DL11 M7800

Noel Chiappa jnc at mercury.lcs.mit.edu
Wed Jun 15 14:16:49 CDT 2016

    > From: William Degnan

    >> For 777560/60 (standard for the console), you want A7/A3 and V4/V5
    >> 'in'.

    > I intend to use a serial terminal to access the console via M912

Got it; that would mean you're wanting the standard console.

    > I believe you're saying to connect A7/A3 and V4/V5

Right, insert jumpers A7 and A3, and also V5 and V4.

    > I still don't understand the pattern.

They specify the device address and vector in binary.

A7 is the 7th bit of the address, i.e. the 0200 bit. And since the DL11 is
'address jumper in for 0', that bit in the device's address is going to be
_0_ when the jumper is in. That would turn 777770 (remember, the device is a
block of 8 bytes, from xxxxx0 to xxxxx7, so you can't set the low 3 bits in
the base address, they must be 0) into 777570. Similarly, A3 is 010, and
turns 777770 to 777760. Put them together, you get 777560.

V5 = 040, V4 = 020, so they become a vector of 060.

    > What would A4, A5, A6 and V7, V6, V3 represent

A4 = 020, A5 = 040, A6 = 0100. V7 = 0200, V6 = 100, V3 = 010.

    > Is there a table with the jumpers and values somewhere?

No, but I'll whip one up and stick it on the Computer History wiki.

    > Specifically something that lists all jumper combos and their
    > corresponding addresses?

Well, _all_ the combinations would be 2^8 combinations (there are 8 address
jumpers), which is pretty sizeable, and I don't feel like listing them all,
but I can list a couple of the most common ones (e.g. console, second line,


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