Re: Question about transistor

"ObamaOrHillary2008" <jgrace5@xxxxxxxxx> wrote in message
news:1188414245.587653.80670@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Aug 29, 12:37 pm, "Dan Coby" <adc...@xxxxxxxxxxxxx> wrote:

So, the transistor functions as a resistor, being shorted out when the
switch is closed.

In this circuit the transistor is functioning as a relay. The base current (from
R1) acts like the control signal. When there is a base current the transistor
is turned on. When the transistor is turned on, it conducts current from the
collector to the emitter. This collector to emitter current path shorts out the
LED. (I.e all of the current through R2 flows through the transistor instead
of through the LED.)

See the circuit diagrams from Jonathan Kirwan in his posting:
news:9tibd31jkhj1sh7palpgo33c43sen58p6v@xxxxxxxxxx


Then we have two parallel routes
to compete for the current, R1 and R2.

Actually the battery will supply current to both R1 and R2.


Since R1 is very large to
restrict current to the base, R2 gets almost all the current
and the LED lights.

No. R1 is usually larger than R2 but that is not necessary for the
circuit to operate properly. R1 is usually selected to give the desired
base current when S1 is open.


However, an open switch allows the current to be drawn into the
collector.

Yes. When S1 is open, there is a base current which allows current
flow into the collector. For there to be a base current, there has to
be about 0.7 volts between the base and the emitter of the transistor.
This voltage (and current) comes from R1 when S1 is open.
When S1 is closed, there will be 0 volts across the base-emitter
junction and as a result there will be no base current.


Then the LED becomes the resistor, where the collector-
emitter
is just like a wire, and it shorts the lighting of the LED.

The LED does not 'becomes the resistor'. LEDs are 'light emitting
diodes'. LEDs emit light when there is a current through the LED.
For current to flow through the LED, there has to be a large enough
forward voltage across the LED. Depending upon the type of the
LED, this voltage is usually 1.4 volts (for red) to about 3 volts (for blue).
A lower voltage results in (almost) no current flow through the LED and
as a result it is dark. (A resistor has a current which is simply proportional
to the voltage across it.)


Is this right? What does the 1/10 volt have to do with it?

Bipolar (your NPN) transistors do not act like a perfect wires when
they are turned on. Instead there is a small voltage (often about 0.3
volts) between the collector and the emitter. Since this voltage is
much lower than the voltage required to light the LED, the transistor
is effectively shorting out the LED.

See previous comment.



.

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