Alt-BEAM Archive

Message #07223



To: beam@sgiblab.sgi.com
From: JVernonM@aol.com
Date: Sun, 31 Oct 1999 16:43:44 EST
Subject: [alt-beam] Re: Craig's chloroplast SE


In a message dated 10/31/99 3:49:48 PM Eastern Standard Time, rigter@cafe.net
writes:

> Truth is the operating voltage is higher! Nothing else is different with
> respect
> to existing SEs!
>
Well, it doesn't dump all of it's load. It retains 5 volts that can be used
for another purpose. It also is much quicker and efficient and doesn't need
to fill the well all the way from the bottom. I would say it is very
different.

See ya,
Jim
http://www.geocities.com/SoHo/Exhibit/8281/beamart.html



7224 Sun, 31 Oct 1999 16:40:04 -0500 [alt-beam] Re: Elementary, My Dear Watson... "Richard Caudle"


content-transfer-encoding: quoted-printable

Here's the circuit description from the article.


"FLIP-FLOP CIRCUIT is built from two saturable amplifiers. In a saturable =
amplifier, as the input voltage increases or decreases, the output saturate=
s at maximum and minimum voltages. The output can be normal or inverted. =
A resistor connects the output of one amplifier to the input of the other; =
it's resistance determines the strength of the connection. The normal outp=
ut terminal can be used to make an excitatory connection. In the flip-flop=
the inverted output terminal is employed instead to make inhibitory connec=
tions. A capacitor and resistor connected in parallel at each input to sto=
re the charge flowing to the terminal and produce an input voltage and to a=
llow the discharge current to flow. If the minimum and maximum outputs are=
+1 and -1 and amplifier A is saturated at +1, B's input will be driven dow=
n and B's output will saaturate at -1. B's output will, in turn be inverte=
d and drive up the inout to A, thus keeping the output of A saturated at +1=
. The reverse situation, in which A is saturated at -1 and B at +1, is als=
o stable. The configuration of the amplifier voltages is represented as a =
point on a two dimensional plane. Each axis represents the output of one o=
f the amplifiers, from -1 to +1. The circuit will always move to one of th=
e two stable points near(+1,-1) and (-1,+1), no matter what the initial vol=
tages were. A third axis represents the value of the computational energy =
E for each voltage configuration. The two stable points appear as valleys =
in the E surface. The edges of the surface rise steeply, because it is imp=
ossible to exceed the minimum and maximum outputs. If an external current =
is given to one of the amplifiers, this will deepen the valley that corresp=
onds to that amplifier's being in the +1 state."

Does that help any? I guessing that the hookups are determined by whether =
you want the next one to be Excitatory or Inhibitory. Let me know if you n=
eed more info.

Richard


content-transfer-encoding: quoted-printable



>
Here's the circuit description f=
rom the =

article.
 

 

"
color=3D#800000>FLIP-FLOP CIRCUIT is built from two saturable =

amplifiers.  In a saturable amplifier, as the input voltage increases =
or =

decreases, the output saturates at maximum and minimum voltages.  The =

output can be normal or inverted.  A resistor connects the output of o=
ne =

amplifier to the input of the other; it's resistance determines the strengt=
h of =

the connection.  The normal output terminal can be used to make an =

excitatory connection.  In the flip-flop the inverted output terminal =
is =

employed instead to make inhibitory connections.  A capacitor and resi=
stor =

connected in parallel at each input to store the charge flowing to the term=
inal =

and produce an input voltage and to allow the discharge current to flow.&nb=
sp; =

If the minimum and maximum outputs are +1 and -1 and amplifier A =

is saturated at +1, B's input will be driven down and B's=
=

output will saaturate at -1.  B's output will, in turn b=
e =

inverted and drive up the inout to A, thus keeping the output of <=
EM>A =

saturated at +1.  The reverse situation, in which A is =

saturated at -1 and B at +1, is also stable.  The =

configuration of the amplifier voltages is represented as a point on a two =

dimensional plane.  Each axis represents the output of one of the =

amplifiers, from -1 to +1.  The circuit will always move to one of the=
two =

stable points near(+1,-1) and (-1,+1), no matter what the initial voltages =

were.  A third axis represents the value of the computational energy =

for each voltage configuration.  The two stable points=
=

appear as valleys in the E surface.  The edges of the surface=
rise =

steeply, because it is impossible to exceed the minimum and maximum =

outputs.  If an external current is given to one of the amplifiers, th=
is =

will deepen the valley that corresponds to that amplifier's being in the +1=
=

state.

size=3D2>"

 

Does that help any?  I gues=
sing that =

the hookups are determined by whether you want the next one to be Excitator=
y or =

Inhibitory.  Let me know if you need more info.

 

Richard






Attachment: sciamflipflop.jpg

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