As i am studying about DC Biasing of transistor, i am just able to fully understand only Base Bias after that i am stuck at Emitter Bias which is also called Dual Supply Emitter bias where we apply voltage Vee to emitter, Now my question is, In Emitter bias we know Ic become independent of Beta when Base voltage become zero, it means as it gets independent of Beta so its operating point will be stable.. and the problem is solved.. than why we move to other bias conditions like Collector feedback bias or voltage divider bias etc etc why we dont end the game at Emitter bias after studying Base Bias. Thankyou.
Electronic – Why we use Dual supply Emitter Bias in Transistor
biasingbjttransistors
Related Solutions
My guess is that you have the Base and Emitter swapped over. Here's an example that is similar to yours. Note how inductor L3 is connected (through R5) to the Emitter, and the erase head is connected from ground (through R3) to the Base.
The circuit is based on a Common Collector Colpitts oscillator. The erase head in combination with C6 and C7 form a tuned circuit. The tap between the two capacitors lowers the impedance on the Emitter side, providing the power gain and positive feedback required to make it oscillate.
You seem to think that because the depletion area is large for the base collector junction that there must be a large field there. That is the opposite of how it works. The voltage difference between the emitter, base, and collector is dropped exclusively in the depletion region. Thus the larger the depletion region, the more spread out the field is and so the less strong the field is. In fact, back injection is one of the reasons for hetero junction transistors. In the non depletion regions, diffusion is the dominant mechanism. Thus by creating a large electric field that essentially acts like a reverse flow barrier, the BJT forces diffusion to occur. The reason the current flows from C->E (NPN) or E->C (PNP) is because the BJT uses different doping levels to create depletion widths for different purposes. The reverse biased C-B depletion region, being large, is dropped almost exclusively on the collector's side. That means there's very little length for diffusion. That small length means that the Collector becomes an excellent source of minority carriers for the base compared to the emitter, whose large electric field is almost insurmountable. When those minority carriers cross, they drive diffusion in the base. Once the base provides that minority carrier to the emitter, it can't get it back, so it must get another one from the collector.
In essence the collector current is dependent only on the amount that flows out of the emitter, not on the field, which is designed to drop almost entirely over the collector.
Please leave a comment on this if you don't understand, because I needed about 10 pictures to get this down in college.
Edit: The BJT picture to rule them all!
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Best Answer
With an BJT transistor in a linear circuit, you need to make the collector current (Ic) not depend on the HFE (used to be called Beta) of the transistor.
The usual method is to make some resistor on the emitter have a lot more voltage drop than the Vbe (Base to Emitter) of the transistor.
This can be done by having a positive and negative supply voltage and keeping the Base near zero volts. Life, however, often doesn't give you two supply voltages. In that case, setting the base at some voltage part way between the positive and negative supplies is much easier than having to make a new power supply circuit. This is the two resistor voltage divider in many designs.
Sometimes you may find your self designing in a case where adding one more component will be objected to by management. A resistor from the collector to the base of a BJT takes fewer parts than the voltage divider.
Sometimes the supply voltage you have is much too low for a normal design. In this cases, you must get very creative about biasing of the transistors.