Help me understand BJT

Discussion in 'General Electronics Chat' started by Evelyn1234, Mar 20, 2015.

  1. Evelyn1234

    Thread Starter New Member

    Mar 20, 2015
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    Hi

    I have read many many articles and write ups on BJT NPN transistor. But one part i am still confused.
    I can understand how electrons flow from emitter to base when Vb is > Ve.

    But what i can't understand is, for example, for BJT NPN to operate in saturate mode, Vb need to be greater than Vc

    If electrons has least barriers from emitter to collector during saturation, how can a higher Vb than Vc (which is a requirement for NPN to operate in saturation mode) make it has less barrier? When Vb is more positive than Vc, how does this help in pushing electrons from base towards collector? Shouldn't it be having a higher Vc than Vb make the electrons move more easily from Base to collector???
     
  2. Papabravo

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    Feb 24, 2006
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    The N layer forms a potential barrier which electrons must traverse by one of two methods. The first method is to have enough kinetic energy to traverse the barrier, and the second method is to tunnel through the barrier. In the case of the first method the height of the potential barrier is lowered so electrons of lower energy may pass over the barrier and roll downhill. You also have to ask yourself which directions are the electrons flowing and remember that the holes are flowing in the opposite direction. Try playing with this animation.

    http://www.learnabout-electronics.org/bipolar_junction_transistors_05.php
     
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  3. Evelyn1234

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    Mar 20, 2015
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    Thanks for replying. I can understand the animation. But it still doesn't answer my questions on how Vbc can help electrons to move from E to C
     
  4. Papabravo

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    As electrons are removed from the base holes are left behind. Electrons from the emitter fill the holes and the holes in the depletion layer. As a result the emitter and the collector are nearly at the same potential. Vce is typically 0.2 V in saturation.

    Try this series of short descriptions.
    http://www.pveducation.org/pvcdrom/pn-junction/conduction-in-semiconductors
     
  5. bertus

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    Apr 5, 2008
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    Hello,

    Have a look at the attached PDF.

    Bertus
     
  6. Evelyn1234

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    Mar 20, 2015
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    Thanks again. If i want to operate the transistor in saturation mode, what is the voltage i need to apply at Vb and Vc; for example?
     
  7. Evelyn1234

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    Mar 20, 2015
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    I only need >0.6v Vb to let the emitter base conducts, to turn fully on the transistor(saturate), i need to apply a voltage at Vc which is lower than Vb., Vc at voltage < 0.6v, such as 0.5v. is this the case? what if i connect Vc to gnd?
     
  8. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    A BJT is not specified with a voltage, it is specified with a current flow. To calculate your current flow, you assume the base is 0.6 volts above the emitter. Then use OHMS LAW to calculate current flow into the base. You will need a voltage that is something above 0.6V (more than emitter voltage) to allow current to flow into the base. Assuming resistors to not limit the current flow, the current flow out of your collector is essentially the current flow into the Base multiplied by the transistors gain (Hfe).

    In Your case, you want the transistor to behave as a switch, by definition, your (current x collector reistor) > (supply voltage - 0.6 volts), where the resistor is placed between collector and supply voltage.
     
  9. crutschow

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    Vbe basically looks like a forward biased diode and that voltage does not change significantly as the transistor goes into saturation.
    The Vce voltage is some higher voltage that goes through a load resistor to limit the collector current when the transistor goes into saturation.
    To insure saturation, it's the base current, not the base-emitter voltage that's of concern. For good saturation the base current should be at least 1/10th of the collector current, thus you typically apply a voltage to the base much higher than Vbe (say 3V or more) and use a resistor in series to establish the base current.

    Edit: Gopher beat me to the punch. ;)
     
  10. MikeML

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    Bertus, I read that, and I am not impressed. I wonder why Beaty thought it was necessary to copyright that nonsense...
     
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  11. bertus

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  12. MikeML

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    This may answer the OP's questions. Here is a simple NPN transistor, such as would be used for a switching application. I have set it up with a constant Base current of 500uA. I am varying the Collector current to see the effect that Ic has on Vbe, when the transistor comes out of saturation, and what its Vce does both while saturated and then when the collector current exceeds the ability of the transistor to stay in saturation...

    Note that the x-axis is the independent variable, namely Ic, plotted on a log plot.
    The upper trace shows the effect of increasing Ic on Vbe.
    The lower trace shows the effect of increasing Ic on Vce. I would say that the transistor comes out of saturation when Ic=~65mA. Since Ib is 500uA, I'll let you figure the Hfe...

    181.gif

    So, after looking at this, with 500uA of base drive, how much load current would you ask this transistor to switch?
     
    Last edited: Mar 20, 2015
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  13. Hypatia's Protege

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    Mar 1, 2015
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    If you'll forgive an observation, you seem to be under some misapprehension -- The bipolar junction transistor is a current operated device...

    With constructive intent
    HP
     
    Last edited: Mar 20, 2015
  14. Hypatia's Protege

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    I don't get it?!?! --- April 1'st is still 12 days off!;)

    Best regards
    HP
     
  15. ian field

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    Oct 27, 2012
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    Typical Vce-sat is about 0.4V for a silicon transistor, you can get optimised low Vce-sat types for special applications that go down to about 0.2V.

    Vbe will be somewhere around 0.7V for saturation, some power transistors need nearly 0.8V Vbe to achieve saturation at the full headline collector current.

    An interesting example of Vce-sat being lower than Vbe, is the Schottky clamped transistor as used in LS-TTl logic devices. A Schottky barrier diode is connected from collector to base - the sum of Vce-sat and the Vf of the Schottky diode is less than 0.7V, so the conducting collector clamps Vbe and prevents the transistor from saturating. This is the basis of the higher speed of LS-TTL.
     
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  16. Papabravo

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    No, no, no. You are looking at things the wrong way. You don't apply any voltage to the collector. In the common emitter configuration the collector is tied through a resistor to the supply voltage Vcc. When the transistor is in cutoff, and there is no current through the collector resistor, the voltage at the collector is Vcc. As soon as collector current flows the value of Vc will be somwhere between Vcc and Ve(GND) + 0.2V. When thr transistor is in saturation the emitter holds Vc at Ve + 0.2V Nobody had to apply any voltage at all to the collector.

    To summarize
    In satuaration:
    Ve is at GND because it can't go anywhere else.
    Vb is at GND + 0.6V because a current source is holding it there to provide base current
    Vc is at GND +0.2V because the emitter is holding it there.
     
  17. crutschow

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    That is a comment that will get a long (and often tedious) argument from certain forum members.
    My view is that the current operated viewpoint works well for large signal considerations (bias point and switching operation, etc.) whereas a voltage operated viewpoint (transconductance operation) often works best for small AC signal calculations.
     
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  18. Evelyn1234

    Thread Starter New Member

    Mar 20, 2015
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    Hi Papabravo!

    Your last answer is spot on to my questions and it helps clarify a lot.
    So when the transistor is in off mode, what is the typical value of Vcc? This is a DC supply voltage correct?

    If i were to set up a circuit just to see how the current and voltage changes at various points of a NPN transistor, i.e from cut off state to active mode until saturation;
    what do i need, is the following correct?

    Vcc at Collector pin via a resistor, Emitter to gnd via another resistor, and applying a varying Voltage at base from zero volt to above 0.6?

    Will i be able to see the changes of Vb, Vc, Ve and Ib, Ic, Ie; which bring the transistor from off to active till saturation?
     
  19. MikeML

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    Sort of like I showed you in post #12.
     
  20. Papabravo

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    You can do things with a simulator that will allow you to answer questions without using actual components.
    Vcc for transistor circuits might be something like 12 Volts.
    For a common emitter configuration
    1. A typical resistor from Vcc to the collector might be 1.2 kΩ
    2. The emitter is connected directly to GND
    3. A base resistor of 10kΩ would be connected to an adjustable voltage source.
    With the adjustable voltage source at 0V, the transistor is in cutoff, the base current is 0, the collector current is 0, and the emitter current is 0.
    As the voltage on the adjustable voltage source is increased, some base current will begin to flow. How much? Ohm's law will tell you.
    Examples
    AV=1.2V, Vb=0.6 ⇒ (1.2V - 0.6V) / 10,000Ω = 6e-9 or 6 nanoamperes
    AV=4.6V, Vb=0.6 ⇒ (4.6V - 0.6V) / 10,000Ω = 400e-6 od 400 microamperes
    Once you know the base current you can use β to compute the collector current.
    At some point the collector current rises to a value that causes all of the supply voltage to be dropped across the collector resistor. How much current is that?
    Ohm's law will tell you. Vcc=12V, Vcesat = 0.2V so ⇒ (12V - 0.2V) /1.2 kΩ = 9.38 milliamperes
    At this point the transistor goes into saturation. You can increase the base current all you want but you cannot increase the collector current beyond the limit imposed by the components in the circuit which are external to the transistor.
     
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