Re: Driving LEDs with a battery pack

On Thu, 9 Jul 2009 03:51:33 -0700 (PDT), fungus
<openglMYSOCKS@xxxxxxxxxx> wrote:

On Jul 8, 12:41 am, Jon Kirwan <j...@xxxxxxxxxxxxxxxxxxx> wrote:

The brightness should also be affected by the base resistor value,
too. Did you try changing it to a lower value -- say about half?

I'll put it together again with my variable resistor see what happens.

Make sure you wire up the transformer, correctly, too. The base side
must _aid_ battery voltage when the transistor is on, not oppose it.
If it isn't working at all (or well), try reversing the leads there.

So *that's* why it didn't work when I put it back together
yesterday... !

hehe.

I tried my variable resistor and it makes a bit of
a difference in brightness. I looked at it with an
oscilloscope and the resistor changes the pulse
frequency.

If you dial it too low the circuit makes a scary
whistling sound. A tiny bit more and it shuts off.

Yeah. The frequency gets well into hearing range.

Can a joule thief power multiple LEDs?

Yes, but don't wire them in parallel. You'd wire them in series.
There is a limit, but with no more than 6 (I think you wrote that) the
required voltage ... at worst ... is 6*3.3V or 19.8V, which I think is
doable. I could play with one here and see, to be sure.

I did six LEDs and it works but quite dim. I tried two batteries
and it's a lot brighter (maybe twice) but still far less bright then
a single LED. In my informal 'point it at the wall' test all six
together were only putting out about the same amount of light
as a single LED at 10mA.

Pulse width goes down as the LEDs stack up. It doesn't get better
that way. But at least it works!

The computation of the peak current actually requires some reading of
the datasheet for the transistor, though you can approximate an overly
high estimate without much information.

I_peakmax = Beta*(N*(V_battery - Vcesat) + V_battery - Vbe) / R_base

Note that the transformer's primary or secondary value in Henrys
doesn't factor into this. What it affects is how long it all takes.
So if you are hearing a terrible racket, that may mean your primary
has too high a value. To fix that, unwind some windings. That will
reduce the inductance (which goes by windings^2.)

You can see that R_base has a definite impact. So does the battery
voltage.

To make much reasoned use of the above equation, N is the ratio of
windings: base winding count / collector winding count. If that is 1,
which it probably is close to, then N=1. You can boost the peak by
increasing that winding ratio a bit. But the risk is that if you wind
the ratio too far, during OFF time for the transistor the base voltage
will be driven even further below the emitter. More than 5V below and
it is likely the transistor will break down. So you need to be
careful here. If you are using a 1.5V battery only, you may be able
to go to N=3 or N=4. If you are using higher voltages, then N=2 might
be a better limit. This is why just keeping N=1 is so safe. It aids
battery voltage, which is good, but not so much as to risk the
transistor.

As a 0th order aproximation, I'd use Vcesat that is equal to Vbe and
beta of 200 (for 2N2222 [which is typical], maybe a little more for a
2N3904.) What happens very near turn-off on the transistor is that
the collector current rises along a clean ramp determined by the
collector winding inductance and the battery voltage (V/L is the
rate.) But meanwhile, the base current remains largely flat (slightly
declining for most of the time.) The ratio of the two, Ic/Ib, is the
beta. At the first of the ON time, that's basically very close to
zero and during which Vce on the transistor is very close to zero
volts. But eventually, as Ic increases, this beta value increases
past 1 and grows towards some transistor limiting value (say 200.)
During this later transition point, Vce also rises. By the time Vce
reaches Vbe, the beta is pretty close to 200 on a 2N2222. Maybe a
little less. But in that area. So let's assume Vbe is about 0.8V and
therefore Vcesat is the same. That should get close.

Assuming N=1 and Vbe=Vcesat, the equation becomes:

I_peakmax = 2*Beta*(V_battery - Vbe) / R_base

With a 1.5V battery (fresh) and a 2N2222, an estimate becomes:

I_peakmax = 2*200*(1.5 - 0.8)/ R_base = 280/R-base

With R_base=2800 ohms, that would be 0.1A peak. But as the voltage
droops, so does the peak current:

I_peakmax = 2*200*(1.1 - 0.8)/ R_base = 120/R-base

In that case, with 2800 ohms, to about 43mA. Less than half.

Note that nowhere in the above equations is the transformer
inductance. That mainly affects frequency. To speed it up, use fewer
windings. To slow it down, more. Too fast and the transistor won't
keep up. Too slow and it will blink or hum.

Higher battery voltages will improve your duty cycle, too, because the
time during which the LED is on is determined by the difference
between the required LED stack voltage and the battery voltage divided
into the inductance. Higher battery voltages reduce the difference
and thereby increase the time. So you might try doing that, just so
long as you stay underneath the LED stack's voltage and DON'T risk
destroying your transistor's base-emitter junction. (Adding some
protection there might be a good idea, now that I'm thinking about
it.)

There is another problem I forget to mention (I'm just a hobbyist.)
The transformer might get saturated. This really becomes a problem
when you are using low frequencies. So one idea is to REDUCE your
windings! See what happens.

Unless this can be improved I'm not sure the JT is the circuit
for me.

It probably can be, though I haven't considered your exact situation
and I probably need to do some playing, myself, before passing along
some ideas. I'm short on these green LEDs (zero), so I need to get
some, first.

(I tried three batteries and the transistor died. I went back
and checked the temp with two batteries and yes, it was
quite warm...)

Ah. Well, there is that. But actually, I'm guessing you may have
zapped the base-emitter junction by the reverse voltage. One
possibility here is to protect that junction by a reverse oriented
diode across it. (Okay, I need to think more about that.)

How important are the number/neatness/type of wire of the windings
on the transformer? (or does it make little difference at 50kHz?)

Within reason, I think you are fine. You cannot use bare wire, for
obvious reasons (shorts itself all over the place.) If you use
insulated wire, your spacing will be pretty wide. Better is 'magnet
wire' which has just enough insulation and not much more than that.

I was using insulated wire because that's all I had.
Do you think the magnet wire fix (or at least help with)
my brightness problem?

To be honest, I'm not sure. I think that mostly affects just how much
inductance you can wind.. but not much else. When I said "better" I
just meant that you have a wider range of choices in winding.

I think 50kHz should be fine for most BJTs. I don't recall the
frequency I measured before, but it seems it was in the 10's of kHz,
so your question is in the right ballpark.

The 60kHz number was from one of the web pages.
I measured about 40kHz.when I had the 'scope
connected.

Okay. That suggests that you are about right, to me. I wouldn't
reduce your windings much, then, unless you do that in conjunction
with reducing your base resistor value. Cutting down the base
resistor value increases the peak current. But getting there takes
longer. Making that take less time, while keeping that lower value
resistor, means cutting down the inductance on the transformer's
collector winding. So the two kind of go together.

Jon
.

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