7 Dec
2007
7 Dec
'07
4:29 p.m.
-----------Original Messages:
Date: Thu, 6 Dec 2007 22:06:49 -0800
From: dwight elvey <dkelvey at hotmail.com>
Subject: RE: Almost OT: Pushbutton switch latching
From: ian_primus at yahoo.com
The main thing I'm trying to achieve is
simplicity.
Nothing fancy. No PIC's, no microcontrollers, etc. I
initially thought of using flip flops and a bunch of
inverters to reset all the non-selected flip flop, but
couldn't work out in my head a good way to do it... I
hadn't taken switch debouncing into the equation - but
aren't flip flops commonly used to debounce switches
anyway?
Hi
The simplest method would be to use the cross coupled
nands as another described. For 8 inputs, you'll need
4 ea 7400's to make the nand latches. You'l need 8 ea 8 input
nands ( forget the 74 number ) and 3 of the hex inverters,
7404s.
For each switch wire one lead to ground. Wire one lead
to a 4.7K pullup resistor and one of the free inputs to the
cross coupled nand pair. Also wire that switch lead to
7 of the 8 input nands ( not the one on this switch circuit. ).
On the remaining free inputs to the cross coupled nands
pair, wire an inverters output. The input of the inverter
goes to the output of the 8 input nand that wasn't one of the
7 connected to this switch.
Repeat this for each of the 8 switches.
If the active level is to be 0 when selected, you can save
one inverter IC. If it needs to be active high, you need the
inverter. in either case, you need to connect to the nand
that has the inverter to the 8 input nand. This will ensure
that multiple switches won't select anything until only
one is selected.
On the 8 input nand, you'll notice that one input isn't
connected. This can be used to reset. Tie all together
with a 1K resistor to +5. Place a diode across this such
that the band is on the +5 ended.
Add a 10uf tantalum to this net with th plus on the
same net and the negative to ground.
This will reset all on powering up. If you want to
one of the outputs to come on when powering up,
you can add a similar capacitor, resistor and diode
circuit to the desired input switch lead. You'd need to
remove the common input to the 8 input nand and
tie it to +5 as well.
I like this better than using some other clocked method
because if multiple switches are selected at the same time,
non will select but when the last switch is still selected,
that output will go active.
There are no ambiguous states and no oscillations. The switches
are debounced as well.
Dwight
------------Reply:
15 ICs and a dozen discretes is "the simplest?"
I think some simpler circuits have been suggested...;-)
As a matter of fact, if you add a diode & cap per switch you could do
the job with 2 IC's for each 8 PBs (7430 & '573 (or 2x'75 if you need
complementary outputs)).
mike
8 Dec
8 Dec
10:06 a.m.
How about this?
8 74xx30's (8 input NAND) with each output feeding one of the inputs
of the other stages. The remaining input of each gate is connected
to a pushbutton whose other side is grounded and the gate side pulled
up to Vcc through an appropriate resistor (say, 2.2K). The circuit
is self-debouncing and the last button released governs the output.
Basically an 8-stable flip-flap-floop...-flop. Initial state is
theoretically random, but probably could be forced with a small
capacitor across one of the buttons.
If you expand much beyond this, note that fanout issues may rear
their ugly heads.
Cheers,
Chuck
1:37 p.m.
From: cclist at sydex.com
How about this?
8 74xx30's (8 input NAND) with each output feeding one of the inputs
of the other stages. The remaining input of each gate is connected
to a pushbutton whose other side is grounded and the gate side pulled
up to Vcc through an appropriate resistor (say, 2.2K). The circuit
is self-debouncing and the last button released governs the output.
Basically an 8-stable flip-flap-floop...-flop. Initial state is
theoretically random, but probably could be forced with a small
capacitor across one of the buttons.
If you expand much beyond this, note that fanout issues may rear
their ugly heads.
Cheers,
Chuck
Hi Chuck
Essencially the same as I described. I think is best meets the
general design criteria. I just used a single throw switch as
the circuit would be more than fast enough to ignore an bounch issues.
If there was a fan out problem, one could do a ripple ring for clearing
but that might require some timed circuits to keep from racing.
Dwight
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1:57 p.m.
On 8 Dec 2007 at 13:37, dwight elvey wrote:
Essencially the same as I described. I think is best
meets the
general design criteria. I just used a single throw switch as
the circuit would be more than fast enough to ignore an bounch issues.
If there was a fan out problem, one could do a ripple ring for clearing
but that might require some timed circuits to keep from racing.
If you wanted to go really retro, implement the NANDS and diode-input
transistor sets. Assuming that the transistors have enough gain and
are beefy enough, you could extend it pretty far.
Or use diode array ICs and an octal buffer.
Anyone ever fool with ternary logic, building flip-flap-flops out of
3-input gates?
Cheers,
Chuck
9:15 p.m.
8 74xx30's (8 input NAND) with each output feeding
one of the inputs
of the other stages. The remaining input of each gate is connected
to a pushbutton [to ground]. Basically an 8-stable
flip-flap-floop...-flop.
Nice idea, but it doesn't work. Draw out three NANDs cross-coupled as
you described. Mark the logic states for a stable state with one
output low. Now consider what happens when each of the inputs goes
low. The only one that does anything is the one feeding the gate whose
output is low.
To put it another way, an input going low doesn't pull the
distinguished state to that gate; it pushes it off that gate. With two
two-input NANDs, there's only one place for it to go, so it's
well-defined; with more, it could stabilize anywhere (or,
theoretically, oscillate forever if the gates have identical
propagation delays).
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11:50 p.m.
On 9 Dec 2007 at 0:15, der Mouse wrote:
Nice idea, but it doesn't work. Draw out three
NANDs cross-coupled as
you described. Mark the logic states for a stable state with one
output low. Now consider what happens when each of the inputs goes
low. The only one that does anything is the one feeding the gate whose
output is low.
Yeah -- I see that. Release a button and the circuit goes into an
unstable state. I wonder what the smallest number of active devices
is for an n-stable circult with n greater than 2.
8 D flip=flops with each switch connected to a PRESET input and a
NAND (or NOR) to edge-clock in a reset state looks to be far from the
minimum. How about some capacitively coupled logic with a few
steering diodes?
Cheers,
Chuck
9 Dec
9 Dec
11:45 a.m.
The only one
that does anything is the one feeding the gate whose
output is low.
Yeah -- I see that. Release a button and the circuit goes into an
unstable state. I wonder what the smallest number of active devices is
for an
n-stable circult with n greater than 2.
Well, you can do a tristable circuit with three _four_-input NANDs.
Cross-couple them, each output driving an input on each other gate;
then, each button drives an input on all but one of the gates.
For example, let the inputs be inA, inB, inC, and the outputs outA,
outB, outC. Then set up
outA = NAND(outB,outC,inB,inC)
outB = NAND(outC,outA,inC,inA)
outC = NAND(outA,outB,inA,inB)
For an N-stable circuit, this technique requires 2*(N-1) inputs on each
gate. For a bistable, or even a tristable, this is tolerable; for an
8-stable, it requires eight 14-input NANDs. You could do it with
8*14=112 diodes (hmm, might need another 8) and a relatively small
handful of resistors and transistors. Doing it with just TTL would
be..somewhat inconvenient.
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1:32 p.m.
On 9 Dec 2007 at 14:45, der Mouse wrote:
For an N-stable circuit, this technique requires
2*(N-1) inputs on each
gate. For a bistable, or even a tristable, this is tolerable; for an
8-stable, it requires eight 14-input NANDs. You could do it with
8*14=112 diodes (hmm, might need another 8) and a relatively small
handful of resistors and transistors. Doing it with just TTL would
be..somewhat inconvenient.
Perhaps something with UJTs and diodes--a UJT would give you a stable
state for each button and some steering logic would get you there.
But UJTs seem to be pretty scarce nowadays.
Better to use a single RS FF per button, with reset being driven by
the OR of all of the other buttons. You could certainly do the OR
with diodes; for an 8-button setup that would be 8 RS flip-flops and
56 diodes (fewer if you want to make a tree of them). That'd be 2
74279s or 4043s with a pile of diodes.
Cheers,
Chuck
3:05 p.m.
On Sunday 09 December 2007 16:32, Chuck Guzis wrote:
On 9 Dec 2007 at 14:45, der Mouse wrote:
You can make your own with a pair of cross-connected transistors.
For an N-stable circuit, this technique requires
2*(N-1) inputs on each
gate. For a bistable, or even a tristable, this is tolerable; for an
8-stable, it requires eight 14-input NANDs. You could do it with
8*14=112 diodes (hmm, might need another 8) and a relatively small
handful of resistors and transistors. Doing it with just TTL would
be..somewhat inconvenient.
Perhaps something with UJTs and diodes--a UJT would give you a stable
state for each button and some steering logic would get you there.
But UJTs seem to be pretty scarce nowadays. Better to use a single RS FF per button, with reset
being driven by
the OR of all of the other buttons. You could certainly do the OR
with diodes; for an 8-button setup that would be 8 RS flip-flops and
56 diodes (fewer if you want to make a tree of them). That'd be 2
74279s or 4043s with a pile of diodes.
A similar application that comes to mind is the "game controller" where each
contestant has a push button and the first one in gets lit and locks the
other ones out. There are a bunch of circuits like that around the web. See
the "4QD" site (in the uk?) for an example of both this and the
how-to-make-your-own UJT particulars.
My initial reply on this went offlist, and I still think it's the simplest.
You take a '374 (well, I have a pile of 'em here... :-) and connect a push
button with a pullup resistor to each input. Then connect a diode to those
points too, with all of the diodes tied together. A single transistor fed
by those diodes will generate the clock...
My application for that was a multiple-battery charging setup. I still have
to finish up the design in figuring out how each select output will choose a
different circuit (different sized batteries) and build the darn thing,
saving me the effort of having to manually hook up all sorts of stuff and
deal with it.
--
Member of the toughest, meanest, deadliest, most unrelenting -- and
ablest -- form of life in this section of space, ?a critter that can
be killed but can't be tamed. ?--Robert A. Heinlein, "The Puppet Masters"
-
Information is more dangerous than cannon to a society ruled by lies. --James
M Dakin
3:35 p.m.
On 9 Dec 2007 at 18:05, Roy J. Tellason wrote:
My initial reply on this went offlist, and I still
think it's the simplest.
You take a '374 (well, I have a pile of 'em here... :-) and connect a push
button with a pullup resistor to each input. Then connect a diode to those
points too, with all of the diodes tied together. A single transistor fed
by those diodes will generate the clock...
You can probably leave out the transistor if you don't need to invert
the diode "OR". i.e. if you need to clock on the falling edge. You
could also fool with pulldowns, N.C. pushbottons, etc.
I was looking for a "valid state guaranteed" circuit where the output
terms would feed back into the input to guarantee that only one
output would ever be valid--even on powerup. You can do it, but the
gate count starts to climb rapidly as the number of states increases,
as der Mouse has illustrated.
There are diode arrays in SIPs and DIPs if you don't like the look of
discretes.
Cheers,
Chuck
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cclist＠sydex.com -
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dm561＠torfree.net -
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