lot of coments have been done about how much light does the LCD absorb (or trasmit), some say it is 50% trasmisive, others 20% trasmisive, I always thought it was less than 9% trsmisive (if you do some math assuming 1 color layer and 2 polarizing filters, as well as 90% fill factor... the result is ideally 9%, thats why i always thought that 10% in the best of cases...)

now the real world test;

I took my striped hami and go to the window with my luxmeter.

placed the luxmeter on the floor, angled as best as posible so the meassuremt is max (there is no lcd yet); 41840 lux is meassured. (it is 16:00 no clouds, and for you that do not know, SPAIN is in north hemisphere, so it is summer )

then just took the hami 8" lcd on my hand and placed it perpendicular to the rays before the luxmeter; 2170 lux is the new meassurement.

So we have 2173/41840=5.2% trasmitance on MY HAMI 8".

My luxmeter is not the best on the world, but if we assume same error desviation, the final error on the % is much smaller. So I believe the 20% trsmitance people trust on, is not true.

If you own a luxmeter, I would like to hear more tests on this as well. (note that the sun test is the best test for this purposse, you know paralell light, no fresnell.. no more variables on the equation...)

Think about it, you are only using narrow bands of red, green and blue from the light source, so the transmittance is bound to be very low.

The first polarizer filter blocks 50% of the incoming light. The other 50% is polarized in a specific direction. Each color filter element then blocks between 30-40% of the light going through it, depending on the bandwidth. So for a completely white screen the lcd should transmit about 0.5x0.35=17.5% of the light. Now for an image this will vary depending on the darkness of the image and colors. But this would probably be a good ballpark for maximum possible lght transmittance through the lcd.

Of course, for our projectors it is even lower. The light source only directs a fraction of the light through the condensor fresnal on axis. Also, each plane of glass/plastic reflects a portion of the light off axis.

"a polarizer blocks, 50% of the light"

can you defend it further? any link...

My ideal numbers are not those you used... but i think my real test is much better than any assumtion we could make

My only question would be if the LCD was powered on when you measured. I'v noted that if my LCD unit is powered off, the light transmission through it is reduced. It requires power in order to come to it's maximum brightness.

For polarizing filters, these can be done blocking very little light. one only needs to look as far as a digital watch, or calculator to find an example of a polarised filter which blocks signifigantly less than 50% of the light passing through it. It's my opinion that the majority of the light blocked by the LCD is in the colour filters, and also in the non-transparent delineation areas between the individual red, green and blue dots. I also feel that by eliminating this kind of inefficiency is how commercial projectors can exhibit more light transmission.

A quick, but informitave reference: howstuffworks.com -- How LCDs work

thanks for the input supra;

the HAMI was not on when i did my test, I thought this should give me the higest trasmitance (full white). But i don´t understan the logic behind your input. Why do you think powering on the LCD does "improve" the trasmitance? I would say it is opposite in any case.

do you think LCDs do opperate "AC" current? I thought they only have the ability to align the LC in one direction (on, aply V to the terminals) or leave them original state (off, do not aply V o the terminals). If you think in TFT techonolly, they are transistors(only 1), they only work in DC. An V+/V- (push pull system) would need double transistors...

What I can say is that if I turn the LCD off on my projector, the image seems darker than when it's on. It's as bright as it gets with no signal input, but when it's off the image is not always at it's "whitest" -- There seemst o be some latent capacitance that is not drained, keeping some of the polarisation in effect when the panel is not powered. Or at least this is the only reasonable explanation that I can provide.

Maybe its an optical illusion.

Duh! every optics engineer knows that. Actually, you can reference me. I am a semi-knowledgable person in how these work. Otherwise, you can look it up in any college optics or electromagnetics textbook, generally junior or senior level.

At any point in time, any light is randomly polarized in any direction. If you freeze time you can visualize the lightwave in a type of corkscrew spiral as it travels forth (let's say z-axis). Now a perfect polarizing filter will allow light polarized in one direction to pass (let's say x-axis)and will block light polarized in a perpedicular direction (let's say y-direction). Any light that is polarized between the x and y directions can be broken into an x component and y component,. The x component simply passes. The y component is blocked.

Since the light is randomly polarized at any given point in time, then statistically just as many portions of the lightwave are polarized in the x-direction as in the y-direction. Therin lies your 50% blockage of light.

The second polarizer is the active matrix crystal. The crystal changes polarization direction depending on the power applied to it. All of the light coming through the crystal will be polarized in one direction (let's say the x direction for consistency) from the first polarizer. If you turn on the active matrix crystal to max power, then the crystal only lets light polarized in the y direction pass. Since no light is polarized in the y direction then the pixel blocks all light and is dark. If you turn on the active matrix to half power then the crystal lets light polarized at an angle of say 45 degrees from the x direction. This will block half the light going through the pixel and show gray. If you apply little or no power the crystal allows light polarized in the x direction to pass, giving you full brightness.

Each pixel element feeds a filter. The software and hardware calculates the orientation of the crystal, depending on the image.

Now, if you would get that chip off your shoulder you might've seen that my post was more or less confirming your measurements. For a pure white screen the maximum light that passes through the lcd is about 17% of the light incident on the lcd. It is easy to envision that reflections and off axis light would reduce the total light that passes through to well below 17%. So lets say your luxmeter measures light incident on the sensor from all angles. Light incident a very steep angles may be read by your lightmeter, but reflected by the multiple screen and polarizers.

the 50% trasmitanceon a polarizer filter is not true. The fact is that a polarizer filter reduces the F stop by 2 steps. thats 25% of light.

anyway let me think about it and I´ll try to defend my "no 50% theory" in terms of math.

I would say the light trsasmitance on a ideal polarizer is close to 36%.

Here it is;
we have a obsortion on Y axiz. If we do a graph, we can draw a sine X function that represents the absortion. Lets draw it from 0 to pi. The area that is under the sine function is absorbed light. This area is not 50% of the 1*pi square area but is the integer of sinx (that is -cosx defined on 0 to pi) this area is equal to 2 [-cos pi + cos 0]. So we have the areas relation 2/pi that is equal to 63.6% absortion or 36.3% trasmitance on a single polarizer film.

now check the polarizer film specs form 3dlens; http://www.3dlens.com/Polarizer.html
"Transmittance: single(38%) ; parallel(30%) ; crossed(0.005%)"

single trasmitance is close to that ideal value. two paraell polarizers would ideally have 38% trasmitance (second would do nothing) but it looks like there is some light lose... finally, two crosed polarizers do block %99.5 of light.

I chalenge you to post a link where a 50% trasmitance is stated on a single polarizer.

Rox can you explain this in a non-mathematical way? I’m having trouble understanding what you are saying.

DJ

Rox - This close enough?

http://www.profluxpolarizer.com/pdf/_TRANS...0MAY%202002.PDF

very interesting information paladin. As far as i understood, there are reflective polarizers as well, so the absorbed light is much less on those kind (because it is being reflected) so I think it is not a good example, here what we are discussing is how much light goes trhow it (trasmitance).

dazzla;

a polarizer film is some kind of component where the molecules are on a certain direction, they have E electric field natively as well (+/- orientation) so the idea is that the E conponent of the light will be absorbed by those E fields that are oriented that way. The light that is oriented 90 degrees to that field, won´t be altered. So the Polarizer film works because some light is absorbed, thats the key.

the X component of the light will go throw, but Y wont, because it is going to be absorbed by those molecules. So ramdonly oriented rays will have sin x distribution on the 0 to pi range. If that area under the sin x function was half of 1*pi (full random polarization input light area), then yes, the light efficiency would be 50%. But this area under the sin x is 63% of the full square area. giving us 36% eficiency of trasmited light.

Click to view attachment

Think of it this way. If I use one polarizer oriented in the x direction and it only blocked 25% of the light. then I used another polarizer oriented in the y direction and it only blocked another 25% of the light. Then you would transmit .75x.75=56% of the light.

However, take any two polarizers and orient them perpendicular to each other and you will actually block 100% (maybe slightly less due to light leakage) of the light.

If it only blocked 25% of the incoming light then an lcd would not work. Your own measurements show that the actual light throughput is significantly lower than 56%. I think you measured 5%.

when I said 25% i was speaking about light trasmision (so 75% absortion=blocked light for you). If you check any polarizer filter for photografy you can check it will decrease F by 2 steps typically.

I see now what you are talking about. Your facts are correct, but your understanding of the process is wrong. What I described was correct, but it did not take into account non-idealities of actual systems. The polarizer from 3Dlens is not an ideal polarizer. So it will tend to absorb and scatter a certain percentage of the light.

It's akin to a transparent glass only transmitting 90% of light instead of 100%. If you use that glass as a substrate for an ideal polarizer then the amount of light transmitted will be 50% of 90% of the incoming light. this would be 45% trannmitted. Now throw on top of that a non-ideal polarizer that may absorb a certain amount of light and you drop the transmittance even lower.

My explanation was a ballpark figure on the maximum light that could be transmitted through the lcd. I think I even stated that an actual system would transmit much less.

I didnt read the whole thread ... and maybe my math is a little shakey ... but lookint at rox's picture of the sine wave ....

I thought to effectively find the work done or magnitude of a sine wave, would be to take the square root of 2 multiplied by the peak value. In this case its 1 .... so that times the square root of 2, would yeald .707 (RMS - root mean square) ....

.707 x pi = .707 x 3.14 = 2.22 ... I believe in the above drawing the area under the sine wave to be 2.22 units. Not 2.

which would chance his final number to 2.22/pi = 2.22/3.14 = back to our magic number of .707

which this would mean that not 36% effeciency .. but rather only a hair more than 30% effeciency

mm, the peak value divided by root square of 2, this is the RMS value, this is not same area under the sine. I wonder what does RMS value help us on.

because in real world physics everything is related as work and power/magnitude.

Labling the area under that sine wave as being a coeffecient to the light passing through an object relative the simple algebraic formula of the area of a sine wave is unrealistic.

thats why RMS exists, it isnt relative to electronics and voltage only.

In the real world that sine wave that you labeled isnt going to stop at pi, as your graph indicates, its going to go on for infinity in the Y direction, and a sine wave is going to have a negative half cycle.

In most cases where physics follows a sine wave pattern the work performed is caluclated with the RMS and not the area under the sine wave. Because one true completion of a sine wave period results in an area of 0 (the postive area plus the equal negative area). However since the work changes direction and work is still performed then you cant have 0 work done. It simply changed direction ... the result is a more realistic coeffecient of .707.

Like I said, I didnt read the whole thread, and I might be wasting server space with this post ..... from your test with your LCD panel ... isnt the lower value effeciency calculation more accurate anyway?

mm, i don´t understand how can we tak about power/work when what we are looking is how much light is there.

I thought if full area represents full input light, the are remaining between the sine and the square area would be the outputing light (that was a personal axiom i took) maybe i was wrong.

The area under sine is not 2.22 in any case but 2 as I said.

And yes, i believe the actual test on the LCD trasmitance with the luxmeter (5.2% trasmitance) is the closest model i could ever get to when it concerns to my hami 8".

Im not coming up with the correct word when I say power/work.

If the full area represents full input light, and the light be blocked is uner the sine wave is your coeffecient.

Im not arguring that because I dont know. What Im argueing is how your stopping at just one sine wave cycle. Does this sine wave continue on the graph towards infinity, and if so, then it would have to have an equally negative half cycle. (that would peak at -1 as it approach 2pi)

a common physical charateristic that ive learned in physics classes is that anything that is responsive in nature to a sine wave response tends to behave more in the RMS fashion

Thats why mathematically the area under a complete sine wave period is 0. However if you were to perform some sort of work( such as current through a wire), then there is no such thing as 0 work. Halfway through the cycle it shifts direction, but work is still performed.

If your stopping at looking at the light transmittance at one given millisecond then your simple algrebriac example will probably hold true. but examing the light as it passes through over a larger period of time.

Again, Im just trying to relate simple mathematic to real world physics. Nothing is perfectly linear .....

I know your not native english speaking (I think), so I might have mis-intrepretted your whole point of the sine wave response .....

, i though my english improved a bit ;D you should see my first posts, you will think i is a new language diferent from english .

well, the sine graph represents the randomly polarized input light; so 1 full cicle is enough to sumarize all posible polarized directions, but i just drow half cicle, symetry would help on it. The sine does represent the atenuation to each correspondent angle, so 0 radian and pi radian light angled is not attenued (blocked) but the points betewwn those two will be attenuated by sine x funcion value. achieveing max attenuation at pi/2 angle (100% block). (sorry i should have explained it on the first post).

by the way, the area under a comlete (0-2pi) sine funtion is not zero. Average value on a sine function is 0 thought, maybe it was what you meant. (note that the misinterpretations because of my bad english are bidirectional , I enjoy discusing with you, thanks).

For the sine wave:

Well, though this is only 1/2 the wave function, it's unimportant because the other 1/2 the unction is exatly the same. (0 degrees and 180 degrees can be treated as identical.)

And since this isn't measuring current, there is no "negative" cycle, as it's simply not possible to have negative light.

However, there's two ways to look at the way a polarising filter works.

The one which you have presented says that it blocks or absorbs the light which is not aligned to pass through the filter, however a polarising filter doesn't have to do that, and it can be also argued that it does not do that.

A polarising filter makes light polar. So where your chart says that it will absorb 71% of the light at 45 degrees to the filter, I could argue that it will instead PASS 71% of that same wave, aligned to the polarising filter. This reverses the graph, and instead means that 63% of the light will pass through the filter.

In that event, a second filter at right angles still blocks out the vast majority of remaining light, because it is not fed a random mix of polar light, but is instead fed light that is polarised perpendicular to the filter. This way also a second polarising filter at 45 degrees passes 71% of the remaining light, resulting in an overall pass of about 45% of the light. (71% of 63%) which seems to be more accurate for what happens with 2 filters at 45 degrees than what your model would predict, which is 29% of 36% or 10% pass.

In the case of an LCD monitor, the colour filters, and the small black area betwen the coloured sub-pixels account for a large amount of the losses in light transmission. In addition, there may be additional light blocking elements involved. After all, for an LCD manufacturer, maximum light transmission may not always be a good thing. Anyone who's used a digital camera in the sunlight can tell you that the image is hard to see in the sun. I can take a piece of glass with a 35% tint film and place it on a sheet of white paper on my desk, and still see that the paper is whte. If that lighting were enough to let me see the white difuser sheet in my LCD monitor, that would be completely unacceptable. Fortunately, I believe that the colour filters and the blacked out spaces between the sub-pixels is enough to ensure that this does not happen. If not, however, the LCD manufacturers would certainly put a tint film on the face of the monitor in order to prevent this from being a problem. Such a film could be cause for certain panels passing more light than others.

you are right supra; the sin X graph (absortion) could be considered as well cos Y graph trasmission... I would say someting was wrong in my assumtion...

look at this paradox;
Lets say i feed 100 lumens of 45 polarized light to my ideal polarizer film polarized at 90 degrees. Then Sin 45=cos 45=0.707 (lets say 70% will be trasmited on Y axiz and 70% will be absorbed by X axiz as well. well, I though i only had 100 lumens . where did those 40 lumens come from? somebody turned on a light?

Someone can feed me a couple of lumens to light my way of understanding this?

oooh! Now theres a debate! I would say all mass is negative light!

Anyway, back to the topic. As you have correctly stated several times, the colour filters and pixel boundaries are a major absorber. Lets say for simple numbers that 1/4 of the LCD area is each red, blue, green, electronics (opaque). For the red 1/4 the full spectrum of light hitting it is reduced from the area under your bulb manufacturer's spectral graph to the area under the red bit! The rest is absorbed - heat.

etc for the other 2 colours.

This is why the projector manufacturers go throught he expence of breaking white first before going through 3 LCDs.

What we really need is a technology where the filter shifts spectrum. This would have better efficiency than 3 panels+splitter hardware. I'm thinking of the colour changing octopus which is a colour change based on chemical reactions.

I'd come up with it myself but it sounds expensive

I figured I'd do a search on spectral changing filters and there it is.

http://micro.magnet.fsu.edu/primer/java/fi...lctf/index.html

Damn, why didnt I think of that? lol

Someone REALLY needs to make an image panel out of this stuff.

just to jump in here, having looked all over the net the last week or two(and following links from others in the anti-glare thread) I have not seen any polarizers (commercially available) that are better than 38% ... even the ones on scientific sites that claim high transmittance..thoug 3m MAY have some that go higher(I'm thinking I saw 43% in their vikuii line but I don't think they are standard polarizers)... also I wonder now where people got the idea of 20% transmission when all the lcd sites I've seen say between 3% (for pva type) t0 at most 9% ... which makes way more sense when you consider 2 polarizers at 38% each, the 2 sheets of glass for the lcd itself PLUS the liquid crystals and mask...actually 5% is pretty good when you add all that up ;-)... and rox...in that other thread you thought that maybe it technically wasn't an increase in brightness when anti-glare was stripped but it made the arc smaller(because less "fuzzy") to the triplet... can you expand upon that if you still think so?

mm, 3% trasmitance for the PVA type of lcds... interesting... any link?

I measured two lcds alrady and one gave me 5,2% trasmission, the other gave 6.6% trasmision. Supra told me his 17" was 8,8% trasmisive and I heard about other guy at diyaudio that measured 5% on his Lilliput.

All those values do 6.4% trasmitance average.

I believe the antiglare layer on the lcd does improve viewing angle and does drop the straight contrast. But does improve overal contrast in all the viewing angle. This antiglare layer is a difusive layer that does 2 things in py opinion; 1 the reflected light from the viewers side is going to be difussed, so virtually no reflection will be seen (antiglare). And the light coming from the lcd is going to be difused as well so viewing angle is better than without the antiglare film.

Supra told me that he was able to project the arc from the lamp on the triplet (paper where the triplet was) very clearly with no lcd in the path. Then with same setup he just introduced the lcd between the frresnells but didn´t change any distance. Then he noticed the arc was much bigger, although the difused arc on the paper was the smallest he could get.

So the trasmitancy on the lcd is what it is. We can´t do anything there i believe. But the difussiveness efect is killing the light efficiency since less light is going to the triplet. Removing the antiglare looks like a huge improvement. I am very interested on meassuring the effects with a luxmeter. Maybe i feel brave enough to strip a lcd one of those days.

Did you read the link in post#13 of this thread?

I think it's only usable with a small parabolic or ellipsoidal reflector though.
But we're working on the latter.

yes, but they didn't mention the total light transmitted(unless they are trying to say 100% of the light input to the polarizer is transmitted, thats impossible) ... now the concept of the reflective polaraizer has been raised(3m has one also) and yes, a new light engine wouldneed to be designed but I don't think this would increase total transmittance(you'd get more lumens yes, because of recylcing the non-polarized light but actual transmittance wouldn't vary)

I agree about the transmittancy, other than removing anti-glare you're pretty much stuck with what it is, but I've stripped a psone lcd of both back and front polarizer and using the non-anti-glare polarizer the image was much crisper and at least seemed to be much brighter(sorry no luxmeter ;-) ..... so unless 3m really has a higher effeiceincy polarizer .... I think a redisigned engine (with please god an elliptical relfector! ;-) that "recycles" the light using a reflective polarizer si our best bet .... now if someone could only find a decent sized elliptical!
and a quick question...someone measured light gain in the anti-glare thread and it averaged to about 24% ...would that mean (for example) that would be 24% of the average 6.4% transmissiveness? so boosting it to > 8% total LCD trans? I get lost sometimes in the math ;-)

someone meassured the gain? (24% light increase???) can you link me?

took me a while to find it, that thread has just exploded, but it has been an exciting couple of weeks... it was oztang on page 40 I think this is ti:

http://www.lumenlab.com/forums/index.php?s...ic=7882&st=780#

thank you very much, I jumped that post, but is one of the most important on the trhead i believe (now that i have read )

the best increase he got was with the tape test, 23% increase on lux reading. Very interesting.

he had 116 lumens output on his standar setup. (I always though the lumenlab average was somewhere 200 lumens).

you're welcome, just a small repayment for all the info you've provided in the forums... yeah I've seen 200-250 quoted a number of places...did you measure yours? and wouldn't it make a difference if it was a 15" or a 17"? (larger LCD to the same size diagonal = more lumens) and how is lumens for a projector calculated anyway? lumens per square inch/cm or a defined area or?

no, mine is not finished yet.

i think the 17" should be sigly brighter but less even because of inverse square law. Brighter because it is larger (more captured light) and the fresnells used typically are the same focal for 15" or 17" (220 and 330).

The luxmeter does read the lux value (lumens per square meter, lm/m^2 ) so since you have 9 lux readings on the screen (that's ansi) you can work out the average value and multiply by the projected area (wide in meters by high in meters) so;

lumens=average lux X meters square area

his image size was 1.6 meters by 1.34 meters. This is 2.144 meters square, So 54 lux average value X 2.144 = 116 ansi lumens

but note he did not use white patern, it will significally improve if white patern is used. (Ansi test is done with white image, RGB at 100%)

ah! thank you, wondered how it was calculated....

How would a 17" be brighter. The center of a 17" is the same as a 15" there should be no difference, the sides should be dimmer due to the inverse square as the edges of a 17" are further than the 15". So a 17" should average out to a dimmer brightness, if the PJ was brought closer to the screen to maintain equal image size then the additional area would capture some more light but then we are dealling with an image that is less of a magnification than a equivalent 15" projected so its not really fair to call the 17" brighter as the configuration has now changed.

well, we should define brightness first and then conclude wich one is brighter;

as far as i can say, brightness is the projector's output in terms of lumens (ansi lumens if you prefer).

So the larger area on the 17" does capture more lumens from the lamp, this is reduced to more lumens on the output, so my final conclusion is 17" is brighter than 15".

Sounds logical, but what about the fact that a larger fresnel and LCD put more material between the source and the screen? More mass should block more light. In addition a bigger LCD means that light is going through the lens at a greater angle. Wide angle lenses are notoriously slow! Im sure many people have dark corners with their lens so is a bigger one brighter?

My CMV 15" was 8.0% transmissive. My 17" LG is 8.8% transmissive.

I'd like to take a stab at the "tape" on the panel to see what that does, but I don't want to sacrifice my 17" panel. I can do that with my 15" panel, though. Nothing to lose, since it's busted.

There is more area and thus more light captured but the center 15" diagonal is the same on both monitors the light you are add in one the sides top and bottom (including corners) they are less bright than the center 15" portion so the average brightness is lower, but if the light is projected in to the same area the projector is closer (less magnification) so it is brighter at the center but given the "dark corners" I think that the average brightness is lower. If both a 15" and a 17" are at the same distance there would be more light spread over a larger area, the center would be the same as the 15" and the corners darker so again the average brightness is lower, the center is the same, the extra light captured gets used up in the larger projection. The problem is a larger LCD will only gain light at the edges which due to inverse square will be darker as defined by a cosine ratio from the center. The center brightness will be better but the corner to center brightness will be a lower ratio. I believe the dark corners will create the illusion that the image is dimmer as our eyes are tuned to differences more than absolute brightness.

inverse law is not cosine function.

mmm, maybe this last rasoning is what you are asking;

the 17" is exactly a 15" and more added outer area. The inside 15" will have same amount of light captured as the real 15". So the outer area will make the only diference in the projector brightness. The 17" should be brighter.

thats what I would say(given the same size projected diagonal for both that is) ..however elken2004 in another thread mentioned he thought the opposite because of pixel density and more mask in the 17" ...thoughts? ..just did quick math..pixel density is different roughly 9300 pixels/sq inch for 17" , 15" is around 7300/sq in...think it makes a big difference? surprised me, figured since pixels were smaller(generally) on 17" it wuld be about same...