How do DVD +/- Recordable disks work?

[raid517]

Hi I am doing a project at college and I need to do a report on how DVD-Rs work. I have heard that there are soon to be new 8GB dual layer DVD-Rs and I am very interested in how this 'double layering' works?

I mean, when we talk about burning a DVD, is that really what happens? Is the process of writing to a DVD a thermal process? And if so and if these DVDs really are dual layered, what prevents the heat used to burn one layer from bleeding into the other and affecting it?

How chemically is it possible to change the properties of varius multilayered surfaces using heat, without the heat source you are using affecting any of the subsquent (and similarly constituded)layers? (I need to figure this out in order to demonstrate a more general principal).

I did search Google - and unfortunately many of the examples I found were either way too technical for the broad audience it is aimed at, or it was not technical enough - in that it provided very little information that would help those who at least had some basic knowledge of the subject to improve their understanding.

Something like a science journal article aimed at skilled lay people would be a good starting point. Or maybe some of you guys can supply your own input?

Any input anyone can offer would be very much appreciated.

Kind regards,

GJ

[red-adair]

I don't know the answers to ure questions but might suggest u look at'dvdrhelp.com'.....big website with lots of info on it.

[jggimi]

There are some good information and links about the six different "burnable" DVD formats currently available starting with Question 4.3 of the official DVD FAQ. (See Doom9's website, "The Basics" page, for the link.)

Doom9 has also tracked news relating to dual layer burnable DVDs ... the most recent news talking about the standard was 20 December 2003 and 11 December 2003.

[raid517]

I'm sorry, I must be blind. I can't see very much about the question I asked there. I didn't see any number 4.3 in the list of FAQ's - and in any case I kind of doubt that someone asking about the fine details about how the new dual layered DVD-Rs work is really likely to be covered in any FAQ. I doubt many people take the time to ask.

Is it possible that anyone has a link to a more detailed source? You mention previous doom9 articles and I have read these, but what I am looking for is diagrams and as best technical explanation I can find for a muti-skilled audience some of whom may have in-depth knowledge of this subject and some of whom will not.

For example I would like to know how thick each of the layers on a muti layered burnable DVD +/- disk are - and if each layer is changed via a thermal process why the heat from the laser burning one layer doesn't affect the subsequent layer.

These questions admittedly might seem a little odd, but I assure you that they do have a practical relevance to me.

Any help anyone can offer would therefore be very much appreciated.

Kind regards,

GJ

[jggimi]

Here's the link, since you had trouble locating it:

http://www.dvddemystified.com/dvdfaq.html#4.3

[raid517]

Yeah thank you for that. But I was really looking for something a little more detailed.

GJ

[jggimi]

There are links in section 4.3 which you should research, and manufacturers' links in section 6.

When it comes to dual-layer burnable discs -- you are asking for technical details of products that are not yet to market. You may get cursory information, but you won't get a great deal of detail from manufacturers' web sites, therefore.

I recommend following the links in the FAQ to the DVD+ and DVD- consortia, as they likely have academic members. Any information that is not trade secret, but is the I.P. of a consortium itself, should be available to consortium members in academia. Your university may already be a member of one of these, or, you may be able to arrange a university membership, or an individual one. Additionally, they may be able to connect you with a colleague at another university who may have permission to divulge what you are looking for under non-disclosure.

[Kedirekin]

jggimi is right. The links in the FAQ do lead to some useful information. There are even some nice cut-away diagrams. I doubt it will provide enough detail to satisfy your appetite though. I'd see if I could find technical data on pressed DVDs were I you. A lot of it should apply, as a DL DVDR has to be compatible.

The primary layer of a DVD5 is sandwiched at the center of two 0.6 mm plastic substrates. It stands to reason that that the semi-transparent layer of a DVD9 is is about half that; 0.2 to 0.3 mm.

In human terms these layers are very close together, but in optical terms they are not. The spot of heat generated by the laser is microscopic and only present on any given location on the disk for a tiny fraction of a second. Yes, the heat will dissipate to the other layer - that's simple thermodynamics. But by the time it dissipates over the 0.2-0.3 mm distance, the temperature will only be a tiny fraction of what it was at the source.

There are several thought experiments you can perform to gain comfort with this. For example, just think about the laser power (a fews 10's of milliwatts) applied to something 0.2-0.3 mm in size, even for an extended period, let alone for a millisecond. It's just not going to get that hot. You could easily apply the laser to the end of your finger continuously and not even feel it.

[ammck55]

raid517--I'm going to move this to the DVD Burning forum, the fellas that hang out there that don't make it to Noobs very often may have some good ideas and help for you.

ammck55

[raid517]

OK well I emailed you a response anyway. Really it is this layered separation between thermal layers that interests me most. Having a transparent insulating layer is another possible option, although by what you have said, this isn't really a factor in this process? I am really very interested in muti-layered surfaces and ways to prevent the heat from a laser from transferring its thermal energy into subsequent layers. The problem I envisage at this time is that 0.3mm in the whole scale of things is really quite thick... Does anyone have any ideas on how one might significantly reduce this thickness, while retaining the ability to affect a specific chemical change between the layers?

I know my questions might seem very odd to some of you guys, but really my questions do have a real relevance. If anyone here is suitably qualified in this field and wishes to learn more, I would be happy to chat to them privately and hopefully seek their advice and guidance - although this might present some slight legal difficulties.

In the mean time if anyone has any further insights to offer, please feel free to let me know.

GJ

Doh! I just got that one, it is the sandwiching material inbetween the layers that is thick, not the layers themselves. So this still begs the question of how thick the individual layers are?

[raid517]

OK I know that you guys are going to want to kill me for asking so many dumb questions - but you see what I am trying to do is chemistry between very thin layers of material. I cannot quite say what kind of chemistry - but it is related to this general field.

What I am wondering is what would the actual calculation of the transport of energy between various thermal layers look like?

I mean you have a couple of variables, such as the chemical/thermal properties of the laser you are using and the frequency/width of the laser source, so how would this equate in real terms? I'm afraid my physics isn't all that strong.

Surely there must be an exact way to calculate this? (Given that one might consider using different kinds of layers to generate different thermal properties).

Just curious...

GJ

[SomeJoe]

If you look up the specifications for DVDs, you'll see that the track pitch on a DVD is 0.74 um. Obviously, the laser focus is accurate enough and power is low enough such that when heating the material to cause the reflectance change, the laser does not affect the adjacent track spiral.

So therefore, if the vertical separation between layers is more than 0.74 um (which is most certainly is), the laser is out of focus and cannot concentrate sufficient energy into the material to affect the reflectance change in the opposite layer.

Now obviously, it's also easier to not affect the adjacent track spiral since it is displaced from the laser's path horizontally rather than vertically. Still, we're talking such infinitesimally small quantities of power that as soon as the laser is out of focus, the power level is insufficient to cause the reflectance change in the dye.

[andyg]

How Chemical do you want to get?
will this do?


"Synthesis, Structural Analysis, and Chemical Properties of Phthalocyanine Derivatives. K. Okada1, N. Sasa1, K. Nakamura2 and S. Okada3, 1Ricoh R & D Center, Yokohama, Japan, 2Showa University, Tokyo, Japan, 3Science University of Tokyo, Tokyo, Japan


Syntheses of the metallophthalocyanic and oxo-bridged phthalocyanine (Pc) dye derivatives were performed and the crystal and molecular structures have been determined by single-crystal X-ray diffraction analysis. The general formula of these Pc derivatives is Si(Pc)(OSiR1R2R3)2: -[Si(Pc)(OSiCH3)2 C7H14]n- (1), Si(Pc)[OSi(C6H13)3]2 (2), Si(Pc)[OSiC8H17(CH3)2]2 (3) and Si(Pc)[OSiC12H25(CH3)2]2 (4). These Pc dye derivatives are prepared by refluxing a mixture of Si(Pc) (OH)2 and SiR1R2R3Cl in pyridine, followed by cooling the mixture slowly, and then drying the resulting precipitates.

The crystal and molecular structures of 1 to 3 have been determined by single-crystal X-ray diffraction analysis. The structures were solved by the application of direct methods. 1 Eight atoms of 1 were located on the special position (z=1/4). All computations were performed on an Compaq personal computer with a DS*SYSTEM3.1

The Pc skeletons of 1 to 3 and the oxo-bridged substituentes are coplanar. The dihedral angle between the mean planes of the Pc skeleton and two single-atom Sime [Sime-O-Sipc-O-Sime] is nearly at right angles. The Sipc-O bonds are shorter than the Sipc-N bonds, and the Sime-O bonds are shorter than the Sipc-O bonds. The polymer chain for 1 is constructed at the siloxy group along the a axis. Two hexyl groups of two siloxy side groups for 2 have the same direction, but one hexyl extends in the opposite direction. For 3 two methyl groups and one octyl group extend in the opposite direction. No readjustment of these siloxy groups is allowed. The crystalline conformations of 2 and 3 are exactly coincident with the potential energy minimum. The energy barriers have unifold patterns, are high and narrow due to steric repulsion caused by the Pc skeleton.

We applied Pc dyes, 2 to 4, mixed with a polymer for the control of the aggregation state to CD-R and DVD-R recording systems. The aggregation state of the recording layer is controlled by choosing the set of axial substituents R1 to R3. The interactions of Pc dyes with polymers are dependant upon the length of axial substituents. We got the best writing contrast and the highest stability with the mixed Pc dye 4. The conformation of axial substituents is very important for the ability of the dye aggregation and the capability to control the aggregation state.


[1] K. Okada and S. Okada, IUCr96, CGA17 Abstr. (1996) PS03.02.10, C-80."

or technical:


"The recording action takes place by momentarily exposing the recording layer to a high power (approximately 8-10 milliwatt) laser beam that is tightly focused onto its surface. As the dye layer is heated, it is permanently altered such that microscopic marks are formed in the pre-groove. These recorded marks differ in length depending on how long the write laser is turned on and off, which is how information is stored on the disc. The light sensitivity of the recording layer has been tuned to an appropriate wavelength of light so that exposure to ambient light or playback lasers will not damage a recording. "


"An optical storage technology in which the disk drive writes data with a laser that changes dots on the disk between amorphous and crystalline states. An optical head reads data by detecting the difference in reflected light from amorphous and crystalline dots. When full a phase-change disk can be erased (or "reformatted") using a medium-intensity pulse to restore the original crystalline structure. CD-RW uses phase-change technology. "

"Magneto-Optical Technology: a rewritable optical storage technology that uses a combination of magnetic and optical methods. Data is written on an MO disk by both a laser and a magnet. The laser heats the bit to the Curie point, which is the temperature at which molecules can be realigned when subjected to a magnetic field. A magnet then changes the bit's polarity. Writing takes two passes. Unlike with phase-change drives MO disks do not have to be "reformatted" when full. See also LIMDOW. "


Wait a minute, this is your homework, not mine!
I'll grab the beer now & you start the search engine if you want to get more technical than this.


ô¿ô

[Kedirekin]

In the first approximation, the calculations are pretty simply. The substrate has a constant heat transfer coefficient, so heat dissipation is linear. You'd just need to know how much heat is being generated and what the transfer coefficient is.

Beyond that though, the calculations get hairy fast. The problem is four dimensional - the heat dissipates in 3 dimensions and is time variant, and on top of that the heat source is moving with respect to the disk. You've basically got a series of point heat sources exploding one after the other along a tangential line, and the wave fronts of heat dissipation from these tiny explosions will impinge on the spot you're interested in a complicated, time varying way. This is not a back of the napkin calculation.

But it doesn't really matter anyway. What SomeJoe says is 100% correct - the dye in the same layer immediately surrounding the heated spot is going to be affected by heat dissipation much more than the dye in other layers would be. If the dye in the same layer is unaffected, you probably don't need to worry about the dye in the other layers.

Of much more concern I think is the focusing of the laser. If you have a beam that focuses on a narrow taper, and you put the layers too close together, the beam will basically be focused on multiple layers at the same time. You'll have the same problem on read as on write - the total reflectivity of the beam will be affected by multiple layers.

[raid517]

Thanks guys, this is all very interesting.

So what you are saying is you can do thermal chemistry between very thin flat layers?

As I said my interest in this subject is inspired by DVD-R technology - but is not entirely related to it. (Although it is still related to computers and data storage).

Again I assumed that there would be some sort of physics type calculation you could do to calculate this? I mean you would have some variables, like the thickness of the layers, the chemical constitution of of the material you were using, the power of the laser etc...

But I really am asking you guys (since clearly you have much more experience of this topic than me) is what exactly would that calculation look like?

I mean how do these variables play in relation to each other in order to give you a final figure about how much energy is transported through any single layer?

All I am looking for is a single equation, which I could then experiment with, by say using different chemical substances and different laser frequencies to give the results I want.

If I then know the thermal properties of certain chemicals, I can then use this equation to narrow down the most likely candidates. Do you see what I am saying?

It's all very vague I know, but I'm looking for a way to make it less vague.

I know Kedirekin made a worded reference to the kind of equatiion I would need, but my maths and physics is weak - and I would really like help with deciding what an actual equation (not worded) like this would look like, given all the variables outlined previously.

Any help you guys can offer would therefore be very much appreciated.

GJ

[raid517]

OK, well possibly you guys think I'm asking too much - but this is really a subject I knew nothing about until you explained it to me. I am very keen not to exhaust any store of goodwill, however I do have one final question.

Specifically I want to know what is the purpose of the the plastic substrate layers separating the recording material on the disks? Is this just for mechanical stiffness, or does it in any way aid the dissipation of thermal energy between the layers?

Again I am very grateful to you guys for teaching me so much - and I hope that you don't mind me asking what might seem like dumb questions to you, but if I could just clear up the last couple of points I will be done for good with this topic.

Any input you can offer would be very welcome.

Kind regards,

GJ

[SomeJoe]

There are two basic equations you need you need to play with to get a physical understanding of heat transfer. One is the heat transfer by conduction equation. It tells you how heat moves through a solid piece of material.

Q   kA(Thot-Tcold)
- = --------------
t         d


Q = Total heat energy transferred through the material, in Joules
t = time, in seconds

(Q/t) is a power term.  Joules/seconds = Watts

A = surface area of the material, in square meters
Thot = temperatue, in Kelvin, of the hot side of the material
Tcold = temperature, in Kelvin, of the cold side of the material
d = thickness of the material, in meters
k = thermal conductivity of the material, in W/(m K).

Kelvin temperature = Celsius temperature + 273


Example:

How much heat is transferred through a 10 cm^2 piece of aluminum 1 cm
thick over 10 seconds, if one side of the aluminum is held at 0
degrees C (freezing point of water), and the other side is held at
100 degrees C (boiling point of water)?  What power rate is this?

Aluminum's k = 205.0 W/(m K).

    tkA(Thot-Tcold)
Q = ---------------
          d

    (10 sec)(205 W/m K)(10 cm^2 * 0.01 m/cm * 0.01 m/cm)(100+273 K - (0+273 K))
Q = ----------------------------------------------------------------------------
                                              0.01 m

Q = 20,500 J

Power = Q/t = 20,500 J / 10 sec = 2050 Watts

The other equation you need to be familiar with is the specific heat equation. It tells you about temperature changes within a material when a specific amount of heat (in Joules) is transferred into it.

Q = cm(delta T)

Q = heat transferred into the material, in Joules
c = specific heat of the material, in J/(g K)
m = mass of the material, in g
(delta T) = change in temperature of the material, in either degrees C or degrees K

Example:

The same piece of aluminum used in the previous example is used
here.  It is initially at 0 degrees C (freezing point of water).  How
much heat is required to raise it's temperature to 100 degrees C
(boiling point of water)?  The density of aluminum is 2.7 g/cm^3,
the specific heat of aluminum is 0.9 J/(g K).

Our piece of aluminum is 10 cm^2 in area, and 1 cm thick.
Thus, we have 10 * 1 = 10 cm^3 of aluminum by volume.

10 cm^3 * 2.7 g/cm^3 = 270 g of aluminum.

Q = cm(delta T)
Q = (0.9 J/(g K))(270 g)(100 - 0)

Q = 24,300 J

Note that it doesn't matter whether the 24,300 J is transferred into
the aluminum within a few seconds or over several days.  Time only
enters the picture when you're talking about the power required to
raise the temperature, not the heat.

So, knowing these two equations, you can now begin to make some estimates of what's happening at the DVD surface during burning.

You can assume some values:

Q/t for the laser ~ 10 mW.
Density of polycarbonate = 1200 kg/m^3
Thermal conductivity of polycarbonate = 0.19 W/(m K)
Specific heat of polycarbonate = 1.3 J/(g K)
Distance between layers on the DVD-R = 50 um
Width of a pit on the surface of a DVD-R = 0.50 um
Length of a typical pit on the surface of a DVD-R = 1.4 um
Rotational speed of a DVD-R at outer edge = 10.5 revolutions/sec
Diameter of disc at outer edge of data = 116 mm

What you need to do with this data is the following:

1. Compute the time that a pit is exposed to the laser, based on the rotational speed, disc dimensions, and pit size.

2. Compute the temperature rise in the pit using the heat capacity equation.

3. Assuming the pit is hot for 1 second after exposure to the laser, find how much heat can be transferred to the 2nd layer using the heat transfer by conduction equation.

4. Assuming the 2nd layer pit is the same size, compute the temperature rise in the 2nd layer pit.


The previous 4 questions are left as exercises for the reader.

[raid517]

Ok thanks man... That was pretty much what I was looking for, and those example questions are great!

Thanks very much!

GJ

[Kedirekin]

Okay, I've been thinking about this. SomeJoe's post kind of makes this post pointless, but I've already taken the effort, so I'm going to post it anyway.

------------------------------

Assume that a laser pulse heats a point within layer 1. For simplicities sake, we'll ignore the duration of the laser pulse, and we'll ignore that the disk is made up of different materials. We'll assume that the energy from the laser pulse is transmitted instantaneously and that this disk as an amorphous homogeneous solid.

After the laser pulse, a cross sectional distribution of temperature might look something like this:

 |*                      
T| *
 | *
E| *
 |  *
M|  *
 |  *
P|  *
 |   *
E|   *
 |   *
R|   *
 |   *
A|    *
 |.....
T|    *.....
 |    *     ....
U|     *        ....      
 |     *             ...
R|     *               ...
 |      *                 ....    
E|      *                     .A..         
 |       *                        .....                   
 |        *                            .....
 |         *                                ......
 |------------------------------------------------
 L1                           L2
                 D I S T A N C E

The upper curve is the temperature distribution a some point shortly after the laser pulse ends. The lower curve is the distribution at a later time, after the 'burned' spot has had a chance to cool and the energy has had a chance to diffuse. Assuming a single laser pulse, the total heat energy under the curve is constant; energy distribution has changed but the total energy remains the same.

I do not know what function describes the curve above. To simplify our calculations, we'll simplify our model and assume the disk material is a thermal superconductor; heat energy diffuses at light speed and within the heat horizon the energy is distributed evenly. This changes our curves as follows:

       
 |                         
T|        
 |         
E|        
 |         
M|***      
 |  *      
P|  *      
 |  *      
E|  *      
 |  *      
R|  *      
 |  *      
A|  *      
 |  *     
T|  *       
 |  *            
U|............................B 
 |  *                         .
R|  *                         .
 |  *                         .  
E|  *                         .            
 |  *                         .                           
 |  *                         .             
 |  *                         .                 
 |------------------------------------------------
 L1                          L2
                 D I S T A N C E

In both graphs, let the time of the lower curve be the time of maximum temperature in layer two (denoted by the distance L2). Point B will always be at a higher temperature than point A (because the total energy under the curve in both cases is the same), so B is an upper bound on the maximum temperature in layer two.

Since the energy distribution is even and completely bounded within the heat horizon, the calculation of the temperature at point B becomes trivial. The volume of heated material is simply a sphere with a radius equal to the layer separation. Using the second equation provided by SomeJoe...

delta T = Q/(c*m)

and

m = 4/3*pi*r^3 * D

where

delta T: temperature (above ambient) at point B
Q: total heat energy provided by the laser pulse (use table of conversions to convert milliwatt seconds to joules)
c: specific heat of the material
m: mass of material within the heat horizon
r: layer separation
D: density of the material

------------------------------------


That's as far as I could take it.

Determining the actual temperature at point A is, in my opinion, non-trivial, and I wouldn't undertake it without a significant government grant.

A special thanks to SomeJoe - you saved me a lot of effort trying to find those equations. It's been ages since college, and most of my texts were rented so I don't have access to them.

[raid517]

Again, spectaucularly good work Dude. That really helps clear things up for me. Thanks very much.

GJ