SSGrower
Well-Known Member
Wait a minute....
I wanted a few 1750 mixed in on my rig, I sense a parts ordering frenzy.
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- Thread starter ChiefRunningPhist
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I wanted a few 1750 mixed in on my rig, I sense a parts ordering frenzy.
ChiefRunningPhist
Well-Known Member
If you can, I'd test it on one of the COBs in your waterfall first. I'm not sure on how much power will/can be generated haha
SSGrower
Well-Known Member
Not since they are epoxied, I remember what happened to the cobs I was looking for now.
I has to remove 6 that were epoxied, 2 didn't make it in what I would consider safe to use condition so they got bined. But I do have a blurple I do proof of concept.
I was able to measure a junction temp of 130F on a cx? 3070 running at 20 watts, so based on the video you posted there is probably enough to run some red and photo red, perhaps charge a cap to run them for eod too.
SSGrower
Well-Known Member
I think it is safe to assume 1.2V @ 170ma per chip running at a junction temperature of 130F. How did I come to this number?
A hot plate stabilized at 120-130 peltier chiller or generator with heatsink and cpu fan running at 5v.
Measurements were as high as 2.2v @ 300ma for the chiller and 2.0v 240ma for the generator. The high readings on the chiller were right after I put it on the hot plate, the generator seemed to stabilize at higher voltage and current.
Given those numbers 1.2v 170ma nominal with a possible peak of 3v 400ma what red leds would fit?
ChiefRunningPhist
Well-Known Member
Good work. That's valuable data. I'd grab a measurement at 150f (?), maybe even higher (?), it could reduce the need for IR and increase peltier? Also, when the hot plate was ~125°, what was the heatsink peltier interface temp (as best you can)? Trying to determine gradient as best as possible. Lastly can you try blowing a fan at the setup from a distance in an effort to simulate a fan in a grow room used to circulate air (or did you use a 5v cpu fan)?
Given the .17A by 1.2V measurement, equating ~0.2w, you could maybe run (1) small 660nm, like an epistar 5730 but you'd want a few to make it worth the while I'd say. You can convert the V & I into whatever is needed, so the wattage or power is what is significant. At 0.2w not much added benefit that I see unless you have several. It would also depend on how many watts it takes to achieve such temps with the COB?
If it's only 0.2w, it could be used for UVB supplementation too. (2) could run (1) UVB chip. But 0.2w at 130f and a decent heatsink with a fan (?) is a little less than earth shattering lol but if it only takes 10w or less per COB to get that temperature it might be significant..
EDIT:
At 20w, a 0.2w realization is only a 1% conversion rate with those particular devices and heatsinks used. Enough to charge an electrolytic for later use, and maybe some other uses too, most mid power 660nm are ~0.2w per chip (from what I've seen).
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SSGrower
Well-Known Member
I don't see converting the voltage to be of much benifit due to additional losses.
I was able to measure temps of about 70-80F on the backside of the heatsink, about 4mm thick aluminum 100W passive heatsink. So it was designed as a passive sink but what I found is without a fan blowing on it, voltage dropped below 1V very quickly. Not sure passive is the way to go here as they rely on a gradual temp gradient, the peltier needs a sharp diference.
Looking at the cxb3590 data sheet it does not even show luminus flux vs drive current for specific Tc for the low drive currents I use, but the different temp lines do converge and looks like between 55c and 85c you would loose myaybe up to 5% relative flux @ 700ma, I am running something like 300-500. If relative flux losses can be held to less than 0.5% due to the increased temp and a supplimental chip efficiency of >50% I feel like we could get ahead of the electrical losses via. generating a more compelete spectra. Thought these parameters my not be wholy quantifyable ie. the smoke is just better/worse.
SSGrower
Well-Known Member
I am always leery of doing current measurements, more so when stoned becease, I have let the smoke out of a meter and a control board on 2 seperate occasions.
I wonder what stacking the Peltiers does? Will try to run that after watering plants.
For now hot plate is occupied.
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ChiefRunningPhist
Well-Known Member
Ya that's what I was saying about a grow room fan. It might be enough air movement if you had a large enough passive to keep the cold side cool. When you stopped air flow the heatsink increased in temp and the gradient suffered so the power generation was less. Idk how big they make the heatsinks, what wattage, but you'd want the biggest bestest baddest one possible (I think), maybe even a 150w or 200w (if they make). If you have to wire a fan up to cool the Peltier enough to use it then it's actually counteracting your power generation. If you can use a source of air flow already present then you don't have to add in power to generate power.
After reviewing the cxb3590 data sheet, using a Peltier/seebeck is not going to increase maximum theoretical μmol/j, but they could still help you possibly achieve better than what you're getting now. If you look at the Tc° plots of "Current" vs "Relative Luminous Flux," you see about a 33% increase in current is needed to maintain intensity between the lowest Tc° data point and the highest Tc° data point, or between 25° and 105°. That's roughly a 33% increase in wattage for the same intensity, and if you only convert about 3% or 4% of the waste heat as electrical energy with the current technology, then it is more effecient to run the COB cooler and use 33% less power than try to recycle 3% or 4% of the heat. So its not going to increase industry μmol/j for COB but it could help you run a UVB supplemental or an FR initiator I think, and depending on your current operating temps. It is not going to increase effeciency better than keeping your chip cool.
Idk, maybe it could help people with heat management issues by running a CPU fan (?), or power a small string of LEDs,.. of course depending on the seebeck devices used, power nessecity of LEDs, heatsink size, and air flow... Or at least it would seem.
On anther note, an OP amp current/voltage conversion cct might be able to achieve the electrical manipulation needed or at least that was my thinking.
EDIT:
- how well a guy can keep 1 side cool using passive heat control or without designed power addition
- the power generated vs "hot side" temp curve of the seebeck device
Heatsink size, & device selection... I think that's what it boils down to..
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SSGrower
Well-Known Member
After running this this thing again today I have some revised numbers. This time I let it sit in passive mode for over 3 hrs. Doing what I could to maintain the hotplate temperature at 190F as this seems to be the point where the peltier kicks it up a notch and stabilizes a bit. The plate temp was hard to keep constant, I think it will be much easier to do with an led and proper dimming driver.
Also notice where those graphs stop (1400 ma), I am running down at 3 - 400 and all the temp lines converge so chip temp differencs have less impact on effeciency.
At any rate after 3hrs of running in passive mode (IIRC it is rated at 100W passive) readings were 1.3v, 160ma, 187F plate temp, 139F backside of heatsink. At that point I turned the fan on and voltage immediately went up, by the time I got probe tips installed it was at 3.9 and falling, curent was over 300ma. It eventually stabilized at 2.35v, 293ma (0.688W !). I had to adjust the hotplate up cause the fan was cooling everything but once it stabilized at 190F the backside of the heatsink was 85F.
One thing that is way outside my comfort zone is designing a circuit to regulate and manage the power swings. I think of things like a current shunt or transistor switch to battery charger that runs the red leds at lights out. So maybe it looks something like a heasink with cob and peltier and some 1W chips driven at 50%, uv during day and red initiator at night (as I don't think the added red would make much of a difference during the day, especially if using a high cri cool color temp chip).
ChiefRunningPhist
Well-Known Member
I should have realized the significance when you pointed that out earlier..
What this is telling me (possibly lol)...
You turned the fan on. The power generation increased because the gradient increased. The fan cooled the heatsink that was attached to the cold side down quickly and while the hot side was catching up transferring the heat across, the gradient was large for a brief time, and as the heatsink pulled more heat away from the hot plate and transferred it into the air the hot plate got cooler, and as a result the gradient decreased which is why there was a spike initially and a subsequent drop. The only thing I can't understand is why it settled to less than the initial spike reading once the hot plate was adjusted up. I was thinking the fan cooled the heatsink to about 85° quickly, and achieved that temp while hot plate was still close to 187° (~100° gradient), so after the increase in hot plate temp to 190° and heatsink at 85° (~100° gradient) I would have thought it would have read similar to the initial power spike.
If my comprehension follows, then it's what I was possibly suspecting earlier, that the reaction would fight itself. As you pull more heat away from the source, the gradient decreases and the reaction slows.
I'm thinking if your fan in your grow room was an oscillating fan, that it would periodically pass air across the heatsink, and this cycling of ON / OFF air flow might give the optimum effeciency. As the fan blows across the heatsink it would cool the heatsink quickly but it doesn't stay on the heatsink for long so it doesn't cause the hot side to lose the heat as much. While the fan blows on the heatsink, the seebeck gradient is greatest, then the fan blows in another direction and the gradient equalizes and the lower power generation is realized.
But I think charging a battery could be a good use. I'd possibly look at how small solar chargers are set up, and then miniaturize it?
0.69 watts could be significant. The relative impact would depend on how many you had per area, and how much power it takes to achieve those gradients per COB?
EDIT:
It also might imply there's a limit to the heatsink size. If your heatsink is too big, I'm now wondering if the gradient will be small because the heatsink will cool everything, not just the hot side. So maybe the biggest bestest baddest heatsink isn't the way to go..
I'm wondering if this is why I've seen some of these seebeck devices with what seems like a greater distance between the plates. I think the proximity of the plates reflects how well a hot side can stay hot without interacting with the cold side? Allowing then for the larger heatsinks or bigger gradients?
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SSGrower
Well-Known Member
I dont look at heat the se way most people seem to. They talk of coolong, where I talk of heat mangment. The universe is cooling, we are not doing anything to facilitate that, we are jist managing where heat goes.
It is like a bucket of water, if I am filling it at a constant rate, eventually it will over top and stabilize. If I punch a large enough hole in it water will escape at a faster rate than it is added, until it reaches the level of the hole.
Edit, also I want to see what happens when I actually put an electeical load on the peltier. I think this will actually increase the temp gradient.
It is like a bucket of water, if I am filling it at a constant rate, eventually it will over top and stabilize. If I punch a large enough hole in it water will escape at a faster rate than it is added, until it reaches the level of the hole.
Edit, also I want to see what happens when I actually put an electeical load on the peltier. I think this will actually increase the temp gradient.
ChiefRunningPhist
Well-Known Member
Haha nice! Thermodynamics! STEM nerds unite! Lol
Like you said, hot & cold are just gradients of thermal energy concentrations. As soon as you use some thermal to produce work the gradient diminishes. The energy is converted from thermal to electrical and when measuring using a multimeter the power draw is nill compared to the source. If you were to power something at the theoretical wattage calculated by the multimeter that means some of the heat is taken away as electrical. The higher the load the higher the % of heat converted or the thermal energy extraction/conversion.
Did you ever stack? I don't think it'd help, but I came across some that were multi-layered?? Seems I have to look into these a bit more, curious, I wonder what's going on..
Like you said, hot & cold are just gradients of thermal energy concentrations. As soon as you use some thermal to produce work the gradient diminishes. The energy is converted from thermal to electrical and when measuring using a multimeter the power draw is nill compared to the source. If you were to power something at the theoretical wattage calculated by the multimeter that means some of the heat is taken away as electrical. The higher the load the higher the % of heat converted or the thermal energy extraction/conversion.
Did you ever stack? I don't think it'd help, but I came across some that were multi-layered?? Seems I have to look into these a bit more, curious, I wonder what's going on..
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SSGrower
Well-Known Member
Stacking only resulted in losses at each peltier, the only benifit would be 2 seperate channels, but there are much more effecient ways of doing that.
I am calling out for help to design the circuit, if you can or know anyone any help in that regard would be appreciated. I can assemble it np, just dont know where to start.
I am calling out for help to design the circuit, if you can or know anyone any help in that regard would be appreciated. I can assemble it np, just dont know where to start.
ChiefRunningPhist
Well-Known Member
Some thoughts on UV...
UV index...
Wikipedia excerpt...
https://en.m.wikipedia.org/wiki/Ultraviolet_index
When looking over what the UV index represents a few things stand out. The scale seems logarithmically weighted towards the shorter WV. 295nm is weighted at 100%, while 305nm only 22%, and 325 only a .3% weight. Refer to chart below...
10mw/m2 = 1μw/cm2
So...
74mw (305nm) + .6mw (295nm)
=
~75mw/m2; ~7.5μw/cm2
It seems a midday summer afternoon could generate approximately 7.5μw/cm2 of UV from 295-305nm. That power UVB is representative of ~10.6 UV index. When referencing the UV index chart I notice 10.6 is actually a pretty high UV forecast, almost 11.
With the UV index weighted logarithmically I've pretty much disregarded 305nm+. With that in mind, it seems I only need to achieve ~7.5μw/cm2 of UVB to realize a relatively high UV index rating.
The light is designed for a 2'× 2', or for 3,716cm2. For the moment lets just assume that the UVB chips are lensed for 0 loss in stray emissions, ie every μw of UV reaches the 3,716cm2. That would equate to roughly ~28,000μw needed in total to cover the 3,716cm2. If the chips I'm looking at are able to produce between 2000μw - 6000μw at a peak of 295nm, an average of 4000μw can be utilized for a mean calculation, and I see that I would theoretically need ~7 chips to achieve a close equivalent of a 10.6 UV index. That's even at only a 1.25% effeciency in chip technology!
(2.8mw/224mw)
Now granted I didn't include 325-305nm in the index measurement, I assumed complete light throw efficiency, and using an arbitrary set of values that I have no way of validating, also not sure what graph they are integrating to get the final value, but other than that its encouraging, lol idk, maybe I'm just trying to see what I want too..
UV index...
Wikipedia excerpt...
https://en.m.wikipedia.org/wiki/Ultraviolet_index
When looking over what the UV index represents a few things stand out. The scale seems logarithmically weighted towards the shorter WV. 295nm is weighted at 100%, while 305nm only 22%, and 325 only a .3% weight. Refer to chart below...
10mw/m2 = 1μw/cm2
So...
74mw (305nm) + .6mw (295nm)
=
~75mw/m2; ~7.5μw/cm2
It seems a midday summer afternoon could generate approximately 7.5μw/cm2 of UV from 295-305nm. That power UVB is representative of ~10.6 UV index. When referencing the UV index chart I notice 10.6 is actually a pretty high UV forecast, almost 11.
With the UV index weighted logarithmically I've pretty much disregarded 305nm+. With that in mind, it seems I only need to achieve ~7.5μw/cm2 of UVB to realize a relatively high UV index rating.
The light is designed for a 2'× 2', or for 3,716cm2. For the moment lets just assume that the UVB chips are lensed for 0 loss in stray emissions, ie every μw of UV reaches the 3,716cm2. That would equate to roughly ~28,000μw needed in total to cover the 3,716cm2. If the chips I'm looking at are able to produce between 2000μw - 6000μw at a peak of 295nm, an average of 4000μw can be utilized for a mean calculation, and I see that I would theoretically need ~7 chips to achieve a close equivalent of a 10.6 UV index. That's even at only a 1.25% effeciency in chip technology!
(2.8mw/224mw)Now granted I didn't include 325-305nm in the index measurement, I assumed complete light throw efficiency, and using an arbitrary set of values that I have no way of validating, also not sure what graph they are integrating to get the final value, but other than that its encouraging, lol idk, maybe I'm just trying to see what I want too..
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SSGrower
Well-Known Member
IIRC this is a 220 watt fixture? So you only need 224mw (at the wall) of uv to get a 10.6 uv index?Some thoughts on UV...
UV index...
View attachment 4323128
Wikipedia excerpt...
View attachment 4323127
https://en.m.wikipedia.org/wiki/Ultraviolet_index
When looking over what the UV index represents a few things stand out. The scale seems logarithmically weighted towards the shorter WV. 295nm is weighted at 100%, while 305nm only 22%, and 325 only a .3% weight. Refer to chart below...
View attachment 4323126
10mw/m2 = 1μw/cm2
So...
74mw (305nm) + .6mw (295nm)
=
~75mw/m2; ~7.5μw/cm2
It seems a midday summer afternoon could generate approximately 7.5μw/cm2 of UV from 295-305nm. That power UVB is representative of ~10.6 UV index. When referencing the UV index chart I notice 10.6 is actually a pretty high UV forecast, almost 11.
With the UV index weighted logarithmically I've pretty much disregarded 305nm+. With that in mind, it seems I only need to achieve ~7.5μw/cm2 of UVB to realize a relatively high UV index rating.
The light is designed for a 2'× 2', or for 3,716cm2. For the moment lets just assume that the UVB chips are lensed for 0 loss in stray emissions, ie every μw of UV reaches the 3,716cm2. That would equate to roughly ~28,000μw needed in total to cover the 3,716cm2. If the chips I'm looking at are able to produce between 2000μw - 6000μw at a peak of 295nm, an average of 4000μw can be utilized for a mean calculation, and I see that I would theoretically need ~7 chips to achieve a close equivalent of a 10.6 UV index. That's even at only a 1.25% effeciency in chip technology!
Now granted I didn't include 325-305nm in the index measurement, I assumed complete light throw efficiency, and using an arbitrary set of values that I have no way of validating, also not sure what graph they are integrating to get the final value, but other than that its encouraging, lol idk, maybe I'm just trying to see what I want too..
Rocket Soul
Well-Known Member
Im not sure if the erythemal action spectrum (how much of what wavelength burns your skinn) can be exactly translated to a plant response. Skin and leaf is not the same. Also, @Randomblame was talking about having the right balance between uva and uvb. Apparently he had loss of yield with too much uvb.Some thoughts on UV...
UV index...
View attachment 4323128
Wikipedia excerpt...
View attachment 4323127
https://en.m.wikipedia.org/wiki/Ultraviolet_index
When looking over what the UV index represents a few things stand out. The scale seems logarithmically weighted towards the shorter WV. 295nm is weighted at 100%, while 305nm only 22%, and 325 only a .3% weight. Refer to chart below...
View attachment 4323126
10mw/m2 = 1μw/cm2
So...
74mw (305nm) + .6mw (295nm)
=
~75mw/m2; ~7.5μw/cm2
It seems a midday summer afternoon could generate approximately 7.5μw/cm2 of UV from 295-305nm. That power UVB is representative of ~10.6 UV index. When referencing the UV index chart I notice 10.6 is actually a pretty high UV forecast, almost 11.
With the UV index weighted logarithmically I've pretty much disregarded 305nm+. With that in mind, it seems I only need to achieve ~7.5μw/cm2 of UVB to realize a relatively high UV index rating.
The light is designed for a 2'× 2', or for 3,716cm2. For the moment lets just assume that the UVB chips are lensed for 0 loss in stray emissions, ie every μw of UV reaches the 3,716cm2. That would equate to roughly ~28,000μw needed in total to cover the 3,716cm2. If the chips I'm looking at are able to produce between 2000μw - 6000μw at a peak of 295nm, an average of 4000μw can be utilized for a mean calculation, and I see that I would theoretically need ~7 chips to achieve a close equivalent of a 10.6 UV index. That's even at only a 1.25% effeciency in chip technology!
Now granted I didn't include 325-305nm in the index measurement, I assumed complete light throw efficiency, and using an arbitrary set of values that I have no way of validating, also not sure what graph they are integrating to get the final value, but other than that its encouraging, lol idk, maybe I'm just trying to see what I want too..
But i still find it and intriguing approach....
whytewidow
Well-Known Member
This is such a good thread. Tyvm for starting is @ChiefRunningPhist and as well as @SSGrower for all his info i out too. Such a good read guys.