Question

Let iE, iC and iB represent the emitter current, the collector current and the base current respectively in a transistor. Then
(a) iC is slightly smaller than iE
(b) iC is slightly greater than iE
(c) iB is much smaller than iE
(d) iB is much greater than iE.

Answer 1

Let iE, iC and iB represent the emitter current, the collector current and the base current respectively in a transistor. Then

(a) iC is slightly smaller than iE

(c) iB is much smaller than iE

Explanation:

Bipolar transistor

• It consists of PN junctions which connects the three terminals and each terminals gives a name to find other two terminals.
• B is the base, E is the emitter and C is the collector.

Common emitter amplifier circuit

• The current flows through the transistor is equal to the current flows through the transistor as emitter.

         Ie = Ic + Ib

• The input impedance is greater, the gain of power and current than that common base configurations and the gain of voltage is lower.
• Lightly doped in the base of the transistor, when the majority carriers because of the forward biasing in the base junction of the emitter, and the batteries repulsive force, it passes through the base region.
• Base is lightly doped and the emitters majority carriers are neutralized, the base current is low.
• The majority carriers which remains will pass through the collector current.
• The base and collector current are slightly smaller than emitter.

To learn more;

1. https://brainly.in/question/5182731
2. https://brainly.in/question/2716991

Answer 2

Let iE, iC and iB represent the emitter current, the collector current and the base current respectively in a transistor. Then  iC is slightly smaller than iE and  iB is much smaller than iE.

Explanation:

• As we cannot understand the meaning, the highlighted sections could not be altered. Kindly look at the logical accuracy’s last line.
• It can be seen that only one option is correct, while there are two correct options in the solution.
• The base of the transistor is doped slightly, the majority of carriers because of emitter base junction’s forward biasing fell the battery’s repulsive force and pass it over to the base region.
• This leads to the formation of emitter current iE.
• The base is lightly doped and is thin, so only little amount of emitter’s majority carriers are at the base neutralized. This leads to the formation of the base current. So, iB is low.
• The emitter’s remaining majority carriers pass to the collector and result in the collector current iC.  So, the relation is given as iE = iB + iC.
• So, the current iC is slightly smaller than iE due to the base.
• Hence, the options (a) and (c) are correct.