Driving LEDs with a battery pack

Maybe I missed it, but I get the sinking feeling that the discussion is overlooking coupling. You both seem focused on inductance, but I don't see where you are considering it (coupling). Jon raised the issue earlier, but we didn't go very far with it.

Ed

Reply to
ehsjr
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I am keeping it in mind and I am DEFINITELY not focused on inductance. It's not a goal, it's the result of other goals.

In the gapped core case and if you read my response to the same post you are responding to here, I wrote, "I guess the secondary wound elsewhere would be linked well enough by the ring of material so that flux changes would induce the desired voltage without the kind of usual poorer linking seen in pure air cases." Note that I brought up linking where it wasn't mentioned, exactly because that is important in this case. The base winding is a help in squeezing out the last bit from a battery.

And although I'm not sure of what I imagine here, I think a gapped core would link just fine. The key is to wind the collector winding over the gap and the base winding over the core, elsewhere. Flux changes in the air gap as energy builds up would be linked readily by the low reluctance path of the core through the base winding, which would have the desired induced voltages. I admit I'm not positive on this point. It just seems 'right' to me. All the core in a gapped situation does is provide a low reluctance pathway between the north and south poles of the air. The spiraling electron charges create linear force lines through the center (a set of fixed magnetic lines induce a force on any electric charge that is always perpendicular to the path of the electron motion, causing electrons to naturally spiral when in motion across a fixed magnetic field... the reverse is also true, that spiraling electrons must create linear magnetic field lines.) These magnetic lines yield an N and S and with the rest of the core, the lines can easily pass to make the necessary closed loop but via the low reluctance core. So any winding on it should experience changes in the flux density created in the air gap, I think.

Jon

Reply to
Jon Kirwan

It seems right to me, too.

Also, since a negligible amount of energy is required from the base winding, it can be ignored for the purpose of calculating energy storage needs.

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Greg
Reply to
greg

I don't see how that helps.

If the voltage across the capacitor is nearly constant, then to get power out of it at a steady rate, the current needs to be nearly constant as well.

I don't immediately see a mechanism that ensures this, because the capacitor voltage will be sitting right about where the LEDs are starting to conduct appreciably, and their voltage-current curve is very nonlinear. So a small change in the capacitor voltage will result in a large current change.

Perhaps the question I should be asking is: How do you calculate the ripple in the LED current when there is no resistor?

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Greg
Reply to
greg

I don't think it was used as a free-running oscillator, but more like a monostable. You trigger it, and it produces a pulse of known length.

I've tracked down a reference that I had in the back of my mind. It's from Digital's "Small Computer Handbook", talking about pulse amplifier circuits used in the PDP-8:

"Modules such as the Type W640 use a transformer-coupled pulse forming circuit to produce standard 400ns and

1us pulses. The time required to saturate the inter- stage coupling transformer determines the output pulse duration."

They also give a circuit, which I'll try to find a way of posting...

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Greg
Reply to
greg

I don't think you need to know the actual air gap, just the average permeability of the core material (taking any explicitly-introduced gap into account).

If I understand correctly, the energy stored per unit volume in the magnetic field depends on the strength of the B field and the permeability of the material it's passing through. So knowing the average permeability and the total volume of the core plus gaps, you can calculate the energy stored per unit of B. Knowing the value of B needed to saturate the core material then tells you how much energy the core can store in total before saturation.

I'm not sure that you really need to involve energy in the calculation for the purpose at hand, though. The goal is to build up a certain current I in a time that's not too short or too long.

Knowing the average permeability and the geometry lets you calculate the inductance for a given number of turns. So you can choose a number of turns giving you an inductance that results in a suitable period. But you need to check that the B produced by those turns at current I doesn't exceed the saturation value.

If the number of turns required to give the desired inductance is excessive, you need to use a core with a higher permeability.

If the core is going to saturate, you need to use one that won't saturate so easily, which probably means using a higher permeability material or adding more air gap.

That will lower the inductance, which will reduce the period. If it makes the period too small, you'll have to add more turns to bring the inductance up again. But that will raise B, bringing the core closer to saturation...

At this point my head is starting to hurt. I think we need an equation linking all these things together, to make the effect they have on each other more apparent.

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Greg
Reply to
greg

Here's the pulse transformer circuit I mentioned earlier.

Haven't quite nutted out exactly how it works yet. Seems to have some similarities to the Joule Thief circuit, but there are some tricky things going on with feedback.

-3V

| .--------------o----------------------------. | | | | V | | - | | | | | | | | .-------o--------o--------. | | | | | | | .--- | | | | | .---. | | --- V .-. .-. *| | | | --- - | | 3k | | 750 C| |C |C | | | | | | | C| |C |C Output |/ | | '-' '-' C| |C |C ---| | === | | | |* | |> | GND | o--------' | | | | | | | '--- | | | - | | | | ^ | | | | | |/ '------o------------------------------------| | | | |>

| | | | | ___ | | ___ | '-----|___|------o--------o---|___|----' | 390 | 10 | |

-15V

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Greg
Reply to
greg

Base unconnected on that transistor?

Jon

Reply to
Jon Kirwan

Yes.

Jon

Reply to
Jon Kirwan

The reason I mentioned that is that the BJT/transformer/resistor (plus battery voltage) set the power and frequency of delivery of increments of it. (Actually, the transformer doesn't have anything to do with the power, just frequency of operation.) The frequency determines how much of it is parceled out per unit time. So if the power is P and the frequency is f, then it is pulsed with energy equal to (P/f.) Meanwhile, the LEDs take their power, as well.

LEDs can be treated as linear for purposes of small changes about a nearby operating condition, so that the power of the LED is P_led(V) = (V/R_led)*(V-V_start), where R_led and V_start are those two modeling parameters I've talked about. The change in voltage on the capacitor, when it is pulsed with a bit of energy from the collector winding, will be tiny because the energy is added to a capacitor that is already packed up with energy. Two states are: V(1)^2*C/2 before the pulse and V(2)^2*C/2, after. The difference is (V(2)^2-V(1)^2)*C/2. Assume the energy delivered is P/f and V(1) = 20V (stack of 6 LEDs at their operating point of 3.3V, rounded up.) Compute V(2), selecting some chosen C value. Call f=50kHz, C=470uF (well, that's what the OP said he was using!) Since they are operating at a presumed 20mA, we must assume that the circuit is designed to deliver 20V*20mA or 400mW.

V(2) = SQRT( 2 * 400mW / (50kHz* 470uF) + 20V^2)

or, 20.00085V

So the voltage went up about 850uV. Not all that much, really.

I just wanted you to think about in energy terms. The ripple on the cap would be huge if the starting voltage of the cap were zero volts

-- almost 185mV change for the first pulse. But once the cap is charged up, ripple is actually quite modest... under 1mV per pulse. This really is NOT going to change the power through the LEDs that much.

Jon

Reply to
Jon Kirwan

The book didn't show the whole circuit, just the output stage. Presumably the base is connected to something which turns the transistor on when a triggering pulse is received.

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Greg
Reply to
greg

Okay, I see what you mean now. You're right, if the capacitor is big enough, the dynamic resistance of the LEDs won't change much over the ripple range.

Also, conservation of energy ensures that the voltage will settle at a point where the average LED current is at a predictable level. Quite a neat system, really. The LEDs function as their own built-in current regulator!

So yes, using a capacitor makes sense if you've got room for it.

There's just one other thing that was bothering me. I had something in the back of my mind about LEDs being more efficient when pulsed. But after reading the discussion about it here:

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it seems that this only applies if the maximum current you have available is a lot less than the optimum rated current of the LED. If you do have enough current, then you'll get the most light by running them continuously at their rated current.

That will be true in our case provided that the battery can supply enough power.

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Greg
Reply to
greg

Well, you don't _need_ a core, at all. An air wound coil works well. (A ferrite core may work better, but that is not the point I'm addressing.)

To make this repeatable for anyone who wants to try, I used readily available material: 1/2 inch PVC pipe as the coil form, and CAT5 cable from which I extracted one of the 4 twisted pairs. The extracted pair gives you bifilar. I wrapped a single layer close wound, 40 bifilar turns around the pvc. I used the "standard" joule thief circuit with a red LED, and changed the base resistor value to

51 ohms. No output cap or catch diode. It makes the red LED glow merrily and runs at ~660 khz. Best of all, while it does glow less brightly as the voltage is lowered, it produces easily visible and useful glow all the way down below .50 v supply. There is a kind of "knee" in the brightness at about .44 volts and the glow is just barely detectable at about .38 volts. I imagine those values will differ slightly with different layout, different NPN (mine is 2n3904), etc. A caveat - there is significant loss in my setup between the power supply and the breadboard, and that contributes positively to the operation. I'm flying in the dark, trying to simulate a weak battery by the seat of my pants. I don't know what the impedance of a "weak" battery is. I do know that if I bypass the supply where it is delivered to the bread board with 100 uF, the glow gets dimmer. I may find time to fiddle with the supply Z and post some numbers on that as well as Ice, but really, it was not the intent to optomize or even fully measure things. The purpose was to devise an easily repeatable experiment that proves that core saturation is not needed. You don't even need a core at all to steal joules. :-)

Ed

Reply to
ehsjr

Thanks, Ed. This confirms my earlier analysis about it, in spades. Unless someone wants to argue that it is the PVC part of it that saturated and made it work. ;) I like the idea of taking cat5 cable (or cat6, which is what I have laying about) and stripping out the twisted pairs for winding. I'm going to try that, despite having so much magnet wire laying about. Can you provide all the dimensions of your air core, though? I'm curious about the approximate inductance

-- for frequency calculations, as I'm wondering if part of the reason it was only 660kHz had more to do with the NPN than the transformer. Oh, and did you lay all of the 40 turns side by side, or stacked in any way?

Thanks again, Jon

Reply to
Jon Kirwan

Jon Kirwan Inscribed thus:

If you recall, in a previous post, I mentioned that I had tried an air core on a 5mm former. I also mentioned that putting a core in improved the behaviour.

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Best Regards:
                Baron.
Reply to
baron

They can argue - but they would be wrong. In spades. To prove that, just take about 8 or 9 feet of the twisted pair. Don't wrap it into a coil - just connect it to replace the transformer, and the thing plays fine.

Frequency is very much voltage dependent. The 660 khz occurs at .4657 volts. Run the voltage up to around .8 and it runs at ~ 200 khz. If you set aside consideration of the value of L and the frequency value, just analyze the circuit, and the reason for the delta F becomes obvious. You have Vsupply + Vinductor = Vtotal and Vtotal lights the LED. Current through the LED discharges L. But the LED stops conducting when Vtotal < Vf causing the pulse in T1. The time taken to discharge depends upon Vtotal, and Vtotal is lower when the battery voltage is lower. Therefore, the frequency rises as Vbatt decreases. This is easily confirmed experimentally. I did it today with a battery instead of my supply to avoid the jury rigged attempt to make the supply look like a "weak" battery.

Regarding the NPN limiting the frequency - it'll run a lot faster in this circuit with different L. I had it running at ~1.3 mhz with one of the setups I played with.

Getting back to your specific question about the inductor, I don't think actual dimensions matter for demonstrating that air cores work, but here they are: PVC form is 1/2" dia sched 40. (The od measures 13/16") The length of the PVC is 3 1/4" and the windings occupy

2 3/4" of that. I drilled little a little hole 1/4" in from each end and passed the wire through - that way, the coil doesn't unwrap. To make it, I stripped 20' of twisted pair out of the CAT5 and close wound it tightly, one layer, around the form. I left 1' of twisted pair sticking out of each end, and had 8' 8" of twisted pair left over. So there's 9' 4" of wire in the coil itself with 1' leads at each end to connect to the breadboard. Each coil's inductance measures 10.96 uH.

Side by side turns, no stacking, just close wound. I can send you a photo if you want.

Ed

Reply to
ehsjr

I was curious about that. What did you use for the form? Finding a ~.2" dia form here would be tough without knowing what to look for. Maybe aquarium tubing? What were the coil dimensions? In the post you mention, you said it would oscillate, but only without a load. Is it the "standard" joule thief circuit but without an LED? Did you try changing the base resistor?

Thanks, Ed

Reply to
ehsjr

In that case where is the energy stored...?

I thought the Joule Thief worked by storing energy somewhere then releasing it when the transistor switches.

/Still waiting for my magnet wire to arrive - I had to order some on eBay.

Reply to
fungus

The energy is stored in exactly the same "place" whether using an air wound or ferrite core transformer: the magnetic field.

When you have a transformer wound on a core of magnetic material the magnetic field is concentrated in the core. With an air wound transformer, the magnetic field is not as concentrated, it occupies a larger volume.

In all cases, the energy is stored in the magnetic field, regardless of where that field is - in a ferrous core or in air.

Ed

Reply to
ehsjr

So there's some energy stored in the air around the wire?

That's too weird for my tiny brain.

Reply to
fungus

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