9 ms·

Every pixel on the panel must work, otherwise the whole panel is useless. Now imagine you have one broken pixel on average for every million pixels. Apple Watch

by 6nf 9y ago

Every pixel on the panel must work, otherwise the whole panel is useless. Now imagine you have one broken pixel on average for every million pixels. Apple Watch screens have 100k pixels so on average 9 out of 10 screens test OK and can be used to make a watch. But the iPhone has about a million pixels, so you end up throwing away about half of all the screens you make. More pixels = lower yield.

kilotaras 9y ago
Probabilities works out even worse. For million pixels roughly 37% of screens would turn out OK.
uptown 9y ago
Could smaller adjacent panels be coordinated to appear imperceptibly as if they’re a single panel?
- cptskippy 9y ago
  Potentially, that's how led billboards are assembled.
nonbel 9y ago
Somethings off. It should work like this: p = 1e-6 n = 10^(1:7) frac_accepted = (1-p)^n Results: num_pixels frac_accepted 1e+01 0.9999900000 1e+02 0.9999000049 1e+03 0.9990004993 1e+04 0.9900498288 1e+05 0.9048373728 1e+06 0.3678792572 1e+07 0.0000453997
- Retric 9y ago
  This assumes each failure is independent. Often a single failure will cause many bad pixels on the same screen. So you might have 98 good screens 1 with one bad pixel and one with 5,000 bad pixels when the per pixel odds say you are going to end up with 1 good screen if you are lucky and 99 bad ones.
  - nonbel 9y ago
    I simulated 100 screens with different numbers of pixels where the probability the first pixel is bad in a screen is p0 = 1e-6 and the presence of each bad pixel increases the probability of the next one by the same amount (ie probability for a second bad pixel is 2e-6 and for last pixel is p0*number_already_bad_px). A screen was accepted if it had 0 bad pixels. The results look pretty much the same: num_pixels frac_accepted 1e+01 1.00 1e+02 1.00 1e+03 1.00 1e+04 0.98 1e+05 0.89 1e+06 0.32 1e+07 0.00 Of course this was just a convenient scheme to simulate. The greatest number of bad pixels was 86k for the 10 million px screen (so chance the last pixel was bad was ~9%). It does show that the independent pixel model isn't necessarily very far off though.
    - Retric 9y ago
      You messed up the math. You need to count (# number of pixels that where bad) / (total number of pixels produced) not simply call the odds for the first bad pixels as the rate of bad pixels.
      - nonbel 9y ago
        You are saying to adjust p0 so I get that the frequency of bad pixels = 1e-6? That makes sense but it would probably now depend on what size screens are being considered... not sure if I'll go back to it.
- 6nf 9y ago
  I didn't actually calculate the probabilities, it was just my off-the-cuff estimates. 37% could be 'about half' :P
baybal2 9y ago
>Every pixel on the panel must work, otherwise the whole panel is useless And that is not so far fetched. The most high yield semi processes can give you defects in single digits for a fully exposed 300mm wafer. The challenge here is that they will have to be grown on a substrate different than silicon for anything bigger than apple watch screen. Growing inorganic semiconductors on glass, organics, non-crystalline materials is what is hard

by 6nf 9y ago

kilotaras 9y ago

Probabilities works out even worse. For million pixels roughly 37% of screens would turn out OK.

uptown 9y ago

Could smaller adjacent panels be coordinated to appear imperceptibly as if they’re a single panel?

cptskippy 9y ago
Potentially, that's how led billboards are assembled.

nonbel 9y ago

Somethings off. It should work like this: p = 1e-6 n = 10^(1:7) frac_accepted = (1-p)^n Results: num_pixels frac_accepted 1e+01 0.9999900000 1e+02 0.9999000049 1e+03 0.9990004993 1e+04 0.9900498288 1e+05 0.9048373728 1e+06 0.3678792572 1e+07 0.0000453997

Retric 9y ago
This assumes each failure is independent. Often a single failure will cause many bad pixels on the same screen. So you might have 98 good screens 1 with one bad pixel and one with 5,000 bad pixels when the per pixel odds say you are going to end up with 1 good screen if you are lucky and 99 bad ones.
- nonbel 9y ago
  I simulated 100 screens with different numbers of pixels where the probability the first pixel is bad in a screen is p0 = 1e-6 and the presence of each bad pixel increases the probability of the next one by the same amount (ie probability for a second bad pixel is 2e-6 and for last pixel is p0*number_already_bad_px). A screen was accepted if it had 0 bad pixels. The results look pretty much the same: num_pixels frac_accepted 1e+01 1.00 1e+02 1.00 1e+03 1.00 1e+04 0.98 1e+05 0.89 1e+06 0.32 1e+07 0.00 Of course this was just a convenient scheme to simulate. The greatest number of bad pixels was 86k for the 10 million px screen (so chance the last pixel was bad was ~9%). It does show that the independent pixel model isn't necessarily very far off though.
  - Retric 9y ago
    You messed up the math. You need to count (# number of pixels that where bad) / (total number of pixels produced) not simply call the odds for the first bad pixels as the rate of bad pixels.
    - nonbel 9y ago
      You are saying to adjust p0 so I get that the frequency of bad pixels = 1e-6? That makes sense but it would probably now depend on what size screens are being considered... not sure if I'll go back to it.
6nf 9y ago
I didn't actually calculate the probabilities, it was just my off-the-cuff estimates. 37% could be 'about half' :P

baybal2 9y ago

>Every pixel on the panel must work, otherwise the whole panel is useless And that is not so far fetched. The most high yield semi processes can give you defects in single digits for a fully exposed 300mm wafer. The challenge here is that they will have to be grown on a substrate different than silicon for anything bigger than apple watch screen. Growing inorganic semiconductors on glass, organics, non-crystalline materials is what is hard