4 ms·

Of course, so let’s assume a 5” display, and assume that the fill factor per LED color is 25% (rather than expected 33%) then 25% of the area of a 5” display wo

by xt00 9y ago

Of course, so let’s assume a 5” display, and assume that the fill factor per LED color is 25% (rather than expected 33%) then 25% of the area of a 5” display would be required to populate one color. So assuming a small dicing kerf let’s assume for a 12 inch wafer you get 8 displays from that for one color. Which feels pretty generous, normally that same vendor would have yielded millions of LEDs from that same wafer and they made say 10000 per wafer, then if you asked them to sell for far less than that, what is their incentive? If they did that they would essentially be crashing the entire LED market in one fell swoop and their own profits. So the die area argument makes sense from apples perspective but doesn’t from the entrenched LED makers perspective. So seems like apple essentially would need to own the actual LED fab as well.

derefr 9y ago
> If they did that they would essentially be crashing the entire LED market in one fell swoop Would they? A one-subpixel microLED isn't going to be able to be used as a regular LED. It's not large or bright enough.
- xt00 9y ago
  interesting point but for a given luminance in nits the area doesn’t matter since it is cd / m^2 so assuming that there is a “cheap variety” of LED wafer material that outputs say 2000 nits for a low current, then a small area — say 20x20um for the subpixel size would save some cost for sure. I would estimate about the area of the subpixel would be about 30 times smaller than most indicator type low cost LEDs — which are in the sub 1mm^2 area range. So we can find the 30X factor but still need around 300X cost reduction still to get a 10000X reduction in price. Ultimately the Display should be around $100 or less, so if you can buy an entire wafer for $800 or so then you have something, but as far as I know most 12 in wafers with zero processing done to them are more like 1k. Then put them thru a fab and add value and run the fab and the cost goes up a ton. And we need 3 different wafers.. so anyway, I’m worried cost is just going to really hurt this approach for a long time.
  - derefr 9y ago
    > the area doesn’t matter Producing the same luminance using a smaller LED is equivalent to running the same current through a thinner wire. At some point you're just going to be burning out the LEDs instead of powering them. Given that you respond to that problem by lowering the current to within the smaller LEDs' reduced tolerance, then you get something too dim to use as an indicator LED.

by xt00 9y ago

derefr 9y ago

> If they did that they would essentially be crashing the entire LED market in one fell swoop Would they? A one-subpixel microLED isn't going to be able to be used as a regular LED. It's not large or bright enough.

xt00 9y ago
interesting point but for a given luminance in nits the area doesn’t matter since it is cd / m^2 so assuming that there is a “cheap variety” of LED wafer material that outputs say 2000 nits for a low current, then a small area — say 20x20um for the subpixel size would save some cost for sure. I would estimate about the area of the subpixel would be about 30 times smaller than most indicator type low cost LEDs — which are in the sub 1mm^2 area range. So we can find the 30X factor but still need around 300X cost reduction still to get a 10000X reduction in price. Ultimately the Display should be around $100 or less, so if you can buy an entire wafer for $800 or so then you have something, but as far as I know most 12 in wafers with zero processing done to them are more like 1k. Then put them thru a fab and add value and run the fab and the cost goes up a ton. And we need 3 different wafers.. so anyway, I’m worried cost is just going to really hurt this approach for a long time.
- derefr 9y ago
  > the area doesn’t matter Producing the same luminance using a smaller LED is equivalent to running the same current through a thinner wire. At some point you're just going to be burning out the LEDs instead of powering them. Given that you respond to that problem by lowering the current to within the smaller LEDs' reduced tolerance, then you get something too dim to use as an indicator LED.