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Configuration

According to my text book,

When we use a transistor as an amplifier, there generally 2 resistors, one is input resister(R in) and the other is output resister(R out). Normally these resistors are connecting as R out is relatively greater than the R in.

  1. I can't understand the reason for Why the R out should greater than the R in and what is the expectation of that?

Characteristic Curves

Output Characteristic

enter image description here

For a concerned IB at the start of the curve, it has a speedy variation (relatively high gradient) and after a while the variation is slow and becoming linear(less gradient).

  1. What is the reason for have two gradients? Starting have a speed variation and then slow gradient... I tried to concentrate it with the Ohm's law but couldn't.(According to the ohm's law the R out should different at the two situations.)

Transfer Charasteristic

enter image description here

  1. I can't understand the reason of why there is a saturated region after the activated region. Why does the activated region can't to be continue.

I expect answers with theoretical descriptions and a deep view of the structure of the transistor.

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I can't understand the reason for Why the R out should greater than the R in and what is the expectation of that?

Roughly, the voltage gain of the CE amplifier is

$$A_v = \beta\frac{R_c}{R_b}$$

So if \$R_b\$ is too large or \$R_c\$ is too small, the gain is low.

I'll answer the 2nd and 3rd questions out of order:

I can't understand the reason of why there is a saturated region after the activated region. Why does the activated region can't to be continue.

If the curves for the active region continued to the left of the y-axis, that would be in the 2nd quadrant of the graph. This would require the transistor to provide power to the rest of the circuit. But a transistor is not a power source, so it can't do that.

I've answered this in more detail in response to past questions: 1 2.

What is the reason for have two gradients? Starting have a speed variation and then slow gradient... I tried to concentrate it with the Ohm's law but couldn't.(According to the ohm's law the R out should different at the two situations.)

Two modes of operation, two different behaviors.

If it's clear the transistor can't continue the active-region behavior into the 2nd graph quadrant, then something has to happen to keep the curve in the 1st quadrant. What happens is what you see in the graph you included in your question.

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  • \$\begingroup\$ Before the last paragraph, what did you mean by "Two modes of operation". \$\endgroup\$ – Osal Thuduwage Oct 7 '18 at 16:54
  • \$\begingroup\$ Forward active, saturation, cut-off, reverse active, and reverse saturation are often called the "modes of operation" of a BJT. \$\endgroup\$ – The Photon Oct 7 '18 at 17:12
  • \$\begingroup\$ On far right of these curves, there is breakdown, internal electric fields so strong the carriers are torn from the atomic matrix and flung to opposite terminals of the transistor. \$\endgroup\$ – analogsystemsrf Oct 7 '18 at 17:22

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