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I took some pictures of a Microchip EPROM die from the late 80s / early 90s (I can't recall the exact part number). The wire bonding pads are surrounded by a comb-like structure. What is the purpose of this structure? eeprom die 1eeprom die 2

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  • \$\begingroup\$ It would be interesting to see if the output pins and address pins look similar. The 24-pin EPROM pinout is pretty standard- the 12 address pins are grouped (with Vcc and Vpp in there) and the 8 output pins are grouped (with Vss in there). \$\endgroup\$ Commented Nov 30, 2014 at 23:21

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They are probably large p-MOS and n-MOS transistors that are used for ESD protection on the bonding pads. Here is a reference that shows various bonding pad designs in detail (in general this information is not easy to come by- IC manufacturers seem to treat ESD protection as a kind of trade secret). Image taken from the above pdf:

enter image description here

I don't recall Microchip ever making memory EPROMs. Is this part of an EPROM microcontroller?

Edit: Just looking at a Microchip PIC16C57, which is probably from a similar era. There are similar patterns on either side of most of the pins (which are I/O) but on only one side of the input-only pins such as T0CKI, /MCLR/Vpp, OSC1. So the structures appear to be drivers on one side and ESD protection circuitry of whatever kind on the other side.

enter image description here

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  • \$\begingroup\$ Certainly the exact details of any specific IC fab process are considered trade secrets, at least until they become commonly known by multiple vendors. It's rare to even see die photos (top layer). \$\endgroup\$
    – MarkU
    Commented Nov 30, 2014 at 21:52
  • \$\begingroup\$ I left the chip at home so at the moment I can't look up the exact part number. I think it was similar to the 27HC64, which Microchip sold around 1990. I believe the chip I have has a few less pins though. \$\endgroup\$ Commented Nov 30, 2014 at 21:55
  • \$\begingroup\$ @ScottLawson Thanks! Obviously, from that datasheet, they did make EPROMs in that era. I was interested in confirming that it was a CMOS process, which it was. \$\endgroup\$ Commented Nov 30, 2014 at 22:02
  • \$\begingroup\$ It would be really interesting to see how you explain how these "MOSFET's" are made in ONLY Metal with no Silicon around. -1 for guessing. \$\endgroup\$ Commented Nov 30, 2014 at 23:49
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    \$\begingroup\$ @placeholder How do you know there is no silicon around- it is obviously on top of a silicon die so you're seeing something else. What would look different to you if those were transistors? \$\endgroup\$ Commented Nov 30, 2014 at 23:57
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At this writing, there are two "answers" that are total guesses - and wrong at that too.

These comb structures are as you might expect to see when you want to induce breakdown at a precise location and into controlled structures rather than else where in the chip. These are in the TOP metal layer, the combs are there to give lots of sharp edges to promote an excessively high ESD event to conduct at that location.

The diode and ESD clamping structures are by necessity in the Silicon.

These are very very far from being the transistor structures which are in the Si at least 3 - 7 metal layers down.

Look at lightning arrestors in the larger world. You will see these exact same things there.

Call it a belt and suspenders approach. Or rather a last chance, the actually ESD structures are rated for much lower voltage events.

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    \$\begingroup\$ -1 for failing to explain why the structures are connected to ground on one side and Vcc on the other. \$\endgroup\$
    – Dave Tweed
    Commented Dec 1, 2014 at 2:51
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    \$\begingroup\$ @DaveTweed guessing again I see. I didn't say they are connected to ground. There isn't enough information to warrant that assumption. What is that pin function? I don't know. \$\endgroup\$ Commented Dec 1, 2014 at 2:56
  • \$\begingroup\$ I think we should all cool down here \$\endgroup\$
    – clabacchio
    Commented Dec 1, 2014 at 22:00
  • \$\begingroup\$ From my POV, when I compare the answers I see one answer claiming that everyone else is wrong, and they are right, because of, well because they are, and another answer that at least tries to back this up with some external sources. \$\endgroup\$
    – PlasmaHH
    Commented Dec 1, 2014 at 22:11
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Those structures are the large transistors required to drive the pins that are used as outputs.

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    \$\begingroup\$ -1 for guessing \$\endgroup\$ Commented Nov 30, 2014 at 23:49

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