DL11 M7800

[william degnan]

On Wed, Jun 15, 2016 at 3:16 PM, Noel Chiappa <jnc at mercury.lcs.mit.edu>
wrote:

>     > 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
>     > CONSOLE ROM.
>
> 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.
>


Thanks, done.  Just about to try it out.


>
>
>     > 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.
>
>
>
OK, this makes sense, thanks.



>     > 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.
>
>
Many would appreciate this I bet.



>     > 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,
> etc).
>
>
Maybe just keep it to the useful combos :-)

Thanks again.

b