Understanding Transistor Polarity and Current Flow in BJT Transistors

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Current in a BJT transistor typically flows from the collector to the emitter when the transistor is on, but it can also flow in reverse if the power supply polarity is reversed, although this is less efficient. The transistor operates with a lower current gain (Beta) in reverse mode, and the reverse breakdown voltage is significantly lower, limiting its effectiveness. When the base current is applied, electrons from the emitter move through the base to the collector, allowing current to flow from the power supply. If the base current is cut off, no current flows, and the electrons return to their original configuration. Understanding the geometry and doping of the transistor regions is crucial for its operation in both forward and reverse modes.
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Using a BJT transistor as an example, can current from the power supply only travel from the collector to the emitter when the transistor is on or can it travel in the reverse direction if the polarity of the DC power supply is reversed?
 
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Bararontok said:
Using a BJT transistor as an example, can current from the power supply only travel from the collector to the emitter when the transistor is on or can it travel in the reverse direction if the polarity of the DC power supply is reversed?

Depending on how you are using the BJT, you can usually get them to work backwards when you reverse the power supply polarity. But they are not as efficient when running backwards -- their current gain Beta is lower in the reverse direction, for example:

http://en.wikipedia.org/wiki/Bjt

.
 
You can make BJT work in reverse direction, but the performance is very different. I am no expert in semi conductor physics, I did design BJT IC before. The collector is very lightly doped. p+ ion is then inject into the collector to form a base region inside the collector. Then the emitter is form by doping with n- ion inside the base region. Therefore the geometry is very distinct and non symmetrical. Also because the doping concentration is so different between the emitter and collector, the reverse breakdown voltage is very different. The reverse break down voltage of B-E diode is much lower, something like 3 to 6 volts only, so you really don't get much of the transistor even if you can live with the low β.
 
The geometry of the transistors regions along with the doping determines how it operates.


tran12.gif




http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html#c5
 
Yes, this is the cross section of an NPN I was trying to describe. They dope and form the collector tub first, then heavily doped the middle and form the base region. Then the last step is to heavily dope the middle of the base to form the emitter.
 
dlgoff said:
The geometry of the transistors regions along with the doping determines how it operates.


tran12.gif




http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html#c5

So if this is the case, then when current is applied to the base, the electrons from the emitter will pass through the base and to the collector and this allows the current from the power supply to pass through the collector? And if the base current is cutoff, do the electrons return to the emitter?
 
Bararontok said:
So if this is the case, then when current is applied to the base, the electrons from the emitter will pass through the base and to the collector and this allows the current from the power supply to pass through the collector?
For a NPN transistor during normal operation, current from the supply enters the collector and exits the emitter as shown here:

tran2.gif



By reversing the supplies (NOT normal operation), current enters the emitter and exits the collector. Of course you'll need a small base current for this to happen.


And if the base current is cutoff, do the electrons return to the emitter?

If there's no base current (cutoff) there will be no current in either case.
 
So this means that the base current simply makes the path from the collector to the emitter conductive using its electrical energy, but of course when the base current is cutoff, the electrons in the transistor will be back in their original configuration with the specified amount of doping returning to its original value in each part of the transistor. Additionally, to pass a current in the reverse direction, the DC bias supply must also have its connecting terminals reversed to reverse the polarity.
 
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