Analogy about Transistor

[Animesh Maurya]

I have developed an analogy about transistor's amplification action.

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.

Thanks

Animesh Maurya

 

[Roger Johansson]

I have a simple explanation and model of how transistors (bipolar)
work.

Current units from the collector input can only pass the emitter if
they are in company of a current unit coming into the transistor
through the base.

But each current unit going through the base-emitter path can take
with it a number of current units coming from the collector.
That number is the amplification factor.

To be more concrete let's say that the current unit is milliamperes
(mA) and the amplification factor is 400.

Each mA we send through the base-emitter path then allows 400 mA from
the collector to accompany it through the emitter.

To control how many mA we want to send through the base-emitter we can
use a base resistor in series with the base.

The voltage at the other end of that resistor controls the base
current, and controls how much collector current can pass.

--
Roger J.

 

[Dana Raymond]

I heard somewhere that "Transistor" comes from Transference of Resistance.
Just a datapoint.

"John Popelish" <jpopelish@rica.net> wrote in message
news:3F53C46E.AC3E2284@rica.net...
Animesh Maurya wrote:

I have developed an analogy about transistor's amplification action.

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.

Thanks

Animesh Maurya

It seems to me that you could replace the transistor with a pair of
resistors (one from emitter to base, and one from base to collector),
and the same analogy would apply just as well. Your model needs more
details before it explains anything. Charge carriers act very
differently in a chunk of semiconductor with little electric field
through it than they do in a chunk containing a large electric field.
Think about which volumes are emptied of almost all charge carriers
and biased with an electric field, and which volumes are flooded with
charge carriers but have little electric field.

--
John Popelish

 

[DJ Bartlett]

It's an amalgam of transducer and resistor

just a datapoint 8 D.

DJ

"Dana Raymond" <draymond@austin.rr.com> wrote in message
news:0BU4b.13423$jt.761221@twister.austin.rr.com...

I heard somewhere that "Transistor" comes from Transference of Resistance.
Just a datapoint.

"John Popelish" <jpopelish@rica.net> wrote in message
news:3F53C46E.AC3E2284@rica.net...
Animesh Maurya wrote:

I have developed an analogy about transistor's amplification action.

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.

Thanks

Animesh Maurya

It seems to me that you could replace the transistor with a pair of
resistors (one from emitter to base, and one from base to collector),
and the same analogy would apply just as well. Your model needs more
details before it explains anything. Charge carriers act very
differently in a chunk of semiconductor with little electric field
through it than they do in a chunk containing a large electric field.
Think about which volumes are emptied of almost all charge carriers
and biased with an electric field, and which volumes are flooded with
charge carriers but have little electric field.

--
John Popelish

 

[Dana Raymond]

Thats surprising since there is no aspect of a transistor that could be
considered a transducer.

"DJ Bartlett" <djbartlett@EvilGeniusTM.com> wrote in message
news:TQ15b.601$1r2.58986668@newssvr13.news.prodigy.com...

It's an amalgam of transducer and resistor

just a datapoint 8 D.

DJ

"Dana Raymond" <draymond@austin.rr.com> wrote in message
news:0BU4b.13423$jt.761221@twister.austin.rr.com...
I heard somewhere that "Transistor" comes from Transference of
Resistance.
Just a datapoint.

"John Popelish" <jpopelish@rica.net> wrote in message
news:3F53C46E.AC3E2284@rica.net...
Animesh Maurya wrote:

I have developed an analogy about transistor's amplification action.

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.

Thanks

Animesh Maurya

It seems to me that you could replace the transistor with a pair of
resistors (one from emitter to base, and one from base to collector),
and the same analogy would apply just as well. Your model needs more
details before it explains anything. Charge carriers act very
differently in a chunk of semiconductor with little electric field
through it than they do in a chunk containing a large electric field.
Think about which volumes are emptied of almost all charge carriers
and biased with an electric field, and which volumes are flooded with
charge carriers but have little electric field.

--
John Popelish

 

[zigoteau]

animesh_m@yahoo.com (Animesh Maurya) wrote in message news:<58eab14a.0308300529.4a2b4204@posting.google.com>...

Hi Animesh,

I have developed an analogy about transistor's amplification action.

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

Now replace the transistor with a hollow cylindrical container.

I'm afraid your analogy is not very good and does not explain or

reproduce any of the characteristic features of a transistor. How does
it explain the rectification of the emitter-base junction and the
base-collector junction? These depend intrinsically on the fact that a
semiconductor has majority and minority carriers, aspects which a
hydraulic model is totally incapable of mimicking.

Cheers,

Zigoteau.

 

[William J. Beaty]

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 Maurya]

Ian Stirling <root@mauve.demon.co.uk> wrote in message news:<biqafh$2ut$2$8300dec7@news.demon.co.uk>...

In sci.physics Animesh Maurya <animesh_m@yahoo.com> wrote:
I have developed an analogy about transistor's amplification action.
snip
Analogies are seldom perfect and at times can be misleading. I don't
know to which extent it is correct.

Ebers-Moll
Google.

Does it resembles with the Eber Molls model ?

Animesh Maurya

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

Thanks

Animesh Maurya

 

[Animesh Maurya]

billb@eskimo.com (William J. Beaty) wrote in message news:<2251b4e6.0309021326.74ff761b@posting.google.com>...

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.

Yes I agree that the model needs some more addition before it can
satisfactorily explain anything. Perhaps it was just an outline. Any
further help will be appreciated.

Animesh Maurya

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 Maurya]

billb@eskimo.com (William J. Beaty) wrote in message news:<2251b4e6.0309021326.74ff761b@posting.google.com>...

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 explanation is very nice, the same I was looking for. It cleared my concept.

Thanks

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/

 

[Ian Stirling]

In sci.physics Animesh Maurya <animesh_m@yahoo.com> wrote:

Ian Stirling <root@mauve.demon.co.uk> wrote in message news:<biqafh$2ut$2$8300dec7@news.demon.co.uk>...
In sci.physics Animesh Maurya <animesh_m@yahoo.com> wrote:
I have developed an analogy about transistor's amplification action.
snip
Analogies are seldom perfect and at times can be misleading. I don't
know to which extent it is correct.

Ebers-Moll
Google.

Does it resembles with the Eber Molls model ?

Very similar

--
http://inquisitor.i.am/ | mailto:inquisitor@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
"Looks like his brainwaves crash a little short of the beach..." - Duckman.