<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>ChuckMcM</author><text>I found this to be a pretty confusing article. I get that they are analyzing the new node, which is great, but the editorializing seems a bit premature to me.&lt;p&gt;I also don&amp;#x27;t think the author understood the TMSC presentation. TMSC clearly said that is used a &amp;quot;model&amp;quot; of a typical SOC of 60% logic, 30% SRAM, and 10% analog. Then they said that for each category of thing, you could expect 1.8x, 1.35x, and 1.2x of shrink. If you do the math, that means an overall shrink for a &amp;#x27;typical&amp;#x27; SOC that conforms to their model would be 1.57x (approximately).&lt;p&gt;That Apple achieved 1.49x would suggest they got 95% of the process shrink effectiveness.&lt;p&gt;Then there is the cost per die and thus cost per transistor discussion. It is important to remember that this is likely the most expensive these wafers will ever be. The reasoning for that statement is that during a process node life-cycle the cost per wafer is set initially to capture &amp;quot;early movers&amp;quot; (who value density over cost). Much like any product where competition will emerge &amp;quot;later&amp;quot; there is a window early on to recapture value which can pay back your R&amp;amp;D and capital equipment investments. As a result the vendor sets the price as high as possible to make that repayment happen as quickly as possible. Once paid back, the price provides profit as long as it can be supported in the presence of other competitors (in this case, I would guess that role is played by Samsung). The GSA uses to publish price surveys of wafers on various nodes over time but they don&amp;#x27;t seem to do that any more. Anyway, so the cost per transistor will go down from this point but how much depends on how much margin is in the current wafer price.&lt;p&gt;So I agree that the cost per transistor is not going down as quickly as it has in the past, and its possible that this node may not get to be as low as the previous node. I&amp;#x27;m curious how it compares when you look at 7nm introduction price per transistor vs todays price per transistor. And if you get the same ramp with the 5nm node what that would mean.</text><parent_chain></parent_chain></comment><story><title>Apple’s A14 Packs 134M Transistors/mm²</title><url>https://semianalysis.com/apples-a14-packs-134-million-transistors-mm2-but-falls-far-short-of-tsmcs-density-claims/</url></story>
<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>topspin</author><text>I really appreciate this analysis and the straightforward top line number 134e6&amp;#x2F;mm^2.  The usual &amp;quot;node&amp;quot; figure is utterly meaningless; an electrical engineer couldn&amp;#x27;t care less about &amp;quot;feature&amp;quot; size (whatever that means.)  What is the count of discrete components in a given area?  There are 40 billion 5nm (the supposed &amp;quot;node&amp;quot; of these chips) squares in a millimeter of area.  That&amp;#x27;s &lt;i&gt;two orders&lt;/i&gt; of magnitude more dense than 134 million.&lt;p&gt;The meaningful achievement is how many discrete electrical components are composed into a given area.  Not some arbitrary dimension of some cherry picked subset of these components.</text><parent_chain></parent_chain></comment><story><title>Apple’s A14 Packs 134M Transistors/mm²</title><url>https://semianalysis.com/apples-a14-packs-134-million-transistors-mm2-but-falls-far-short-of-tsmcs-density-claims/</url></story>