-2
\$\begingroup\$

Tell me if i am correct or not.

In transistor as an amplifier, we have n-p-n transistor (base in the middle and emitter and collector at the sides). The collector has electrons in majority because it is n-type. Similarly, holes are in majority in base but as it is lightly doped so it has very little amount of holes.

Now, when a small signal of alternating current is supplied to the base, the base current changes and this change is called input current. Most of this input current moves into the collector (base has less holes so most of the current moves on when it fills the holes).

Now, the collector has its own majority charge carriers (electrons) and some more electrons are added to it by the input signal. So, when collector gives current to the load, it will be the combination of input signal and the already present electrons. Hence, small signal is amplified. Am i right?

|improve this question
\$\endgroup\$

closed as not a real question by Leon Heller, Dave Tweed♦, Anindo Ghosh, zebonaut, Nick Alexeev♦ Jan 24 '13 at 22:56

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Please fix your capitalization. It is harder to read and the part at the end will likely be seen as shouting. \$\endgroup\$ – Brian Carlton Jan 22 '13 at 20:20
  • \$\begingroup\$ i hope there isn't any problem now. \$\endgroup\$ – Muhammad Rafique Jan 22 '13 at 20:27
  • 1
    \$\begingroup\$ It's easier to read now, but still not really clear what's being asked. It seems to boil down to "how does a transistor work?", and that can be answered in so many ways. We can discuss the physics, we can discuss mathematical models, intuitive models, and so on. It's been the topic of entire books, so it's difficult to know how to begin with an answer here. \$\endgroup\$ – Phil Frost Jan 22 '13 at 21:20
  • \$\begingroup\$ i am not asking actually. i just want to know if i am right or not about the concept. \$\endgroup\$ – Muhammad Rafique Jan 23 '13 at 5:01
4
\$\begingroup\$

We don't have to understand why the transistor exhibits current gain in order to understand amplification. Those topics can be understood separately. To understand the amplifier, we can regard the transistor as a "black box" which has certain behaviors which can be described with math functions and plotted as graphs.

If you approach an amplifier circuit from the perspective of majority carriers and doped regions in the transistor, it will be unnecessarily confusing.

Also, to understand the transistor circuit, we should delay thinking about AC and first concentrate on DC. There are DC amplifiers, which should be understood first! The AC amplifier essentially works by superimposing back-and-forth disturbances on the quiescent point of a DC amplifier. (But the AC amplifier is optimized toward AC operation, which is necessary because AC signals "see" different impedances in a circuit from DC. The AC amplifier has a different load line and gain from the DC amplifier which it is based on. These are complications which are easier to understand once you have a grasp on the DC amplifier.)

Voltage amplification takes place when the output is taken from a collector-side load resistance. Changes in the base voltage cause corresponding changes in the much larger current through the collector. The fluctuations in the collector current are multiplied by the load resistance on the collector side to create a fluctuating voltage (V = IR). If the amplifier is biased right, that voltage is larger than the input voltage.

If the output is taken from the emitter resistor instead, then there is no voltage amplification: the amplifier is a voltage follower. But in that case, still the transistor provides a boost to the signal. Even though the output voltage tracks the input voltage, the transistor can supply more current, so it can drive heavier loads (lower impedances). It is is a power amplifier.

Why does the NPN transistor exhibit current gain? The simple explanation "for dummies" is that the behavior occurs because the base region is very thin. Electrons are driven by the forward bias of the base-emitter diode to flow from the emitter terminal to the base terminal, which requires them to traverse the emitter and then the base. Because the base is lightly doped, the electrons have a hard time reaching the base terminal. They have to "go slow". And because the base is thin, the electrons are forced to travel across the base close to the collector. It's as if the electrons are asked to walk on a rope. So, most of them "slip and fall" into the collector instead before they are able to escape to the base terminal. Hence, the emitter has to supply many extra electrons in order to sustain the base current, and most of the emitted electrons go into the collector.

|improve this answer
\$\endgroup\$
  • 1
    \$\begingroup\$ I wouldn't call an emitter follower a power amplifier any more than most transistor configurations. I'd call it a current amplifier, which is more specific. \$\endgroup\$ – Phil Frost Jan 22 '13 at 22:08
  • \$\begingroup\$ Thanks. I hesitated over that. The problem is that another meaning for current amplifier is: driving an amount of current into the load that is proportional to the input voltage. \$\endgroup\$ – Kaz Jan 22 '13 at 22:40
  • \$\begingroup\$ +1 for decomposing into macro and micro and offering an explanations at each level. \$\endgroup\$ – JRobert Jan 22 '13 at 23:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.