Is the LED pulsed? If so, the boost can go directly into the LED without an intermediate DC voltage. The associated inductor value might be embarassing.
A charge pumpy thing could do the same, working off a 5 volt supply and boosting the 24 to 29ish.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Den tirsdag den 4. november 2014 22.04.26 UTC+1 skrev DecadentLinuxUserNumeroUno:
700mA@28V isn't exactly very low consumption-Lasse
SEPIC converters are pretty friendly. The input current is a triangular waveshape that is easier to filter out noise than a simple boost converter. LT has a few SEPIC controllers.
-- Mark
It's a 20 watt LED (28V @ 700 mA). So I need a real power supply.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Looks like it might be going away--high prices, very little stock anyplace AFAICT.
Thanks
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
While you are considering options, don't forget changing to a 28 volt supply and finding a 20 watt led that only requires 24 volts. Both of these solutions do not make the box more complicated.
Dan
The measurement stability depends on the LED current waveform being predictable and pretty close to square. My first choice is to jack up the rail a bit and then run the LED from a switched current source.
I'm leaning towards an LT3580 boost with two sections of RC decoupling on the supply side (0.51 ohm pulse-rated resistors and 3300 uF caps). That gets rid of most of the 1-2 kHz current ripple on the input supply, and as a bonus reduces the 400 kHz switching junk to a very low level. Total cost probably $7 in 100-piece quantities.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
All I need in order to do that is a different customer. ;)
Cheers
Phil Hobbs
>-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Are you going to build 100 egg inspection machines?
Lotta eggs.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
That was my thought... well I was only thinking about a 24V 20W led.
George H.
One of these machines can keep up with a few hundred thousand hens. However, there are a _lot_ of people eating eggs. ;)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
OK, for a single-inductor converter that's right.
Just now I'm designing a laser driver that runs from +5, and I need a bit of +8 for auxiliaries. In that case, it makes sense to buy a little half-sip 5-to-3.3 converter and stack its output on top of +5.
-- John Larkin Highland Technology, Inc jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
On Tue, 4 Nov 2014 13:11:29 -0800 (PST), Lasse Langwadt Christensen Gave us:
Well, the things should be able to be fired from the lower voltage as there are no 28V LEDs. Perhaps LED arrays, or some OTS part, but not individual discreet elements.
So, this must be some already made, already bought subassembly that simply will not "come up" and operate without the added voltage.
Otherwise, there is no reason less than 25W worth of LED load could not be driven by a cheap circuit being fed the 24V value. Wow... it would pull nearly an amp.
I still do not see the reasoning for this at all.
I mean, I made a supply where I had to make 5V from a 3 volt source, in order to make the op amps work right, but that wasn't a big loading, and was easily done with a small multiplier.
How about just stacking a 5v dc dc converter on top of the 24v supply? Most are isolated and a 5v is fairly inexpensive.
Cheers
Whoa! You only need regulated-voltage power if the pulse has to be a square-top regulated voltage.
It's pretty easy to ramp up an inductor to 700 mA and let it ramp down through the LED. Heck, I've got vacuum-tube strobes that do more current (and higher frequencies) into worse loads.
Sure, but this is a high precision measurement gizmo, not a light bulb. I care about the phase relationship between the illumination and the drive waveform. Also the light output of these LEDs is peak-current limited, which I also care about.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
If the duty cycle is low enough a current diverter may do the trick. They came from core memory designs.
?-)
It's a replacement for a flashlamp/PMT design--brown eggs are pretty dark, and the measurement principle is spectral differencing between two
4-nm wide bands on either side of the huge 589-nm absorption feature of protoporphyrin (which is what makes brown eggs brown).The oxyhaemoglobin peak is at 578 nm, which sounds fine until I tell you that both peaks are 10-20 nm wide, so they overlap considerably. The haemoglobin absorption can easily be an almost imperceptible 0.02-AU shoulder on a ruddy great 2-AU protoporphyrin peak. (AU = absorption unit, i.e. 10 dB optical / 20 dB electrical.)
The solution is to find another point on the other shoulder of the protoporphyrin line that goes up and down at about the same rate as 578 nm, which in this case happens at about 598 nm. The error between the two absorption slopes is taken out by a gain and offset term in absorption units (i.e. a ratio and power law):
A(HG) = A(578 nm) - A(598 nm)*(1+epsilon) + offset
This works to within a percent or so over a 50-dB range (electrical), which I'm pretty happy with. (This sort of measurement isn't original with me.)
It's sort of a fun project--I haven't played with Easter egg dye since my kids were small. ;)
Cheers
Phil Hobb
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Awww... I here I thought that it was because they were *local* eggs!
^
And. Doh.
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