Jeff:
I couldnt agree with everything in your post and im not sure
what you meant by
some things so i rewrote a few things with the hope that this
would help clear
up some issues. Also, keep in mind this
is in reference to the Brinkmann circuit,
not the circuit in PeLu's post ok?

As you probably already know, in these circuits the transistor
acts more or less
like a switch. When current flows base to emitter, it causes the
collector
emitter resistance to fall to a very low value. This decrease in
resistance
makes the collector emitter look like the two connections of a
simple
mechanical switch that is controlled by the base current. When
base current
flows, the switch is turned on. When base current stops flowing,
the switch
is turned off. There is an exception to this however, and that is if
the
current through the collector emitter rises high enough when the
transistor
is turned "on", the transistor will appear to turn off even though
base current
still flows. This action is what starts Q2 turning off.
The base drive sets the max level of collector emitter current:
more base drive means the transistor collector emitter current
will rise to
a higher value before starting to turn off. This means R2 will
partly determine
the operating frequency as well.

As you probably already also know, when you turn on Q2 you are
effectively
putting a dc voltage across the coil, and when this happens
current starts to
rise in the coil according to:

i=v*t/L
where
i is current(ramping up)
v is voltage
t is the voltage pulse time
L is inductance

This of course means the longer the pulse time, the higher the
current ramps up to.
The current ramps up untill the Q2 transistor can no longer stay
turned on,
and then it starts to turn off. When it turns off, the voltage
across the
inductor suddenly rises because it tries to keep the same current
flowing
through the transistor. Since the transistor resistance is
increasing at this
time, the voltage keeps increasing untill it reaches the level at
which the
LED first turns on, and then the LED resistance starts to limit
the voltage
that the inductor will rise up to. Feedback though the cap turns
off the two
transistors so that the only resistance left that absorbs L's
energy is now the LED.
The voltage is thus clamped by the LED.
The initial current is about the same as that which was flowing
before the
transistor just started to turn off, only somewhat lower because
the transistor
does absorb some energy while turning off. The current then
ramps down
as the energy from the inductor dissipates through the LED.

The reason the voltage can rise across the inductor is because
one important
property of an inductor is that it tries to maintain a constant
current
(over very short periods of time) by reversing (and raising) its
own terminal
voltage when the resistance (Q2 collector emitter junction) in
series with
the inductor increases. This occurs because the quickly colapsing
field in
the inductor generates a reverse voltage which tends to keep the
current at
the same level and flowing in the same direction as it was before
the
resistance increased.

Here is your post with some modifications and some other notes:

1. Current flows Q1(eb) to ground through R1, and R1 limits the
current flow.
2. This causes current flow through Q1(ec), so current starts to
flow through Q2(be)
through R2, and R2 limits the current flow through Q2(be).

3. This causes Q2(ce) to conduct current, so current starts
flowing
through L1 which builds up a field in L1's core. As L1 charges,
more current
flows though L1 as time goes by. The more time that passes, the
more current
flows through L1. As the current rises, Q2(ce) is eventually
forced out of
saturation.

4. Eventually Q2(ce) reaches a level where the Q2(ce) doesnt look
like a short anymore.
This happens when Q2 comes out of saturation. When this
happens, Q2(ce) voltage
starts to rise. This happens because Q2(be) receives only a small
current through
its base resistor. Q2 starts to come out of sat approximately
when the current
in the coil exceeds Q2's base current times its beta ( ib * B ).

5. This eventually breaks the current flow through Q2(ce), and L1
being an inductor,
needs a place to dump its stored energy and so it creates a high
voltage across
it in a direction that will try to keep the current flowing in the
same
direction that it was before Q2(ce) opened up.

6. Since the voltage shoots up to several volts, its enough to
turn on the LED.
Once the LED turns on, current starts to flow through it from
the inductor.
The voltage will rise to any level that will keep the current flowing
through it
and thats what turns the LED on. Since the inductor had
amount of energy during its charge, it now dumps that energy
into the LED.
Theoretically without considering the max flux density in the coil,
the voltage
would rise to any level meaning you could wire 100 LED's in
series and when
that inductor "kicks back" (when Q2(ce) opens up) they would all
light up, but
because the series voltage would be over 300 volts the pulse
would only last
for a very short time, and there would be other practical
considerations such
as the voltage rating of the transistor. Considering the max flux
density
in the coil is limited in any practical coil, the maximum level of
voltage
attainable will be limited by whatever the coil can sustain without
saturating.

7. Once Q2 turns off, its collector voltage rises and that is
coupled to
the base to Q1 through the capacitor. This turns Q1 off which
then turns
Q2 off completely. The capacitor then discharges to ground
through R1.

8. Once the cap discharges, Q1 starts to turn back on and the
cycle repeats.

Notes:

a. The time it takes for C1 to discharge and the time it takes for
Q2 to come out of
saturation approximately determines the operating frequency.
b. R1 helps to bias Q1 as an inverter, as well as sets the
discharge time of the cap.
R2 sets the current level that the inductor reaches before forcing
Q2 out of sat.
c. The energy to light the led comes primarily from the L1 field
collapse when the circuit opens.
(Yes, thats true). Without the inductive kick back, the voltage
couldnt rise higher
then the battery voltage. This is the main idea behind the boost
converter as well
as many other types of converters. The inductor is charged up
with a constant voltage,
and when the circuit is opened the inductor generates a high
voltage which is
then used to drive a higher voltage device such as an LED.

Scale up proposals for consideration:

1. This is what im looking at next. Im going to try to answer the
question:
"if we could change anything about the circuit in order to drive
any number
of led's, what would be the best changes to make for 1 led, 2
led's, 3 led's,
etc." There are going to be some things that are better to
change then others
while keeping efficiency in mind. My guess is that something will
have to
change about the circuit in order to accommodate a particular
number of LED's.
Two leds instead of one means either twice the current(parallel)
or twice the voltage(series).

Some of the constraints are:

1. Keeping the duty cycle reasonably high (close to 50%)
because the higher
the duty cycle the higher the overall efficiency due to the true
nature
of the current vs illumination curve of the led. 50% duty cycle
seems to
be a good tradeoff between getting good efficiency and building a
practical
circuit. If you have to bang the LED's with a 10% duty cycle, you
have
to raise the pulse current considerably higher then at 50% duty
cycle
to get to full brightness, so that would reduce efficiency too
much.
Also, a low duty cycle creates a problem in trying to get full
brightness
from an LED without going over the manufacturers
recommended max pulse
current rating.
2. Being able to easily modify the circuit to operate on either one
or two
alkaline cells, or one or two NiCd cells.
3. Low parts count.
4. Very low cost so that it doesnt contribute significanly to the
cost of
any flashlight, especially flashlights with only one or two LED's.

Future ideas:

1. Being able to modify the circuit to provide a constant current
through the
LED's for the full life of the battery. Some people really want this.
Im not sure which i like better, constant current or allowing the
brightness to fall off a little as the batteries wear down. The
falling
brightness gives you an indication that the batteries are wearing
down
rather then have the light suddenly go dim. Im not sure what i
like
better, but i would like to provide the option of adding constant
current
control if desired.

Some undesirables:

1. Dont add a cap in parallel with the LEDs though, because the
transistor will have
to discharge it and that will generate unwanted heat as well as
waste energy.

2. Also, for driving LED's it should be unnecessary to require a
Schottkey diode
in the output circuit, but i wont rule it out completely when more
then one
LED can be wired in series. With only one white LED, the lost
efficiency that
With two LED's
wired in series that drops to about 5 percent. With three in
series that drops
even lower to 3.33 percent. Of course you can only wire them in
series with
a circuit that can provide both the required voltage and current
to the LED
at a decent duty cycle.

Good luck with your LED circuits,
--Al

--------------------

LED's vs Bulb's, the battle is on.

http://www.nteinc.com/Web_pgs/device_list.html

From what i see there, it looks like the
NTE11 transistor is a good transistor
for this kind of circuit :-)
From what i saw of the 2N3055 transistor,
the gain is low and so is the frequency,
so i would rather use another type, unless
i didnt have anything else laying around.
It will oscillate but with a much lower
value of R2 and only at a lower frequency.
After all, its original app was linear audio,
not switching. The NTE11 is about 10 times
better except of course the collector current
rating is lower (but still plenty high
enough for this app).

Oh by the way, five of the most important
transistor characteristics:
HFE=dc current gain
Vceo=max voltage across collector emitter
IC=max collector current
FT=transition frequency
Vsat=collector emitter saturation voltage

HFE is the current gain in a dc circuit.
For example, if you have 2ma going into the
base and 200ma flowing through the collector,
you have an HFE of 100 because 2ma times 100
equals 200ma.

Vceo is just the breakdown voltage of the
transistor across collector and emitter.
If you exceed this voltage, the transistor
will break down and be permanantly damaged.

IC is just the maximum current that can
flow through the collector before the
transistor breaks down. Sometimes there
are two ratings for this:
continuous current and max current.
The continuous current rating is just that,
while the max current is the max for a pulse
of current.

FT is just the ac frequency at which the transistor gain goes
down to 1. Since
the gain goes down as frequency goes up,
at a high enough frequency the transistor
isnt useable any more because you cant get
any gain out of it, and thats what makes the
transistor useful in the first place:-)

Vsat is just the voltage drop across the
collector and emitter when the transistor
is turned fully 'on'. There is usually a
collector current spec with this (like
200ma or something) so you know what current
is flowing through the collector at the time
the Vsat is spec'd. If you go above this
current, the vsat goes up too.
The Vsat spec is important for high efficiency and for app's that
have to
work at very low voltage (like 1.5v).
The lower this spec, the better the transistor, but keep in mind
most NPN
transistors like we are using all have
Vsat below 0.4 volts for currents around
150ma or so.

Oh as far as the parts go, i would say if
you like building these circuits up then
you should probably get more parts to
try out. You can get packs of 50 resistors
with an assortment of values. Do you order
on the web too?

Well good luck with your LED circuits,
and ill be posting more circuits on my