Re: Question about transistor

ObamaOrHillary2008 wrote:
On Aug 28, 4:36 pm, Jonathan Kirwan <jkir...@xxxxxxxxxxxxxx> wrote:

On Tue, 28 Aug 2007 13:11:24 -0700, ObamaOrHillary2008





<jgra...@xxxxxxxxx> wrote:

On Aug 27, 2:06 pm, Eeyore <rabbitsfriendsandrelati...@xxxxxxxxxxx>
wrote:

ObamaOrHillary2008 wrote:

In the diagram below:

3V
|
----------
NPN LED
---
-

from my electronics learning kit, the NPN is in parallel with the led,
with its base connected to a switch. When you press the switch, the
LED, previously off, suddenly turns on. I've left out the resistance
stuff.

My question if, since the voltage is in parallel with the LED, why
can't the current just go through the LED and not the transistor. Why
does the lack of current in the transistor keep the LED off until the
base is activated?

It makes no sense without the resistors.

Graham- Hide quoted text -

- Show quoted text -

Thanks for the responses. I don't have the schematic with me (I'm at
work). I gather that it all has to do with the voltage and current
equations for the particular circuit. When you say it makes no sense
without the resistors, I gather that there is some threshold of
current and/or voltage that makes the LED come on, even though it is
connected to the battery. The problem with electronics learning kits
is they don't fully explain why there is a resistor here, a capacitor
in the corner there, just a bit about the main capacitor and such.

Was the circuit something like this (view using fixed-spaced font):






: ,--------------------+----------,
: | | |
: | | |
: | \ |
: | / R2 |
: \ \ |
: / R1 / |
: \ | |
: / | |
: | | |
: | ,--------+ |
: | | | ---
: | | | - BATTERY
: | |/c Q1 | ---
: +---------| NPN | -
: | |>e | + |
: | | --- |
: | | \ / LED |
: o | --- |
: / S1 | | |
: / push | | |
: o button | | |
: | | | |
: | | | |
: '-----------+--------+----------'

If so, then pushing the button to turn on the switch will turn off the
transistor. When the transistor is off, the LED will come on. That's
because the transistor, when on, bypasses the LED and doesn't allow
current sufficient to light it to pass through it. When the switch S1
is not engaged and is off, then current in R1 supplies the needed base
drive for Q1 to turn Q1 on. When the switch S1 is pressed and is on,
then S1 forces Q1's base to the negative side of the battery and this
causes Q1 to be off. When Q1 is off, you can just imagine removing Q1
entirely from the circuit to see what is happening. When Q1 is on,
think of it as a switch that is turned on and tying the (+) side of
the LED to its negative side and the negative terminal of the battery
-- forcing the LED off.

Jon- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -


Thank you very much for the diagram, which is what I had in mind.

Actually, what would be really helpful would be, in case a and b, do
draw
arrows in the circuit. If I am a positive charge, starting at the +
side of the battery,
and I want to go home to the negative side, what is my route when the
switch is off?
Do I go through the NPN, do I go through the LED and THEN to the NPN?
What about
in case b? For example:

,--------------------+----------,
| <------ | <------ |
| | |
| \ | ^
| / R2 | |
\ \ | |
/ R1 / | |
\ | |
/ | |
| | ----> | |
| | ,--------+ | start
| | | || ---
| \/ | || - BATTERY
| ---> |/c Q1 |\/ ---
+---------| NPN | -
| |>e | + | end
| | --- |
| | \ / LED | ^
o | --- | |
/ S1 | | | |
/ push | | | |
o button | | | |
| | | | |
| | | \/ |
'-----------+--------+----------'



With S1 open:
When a small current from the (+) of the battery and through
R1 enters the base of the transistor, it flows out the emitter
and back to the (-) side of the battery.

That small current enables the transistor to conduct a much
bigger current through the transistor as follows: from the
(+) side of the battery, through R2 to the collector, from
the collector to the emitter and back to the battery.

The current through R2 causes a voltage drop (V=I*R) which
makes the voltage at the top of the LED in your diagram
lower than the voltage needed to make the LED conduct, so
there is no current through the LED. Therefore, it does not
glow.

When S1 is closed:
No current can enter the base of the transistor, because
the bottom of the resistor (and therefore the base) is
connected to the (-) of the battery, so there is no (+)
to go into the base.

Since there is no current from the base to the emitter
(there cannot be, because there is no potential difference
between them), there can be no current from the collector
to the base. Therefore, the transistor cannot cause a
voltage drop across R2. Since the transistor causes no
voltage drop across R2, the bottom of it is (+) and
current can flow from the battery (+) through R2, through
the LED and back to the battery (-). There will be a
voltage drop across R2, but it will be smaller than the
voltage drop caused by the transistor because the transistor,
when conducting, draws more current than the LED does. So
the voltage at the top of the LED will be sufficient for
current to flow through it, and it will glow.

Ed
.

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