Discussion in 'Electronic Basics' started by ObamaOrHillary2008, Aug 27, 2007.

1. ### ObamaOrHillary2008Guest

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?

2. ### BobGuest

Is the collector-emitter of the transistor really in parallel with the led
or is it in series?

If the transistor is really in parallel with the led then the led will be
OFF when the transistor is ON. This will happen because when the transistor
is ON, the voltage from the emitter to collector (and thus also across the
led) will be below the level that the led will draw any appreciable current.
When the transistor is off then voltage across the led will be allowed to
seek its own level (as if the transistor weren't even in the circuit).

If the transistor is in series with the led then the led will be ON when the
transistor is ON.

Bob

3. ### John PopelishGuest

Leaving out the resistors makes it a lot harder to
understand the circuit, because the voltage drops across
those resistors are a big part of what makes this work.

LEDs require some minimum voltage across them before they
conduct enough current to show a visible output. The
voltage varies with the color (energy) of the photons
3 volts. Note that this circuit runs of a very low voltage,
so there is barely enough to light an LED.

The problem with switching the LED by putting a transistor
in its current path is that this circuit has very little
extra voltage available to waste while that current passes
it seems that the designer has chosen to turn the LED off by
detouring its supply of current around the LED with the
transistor. As long as the transistor can carry this
current while dropping less than the voltage it takes to
light the LED, it works. The disadvantage of this method of
control is that the current through the LED supply resistor
is higher when the LED is off (and the transistor is on)
than when the LED is on (and the transistor is off).

4. ### EeyoreGuest

It makes no sense without the resistors.

Graham

5. ### ObamaOrHillary2008Guest

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.

6. ### Jonathan KirwanGuest

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

7. ### ObamaOrHillary2008Guest

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?
in case b? For example:

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

8. ### Dan CobyGuest

In your circuit, he current which enters the base of the NPN transistor
flows
through to the emitter and then back to the battery. It does not go out
through
the collector. The current through the base-emitter junction allows a
current
to flow into the collector of the transistor. Please note that the collector
current
is 'into' the collector which is the opposite direction from the arrow which
you

The LED is on when switch S1 is closed. Closing S1 shorts out the
base-emitter
junction of the transistor. As a result there is no base-emitter current and
no
collector current flow in the transistor. There is a current flow from the
positive
side of the battery through R2 to the LED. This current lights the LED.

When S1 is open, there is a base current which comes from the battery via
R1.
The base current allows current to flow into the collector. This collector
current
comes from the battery via R2. Typically for this type of circuit, the
values for R1
and R2 are chosen so that current flow into the collector will be large
enough to
pull the collector voltage down to a few tenths of a volt. (The voltage at
the collector
is the battery voltage minus the I*R drop across R2.) Since the collector
voltage
and the LED voltage are the same (they are connected together), the voltage
across the LED will only be a few tenths of a volt and the LED will not be
lit.

9. ### ObamaOrHillary2008Guest

So, the transistor functions as a resistor, being shorted out when the
switch is closed. Then we have two parallel routes
to compete for the current, R1 and R2. Since R1 is very large to
restrict current to the base, R2 gets almost all the current
and the LED lights.

However, an open switch allows the current to be drawn into the
collector. Then the LED becomes the resistor, where the collector-
emitter
is just like a wire, and it shorts the lighting of the LED.

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

10. ### Jonathan KirwanGuest

I the case where the button is NOT PRESSED, you have:

I the case where the button is PRESSED, you have:
Hope that helps.

Jon

11. ### ehsjrGuest

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

12. ### Dan CobyGuest

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:

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

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.

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.

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.)

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.

13. ### ObamaOrHillary2008Guest

Thank you again for the drawings and responses; I understand it a lot
better now.

I'd like to take the questions a step further. I was working with my
electronics set
yesterday. The following diagram (hope you can read it ok).

+3V
-----------------------------------
10k 10k
| |
-- C 4.7k 4.7k C-----LED
B - x - B |
E | | E __ |
| s1 s2 |
------------------------------
-------------------
-----

This is supposed to be an RS flipflop, and when I connect the 3v, the
led is off.
Pressing s2 and releasing turns on the led and it stays on. Pressing
s1 and releasing
turns off the led again.

My question is, it seems like the NPN on the left is interfering with
the NPN on the
right. Why is this, even though the switches as released?

14. ### Dan CobyGuest

Your circuit diagram is distorted (even with a fixed size font) so I am not
exactly sure about what you have. My assumption is that you have some
form of what is called a 'bistable multivibrator'. See the wikipedia entry:

http://en.wikipedia.org/wiki/Multivibrator#Bistable_Multivibrator_Circuit

states for this circuit. In each stable state, one transistor is turned on and
the other is turned off. In one stable state, the left transistor is on and the
right is off. In the other state the left is off and the right transistor in on.

The transistors act in a feed back loop. For instance if S1 is used to
short out the base of the left transistor, Then there will be no base current
into that transistor. The lack of base current will turn off the left transistor.
As a result the collector current of the left transistor will stop. With no collector
current, the collector voltage will increase. The collector voltage is being
pulled up by the left side 10K resistor. With a high collector voltage on the
left transistor, there will be a current through the 4.7K resistor into the base
of the right side transistor. This will turn on the right transistor. This gives us
a collector current into the right transistor. The collector current will cause a
voltage drop across the right 10K resistor. This will pull the collector voltage
of the right transistor down to about 0.3 volts. With this low voltage there will
be no current flow through the 4.7K resistor which is connected to the base
of the left transistor. This keeps the left transistor turned off even if S1 is then
opened. So the feed back loop is that S1 turns off the left transistor. This
turns on the right transistor which keeps the left transistor turned off. However
closing S2 will switch to the other stable state with the right transistor off and
the left transistor on.

15. ### Guest

I came across this post and was curious why someone would want to
illuminate an LED this way. Is there an inherent advantage to running
the load in parallel with the collector and emitter, as opposed to in
series? I've seen some (simple) audio amplifier circuits that do the
same thing, too.

Andy

16. ### Jonathan KirwanGuest

I read the original post and came up with the schematic by "reading
between the lines" and accepting the OP's ability to write accurately
about whether or not the switch was pressed when the LED was on or
off. It seemed the simpler circuit that would not be inconsistent
with the OPs rather vague writing about it.

One of the insights that helped me accept the OP's writing was the
phrase, "electronics learning kit." I don't think someone would
actually do this as a practical matter, but since it was in some
learning kit I had no problem accepting the circuit idea that seemed
to fit, as a possible explanation for what he wrote.

Jon

17. ### Guest

That makes sense. His wording does seem to lend itself to the diagram
you came up with. I guess the schematic seems viable from an
educational standpoint, as you pointed out. Thanks!

18. ### Ken MoffettGuest

Jon,

Actually I did this as a "practical" matter. I thought this
schematic looked a little familiar, so I went back looking at
some of my circuits from the past. This was a method I
used of pulsing a laser diode fed by a switchable constant
current source.

Ken

.----.
| |
+10v +------+------+-|7805|-+--------------+---+---+
| | '----' | 18 | 33| 82|
| 0.1 -+- | --- .-. .-. .-.
.-. --- | --- 0.1 | | | | | |
| | | | | | | | | | |
| | +---+----+ '-' '-' '-'
'-' | | | |
390 | | .--+ | |
| | | o o /o
| | | /
| | | /
| | | o
| | | |
| | . | |
| +----------------+------+
- | | |
| | | | |
| | | |/ |
| | +----------------------+-+| TIP29 V
| | | |> -
| | 4.7K 4.7K | | | Laser Diode
-- --- ___ ___ |/ | |
Signal +--------+-|___|+--+-+-|___|--| | |
| |> | |
- | | |
1N4148 ^ 2N3904 | | |
| | | |
COM +------------------+------------+----+------+
Pulsed Constant Current Laser Diode
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

19. ### Jonathan KirwanGuest

Makes sense. I actually have thought about doing just about that, as
well. Just that it didn't come to mind when thinking about a push
button and a simple LED. In general, though, you are right to point
out that there are some uses for "current steering."

Thanks,
Jon

20. ### ehsjrGuest

Maybe for an Off indicator? Something like this
comes to mind:

+12 -----+-----+
| |
[D1] [Rly]
| |
+-----+--{{------+
| |
/c [1K]
---[R]---| |
\e [LED] Off
| |
Gnd ---------+--{{-------+

You might have an existing circuit like the relay circuit
above, where you could add a resistor and LED to indicate
that the relay is not energized.

Ed