Yes, but we usually treat BJT as a current controlled device (current controlled current source Ib = hfe * Ic) or as a voltage controlled device, a voltage controlled (Vbe in this case) current source (Collector current).
    The word "drawn" is not proper here, because voltage do not flow, only current can flow not the voltage.
    Vce = Voltage between collector and emitter is equal to:
    Vce = Vcc - VRc = Vcc - Ic*Rc (Rc resistor convert collector current into voltage)

    Do not disregard the emitter as a output. Because emitter follower is a very useful circuit.
     
  8. Nov 11, 2014 #7
    I prefer calling it current controlled device rather than voltage controlled device, since Vbe will stay the same (0,6V). Meanwhile Ic or Ie will change with regard of change in Ib.
    Yea, voltage doesn't flows. But, Vce value will never get bigger than Vcc value. Therefore, it "look like" Vce value generated with Vcc as reference. The value is controlled by Ic and Rc but it can't get bigger than Vcc (I'm imagining Vcc as a water tank). Still, i agree that fundamentally Voltage doesn't flow. Sorry for my bad wordings.
    Does Ve always Vb-Vbe? then do you mean Ve always follow Vb value?
    Could you give me example of emitter follower usefulness?
     
  9. Nov 12, 2014 #8
    Yes, always, because the base-emitter junctions form p-n junctions and p-n junction form a simple diode.

    Emitter voltage follows the base voltage (Ve = Vb - Vbe )

    Emitter follower is a very simply circuit. The output voltage is always 0.6V lower the the input voltage.
    See some examples
    Once again look at this examples
    ef.PNG
    Notice that the base current is (β+1) smaller then emitter current (load current).
    So our base current source B1 see our load ( Re resistor) not as 100Ω resistor. But B1 see (β+1)*Re load.
    Here you have anther example.
    12a.PNG
    Spouse we have a 1K resistors voltage divider supply from 10V battery.
    Without any load connect to the output terminal, voltage divider output voltage is equal 5V. Now if we connected a load resistor (100Ω) across the output terminal, our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.
    To fix this issue we add a buffer (emitter follower)
     
  10. Nov 12, 2014 #9
    That's some nice collection of example pictures you got there :D
    On the buffer circuit picture, how can we know the value of Vb=4.79V ??
    which one got determined first? Ve or Vb??
     
  11. Nov 12, 2014 #10
    I use a circuit theory and solve the circuit.
    a2.PNG

    From the II Kirchhoff's law we can write

    Vcc = I1*R1 + I2*R2 (1)

    I1 = Ib + I2 (2)

    I2*R2 = Vbe + Ie*Re (3)

    And Ib = Ie/(β+1) (4)

    And we can solve this for Ib.

    [tex]Ib = \frac{R2 Vcc - Vbe(R1+R2)}{(\beta+1) Re (R1+R2) +(R1*R2)}=419.047619\mu A[/tex]

    I assume β = 99 and Vbe = 0.6V

    Knowing Ib we can easy solve for Ie, Ve and Vb

    But there is also a simpler way to solve this circuit by using thevenin's theorem.
    We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit

    Vth = Vcc * R2/(R1+R2)

    Rth = R1||R2 = (R1*R2)/(R1+R2)


    2a.PNG
    And now we can solve for Ib

    Vth - Ib*Rth - Vbe - Ie*Re = 0


    And we also know that

    Ie = Ib*(β +1)

    Or

    Ib = Ie/(β +1)

    so we end up with

    Vth - Ie/(β +1)*Rth - Vbe - Ie*Re = 0

    [tex]\Large Ie = \frac{Vth - Vbe}{\frac{Rth}{\beta +1} +Re}[/tex]

    And this is the end.

    PS. Notice that Ie current is 41.9mA not 4.19mA
    Ve = 4.19V ----> Ie = 4.19V/100ohm = 41.9mA
    Sorry for stupid mistake
     
    Last edited: Nov 12, 2014
  12. Nov 13, 2014 #11
    Joney gave a great run down on this. Remember this is about the BJT being in the Active mode.

    BJT can be in Cutoff or Saturated as well as other modes. As you move forward, make sure you use proper model for mode it is in.
     
  13. Nov 16, 2014 #12
    This part kinda get me confused. This was my steps on breaking down this function:
    1. Vth - Ie/(β +1)*Rth - Vbe = Ie*Re
    2. (Vth - Vbe - Ie/(β +1)*Rth) / Re = Ie
    3. Vth/Re - Vbe/Re - (Ie/(β +1)*Rth*Re) = Ie
    4. (Vth - Vbe)/Re = Ie - Ie/(β +1)*Rth*Re

    and i stuck at this point. . . . can someone please point out my mistake?
     
  14. Nov 17, 2014 #13
    Hi, you made a error in step 3. But why you solving this equation is such a strange way ?

    Vth - Ie/(β +1)*Rth - Vbe - Ie*Re = 0

    Vth = Ie * Rth/(β +1) + Vbe + Ie*Re

    collect Ie terms

    Vth = Ie * (Rth/(β +1) + Re) + Vbe

    subtract Vbe

    Vth - Vbe = Ie * (Rth/(β +1) + Re)

    divide by (Rth/(β +1) + Re)

    (Vth - Vbe)/ (Rth/(β +1) + Re) = Ie

    And we done.
     
  15. Nov 17, 2014 #14
    Here is a pop quiz answer key by Dr. Ruben Kelly.

    Note you can get all of your equations by Inspection. This works on more difficult circuits as well including feedback.


    Old material, but still works well. EECS321 Pop quiz7 answer key.jpg

    Dr. Kelly was pretty picky on us kids using our calculators properly. Rounding numbers would make you fail class.

    Last time I looked his lecture notes were available on Amazon.
     
  16. Nov 17, 2014 #15
    Thank you so much Jony130 for explaining this in a very clear manner :D
    This picture is interesting, the Ibq function is same as previous equation explained by Jony130.
    Does it mean this Ibq function is a common knowledge that can be applied anywhere (w/o feedback)?
     
  17. Nov 18, 2014 #16
    Yes, for all BJT circuit which contain a voltage divider and Re resistor.

    Please try solve this circuit

    63.png

    For Vcc = 10V ; Rb =10K; Rc = 1K; Re = 100Ω ; and Vbe = 0.6V and Hfe = 100
     
  18. Nov 18, 2014 #17
    Ve = Vcc - Vbe
    Ve = 10V - 0.6V
    Ve = 9.4V

    Ie = Ve / Re
    Ie = 9.4V / 100Ω
    Ie = 94mA

    Ib = Ie / (Hfe+1)
    Ib = 94mA / (100+1)
    Ib = 0.93mA

    Ic = Ib * Hfe
    Ic = 0.93mA * 100
    Ic = 93mA

    Vc = Ic * Rc
    Vc = 93mA * 1KΩ
    Vc = 10V ?? (93V bigger than Vcc)

    Vce = Vc - Ve
    Vce = 10V - 9.4V
    Vce = 0.6V
     
  19. Nov 18, 2014 #18
    How can this be true ? Do you forget about base current and voltage drop across Rb resistor ?

    63.png

    And the KVL for this circuit look like this

    Vcc = VRb + Vbe + VRe

    Vcc = Ib*Rb + Vbe + Ie*Re
     
  20. Nov 19, 2014 #19
    WHOOPS!! what a mistake :))
    sorry for being amateur.
    When doing KVL with lot of resistors i don't have any problem but, when it comes to applying KVL/KCL with a real circuit, my mind goes blank. . . .

    Vcc = Ib*Rb + Vbe + Ie*Re
    10V = Ib*10KΩ + 0.6V + Ie*100Ω
    9.4V = Ib*10KΩ + Ib*(β+1)*100Ω
    9.4V = 10KΩ*Ib + 10.1KΩ*Ib
    9.4V / 20.1KΩ = Ib
    Ib = 0.467mA
    Ic =
    β*Ib = 100 * 0.467mA = 46.7mA
    Ie =
    Ic + Ib = 46.7mA + 0.46mA = 47.16mA

    Vb = Vcc - Ib*Rb
    Vb = 10V - 0.467mA*10KΩ
    Vb = 10V - 4.67V
    Vb = 5.33V

    Vc = Vcc - Ic.Rc
    Vc = 10V - 46.7mA*1KΩ
    Vc = 10V - 46.7V ???
    Vc = -36.7V??

    Ve = Vb - Vbe
    Ve = 5.33V - 0.6V
    Ve = 4.73V
    or
    Ve = Ie*Re
    Ve = 47.16mA*100Ω
    Ve = 4.72V (close enough :w)

    Vce = Vc - Ve
    Vce = -36.7V?? - 4.72V
    Vce = -41.42V???

    damn, i still having tough time doing this simple problem.
    do i get it right this time?
     
    Last edited: Nov 19, 2014
  21. Nov 19, 2014 #20
    Believe it or not but your solution is correct.
    Because what we have done first is that we assumed that the BJT works in active region, which is not always the case.
    And this is why this strange results (Vce = -41V) give as one important hint.
    Our BJT is saturated and Ic = Hfe*Ib don't hold anymore.
    Because if Ic = Hfe*Ib = 46.7mA is larger than Ic_max = Vcc/(Rc+RE) =10V/1.1K = 9mA transistor is in saturation region.
    So to solve this circuit we need to assume Vce_sat voltage and use this equation
    Ie = Ib + Ic.
    Let as assume Vce_sat = 0.1V and Vbe = 0.7V.
    Ie = Ve/Re (1)
    Ib = (Vcc - Vbe - Ve)/RB (2)
    Ic = (Vcc - Vce_sat - Ve)/Rc (3)
    And now if we solve this for VE we have this
    [tex]\Large Ve = (\frac{Vcc - Vbe}{RB} + \frac{Vcc - Vce_{sat}}{RC}) * RB||RC||RE = 0.975676V [/tex]
    So
    Ve = 0.975676V
    Vb = Ve + Vbe = 0.975676V + 0.7V = 1.675676V
    Vc = Ve + Vce_sat = 1.075676V
    And the currents
    Ie = Ve/Re = 9.75676mA
    Ic = (Vcc - Vc)/Rc = 8.924324mA
    Ib = (Vcc - Vb)/RB = 0.8324324mA
     
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