Driving LEDs with a battery pack

That's the nature of the millicandella rating though. It is easy to make a poor performer sound good simply by narrowing the focus of the spot of light.

The same total flux of light from a led, is brighter when the light energy is concentrated in a smaller beam. The led chip and total flux output can be the same, and the manufacturer just offers different beam angles by changing the shape of the lens.

You, no doubt, have an idea of the size of the area you want to illuminate. You select based on the distance from the source and illumination needed and choose the angle and candela output to match.

I have some 25 leds up on a pole illuminating a ramp I use to load my kayak in the dark - narrow angle, high output leds work better for that application.

I don't think so. I can do the calculations, and achieve accurate results from them, without taking saturation into account. I'll try and give a more detailed account than I have, if interested.

Jon

"Breaking down" and going bzzzzt internally because the voltage is too much. It doesn't make the transistor better. ;)

Jon

Obviously...

So you think the LEDs this guy sells are just narrower focus than the ones I have, ie. total number of photons is approx. the same for all the 2.0V @ 20mA LEDs out there?

Okay. Here's another shot at the general idea. I'll write this in ASCII characters. You'll need to use a fixed-spaced font to read it, such as Courier or Courier New or Prestige Elite. Notepad will do that, so will many newsgroup readers. Even google can do it, if you look around for the option on the page.

I'll assume 6 LEDs and 3 batteries:

Tweek your transformer to get in the vacinity of 20-50kHz. The LED current will vary somewhat less, closer to 50kHz, but down at 20kHz it should look fine, as well. C1 should be at least 35V, but 50V is safer.

The value for R1 can be as you see fit. I'd stay at or above 1k and you can try it upwards of 3.3k. I tweeked things so that 2.7k appears to work fine.

LED current will definitely droop as the battery voltages also droop. Nothing in there to help that problem. As battery voltage goes down, so does base current. Lower base current, lower peak collector current. Lower peak collector current, lower energy stored and transferred. Etc. But I think it will move down by the ratio of voltages, so no worse than 1.1/1.5 or about 3/4ths.

What you need to add is C1, R2 (not strictly necessary, I just added it so you could play with some values there), and D8. Some folks may recommend a schottky diode for that, because they present a somewhat lower voltage and they switch fast. But probably you can use what you have available.

Oh, and D7!! I added that to protect your Q1 base-emitter from reverse voltages. That will help protect your poor BJT.

If you try this, let me know what happens!

Jon

Tweak!

Jon

If you read 'default' more closely, you'll see him write, "They also have some impressive mcd numbers ALONG with fairly large beam spreads." Note that he includes "___ALONG with___" here and describes the idea of large beams. So, actually, I think 'default' is saying that these seem to be pretty impressive not just for having high numbers but also for having those numbers while also having wide spreads.

So to the opposite, I think. I don't think he thinks they get their numbers from a narrower focus. He feels pretty positive, I gathered.

Jon

I think you _are_ mistaken.

No. Saturation of the core isn't required. This would work on an air core transformer, I believe.

....

The base current starts out unaided by any induced voltage on the base winding, but still turns the BJT on. Once that happens, though, the collector drops immediately to near zero (there is no collector winding current to speak of, at that point.) This places a near-fixed voltage across the collector winding, which allows the collector current to rise according to V/L. Almost from the first moment this takes place, there is an induced voltage in the base winding due to the rate of flux change on the collector winding which adds to the battery voltage (if wired with the correct orientation, of course) and this increases the base current to it's initially 'highish' value of (2*V_bat-Vbe)/R.

At first, the Vce of the BJT remains very close to zero because the Ic/Ib is well below 10. (The BJT is severely saturated.) But as Ic rises along the ramp of V/L, while at the same time Ib remains close to fixed, it eventually reaches the point where Ic/Ib goes over 1, then goes over 10, then goes over 20, etc. 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. That reduces the base current, but does so exactly at the point when higher Ic requires more, not less, Ib. In other words, the aiding voltage by the base winding is falling and _reducing_ base current right at the point where Vce is rising and reducing the voltage across the collector winding.

Suddenly, the beta just isn't enough and the BJT attempts to reduce Ic. As soon as it 'tries' to do that, though, the collector winding immediately responds by reversing its polarity as the only possible response to allow a reduction in Ic (V/L must reverse its sign.) But that immediately causes the base winding to also reverse its polarity and __oppose__ the battery voltage that is struggling to drive current into the base. The whole thing collapses with the battery voltage opposed by an overwhelming reverse polarity, base current goes to zero, base voltage goes below ground, etc. The BJT is off at this point.

After the collector winding reverses voltage, current is driven now through the LED as the reversed-sign V/L allows the current through it to decline along a ramp (the LED maintains a fairly fixed, but gradually declining V/L.) At some point, the LED is no longer able to accept much current (non-linear decline, as well) and the collector winding's field entirely collapses and has no energy remaining. It's voltage goes to zero, so does the induced voltage on the base winding, the battery is now able to generate some current into the base of the BJT, the BJT turns back on, a voltage is applied to the collector winding, the collector winding induces a renewed aiding voltage on the base winding, the base current rises a bit, and the whole cycle repeats.

No saturation of a core invoked here.

But I disagree. And, it appears, so does LTSpice where I don't have to add any saturation effects to get it to oscillate just fine.

Hope that helps. Or, if you find good fault with my reasoning, it will help me. Either way, it's all good.

Jon

I added D8 and a big scary electrolytic capacitor* I found in my box at C1 (I don't exactly have a big selection of components to play with and that's the only capacitor which looked likely to do anything...)

The capacitor adds a cool effect - the LEDs fade up/down when I switch the thing on/off.

But ... it works! I think I'm pretty much at full brightness with the capacitor in there.

Looking on the scope, the voltage at the first diode is perfectly flat (to be expected).

Best of all ... I measured the output current and with 4.2V input it was 106mA for six LEDs, that's an average of 17.6mA each - right on target.

With some half-depleted batteries I got 104mA across the LEDs from 3.8V input.

With two half-depleted batteries I got 70mA from 2.5V

- still quite respectable.

I'll be thinking of him...

So far so good... now I want to work on the size of the ferrite bead (I want this to fit in a more discrete package).

I'm definitely interested... :-)

Which leads to another question before I go parts-shopping tomorrow ... what sort of capacitor is best for this? There's a zillion different types.

Electrolytic for cheapness, 10uF or bigger, 35V or higher. I don't think it matters that much, otherwise. You can strip something out of some electronic device (the same one you got the transformer core out of?) Don't shop if you don't have to....

Jon

Yes, that's right. Which means you need to keep the frequency low enough so that the switching time is small compared to the on time.

If you have a scope, you could look at the voltage between the emitter and collector. It should be close to a square wave, with steep slopes on the rising and falling edges, for best efficiency.

What part of the PC? If it was somewhere in the power supply, it's probably okay.

It means that there is a high enough reverse voltage being applied to the base-emitter junction for it to break down like a zener diode, which could damage the transistor.

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

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.

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

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

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.

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!

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.

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

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?

There's dozens of capacitors within easy reach but they're all soldered to PCBs and only have 1mm legs... :-(

Wasn't from the PSU. I just opened a PSU and pulled a couple of smaller ones out. They're made of blue ceramic or something and they stick a lot harder to a magnet than my big one does. I'll give them a try when I can get some better wire.

PS: Hard disk magnets ... crazy strong. I nearly lost a finger.

No. The leds this guy sells specify beam angle, and has a range of beam widths.

I'm very happy with the tricolor leds I got for a project. I'm using them for power indicators and the wide beam lets me see them from anywhere in the room with any ambient light level.

His price was 60 cents a led versus $2.80 from a US importer of the same part.

Only negative comment I wish he'd link to the actual data sheets for the Leds instead of the abbreviated ones he provides.

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.

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.

I think in a profoundly different mode. But yes.

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.

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

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

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