11 Apr
2000
11 Apr
'00
11:50 p.m.
For the heck of it, I've tried to design an electro-mechanical binary
adder that can automatically do the carry of the one that arises in 1 +
1 = 10. Of course, you reply, "George Stibitz's already did that in
1937 as you see here..."
http://www.toronto-montessori.on.ca/bsutherland/electricity/stibitz.html
But hey: this is retrocomputing, after all, and I'm not trying to
duplicate Stibitz--my aim is to come up with an electro-mechanical
binary adder architecture that's even simpler than his 2 switches, 2
bulbs, 2 batteries, and 2 relays version.
I think I've done it using 3 switches, 4 bulbs, and 1 battery. Yes,
still eight components but because it lacks the relays, I think it's
definitely simpler and therefore a kind of retro-breakthrough. And yes,
mine does carry the one for 1 + 1 = 10. Feel free to try building this
(or better it if you can with even fewer components).
But I do have a stupid question for everyone that exposes the holes in
my education: if my adder can successfully do all of the below
calculations but no more
0 + 0 = 0
0 + 1 = 1
0 + 10 = 10
0 + 11 = 11
1 + 0 = 1
1 + 1 = 10
1 + 10 = 11
10 + 0 = 10
10 + 1 = 11
11 + 0 = 11
have I constructed a 1-bit binary adder or a 2-bit binary adder?
Thanks,
S.F. Hall
11 Apr
11 Apr
11:25 p.m.
have I constructed a 1-bit binary adder or a 2-bit
binary adder?
1-bit. A 2-bit adder has three bits of output and handles six more cases:
01 + 11 = 100
10 + 10 = 100
10 + 11 = 101
11 + 01 = 100
11 + 10 = 101
11 + 11 = 110
However, it still sounds like fun!
Logically a 2-bit adder can be decomposed into two one-bit full adders,
each with the following truth table:
A B Cin Cout Q
-------- ------
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
The least significant stage of the mutli-bit adder can be simpler since
there is (usually) no carry-in:
A B Cout Q
---- ------
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0
This is exactly what George Stibitz's Model K does.
Note that this can be implemented with no relays if
the inputs are only from switches (not the outputs of
other circuits), by using three-pole double-throw switches.
(I'm not going to try to draw ascii art unless someone
requests it.)
So an interesting challenge would be to minimize the
circuit for the full adder, given that the A and B
inputs are from switches, but the C input is an
electrical input from the lower-order stage.
I think at least one relay is needed, to act as an
inverter for the carry-in stage.
Eric
12 Apr
12 Apr
1:49 p.m.
New subject: More disk problems
Hi
I was going to trace the write gate back from the
drive I have in my NIC-80 system but came across
a problem. The drive is a Shuggart 900-1. It
has a 44 pin connector instead of the more standard
50 pin.
Does anyone have the pinout for this odd one
around some place?
Thanks
Dwight
7:26 a.m.
What you have done is a 1bit full adder with carry in. It's not quite two
bits and not limited to one bit pair.
Allison
On Tue, 11 Apr 2000, Scott F. Hall wrote:
For the heck of it, I've tried to design an
electro-mechanical binary
adder that can automatically do the carry of the one that arises in 1 +
1 = 10. Of course, you reply, "George Stibitz's already did that in
1937 as you see here..."
http://www.toronto-montessori.on.ca/bsutherland/electricity/stibitz.html
But hey: this is retrocomputing, after all, and I'm not trying to
duplicate Stibitz--my aim is to come up with an electro-mechanical
binary adder architecture that's even simpler than his 2 switches, 2
bulbs, 2 batteries, and 2 relays version.
I think I've done it using 3 switches, 4 bulbs, and 1 battery. Yes,
still eight components but because it lacks the relays, I think it's
definitely simpler and therefore a kind of retro-breakthrough. And yes,
mine does carry the one for 1 + 1 = 10. Feel free to try building this
(or better it if you can with even fewer components).
But I do have a stupid question for everyone that exposes the holes in
my education: if my adder can successfully do all of the below
calculations but no more
0 + 0 = 0
0 + 1 = 1
0 + 10 = 10
0 + 11 = 11
1 + 0 = 1
1 + 1 = 10
1 + 10 = 11
10 + 0 = 10
10 + 1 = 11
11 + 0 = 11
have I constructed a 1-bit binary adder or a 2-bit binary adder?
Thanks,
S.F. Hall
9:14 a.m.
Doesn't a full-adder have 3 inputs and two outputs any more? After all, the
three bits can't generate a sum of more than two bits.
Dick
----- Original Message -----
From: <allisonp(a)world.std.com>
To: <classiccmp(a)classiccmp.org>
Cc: <bluebike(a)linkny.com>om>; <cruiksdb(a)alfredtech.edu>
Sent: Wednesday, April 12, 2000 6:26 AM
Subject: Re: stibitz.adder
What you have done is a 1bit full adder with carry in.
It's not quite two
bits and not limited to one bit pair.
Allison
On Tue, 11 Apr 2000, Scott F. Hall wrote:
For the heck of it, I've tried to design an
electro-mechanical binary
adder that can automatically do the carry of the one that arises in 1 +
1 = 10. Of course, you reply, "George Stibitz's already did that in
1937 as you see here..."
http://www.toronto-montessori.on.ca/bsutherland/electricity/stibitz.html
But hey: this is retrocomputing, after all, and I'm not trying to
duplicate Stibitz--my aim is to come up with an electro-mechanical
binary adder architecture that's even simpler than his 2 switches, 2
bulbs, 2 batteries, and 2 relays version.
I think I've done it using 3 switches, 4 bulbs, and 1 battery. Yes,
still eight components but because it lacks the relays, I think it's
definitely simpler and therefore a kind of retro-breakthrough. And yes,
mine does carry the one for 1 + 1 = 10. Feel free to try building this
(or better it if you can with even fewer components).
But I do have a stupid question for everyone that exposes the holes in
my education: if my adder can successfully do all of the below
calculations but no more
0 + 0 = 0
0 + 1 = 1
0 + 10 = 10
0 + 11 = 11
1 + 0 = 1
1 + 1 = 10
1 + 10 = 11
10 + 0 = 10
10 + 1 = 11
11 + 0 = 11
have I constructed a 1-bit binary adder or a 2-bit binary adder?
Thanks,
S.F. Hall
12:55 p.m.
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2:08 p.m.
Hi
Here is a full adder that I did in relays a while
back. All of the relays are double throw and relay
Y has three sets of contacts. This is a DC relay
design and uses two diodes, D1 and D2, to form a
blocking OR function. This could have been done
with more contacts. Parts of the same toggle
in a double throw pair are indicated by the same
number, such as Ci1 being in two locations.
One is normally close and the other normally open.
The common wire is the common in the relay contact.
I hope you like the ascii art because it took me some time to do.
X1 Y1
R ---|/|----| |-----+
| Ci1
+-----+---|/|----+
X1 Y1 | | |
R ---| |----|/|-----+ | |
| +----- Z ( sum )
X2 Y2 | |
R ---|/|----|/|--------+ | |
| | Ci1 |
+--)---| |----+
X2 Y3 D1 | |
R ---| |----| |--+-|>|-+ | Ci2
| +---| |----+----- Co ( Carry out )
| D2 |
+-|>|---------------+
R = rail voltage
--| |-- = normally open contacts
--|/|-- = normally open contacts
|
-)- = no connection
|
|
-+- = connection
|
Coils:
___
X --|___|-- GND
___
Y --|___|-- GND
___
Ci--|___|-- GND
I would like to see anyone suggested changes? Remember,
switches are not valid unless you also show the robot
that controls it. Four pole/double throw relays can be used
but they are harder to find. I'm always looking for
ways to use more diodes and less relays.
Dwight
9027
days inactive
9028
days old
6 comments
6 participants
participants (6)
-
allisonp＠world.std.com -
ard＠p850ug1.demon.co.uk -
edick＠idcomm.com -
elvey＠hal.com -
eric＠brouhaha.com -
hallsf＠alfredtech.edu