Consider an PNP transistor in Common-Base configuration.
Emitter-Base region is forward biased and that of Collector-Base
region is reverse biased.

______________________
| | | |
-------| P | N | P |-------
| | | | | |
| _________ __________ |
| | |
|+ =======>> | - |
----- | ---
--- | -----
|- | +|
| | <<======== |
| | |
------------------------------------


Now replace the transistor with a hollow cylindrical container.

Place two porous membranes vertically opposite to each other at the
centre of the cylinder and call the enclosed region as base.

Distance between these two membranes is considered small as compared
to the length of the cylinder due to the fact that base region is
small in a transistor.

A small hole in made on the body of the cylinder in the base region.
Diameter of this hole is small than that of emitter & collector.

Connect this assembly using tubes in Common-Base configuration and put
pumps in place of a battery.
Fill up the tubes with water. Assume that water flows in the same
direction as that of the conventional current of battery (i.e. form
+ve to ?ve).

Now put on the pumps and see what happens.

Let us first highlight the Emitter-Base region.

Water flowing form the emitter reaches the base and will start
dividing. But majority of water will be transferred to the collector,
as the base opening in very small.

Moreover in Collector-Base region the pump is applying force which is
just opposite to that of the base and thus making base current less
and less, which in turns facilitates large collector current. This
causes amplification.

Analogies are seldom perfect and at times can be misleading. I don't
know to which extent it is correct.

Also one major drawback in that I cant explain amplification in case
of a NPN transistor, if the above assumption are unaltered.

Animesh Maurya

Water flowing form the emitter reaches the base and will start
dividing. But majority of water will be transferred to the collector,
as the base opening in very small.

Moreover in Collector-Base region the pump is applying force which is
just opposite to that of the base and thus making base current less
and less, which in turns facilitates large collector current. This
causes amplification.

animesh_m@yahoo.com (Animesh Maurya) wrote in message news:<58eab14a.0308300329.7c06035a@posting.google.com>...
Water flowing form the emitter reaches the base and will start
dividing. But majority of water will be transferred to the collector,
as the base opening in very small.

Moreover in Collector-Base region the pump is applying force which is
just opposite to that of the base and thus making base current less
and less, which in turns facilitates large collector current. This
causes amplification.

Your analogy seems to be missing the entire transistor action. In
the above design, tiny changes in the base circuit will not produce
large changes in the collector circuit. This fluid analogy simply
acts like two resistors and two power supplies.

The core concept in transistor or vacuum tube amplification is
"controllable valve action," where a tiny amount of energy is
used to open/close some sort of fluid valve. Or in other words,
small energy expended per second is used to control a large amount
of energy per second being produced by some sort of power supply.
Perforated plates essentially behave as resistors, not as
controllable valves.

In transistors, the base/emitter junction acts as a genuine
"electricity valve" which determines the value of current
in emitter and collector leads. To understand transistors, don't
use the common-base configuration. Instead imagine an "ideal"
transistor where the base width is zero and the base current is
zero, yet where Vbe still determines Ic (still like a diode, but
a diode where the Vf forward voltage is applied between two terminals,
while the main current path is through two entirely DIFFERENT
terminals.)


PS
Any analogy which tries to explain why base current can control
the collector current is doomed to failure. Why? Because in real-
world transistors, base current doesn't control collector current!


Here's someone's simple but excellent analogy:

http://www.satcure-focus.com/tutor/page4.htm


See? In the above analogy, base current doesn't control
collector current. Instead, the pressure applied to the
base will open the "collector valve" as well as causing
fluid leakage through the base, determining both the Ib and
the Ic. The central concept is "Vbe controls Ic."
Understand that, and you understand transistors. Oh, yeah,
by the way, Vbe also controls Ib, so Ib just happens to be
proportional to Ic; a useful phenomenon for simplifying design
calcs, but for actually UNDERSTANDING transistors, the whole
"current gain" concept sends you up a dead-end road. There
is no mechanism in bipolar transistors whereby Ib can directly
determine Ic.



Here's my own oversimplified explanation (w/lots of ASCII art!) :

http://amasci.com/amateur/transis.html


((((((((((((((((((((((( ( ( (o) ) ) )))))))))))))))))))))))
William J. Beaty Research Engineer
beaty@chem.washington.edu UW Chem Dept, Bagley Hall RM74
billb@eskimo.com Box 351700, Seattle, WA 98195-1700
ph206-543-6195 http//staff.washington.edu/wbeaty/

animesh_m@yahoo.com (Animesh Maurya) wrote in message news:<58eab14a.0308300329.7c06035a@posting.google.com>...
Water flowing form the emitter reaches the base and will start
dividing. But majority of water will be transferred to the collector,
as the base opening in very small.

Moreover in Collector-Base region the pump is applying force which is
just opposite to that of the base and thus making base current less
and less, which in turns facilitates large collector current. This
causes amplification.

Your analogy seems to be missing the entire transistor action. In
the above design, tiny changes in the base circuit will not produce
large changes in the collector circuit. This fluid analogy simply
acts like two resistors and two power supplies.

The core concept in transistor or vacuum tube amplification is
"controllable valve action," where a tiny amount of energy is
used to open/close some sort of fluid valve. Or in other words,
small energy expended per second is used to control a large amount
of energy per second being produced by some sort of power supply.
Perforated plates essentially behave as resistors, not as
controllable valves.

In transistors, the base/emitter junction acts as a genuine
"electricity valve" which determines the value of current
in emitter and collector leads. To understand transistors, don't
use the common-base configuration. Instead imagine an "ideal"
transistor where the base width is zero and the base current is
zero, yet where Vbe still determines Ic (still like a diode, but
a diode where the Vf forward voltage is applied between two terminals,
while the main current path is through two entirely DIFFERENT
terminals.)


PS
Any analogy which tries to explain why base current can control
the collector current is doomed to failure. Why? Because in real-
world transistors, base current doesn't control collector current!


Here's someone's simple but excellent analogy:

http://www.satcure-focus.com/tutor/page4.htm


See? In the above analogy, base current doesn't control
collector current. Instead, the pressure applied to the
base will open the "collector valve" as well as causing
fluid leakage through the base, determining both the Ib and
the Ic. The central concept is "Vbe controls Ic."
Understand that, and you understand transistors. Oh, yeah,
by the way, Vbe also controls Ib, so Ib just happens to be
proportional to Ic; a useful phenomenon for simplifying design
calcs, but for actually UNDERSTANDING transistors, the whole
"current gain" concept sends you up a dead-end road. There
is no mechanism in bipolar transistors whereby Ib can directly
determine Ic.



Here's my own oversimplified explanation (w/lots of ASCII art!) :

http://amasci.com/amateur/transis.html


((((((((((((((((((((((( ( ( (o) ) ) )))))))))))))))))))))))
William J. Beaty Research Engineer
beaty@chem.washington.edu UW Chem Dept, Bagley Hall RM74
billb@eskimo.com Box 351700, Seattle, WA 98195-1700
ph206-543-6195 http//staff.washington.edu/wbeaty/