Please identify this circuit board

[Christian Corti]

[Exceptionally top-posting...]

There are two modules each containing eight pulse transformers. That's it. 
The left module has the cover removed.
No memory, no register. This board is from 1971, so it would contain some 
TTL FFs or RAM (e.g. 7489) for registers.

Christian


On Wed, 1 Feb 2017, dwight wrote:
> I think it is a register board. Something
>
> to hold data through a power down.
>
> Dwight
>
>> On Feb 1, 2017, at 1:02 PM, Paul Koning <paulkoning at comcast.net> wrote:
>>
>>
>>> On Feb 1, 2017, at 11:57 AM, Rick Bensene <rickb at bensene.com> wrote:
>>>
>>> Mr. Havermout wrote:
>>>
>>>> Can someone identify this circuit board? It's some sort of magnetic core
>>>> memory. I've had this for ages and I've always wondered what it is and
>>>> where it comes from.
>>>>
>>>> http://lookpic.com/O/i2/366/iAFq4mLF.jpeg
>>>>
>>>
>>> Whatever this was, it appears that it has been scavenged for parts over time.  Many components appear to be missing.
>>>
>>> It seems  to me from looking at the board carefully  that it could be a small wire-rope ROM, or, perhaps the cores (one module has its cover removed exposing the large ferrite rings) serve as pulse transformers for magnetic core that resides on another circuit board.
>>>
>>> If it is a small wire-rope read-only memory, it appears that it could be a 16x16 ROM, perhaps hard wired for some kind of small bootstrap loader or the like.
>>> It looks like there are 16 wire drivers (two of the driver transistors are missing), and 16 sense amplifiers.
>>
>> If it's a core-rope ROM, which seems plausible, then it would be 256 x 16.  Core rope uses 2 select ("inhibit") lines/drivers per address bit.  So 16 drivers means 8 address bits, i.e., 2^8 words.
>
> Never mind, that was all wrong.  It takes only 2*n wires to select one of 2^n cores in a core-rope ROM, so far that was correct.  But the number of ROM words equals the number of cores (obviously).  So if there are 16 cores that would mean 16 words, but it also means that only 8 drivers are needed (4 address bits).  So there's a puzzle.
>
>        paul