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post #12631 of 16193 Old 11-02-2015, 05:43 PM
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Originally Posted by fafrd View Post
while LG focuses on actually productizing the 77" (meaning a simple, flat version with no bells and whistles and a back-down-to-earth price)...

Fingers crossed!
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post #12632 of 16193 Old 11-02-2015, 06:10 PM
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Fingers crossed!
Well, I assume you saw that they are ditching 1080p next year, so they have got to do something to enlarge the product line beyond a 55" and a 65" 4K OLED available in flat or curved configuration...

The 2080p M1 line may also be repurposed to manufacture the new-Flagship 99" .
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post #12633 of 16193 Old 11-02-2015, 08:04 PM
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Originally Posted by fafrd View Post
Well, I assume you saw that they are ditching 1080p next year, so they have got to do something to enlarge the product line beyond a 55" and a 65" 4K OLED available in flat or curved configuration...

The 2080p M1 line may also be repurposed to manufacture the new-Flagship 99" .
I think I need a 77" at least to upgrade. I just don't think I can get rid of an awesome 65" plasma and not go bigger. I am starting to look at PJs as another option.
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post #12634 of 16193 Old 11-03-2015, 06:55 AM
 
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I meant cd/m2 per pixel.
Please rephrase. There is no cd/m² per pixel metric that makes sense to me. A cd/m² is a cendela per square meter. To make a per square meter per pixel unit doesn't make sense.

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Originally Posted by stas3098 View Post
OLEDs are current-driven, (..........snipped.........)
Hence the start of the original argument I recall between you two, (if I recall correctly). Everything you said is how we've always approached the operation of OLED because it seems the most sensible at first. Amps are a measurement of electrons per second, electron-hole-recombination creates the light, more amps yield more light, etc.

And at a certain level that's nominally correct.

But I completely revisited this with xrox recently by PM and it's still the case that the "switch" for effectively starting that process is voltage based. I have a far better than average understanding of physics and electrical processes myself and it took a bit of time for me to get ahold of how that really works.

However, I can't speak for xrox: If he wants to chime in, I'll let him take control on this subtopic----I'm bowing out because I don't yet have a good way of describing it that I am comfortable with. I do have to say though that as valid as xrox's point is, at first it's somewhat counter-intuitive. For me it still falls in that section of physics where you can say with complete honesty "well, you're mostly right because that is how it seems, but this is really what's going on".

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post #12635 of 16193 Old 11-03-2015, 11:43 AM
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Originally Posted by tgm1024 View Post
Please rephrase. There is no cd/m² per pixel metric that makes sense to me. A cd/m² is a cendela per square meter. To make a per square meter per pixel unit doesn't make sense.



Hence the start of the original argument I recall between you two, (if I recall correctly). Everything you said is how we've always approached the operation of OLED because it seems the most sensible at first. Amps are a measurement of electrons per second, electron-hole-recombination creates the light, more amps yield more light, etc.

And at a certain level that's nominally correct.

But I completely revisited this with xrox recently by PM and it's still the case that the "switch" for effectively starting that process is voltage based. I have a far better than average understanding of physics and electrical processes myself and it took a bit of time for me to get ahold of how that really works.

However, I can't speak for xrox: If he wants to chime in, I'll let him take control on this subtopic----I'm bowing out because I don't yet have a good way of describing it that I am comfortable with. I do have to say though that as valid as xrox's point is, at first it's somewhat counter-intuitive. For me it still falls in that section of physics where you can say with complete honesty "well, you're mostly right because that is how it seems, but this is really what's going on".
I meant luminance per pixel (luminance of a pixel). I used cd/m2 (luminance) there for convenience, I guess...

All I meant in that post is that if the green and the red sub-pixels in a pixel are turned off and don't emit any light at all then the blue sub-pixel effectively "turns" into a pixel (for all intents and purposes). This is the fact on which all of my previous posts were predicated.

P.S. The way science seems to work: the deeper you go, the more complicated it gets.

....
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post #12636 of 16193 Old 11-03-2015, 12:27 PM
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Originally Posted by tgm1024 View Post
Please rephrase. There is no cd/m² per pixel metric that makes sense to me. A cd/m² is a cendela per square meter. To make a per square meter per pixel unit doesn't make sense.
Looks like I didn't word the question he was responding to that well. I should have said sub-pixel for one.

The way I meant the question was whether he meant the cd/m2 only counting the area the blue sub-pixel used, or the whole pixel area including dark space. Put another way, in my example of shutters blocking 90% of each blue sub-pixel, if the blue sub-pixels themselves were putting out 70 cd/m2 and the fill ratio was 10% just considering blue then the screen would have 7 cd/m2 worth of blue. If 90% of the sub-pixel area was blocked and those sub-pixels were uniform then the leftover part of the blue sub-pixels would still be putting out 70 cd/m2, but the screen would have only 0.7 cd/m2 of blue because the effective fill ratio would now be 1%.

Does that make sense?

--Darin

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post #12637 of 16193 Old 11-03-2015, 12:52 PM
 
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Originally Posted by darinp2 View Post
Looks like I didn't word the question he was responding to that well. I should have said sub-pixel for one.

The way I meant the question was whether he meant the cd/m2 only counting the area the blue sub-pixel used, or the whole pixel area including dark space. Put another way, in my example of shutters blocking 90% of each blue sub-pixel, if the blue sub-pixels themselves were putting out 70 cd/m2 and the fill ratio was 10% just considering blue then the screen would have 7 cd/m2 worth of blue. If 90% of the sub-pixel area was blocked and those sub-pixels were uniform then the leftover part of the blue sub-pixels would still be putting out 70 cd/m2, but the screen would have only 0.7 cd/m2 of blue because the effective fill ratio would now be 1%.

Does that make sense?

--Darin
I suppose, but what matters is that he seems to be backing away from his original assertion, or at least holding up the "let me think for a second" flag, either of which are perfectly fine.
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post #12638 of 16193 Old 11-03-2015, 03:50 PM
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Originally Posted by darinp2 View Post
Looks like I didn't word the question he was responding to that well. I should have said sub-pixel for one.

The way I meant the question was whether he meant the cd/m2 only counting the area the blue sub-pixel used, or the whole pixel area including dark space. Put another way, in my example of shutters blocking 90% of each blue sub-pixel, if the blue sub-pixels themselves were putting out 70 cd/m2 and the fill ratio was 10% just considering blue then the screen would have 7 cd/m2 worth of blue. If 90% of the sub-pixel area was blocked and those sub-pixels were uniform then the leftover part of the blue sub-pixels would still be putting out 70 cd/m2, but the screen would have only 0.7 cd/m2 of blue because the effective fill ratio would now be 1%.

Does that make sense?

--Darin
Yes, it makes sense. You got the idea behind it down. Now, all you have to do is learn the proper terminology.

The blue subpixels' luminous intensity would be the same 70 candelas (even when 90% of their light output is blocked by shutters). Luminous intensity is proportional to current intensity, and thus If you were to decrease the current intensity/strength (i.e amperage) by 90 percent then the luminous intensity would, consequently, decrease by 90 percent.

But, the blue subpixels' luminance (measured in candelas per square meter) would decrease by 90 percent when 90 percent of their area was blocked.

But, there are some caveats there which have to do with the human visual system, though... I'll try to shed some light on that one a bit later.

....
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post #12639 of 16193 Old 11-03-2015, 04:04 PM
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Originally Posted by stas3098 View Post
But, the blue subpixels' luminance (measured in candelas per square meter) would decrease by 90 percent when 90 percent of their area was blocked.
Of the whole sub-pixel area. whole pixel area, or whole screen area I would agree. However, for just the part of a single sub-pixel that is still showing through the luminance wouldn't change though, would it? That is, if you could measure the light and area for something that small (maybe around one 10 billionths of a meter squared if the math in my head is right).

--Darin

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post #12640 of 16193 Old 11-03-2015, 07:48 PM
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The blue subpixels' luminous intensity would be the same 70 candelas (even when 90% of their light output is blocked by shutters).
Going back to this part, I don't believe that is correct unless you want to count the light being blocked by the shutters. As long as 1/10th of the pixel area left is sending the light over the same number of steridians as the unblocked sub pixel was the luminous intensity from each sub-pixel is now 1/10th of what it was, whether using candelas or lumens.

Imagine a projector with a small imaging chip and the zoom lens left in one position so the beam size and angle don't change. Now put the projector in a mode where all the pixels are white and another mode where 10% are white and 90% are black. Would you claim that the lumens or candelas of light from the projector are the same in both cases?

Another way to look at it is to go back to the 70 candelas and instead of putting a shutter over 90% of each sub-pixel put a lens that focuses that light toward a side wall. The lensed part of the sub-pixels would send out a certain number of candelas and the unlensed part would too. You wouldn't say that those would now add up to more than 70 candelas, would you? If not, then the unlensed part would have to send out less than 70 by definition if the lensed part sent out any.

Do you disagree that the luminous flux outside the shutters goes down by 90% if shutters cover 90% of the sub-pixel area, for that example from earlier?

How about if shutters went all the way to blocking 100%? Would that change the luminous intensity or luminous flux?

Thanks,
Darin

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post #12641 of 16193 Old 11-04-2015, 06:17 AM
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Originally Posted by darinp2 View Post
Going back to this part, I don't believe that is correct unless you want to count the light being blocked by the shutters. As long as 1/10th of the pixel area left is sending the light over the same number of steridians as the unblocked sub pixel was the luminous intensity from each sub-pixel is now 1/10th of what it was, whether using candelas or lumens.

Imagine a projector with a small imaging chip and the zoom lens left in one position so the beam size and angle don't change. Now put the projector in a mode where all the pixels are white and another mode where 10% are white and 90% are black. Would you claim that the lumens or candelas of light from the projector are the same in both cases?

Another way to look at it is to go back to the 70 candelas and instead of putting a shutter over 90% of each sub-pixel put a lens that focuses that light toward a side wall. The lensed part of the sub-pixels would send out a certain number of candelas and the unlensed part would too. You wouldn't say that those would now add up to more than 70 candelas, would you? If not, then the unlensed part would have to send out less than 70 by definition if the lensed part sent out any.

Do you disagree that the luminous flux outside the shutters goes down by 90% if shutters cover 90% of the sub-pixel area, for that example from earlier?

How about if shutters went all the way to blocking 100%? Would that change the luminous intensity or luminous flux?

Thanks,
Darin
Let's start from scratch.

If you have one light emitting molecule and you apply 0.1 micro-amperes to it it is liable to produce, say, 1 billion photons per unit time.

And if you have two similar light emitting molecules and apply the same 0.1 micro-amperes to them, they are still liable to produce 1 billion photons per unit time, but now each light emitting molecule produces only half a billion photons. In this case, the luminous intensity of each molecule halved, but the net luminous power/flux of the two light emitting molecules ( i.e. the number of photons produced per unit time) stayed the same as in the first example.

And if you were to put ten similar light emitting molecules together and feed them an amperage of 1 micro-ampere (which would yield 10 billion photons per unit time in total) and then were to cover up nine of them, the uncovered light emitting molecule would produce 1 billion photons, and among the nine covered light emitting molecules each light emitting molecule would also produce 1 billion photons, because all ten light emitting molecules would have the same current density of 0.1 micro-amperes, but the total luminous power of the ten molecules would decrease by 90 percent (when nine of them are covered up).

Now let's get back to the projector example. If you were to cover 90 percent of the lens with a shutter then its luminous flux would decrease by 90 percent (or in other words, the amount of photons coming out the lens would fall by 90 percent if 90 percent of it was blocked), but the luminous intensity of the remaining 10 percent would stay the same.

And if you were to cover 90 percent of each pixel of the projector then the luminance of the entire screen on which the projector was projecting light would decrease by 90 percent, however the remaining 10 percent of each pixel would still maintain the same luminous intensity as before (i.e. if the projector was OLED-based then each light emitting molecule would still produce the same amount of photons per unit time).

I hope this makes things a bit clearer.

....

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post #12642 of 16193 Old 11-04-2015, 06:42 AM
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Originally Posted by darinp2 View Post

Do you disagree that the luminous flux outside the shutters goes down by 90% if shutters cover 90% of the sub-pixel area, for that example from earlier?

How about if shutters went all the way to blocking 100%? Would that change the luminous intensity or luminous flux?

Thanks,
Darin

Do you disagree that the luminous flux (the total amount of photons coming out of the subpixel) outside the shutters goes down by 90% if shutters cover 90% of the sub-pixel area, for that example from earlier?



No, I do not disagree.

And if 100 percent of the subpixel's area was blocked then the luminous flux from that subpixel would be zero.

....
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post #12643 of 16193 Old 11-04-2015, 12:10 PM
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Originally Posted by stas3098 View Post
And if 100 percent of the subpixel's area was blocked then the luminous flux from that subpixel would be zero.
And so would the luminous intensity for that subpixel.

I can go back and address the projector stuff, but for the moment I'll concentrate on one thing.

It seems like you are assuming some uniformity is required for luminous intensity or are changing the area even though luminous intensity doesn't have area (it has angle). At this level the area the pixel takes up doesn't really matter, just the amount of light it puts out per steradian angle.

If we consider each subpixel in the shutter blocking 90% case as uniform with each 1/10th of the subpixel being something I'll call a block for now, each block has the same luminous flux and luminous intensity before the shutter is put in place. After the shutter is put in place 9 out of 10 have both of those dropped to effectively zero, while 1 of them stays the same for both. So, the average luminous intensity for the whole sub-pixel has changed even though the luminous flux and luminous intensity for one of the blocks hasn't.

If you look at the light for one steradian angle it has changed from one subpixel (10 blocks) between the shutters being open and closed.

To put it simply, when 90% of the subpixel area is blocked the perceived power per unit solid angle to viewer's goes down.

Do you disagree with that?

If we agree there then we can go back to your original position about how hard things are driven where I think I can see your point if we define how hard things are driven by the current. By the way you describe OLED I think a person could say that driving a very small subpixel at 1 amp is driving it harder than driving a larger subpixel at 1 amp, and therefore larger subpixels can be driven less hard for the same luminous flux and luminous intensity, but that is largely semantic.

Thanks,
Darin

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post #12644 of 16193 Old 11-05-2015, 06:27 AM
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Originally Posted by darinp2 View Post
And so would the luminous intensity for that subpixel.

I can go back and address the projector stuff, but for the moment I'll concentrate on one thing.

It seems like you are assuming some uniformity is required for luminous intensity or are changing the area even though luminous intensity doesn't have area (it has angle). At this level the area the pixel takes up doesn't really matter, just the amount of light it puts out per steradian angle.

If we consider each subpixel in the shutter blocking 90% case as uniform with each 1/10th of the subpixel being something I'll call a block for now, each block has the same luminous flux and luminous intensity before the shutter is put in place. After the shutter is put in place 9 out of 10 have both of those dropped to effectively zero, while 1 of them stays the same for both. So, the average luminous intensity for the whole sub-pixel has changed even though the luminous flux and luminous intensity for one of the blocks hasn't.

If you look at the light for one steradian angle it has changed from one subpixel (10 blocks) between the shutters being open and closed.

To put it simply, when 90% of the subpixel area is blocked the perceived power per unit solid angle to viewer's goes down.

Do you disagree with that?

If we agree there then we can go back to your original position about how hard things are driven where I think I can see your point if we define how hard things are driven by the current. By the way you describe OLED I think a person could say that driving a very small subpixel at 1 amp is driving it harder than driving a larger subpixel at 1 amp, and therefore larger subpixels can be driven less hard for the same luminous flux and luminous intensity, but that is largely semantic.

Thanks,
Darin
No, I do not disagree with any of your points here.

But with one correction, though:... and therefore larger subpixels can be driven less hard for the same luminous flux, but at a lower luminous intensity...

Luminous intensity was designed this way (the way I described it), because it needs to be easily convertible from radiant intensity, but the difference between those two is mostly pragmatic.

And by the way, the candela (or standard candle) is defined as the luminous intensity weighted with the response of the human eye emitted from 1/60 of 1cm2 projected area of an ideal blackbody radiator, in any direction, at a temperature of 2045 K.

....

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post #12645 of 16193 Old 11-05-2015, 07:40 AM
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But with one correction, though:... and therefore larger subpixels can be driven less hard for the same luminous flux, but at a lower luminous intensity...
What is the correction exactly? Is it the area thing? In this case I believe the subpixels are less than 1/60th of 1 cm2 in both cases.

It seems like you are claiming the watts go down, but the watts per steradian do not in the blocking example. Is that right? Are you claiming the radiant intensity for a whole subpixel stays the same in my example of blocking 90% of that subpixel?

Edit: I'm confused about whether you have changed your mind from earlier when you said the luminous intensity for a blue subpixel stays the same when 90% of it is blocked.

Thanks,
Darin

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Originally Posted by darinp2 View Post
What is the correction exactly? Is it the area thing? In this case I believe the subpixels are less than 1/60th of 1 cm2 in both cases.

It seems like you are claiming the watts go down, but the watts per steradian do not in the blocking example. Is that right? Are you claiming the radiant intensity for a whole subpixel stays the same in my example of blocking 90% of that subpixel?

Edit: I'm confused about whether you have changed your mind from earlier when you said the luminous intensity for a blue subpixel stays the same when 90% of it is blocked.

Thanks,
Darin
Luminous intensity and luminous flux can be equal at a molecule level, in a subpixel there will be millions of molecules and so the luminous power/flux from one molecule and millions of molecules in a subpixel would not match up. I hope you can see this...

....

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post #12647 of 16193 Old 11-05-2015, 08:23 AM
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Originally Posted by darinp2 View Post
What is the correction exactly? Is it the area thing? In this case I believe the subpixels are less than 1/60th of 1 cm2 in both cases.

It seems like you are claiming the watts go down, but the watts per steradian do not in the blocking example. Is that right? Are you claiming the radiant intensity for a whole subpixel stays the same in my example of blocking 90% of that subpixel?

Edit: I'm confused about whether you have changed your mind from earlier when you said the luminous intensity for a blue subpixel stays the same when 90% of it is blocked.

Thanks,
Darin
If a molecule has a luminous intensity of 0,000001 cd then its luminous flux is 0,000001 lumens. But if there are a million molecules in a pixel and each of them has a luminous intensity of 0,000001 cd then the pixel's luminous flux is 1 lumen.

I hope you can see now that you have to add up all the molecules in a pixel to get the luminous flux.

....

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Originally Posted by darinp2 View Post

Edit: I'm confused about whether you have changed your mind from earlier when you said the luminous intensity for a blue subpixel stays the same when 90% of it is blocked.

Thanks,
Darin
Let's say, that the blue sub-pixel in question had one million molecules with a luminous intensity of 0,000001 cd and thus it had a luminous flux of 1 lumen (i.e each molecule in the blue sub-pixel had a luminous flux of 0,000001 lumens but all of them combined together had a luminous flux of 1 lumen).
What do you think would happen if you could remove 90 percent of the molecules from the blue subpixel if the act of the removal of those molecules did not result in a change in luminous intensity for the remaining 10 percent of the molecules. What would now the luminous flux from the blue sub-pixel be?

....

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If a molecule has a luminous intensity of 0,000001 cd then its luminous flux is 0,000001 lumens. But if there are a million molecules in a pixel and each of them has a luminous intensity of 0,000001 cd then the pixel's luminous flux is 1 cd.
and the luminous intensity of the pixel is 1 cd. Do you disagree?

Do you disagree that the units of luminous intensity are watts per steradian? Can a candle have a luminous intensity? What would it be for no lens and a candle with luminous flux of 1 candela?
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I hope you can see now that you have to add up all the molecules in a pixel to get the luminous flux.
Same thing for getting the luminous intensity of the pixel.

--Darin
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You seem to be confused by the word "intensity" and are taking it as the concentration of light within the pixel. It is not. It is the concentration of light emitted by the pixel (away from the display). Otherwise it could not be easily measured.

If you think about how you would measure the luminous intensity for a pixel it might make more sense to you. If you still believe you need to know how many molecules are in a pixel in order to figure out the luminous intensity of the pixel please explain how you think luminous intensity is measured.

Here is what I consider a clue. Do you have to know how big the flame is on a burning candle to measure the luminous intensity of the light from the candle in a particular direction?

--Darin

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post #12651 of 16193 Old 11-05-2015, 09:26 AM
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Stas,

It seems that your latest position contradicts a passage you agreed with earlier, where you quoted:

"Luminous intensity should not be confused with another photometric unit, luminous flux, which is the total perceived power emitted in all directions. Luminous intensity is the perceived power per unit solid angle. If a lamp has a 1 lumen bulb and the optics of the lamp are set up to focus the light evenly into a 1 steradian beam, then the beam would have a luminous intensity of 1 candela. If the optics were changed to concentrate the beam into 1/2 steradian then the source would have a luminous intensity of 2 candela. The resulting beam is narrower and brighter, though its luminous flux remains unchanged."

Notice how the number of molecules is not needed to determine the luminous intensity?

--Darin
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Originally Posted by darinp2 View Post
Stas,

It seems that your latest position contradicts a passage you agreed with earlier, where you quoted:

"Luminous intensity should not be confused with another photometric unit, luminous flux, which is the total perceived power emitted in all directions. Luminous intensity is the perceived power per unit solid angle. If a lamp has a 1 lumen bulb and the optics of the lamp are set up to focus the light evenly into a 1 steradian beam, then the beam would have a luminous intensity of 1 candela. If the optics were changed to concentrate the beam into 1/2 steradian then the source would have a luminous intensity of 2 candela. The resulting beam is narrower and brighter, though its luminous flux remains unchanged."

Notice how the number of molecules is not needed to determine the luminous intensity?

--Darin
I don't know, I need to take some time to think things through,

....
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post #12653 of 16193 Old 11-05-2015, 02:03 PM
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I don't know, I need to take some time to think things through,
Okay. One more thing in case it helps. A candela is equal to the amount of energy from a fixed area of that black body you mentioned, but that doesn't mean that luminous intensity has an area component. The energy and luminous intensity for other things (like pixels) don't require the area. That 1cm2 was just for making a standard unit of energy for a known case. If you have the luminance you can solve for luminous intensity by multiplying by the area, but measuring luminous intensity doesn't require measuring luminance or area of the source.

If you know how much visible energy something like the sun emits per square acre for instance you can determine how much luminous intensity is emitted based on area, but then the luminous intensity scales with area just like luminous flux. For instance, 2 square acres of the sun's surface emits twice the luminous intensity of 1 square acre even though it is the same cd/m2 when measuring at the sun's surface. The 2 square acres of the sun's surface illuminates the earth's surface twice as much 1 square acre of the sun's surface does because it has twice the luminous intensity.

--Darin

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post #12654 of 16193 Old 11-06-2015, 07:10 AM
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Hi everyone,

Well I still wait on new OLED TV's from Samsung.
There must be an other reason why Samsung is that slow compared to LG.

Samsung display is the biggest as you all know.

2 weeks ago I was at a Samsung event. (I work for SamMobile) A Samsung employer told me that all OLED TV's from LG are having a oled screen burn-in issue. I was surprised and I do think he's wrong maybe he is more jealous, don't know .

One thing is sure: Samsung never talks in public about their new display technology. SUHD came as a surprise as well.

I think that Samsung should announce their new OLED line-up at CES 2016.
There are a couple of marketing reasons.

2016
European football cup 2016.
Rio Olympics & Paralympics 2016. (Samsung is official partner)
This could be a win-win situation...

If Samsung won't announce any new OLED TV at CES 2016, I as Samsung fan will switch to LG's amazing OLED TV.

PS, sorry for my bad English.
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post #12655 of 16193 Old 11-06-2015, 02:40 PM
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The discussion about larger blue subpixels and perception is a really good study on photometry IMO. My good friend is a photometry/OLED professor and researcher and I will love to ask him this when I get a chance.

I’m not in his league but just some thoughts of mine below (IMO only).I think you guys shouldn’t be concerned with any metrics other than luminance(cd/m2).

  • Human eye only sees luminance (think of it as a luminance meter)
  • Luminance is independent of distance from the source or size of the source.
  • Our eye response (pupil/exposure/adaptation) is to the average luminance in the visual field.
  • Enlarging the blue pixel and reducing the current density will produce a lower luminance out of a larger area. However, the lifetime is increased.
  • The average luminance in our visual field remains the same

Over thinking, over analyzing separates the body from the mind

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post #12656 of 16193 Old 11-06-2015, 04:23 PM
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Originally Posted by Danny Dorresteijn View Post
Hi everyone,

Well I still wait on new OLED TV's from Samsung.
There must be an other reason why Samsung is that slow compared to LG.
They have no means of manufacturing OLED TVs. That's why they are "slow". Company size could not be less relevant. GM is bigger than Cadbury, but don't buy a chocolate bar from GM.
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2 weeks ago I was at a Samsung event. (I work for SamMobile) A Samsung employer told me that all OLED TV's from LG are having a oled screen burn-in issue. I was surprised and I do think he's wrong maybe he is more jealous, don't know .
Shocked, shocked to learn a low-level Samsung employee would lie about the competition.
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One thing is sure: Samsung never talks in public about their new display technology. SUHD came as a surprise as well.
SUHD is not technology, it's marketing. Companies don't pre-announce marketing. Samsung *regularly* demonstrates technology in the early development stage.
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I think that Samsung should announce their new OLED line-up at CES 2016.
If they have a licensing agreement with LG (unannounced but rumored to be in the works), they still could not produce OLED TVs in 2016 on their own unless they have secretly built a fab using technology they could not legally use until recently (or still can't legally use).
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If Samsung won't announce any new OLED TV at CES 2016, I as Samsung fan will switch to LG's amazing OLED TV.
Even if they do announce something, you should switch to LG. LG now has been making these since 2013 or so. They are advancing production, lowering costs, expanding SKUs.
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PS, sorry for my bad English.
No need to apologize for not being a native English speaker You're thoughts are totally easy to understand. I don't totally agree with them (as I outlined above), but they are well expressed.

There's a saying about "everything in moderation". If only it was applied to well, you know...
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post #12657 of 16193 Old 11-06-2015, 05:30 PM
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The discussion about larger blue subpixels and perception is a really good study on photometry IMO. My good friend is a photometry/OLED professor and researcher and I will love to ask him this when I get a chance.

I’m not in his league but just some thoughts of mine below (IMO only).I think you guys shouldn’t be concerned with any metrics other than luminance(cd/m2).

  • Human eye only sees luminance (think of it as a luminance meter)
  • Luminance is independent of distance from the source or size of the source.
  • Our eye response (pupil/exposure/adaptation) is to the average luminance in the visual field.
  • Enlarging the blue pixel and reducing the current density will produce a lower luminance out of a larger area. However, the lifetime is increased.
  • The average luminance in our visual field remains the same
Perhaps, these questions are stupid and don't make much sense, but I still feel like I have to ask them:

But if the human eye is a luminance meter and reducing the current density (by enlarging the blue pixel) produces a lower luminance, then wouldn't it make the blue pixel appear dimmer to the human eye/luminance meter were it to be made larger and yet still fed the same amount of electric current?

Why would the average luminance in our visual field remain the same if the area the blue pixel now occupies had increased and its net luminance decreased?

P.S I see how the sun example might make sense and that I conflated the terminology there, a bit. But, I am still apprehensive of it. After much thought, I have now seen this entire thing in a whole new light, but it's still a wee bit unclear to me which side of this thing is up and which one is down and until I figure this one out I feel a tiny bit unconformable to proceed further with the discussion, but, at least, I feel like I am making some progress here.

....
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post #12658 of 16193 Old 11-06-2015, 05:41 PM
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I think you guys shouldn’t be concerned with any metrics other than luminance(cd/m2).
I think that is fair as long as we keep in mind that this averaging works for small angles. Put another way, if you are at a normal viewing ratio with a high resolution display then blocking 90% of each sub-pixel reduces the cd/m2 that the human eyes sees. Blocking 90% of the screen with a blanket is different since human vision then sees one part that didn't change and one part that did. Move hundreds of yards away so that the whole screen looks about the same size as a sub-pixel at normal viewing ratio and human vision will now see the 90% covered screen as dimmer than the uncovered screen, like a blinking light without discernable size.

Kind of like how covering one half of the moon would be different than covering half of a star.

--Darin

This is the AV Science Forum. Please don't be gullible and please do remember the saying, "Fool me once, shame on you. Fool me twice, shame on me."
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But if the human eye is a luminance meter and reducing the current density (by enlarging the blue pixel) produces a lower luminance, then wouldn't it make the blue pixel appear dimmer to the human eye/luminance meter were it to be made larger and yet still fed the same amount of electric current?

Why would the average luminance in our visual field remain the same if the area the blue pixel now occupies had increased and its net luminance decreased?
The way I think of it is that the human eye has millions of detectors all fed by a single set of optics.

The display can be thought of as a compilation of millions of point sources that we average into one large extended source. The abundance of black between pixels contribute to the average.

If we move close enough that the single blue subpixel is its own extended source than we would perceive the fluctuation in luminance. But as an array of point sources we average the luminance within the field.

Assuming the absence of glare, if the sun were to move further away at what point do we start to say it looks less bright? I think the threshold is when it changes from a perceivable extended source into a point source.

Over thinking, over analyzing separates the body from the mind
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post #12660 of 16193 Old 11-07-2015, 02:30 AM
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blocking 90% of each sub-pixel reduces the cd/m2 that the human eyes sees.
That's correct, but even if luminous flux is reduced by 90%, luminous intensity doesn't change (it depends only by the current, not by the emissive area)! That's where all the confusion comes from.
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