Re: Driving LEDs with a battery pack

On Sat, 11 Jul 2009 20:50:17 GMT, ehsjr <ehsjr@xxxxxxxxxxxxxxxxx>
wrote:

Jon Kirwan wrote:
On Sat, 11 Jul 2009 22:46:00 +1200, greg <greg@xxxxxxxxxxxxxxxxxxxxx>
wrote:


Jon Kirwan wrote:


During this time, the Vce
rises, too. As that happens, the induced voltage on the base winding
declines due to a falling rate of flux change.

Interesting -- there appear to be more subtle things
going on in this circuit than I thought!


Actually, it's a beautiful piece of work. The Joules per unit time
(power) is independent of the inductance of the transformer, so you
don't mess things up if you wind it too many times or too few. All
that affects is the frequency of operation. A simple resistor sets
the power. Just a very few components, too. It's sweet.


But it seems like saturation would have much the same
effect -- the flux suddenly stops rising, causing the
base voltage to fall to the point where transistor
can't sustain the current and a flip-over occurs.


The point is that saturation isn't required and will actually waste
power. So I think that saturation would be a problem, actually.

Assume the collector winding saturates the core. This happens as Ic
is rising on a V/L ramp and, for purposes you are putting it to,
occurs by definition well __before__ Ic/Ib gets anywhere near the beta
limit of the BJT. So let's say this happens when Ib=Ic, just to keep
it easy. So the BJT is in deep saturation, still. And the core
suddenly decides "that's it, I'm tapped out!" At that point, it isn't
the BJT, but the core that decides the voltage across it must cease.
So the collector winding voltage goes to zero, suddenly, which means
Vce on the BJT suddenly rises to the fully battery voltage -- meaning
VERY BAD dissipation. You are right. The base winding also goes to
zero volts and the base current takes a hit. But it goes to about 1/2
the earlier value. Since Ic doesn't change, but only Vce did, the
beta is now 2. Which means the BJT can still support the Ic required
(which hasn't yet changed -- the collector winding is zero volts, not
negative, just the Vce has jumped up.)

At this point, we see the BJT with beta 2 (which is still deep
saturation) and a base current that is 1/2 of what it just was but is
still well more than enough. So the BJT's Vce times Ic gets
dissipated by the BJT, which is now heating up big-time. (Beta
capability actually rises with temperature increases, memory serving.)

If inductors were perfect (superconducting?), I suspect things would
just sit there with beta=2, the BJT heating up until it reaches some
stable point, Ic remaining fixed, Ib remaining fixed, and that would
be that. But the resistance in the inductor is instead gradually
(slowly) eating up the field's energy, causing a voltage reversal...
probably on the order of timing less than a second but not nearly at
the design frequency rate. This opposes the battery voltage to the
base, of course, and the whole thing does wind down. But I suspect
that saturation is NOT a good thing here because of all this. In
other words, it is not "a friend" to the process. It's to be avoided.

But I've only two or three weeks ago started studying magnetics design
for the first time. And I may have something wrong. But that's the
way it looks to me.


So the circuit will still work if saturation occurs,
just in a slightly different mode.


I think in a profoundly different mode. But yes.


It might even work slightly better in that mode, if
it causes the transistor to cut off more sharply.


I don't think so. Read what I wrote and see if you can find fault
with it. I'm curious about learning this stuff better.


In any case, it seems like it would operate more
predictably, since the period and maximum current
would depend on the saturation flux of the core
rather than some rather uncertain transistor parameters.


Well, I didn't take your starting point so conclusions from it still
don't flow from it, for me.


If you just pick some random transistor, then it's only
by luck that you avoid overdriving the LED.
Seems like a rather hairy way to design a circuit to me!


Actually, I think it is an incredible arrangement.

Look. You can design this entirely with the idea of just knowing how
much current on the other end you require. Knowing the current, you
can compute the peak current you want in the collector winding. It is
just:

Ipeak = Iout*[2*(Vout+Vd-Vceon)/(Vbattery-Vceon)]

(These assume that diode I suggested to the OP, so Vd is the forward
voltage of it during BJT-off times. Vceon and Vceoff will be rough
numbers used and aren't all that critical. I use Vceon=0.1 [as an
average value between 0.0V and 0.2V during on-time] and use Vceoff of
between 0.4V and 0.7V depending on just how big Ic happens to be.)

Notice that there is no inductance here, no base resistor value, etc.
There is a pure number ratio (bracketed) times the desired Iout. Very
simple and independent of a lot of stuff you don't want to have to
worry about. This makes design easy.

Then, you compute the inductance of the collector winding. At this
point, you need to know some idea about the desired frequency. The
beauty of this is that you can now worry about volt-seconds and that
saturation problem. To avoid volt-second problems, choose a faster
frequency. The frequency will have NO effect on the power/Iout that
is delivered. You get to choose it independently and not worry about
its effects on Iout. Wonderful, for designing. So to avoid huge
volt-second figures, pick a frequency that is high. But to avoid
dealing with BJT capacitance and charge mobility issues, choose a
frequency that is low. Between these is a nice region in the tens of
kHz... so I like 50kHz as a good place to be. At this point, compute
L:
(Vbattery-Vceon)*(Vout+Vd-Vbattery)
L = -------------------------------------
(Vout+Vbattery-Vceon)*Ipeak*frequency

At this point, Rbase can also be calculated:

Rbase = beta*[N*(Vbattery-Vceoff)+Vbattery-Vbeon]/Ipeak

Here, N is the winding ratio, with higher values occuring when the
base winding has more turns. Normally, N=1. Also, beta is selected
from the datasheet at the value for Ic=Ipeak, roughly. Also, Vbeon is
picked up from the datasheet (or estimated.)

Your peak base current isn't simply the Ipeak divided by the chosen
beta, here. It's more like:

Ibaseavg = [N*(Vbattery-Vceon)+Vbattery-Vbeon]/Rbase

Rbase is determined using a higher Vceoff calculation because the
value of beta won't apply well, if applied using the lower Vceon.


Another thought: For driving several LEDs in series,
maybe it would help to use a separate secondary winding
with more turns for the output. That would give a
current step-down relative to the transistor current
and allow the output duty cycle to be increased for
the same output current.


But with the freewheeling diode and a capacitor, there is no need. And
that arrangement provides a nice smooth current. The issue with
pulsing is that the LED voltages rise with high currents flowing and
this wastes energy uselessly. What the diodes really want is a steady
DC at the right level, not huge pulses of current. Best efficiency is
not necessarily at the nominal current for the LEDs, but it usually
isn't too far from it. Pulsing with duty cycles not unlike 1:10 means
10-fold current increases and that is way past their efficiency curve
peak. It wastes power. I think the diode/cap almost suggests itself.


E.g. suppose you have 4 LEDS in series and want to
drive them at 20mA max. If you use a 4:1 turns ratio
and arrange things so that the primary charges up
to 80mA over an on-time of t, it will then deliver
20mA initially to the secondary, ramping down to 0
over a time of 4t before the stored energy runs out.
So the LEDs are driven at a duty cycle of 80%.

In contrast, without the current step-down, the peak
transistor current has to be limited to the maximum LED
current, and the LED duty cycle falls in proportion to
the number of LEDs -- which means you can never get more
light out of series LEDs than you could from a single LED.

Does any of that make sense?


On the surface, it seems easier and better to just add a diode and cap
than to wind another winding in a tiny bead.

Jon

My recollection of messing with the joule thief a few
years ago was that it is transistor saturation, not
core saturation, and that Vbatt is the key.

Which is ALL I've been saying here. That's why I disagree with greg
on this.

In essence,
the base wants to drive the transistor harder, the transistor
is capable, but Vbatt prevents any further increase in Ic.

Actually, Ic rises until Ic/Ib exceeds the BJT beta at that Ic.

BTW - I made one using an air core (we can eliminate core
saturation from that one), and several using ferrite cores
where core sat was perhaps possible. But again, my recollection
is that it was transistor saturation due to the low battery
voltage. I suppose that may be too simplistic - you could get
into the Vdrop within the battery's internal resistance (Rbatt)
and Rbatt limiting the current through the coil ...

If you read some of my longer postings here (it seems you have not),
you will see a VERY DETAILED description. Not to mention the
equations I've provided. So I have a pretty thorough understanding
I'm applying here.

Jon
.

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