Very basic NOT gate question

[Sharms]

1) Voltage---------WWWWW(resistor)----------Switch--------ground
| |
| |
| |
LED1 LED2

2) Voltage---------WWWWW(resistor)----------Transistor-----ground
| | |
| | |
| | |
LED1 LED2 Base Input


The above two circuits are supposed to represent the same circuit, where
a transistor is used in circuit 2 in place of the mechanical switch in
circuit 1. The circuits are discussed at this web page:
http://www.spsu.edu/cs/faculty/bbrown/web_lectures/transistors/index.html.
When the switch is open, both LED's are supposed to light up, meaning
there is both a voltage and current across them. When the switches are
closed (or there is satisfactory input at the transistor's base), LED1
lights up, but LED2 does not. Understanding why this happens seems
pretty critical to understanding how basic logic gates work. Can someone
please explain this to me? Is it because closing the switch causes high
current to flow through the resistor, such that the voltage drop across
that resistor is so high that there is not enough voltage to the LED2 to
light it up? I really appreciate anyone's help.
Sharm

 

[Michael Black]

Sharms (ajdsharma@netscape.net) writes:

1) Voltage---------WWWWW(resistor)----------Switch--------ground
| |
| |
| |
LED1 LED2

2) Voltage---------WWWWW(resistor)----------Transistor-----ground
| | |
| | |
| | |
LED1 LED2 Base Input


The above two circuits are supposed to represent the same circuit, where
a transistor is used in circuit 2 in place of the mechanical switch in
circuit 1. The circuits are discussed at this web page:
http://www.spsu.edu/cs/faculty/bbrown/web_lectures/transistors/index.html.
When the switch is open, both LED's are supposed to light up, meaning
there is both a voltage and current across them. When the switches are
closed (or there is satisfactory input at the transistor's base), LED1
lights up, but LED2 does not. Understanding why this happens seems
pretty critical to understanding how basic logic gates work. Can someone
please explain this to me? Is it because closing the switch causes high
current to flow through the resistor, such that the voltage drop across
that resistor is so high that there is not enough voltage to the LED2 to
light it up? I really appreciate anyone's help.
Sharm

You're looking for way too complicated an answer.

Close the switch, and that second LED can see no voltage, because
the point is shorted to ground.

Since there is that resistor in series with the two LEDs, the first LED
does not see a short to ground, but a current drop (the value depending
on the resistor value). At most, that first LED will drop in output
slightly.

Shorting the resistor cannot cause "high current to flow through the
resistor" because it is the resistor that determines how much current
flows. The current through the resistor does not change whether the
switch is open or closed (well, leaving out the voltage drop in the
LED which will lower the current since the voltage seen by it is lower).
The switch merely shunts that current to ground rather than letting it
flow through the second LED.

I don't think this is a good example for explaining logic circuits.
If it's supposed to show a simple inverter, it's complicated by that
first LED, which does not make it as clear as an example as it could
be.

An inverter simply brings the output low as the input goes high.

Take out that first LED. Now with no voltage on the base of
the transistor, the LED is on. The collector of the transistor
is not conducting. Now, apply voltage to the base of the transistor.
The collector goes low, the signal is inverted from the input, and
the LED goes off.

Michael

 

[John Popelish]

Sharms wrote:

1) Voltage---------WWWWW(resistor)----------Switch--------ground
| |
| |
| |
LED1 LED2

2) Voltage---------WWWWW(resistor)----------Transistor-----ground
| | |
| | |
| | |
LED1 LED2 Base Input

The above two circuits are supposed to represent the same circuit, where
a transistor is used in circuit 2 in place of the mechanical switch in
circuit 1. The circuits are discussed at this web page:
http://www.spsu.edu/cs/faculty/bbrown/web_lectures/transistors/index.html.

The first detail you need to understand is that LEDs have a very
nonlinear voltage current relationship. But their current to
brightness relationship is nearly linear. As you apply a rising
voltage to an LED, it conducts almost no current till the voltage is
very near the normal operating voltage (that varies with LED color).
But then very small increases in voltage produce very large increases
in current. So LEDs are almost never connected directly to voltage
sources, but are driven by regulated current sources or with series
current limiting resistors.

When the switch is open, both LED's are supposed to light up, meaning
there is both a voltage and current across them. When the switches are
closed (or there is satisfactory input at the transistor's base), LED1
lights up, but LED2 does not. Understanding why this happens seems
pretty critical to understanding how basic logic gates work. Can someone
please explain this to me? Is it because closing the switch causes high
current to flow through the resistor, such that the voltage drop across
that resistor is so high that there is not enough voltage to the LED2 to
light it up? I really appreciate anyone's help.
Sharm

Start out by adding a series resistor to each LED, so that they act
more like visual volt meters. If you are using a 9 volt battery for
your experiments, 1000 to 10,000 ohms might be good. With the switch
(or transistor) off, you would expect both LEDs to light, though the
left one would be brighter, because the resistor between them will
drop some voltage. You should also use a resistor in this range
between them (as shown on the schematic).

When the switch (or transistor) is on, you should see the right LED go
off, but the right one should still be on.

--
John Popelish

 

[Sharms]

Thanks for your help. . . I think I just didn't realize that allowing
current to flow through the transistor (switch) creates a short circuit.
Does anyone know where on the web I can find a good explanation of how
logic gates are physically implemented in chips? All I have been able to
find on the web are discussions of Boolean logic and stand-alone gates.
Thanks again for all the help.

 

[R. Steve Walz]

Sharms wrote:

Thanks for your help. . . I think I just didn't realize that allowing
current to flow through the transistor (switch) creates a short circuit.
Does anyone know where on the web I can find a good explanation of how
logic gates are physically implemented in chips? All I have been able to
find on the web are discussions of Boolean logic and stand-alone gates.
Thanks again for all the help.
-----------------

In TTL logic schematics, such as this one,

http://www.armory.com/~rstevew/Public/Tutor/LogicCcts/ttl.gif

and these,
http://www.armory.com/~rstevew/Public/Tutor/LogicCcts/TTL_NAND_totem.gif
http://www.armory.com/~rstevew/Public/Tutor/LogicCcts/TTLinside.gif
and others, http://www.armory.com/~rstevew/Public/Tutor/LogicCcts/

just replace each transistor with a switch, and imagine it controlled
by its base as the switch lever, ON when current can flow into it,
and then do the simple Ohms Law analysis for each case, and you'll
see immediately how logic gates are produced using transistors.

The weird input transistor in these circuits is really only a use of
a multi-emitter transistor only as a set of diodes, and because it
is easy to fabricate on silicon, and you can use transistors in
parallel to emulate this weird transistor construction, tying their
collectors and bases together and leaving each emitter separate.

The grounding of these multiple emitter inputs, routes current to
ground which flow then starves the next section's transistor's base,
thus turning it off. If an emitter is grounded, then the current in
that input transistor flows base to collector, per that transistors
"two-diode nature" and the current keeps the next section's transistor
turned ON.

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public