Why Does Collector Current in NPN BJT Depend on Base Current in Common Base Mode?

Dear All,

Please guide me i am little bit confuse now. As we know

N-P-N transistor combination of two junctions
1-Emitter- Base
2-Collector Base.

In common Base configuration we have to make forward bise the Emitter- Base junction and Reverse Bise the Collector- Base junction. but i am unable to understand these two things mention below.

1- Let say i make open circuit of V(eb) and I(b) will be zero and collector- base junction connects in reverse bise configuration then why no current flows? why collector current depends on base current?

2- If i add a voltage source V(eb) and make Emitter- Base junction as forward bise then current will flow from emitter to base ok i agree because both are connected in forward bise, and we connect the collector- base junction in reverse bise then "how current will flow from base to collector as we know they are connected in reverse bise"????
*

This question starts by understanding the physics and chemistry of the transistor and best understand this before getting too involved in the reason.

It all has to do with the doping of the NPN material and the width of the base junction compared to the emitter and collector junctions. As the current flows through the base emitter junction it fills the base P region with N creating simply a cross over path between the collector and emitter. The more current through the base the more the base region is filled allow a larger crossover path for the collector emitter junction. The base material is very thin to create this affect.

There is also an proportional affect for the base emitter junction but the relationship is easier to understand with the collector controlled by the base.

2 - As the voltage is applied to the collector the depletion region between the base an collector get larger hence no current flow, but instead create the flow across the emitter collector region.

I am not the best in explaining this process as I have only learn't it myself but always assumed this while as a technician. Now as a student engineering it is a different story.

Corrections welcomed

In case 2, the current doesn't flow from the collector to the base, it flows from the collector to the emitter. The only current that flows from the collector to the base is reverse leakage current, which for "normal operating range" temperatures, is so small as to be negligible.

And regarding the actual solid-state physics that makes it possible for a small emitter-base current to control a larger Emitter-Collector current, I suspect most engineer's experience is the same as mine. Way back in engineering school, we struggled with the theories that explain this, but since most of us didn't major in solid-state physics [and thus we didn't really get into quantum energy states or whatever the heck other QP* voodoo is necessary to really understand what is going on [assuming anyone _really_ understands this], the bipolar transistor remains a black-box and we all just take it on faith that the thing works ;)


.* Quantum Physics and the whole science of really tiny things!

Also, it doesn't matter what circuit configuration the transistor is in (common-base, emitter-follower, common-emitter, etc.), the same law applies: a small emitter-base current controls a larger emitter-collector current [assuming the transistor is biased into the "active region"]. Too little forward current, or no current flowing into the emitter-base junction will inhibit all but leakage current from flowing from the emitter to the collector. In this case, the transistor is considered "turned off" or in "cutoff". Then there is the saturation state, where the forward current through the emitter-base junction causes the transistor to turn so far on, that there is little change in the emitter-collector current for changes in the emitter-base current. In this case, the transistor is considered "on" or "saturated".

In the case of a "switching transistor", and in terms of attempting to achieve the ideal state of zero voltage drop in the on state, the "saturation voltage" of the emitter-collector is made as low as possible by driving the emitter-base junction way beyond the beta required for operation in the active region. Typically a beta of around 10 is used to achieve this (where the beta might be around 150 for a transistor biased as a class-A audio preamplifier, for instance). But, there is a tradeoff: the further the transistor is driven into saturation, the longer it takes to pull it out of saturation, thus the switching speed suffers. Thus, for speed, bias the transistor at the edge of saturation and sacrifice some power due to higher saturation voltage -- or just use a MOSFET ;)

Believe me, it took a while for me to get all of that sorted out! Thus my eagerness to share it with those struggling with these concepts.

I really think it is best seen with a transistor model. In the picture you provided, it appears the cicuit is a common emitter (CE) configuration, so a comon emitter model would be appropriate. The collector current is a function of the base-emitter voltage as stated by the relationship: Collector Current I_C = Base-Emitter Diode Saturation current I_ES * e(V_BE/V_T). However, the collector current can also be expressed as a function of the base current, as you are asking. It's more difficult to see on the screen, so I'd recommend you write it on paper or something similar. The relationship written above will not yield any positive I_C without a positive, non-zero base-emitter voltage V_BE. The V_BE voltage we usually see being used is 0.7 V. 0.7 produces a value of current that can be readily seen, or measured. Anything lower, say 0.6, 0.5, 0.4, will mathematically produce values, but these values are too low to measure. V_T, the thermal voltage, is approximately 0.025 V for a small transistor, and I_ES is the saturation current of the base-emitter diode junction which is approximatley 10^-12 Amperes. Adding these into the relationship it is not too difficult to see the following response of I_C to changes in V_BE:

I_C = 10^-12 * exp (0.7/0.025) ) = 1.44 A
I_C = 10^-12 * exp (0.6/0.025) ) = 0.026489 A
I_C = 10^-12 * exp (0.5/0.025) ) = 0.000485 A
I_C = 10^-12 * exp (0.4/0.025) ) = 0.000009 A

The attached picture contains the CE NPN transistor model. The same model is shown twice. The only difference between them is how the current source is expressed. I_C is driven by the voltage-dependent current source on the right. The dependent current source depends on V_BE (V_BE is the only variable in the relationship above). You can leave it like this, or express I_C as a function of I_B. To do this, use the equation I_C = Beta * I_B. Keep the right-hand side of this equation and place it next to the dependent current source. This simple change takes it from a voltage-dependent current-source, to a current- dependent current source. This current-dependent current source drives the Collector current I_C. It is now easier to see that I_C depends on I_B.

I appologize for the hand writing. I hope by adding some numbers and a transistor model with some functions (known as the large signal model for the NPN BJT) it will help show the dependency of I_C on I_B. I wish I could qualitatively explain the solid-state physics behind it (I really cannot), but I think the responses before me have done a great job of that.