Wow...

I didn't realize the LED color might make a difference, or maybe better put, I didn't realize that the Vf would make that big a difference.

My first thought is to use 1 red and 1 green Luxeon III LED.
Green: http://ledsupply.com/lxhl-lm3c.php (Vf 3.9)

Where did you read the Vf=3.9?
I'm seeing Vf(typ) = 3.7 @ 700mA on page 6. Oh, were you reading the Vf @ 1000mA? Yes, that makes a difference.

Red: http://ledsupply.com/lxhl-ld3c.php (Vf 3.2)

OK, the red Luxeon LED shows a typical Vf of 2.95v @ 1400mA; it'll be even lower at 700mA.

But then I was pointed toward these:
Green: http://ledsupply.com/creexre-g.php (Vf 3.6 @ 700mA)

The datasheet indicates that Vf is 3.5v @ 700mA for green LEDs on page 4 (by default)

Red: http://ledsupply.com/creexre-r.php (Vf 3.2 @700mA)

This 2nd link shows the same datasheet; red is not an available color. Cool White, Neutral White, Warm White, Royal Blue, Blue and Green are the only colors listed in that datasheet.

 

This link is related; might be helpful 'though dated: http://bikeled.org/elec_LED.html

 

That link is pretty interetsing...I'll have to read through it more carefully tomorrow!

 

I don't know if you've planned for heat dissipation from the LEDs.

Basically, power in Watts = voltage x current.

So, if you have an LED with a Vf of 3.5v that you're driving with 700mA current, the LED will be dissipating 3.5v x 0.7a = 2.45 Watts of power.

Now since you're planning on these LEDs to only have a 50% duty cycle, the average power dissipation will be 1.225 Watts. But still, you'll have to do something with that heat, or the LEDs will self-destruct.

 

I hadn't thought of the heat sinking for the LEDs, should have.

I think you're wrong about how they work, though I can't prove it via the data sheet. The data sheet does say this...

I suspect the only way we'll prove or disprove this is by talking with someone that has hands on.

If I'm right then the conversion would look something like this...

3.6V @ 700ma = 5V @ 504ma (100%)

Figuring 10% loss this number would become:

504ma / 90% = 560ma

Time will tell. Maybe the OP will tell us the current draw when he builds it.

 

OK, so we're in trouble anyway.

"Due to the nature of the buck regulator, the input voltage must always be higher than the total forward voltage drop of the LED junction(s) connected in series, 2.0v for DC models..."

With two CR123 batteries @ 750mA draw, the input voltage will be 5v; 5v-2v=3v - and TJ is looking at LEDs with Vf's of 2.95v and 3.5v-3.7v. The red LEDs will likely work while the batteries are fresh (unless he happens to get one of the higher Vf LEDs), but there's not enough headroom for the green LEDs.

 

I'd be tempted to try it as is, but it would be safer to use a 3rd battery (damnit).

 

I've written to Luxeon's service department to get their answer to the 737mA vs 560mA debate.

Now the question is, do you prefer roasted, baked, or pan-fried?

 

If they could come back with that overhead question it would be nice too. I suspect two batteries would work, but 3 would let you suck em dry before you throw them away. A lot of parts vendors tend to be conservative, they have to be.

***************

You know, I may redo this project on another thread, to develop an alternate to buck puck. Reason I would go with a different thread is because it would take a little while to do it right. A current sense, a digital MOSFET, op amp and 555 might do it. No coils, but I think caps would be needed.

 

Gee, the 2v overhead requirement for DC models (4v for AC models) is already stated on the website and in the datasheets. Do we really need to confirm it?

I'm really rather inclined to take the website and datasheets' word for it. It might give near the rated output at less headroom, but I'm pretty certain it won't be up to specifications. The red LED's Vf is really rather close to the 2v headroom specs; just 100mV (or so) to spare if the one TJ gets falls into the "typical" range; but you can't specify the maximum Vf in your order or EVERYONE would specify the lowest Vf, and the higher Vf LEDs would sit lonely in a huge bin somewhere.

Take a look at the link in my reply #42.

 

The message I sent:

Dear Luxeon Representative,

A friend has asked me to confirm the current input requirements for the BuckPuck 3023-D-N-700.

LED being driven: Luxeon Star III, 2.95v @ 1400mA, but will be driven @ 700mA.
Power supply: 5V DC.
The minimum 2v headroom is satisfied.

From your website, the stated efficiency of the BuckPuck 3023-D-N-700 is 95% when Vin is 5v.

I calculated that the input current will be approximately 700mA / 95% = 736.8mA.

A colleague is suggesting that the input current will be closer to 560mA; but that simply does not make sense to me.

Would you please confirm the input current requirements for the 3023-D-N-700 from a 5v supply when driving a 2.9vf LED @ 700mA?

Thank you for your attention.

...And the word is:

Your calculation is correct. you will need at least 736mA input current. Ideally you should use 750mA to 800mA.

Keep in mind that the forward voltage of the Luxeon III LEDs can vary from 2.31V to 3.51V, so ideally you should be using a 6V 800mA power supply to ensure reliable operation.

Cheers!

Ron Warris
Luxeon Star LEDs

 

OK, NOW this is bugging me.
The specs say 95% efficiency with a 5v input.
But the Luxeon rep is suggesting 750mA to 800mA current.
If the input current is 750mA, that's 93.3% efficiency.
If the input current is 800mA, that's 87.5% efficiency.

The 800mA suggestion may just be for the "start up" overhead.

 

Here's the part that bothers me, if it drops the voltage, but the current is the same, that should become heat. I don't believe these parts become hot, they don't look designed for it. With conversion the heat issue took care of itself.

 

Here's the part that bothers me, if it drops the voltage, but the current is the same, that should become heat. I don't believe these parts become hot, they don't look designed for it. With conversion the heat issue took care of itself.

OK, so that's where you're getting confused.

Basically, these BuckPucks are just little "buck"-type switching current supplies.

Imagine if you will, a PWM circuit powered by 6v-9v used to control the current through a standard red 2v @ 20mA LED. Without current limiting (like using a resistor), there is no current flow through the LED on part of the cycle, and the other part of the cycle the current is WAY too high.

An inductor is added to the current path. The inductor resists any change in current. If the PWM frequency is high compared to the inductance value of the coil, the current output is much more constant than if the inductor were not present.

The inductor does dissipate a small amount of heat, as does the switch (MOSFET) and the current sense element. However, these losses are minute compared to a linear regulator or plain resistor.

In a standard "buck" regulator circuit, a "flywheel" diode is used across the load and inductor to provide a current path from one end of the coil through the load and diode when the current source is cut off. In a synchronous buck regulator, a MOSFET is used as an "ideal diode" to minimize the losses in the "flywheel diode"

The current flow is controlled by measuring the voltage across a small value resistance. When the voltage drops too low, the current source is switched on; too high and it's turned off.

 

I understand this, but we are still dealing with black box electronics. The excess has to go somewhere. If a box is passing same current in and same current out with DC, and dropping X amount of volts, this will be heat.

The PuckBuck is a black box is dropping a voltage (that also sets the current through the LED). If the current is the same on both sides, then you will get heat, this is DC on the outside (and I'm not interested in the innards for the sake of this arguement).

The way heat is avoided in switchers is conversion. The current from the power supply is less than the output current, because the total wattage is approximately the same and the power is passed through. There is no excess power in the system to disappate.

With any constant current source, mode of operation doesn't matter (black box), if it isn't doing a straight through transfer of power, if there is a mismatch, it has to show up as heat. In this system 5V X .7A is 3.5W, the LED is 3.6V X .7A (2.5W), there is a watt that has to show up somewhere.

 

I understand this, but we are still dealing with black box electronics. The excess has to go somewhere. If a box is passing same current in and same current out with DC, and dropping X amount of volts, this will be heat.

That is definitely the case with linear current or voltage regulation; the power is dissipated as heat.
That is not the case with "buck"-type regulation. In a buck regulator, much of the losses are due to the sensing of the current and driving the gates of the MOSFETs.

The PuckBuck is a black box is dropping a voltage (that also sets the current through the LED). If the current is the same on both sides, then you will get heat, this is DC on the outside (and I'm not interested in the innards for the sake of this arguement).

Then answer this for me;
Why does a linear regulator like an LM317 get hot, but a PWM circuit stay cool, if both are passing the same average amount of current?

The way heat is avoided in switchers is conversion. The current from the power supply is less than the output current, because the total wattage is approximately the same and the power is passed through. There is no excess power in the system to disappate.

With any constant current source, mode of operation doesn't matter (black box), if it isn't doing a straight through transfer of power, if there is a mismatch, it has to show up as heat. In this system 5V X .7A is 3.5W, the LED is 3.6V X .7A (2.5W), there is a watt that has to show up somewhere.

Actually, it's just the 5% inefficiency of the buck puck that shows up as heat. The other 95% shows up as heat in the LED. The current drain on the battery isn't a flat 700mA; it's more like 737mA.

I see your point about the "Mysterious Missing Watt". However, the power isn't being dissipated in the regulator (5% inefficiency aside) - the power is being "chopped up" and transferred via "bucket brigade".

 

Then answer this for me;
Why does a linear regulator like an LM317 get hot, but a PWM circuit stay cool, if both are passing the same average amount of current?



Actually, it's just the 5% inefficiency of the buck puck that shows up as heat. The other 95% shows up as heat in the LED. The current drain on the battery isn't a flat 700mA; it's more like 737mA.

I see your point about the "Mysterious Missing Watt". However, the power isn't being dissipated in the regulator (5% inefficiency aside) - the power is being "chopped up" and transferred via "bucket brigade".

The battery will show a reduced load similar to what I was discribing. If the battery is sending 3.5 watts down the line, then there will be 3.5 watts disappated somewhere. PWM is a form of conversion, the current and voltage through the load does not necessarily match the current and voltage from the battery. The battery isn't AC and reactance doesn't apply (and I'm not interested in the innards at this point), the math is much simplier as seen from the battery terminals. The power a battery or any DC source sends out must all be accounted for, though not all at the same point.

While this does apply for the OP project, I think we've stretched the edges of the threads purpose. This does give me another slant to add for the PWM article I've yet to write elsewhere.

 

So to be safe...should I use 3 batteries? Which actually doesn't kill me at all in terms of form factor.

Thanks guys! Reading this has been an education. I now know that I know even less than I thought I knew when we started this whole thing!

 

I would, if it isn't a show stopper. There is also an added advantage that when the power dies it will be all at once, and the batteries will be completely discharged.

I think it would work with two, but it would be much weaker, and die before the batteries were truly exhausted. You look at the current discharge shown in post #31 and you will see these batteries give their all up to the last, if allowed.

It must be obvious you've touched on some technical issues Wookie and I both need to resolve. Part of the reason I do this is you never stop learning, and sometimes the most innocuous projects have something to teach.

 

Cool. I'm glad I sparked the converstaion. And I'm glad it wasn't super easy because then I would have felt even more electronics-stupid than I already do.