Chip with simple program for Toy

[Terry Pinnell]

"John" <John.Mason1212@hotmail.com> wrote:

Yes, youve got it. If input goes high during the time period then the output
should go high after the end of the time period. If input is low after the
time interval then output should also be low. After the time period if the
input goes high the output should immediatley go high.

The output should be the same as the input - with the exception that once
input goes low it remain low for 5 mins regardless of what the input does.
After the time period it follows the input again. i.e stays low untill input
goes high (then output goes high), or goes immedialy high is input is high.
When input falls low then time starts/cycle starts again.

The 5 mins isnt critical if it varys it doesnt matter as long as it is say
within 1min either way. (i.e 20% tollerance)

I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[mike]

Tim Wescott wrote:

Walter Harley wrote:

"John" <John.Mason1212@hotmail.com> wrote in message
news:VqRhc.1214$Aa7.1019@pathologist.blueyonder.net...

I have a output that provides a voltage of 5v when high/0v when low.

This output can go repeatedly high & low. I want to smooth the
output, so
when in goes low it cannot go high again for at least 5 mins.

The timer needs to start when the input goes low, and not start the
timer
again if the input goes high/low during the "off" time.

If the input is high but the output is low (off time), I want the output


to

go high again at the end of the time period.



I think what you're looking for is a "non-retriggerable monostable
multivibrator" (Google for that). Use the input signal as the input
to the
multivibrator (you might need to invert it first); then use the output of
the multivibrator to gate the input signal with an AND gate (again, you
might need to use an inverter).

In other words, the logic is that when the input signal goes low you
trigger
a 300-second pulse that can't be retriggered until it's done; whenever
that
pulse is high, you force the output to be low, and otherwise you let the
output be a copy of the input.

You should be able to do this with two chips: they make CMOS
non-retriggerable multivibrators, e.g. the 74HC221, and then you'd use a
quad NAND gate to get all the rest of your logic (tie the inputs of a
NAND
together to get an inverter; use an inverter on the output of a NAND
to get
an AND; and so forth). The only question is whether you can get a stable
timing interval as long as 300 seconds; if not, your other recourses are
going to be more complicated.



For a five minute time interval you really ought to be looking at
something with a counter. For something with a counter the smallest
thing by far that you'll find is a PIC or an AVR microcontroller.

If you just can't stand the thought of an 8-pin microcontroller then
consider a 555 for timing, a 4040, 4020, 4060 for a divider, a 74HC76
for tracking the state, and some glue logic. That's only four chips and
40 pins...

This is an excellent suggestion.
You can fit a PIC or AVR and a battery to run it in less space
and at less cost than the capacitor you're gonna need to make a
555 work at 5 minutes, if at all. A 555 definitely won't do
5 minutes with an electrolytic cap...at least not one you can afford.

If you want something really arcane, I once built a Ham Repeater
10-minute ID timer with an EPROM and a dual 555. Used the EPROM
as the counter and as the place to put the CW ID. Think of it
as a really big state machine with dual-mode clock.
Today, I'd use a PIC.
mike

--
Return address is VALID.
Bunch of stuff For Sale and Wanted at the link below.
Toshiba & Compaq LiIon Batteries, Test Equipment
Honda CB-125S $800 in PDX
Yaesu FTV901R Transverter, 30pS pulser
Tektronix Concept Books, spot welding head...
http://www.geocities.com/SiliconValley/Monitor/4710/

 

[John Fields]

On Thu, 22 Apr 2004 12:57:59 -0700, mike <spamme0@juno.com> wrote:

A 555 definitely won't do
5 minutes with an electrolytic cap...at least not one you can afford.

---
Sure it can!

Even a bipolar 555 with 250nA of threshold current will let you put a
1M timing resistor in there and, depending on the cap's leakage
current, about a 300ľF cap to get the 300 seconds, and 300ľF 6V
aluminum electrolytics are cheap.

A CMOS 555 (7555) with about 100pA of threshold current is 2500 times
better and will let you use a smaller capacitor. Again, depending on
the capacitor's own leakage current. Something reasonable might be 10
megohms and 30ľF which will get you a _cheap_ aluminum electrolytic
or, if you need lower leakage, a tantalum electrolytic.
---

--
John Fields

 

[mike]

John Fields wrote:

On Thu, 22 Apr 2004 12:57:59 -0700, mike <spamme0@juno.com> wrote:


A 555 definitely won't do
5 minutes with an electrolytic cap...at least not one you can afford.


---
Sure it can!

Even a bipolar 555 with 250nA of threshold current will let you put a
1M timing resistor in there and, depending on the cap's leakage
current, about a 300ľF cap to get the 300 seconds, and 300ľF 6V
aluminum electrolytics are cheap.

A CMOS 555 (7555) with about 100pA of threshold current is 2500 times
better and will let you use a smaller capacitor. Again, depending on
the capacitor's own leakage current. Something reasonable might be 10
megohms and 30ľF which will get you a _cheap_ aluminum electrolytic
or, if you need lower leakage, a tantalum electrolytic.
---

I'll admit I didn't search very hard, but just digging thru the
Panasonic caps in the digikey catalog, a typical electrolytic
leakage spec is i<.01CV or 3uA whichever is greater. at 20 degrees C.
300uF * 6V = 1800 *.01 = 18uA.
VCC can't be over 9V with a 6V cap. So, you've got 3uA of current to
supply the 555 input current plus 18uA of cap leakage.

So, tell me again how this is gonna work with a 1Meg resistor.

mike

--
Return address is VALID.
Bunch of stuff For Sale and Wanted at the link below.
Toshiba & Compaq LiIon Batteries, Test Equipment
Honda CB-125S $800 in PDX
Yaesu FTV901R Transverter, 30pS pulser
Tektronix Concept Books, spot welding head...
http://www.geocities.com/SiliconValley/Monitor/4710/

 

[Ben Bradley]

In sci.electronics.design, Tim Wescott <tim@wescottnospamdesign.com>
wrote:

For a five minute time interval you really ought to be looking at
something with a counter.

Agreed - no one has discussed this. An earlier post's link shows a
470uF cap and 1 meg resistor, which may be correct values for the
proper R-C time constant, but leakage current in the cap is greater
than the 1 meg can supply.

...

If you just can't stand the thought of an 8-pin microcontroller then
consider a 555 for timing, a 4040, 4020, 4060 for a divider, a 74HC76
for tracking the state, and some glue logic. That's only four chips and
40 pins...

Wasn't there a chip, maybe 4xxx CMOS, that's a 'combination
monostable,' an R-C oscillator, several-stage divider and glue logic,
meant especially for extra long cycle times like this?

 

[Tim Wescott]

mike wrote:

John Fields wrote:

On Thu, 22 Apr 2004 12:57:59 -0700, mike <spamme0@juno.com> wrote:


A 555 definitely won't do
5 minutes with an electrolytic cap...at least not one you can afford.



---
Sure it can!

Even a bipolar 555 with 250nA of threshold current will let you put a
1M timing resistor in there and, depending on the cap's leakage
current, about a 300ľF cap to get the 300 seconds, and 300ľF 6V
aluminum electrolytics are cheap.
A CMOS 555 (7555) with about 100pA of threshold current is 2500 times
better and will let you use a smaller capacitor. Again, depending on
the capacitor's own leakage current. Something reasonable might be 10
megohms and 30ľF which will get you a _cheap_ aluminum electrolytic
or, if you need lower leakage, a tantalum electrolytic.
---


I'll admit I didn't search very hard, but just digging thru the
Panasonic caps in the digikey catalog, a typical electrolytic
leakage spec is i<.01CV or 3uA whichever is greater. at 20 degrees C.
300uF * 6V = 1800 *.01 = 18uA.
VCC can't be over 9V with a 6V cap. So, you've got 3uA of current to
supply the 555 input current plus 18uA of cap leakage.

So, tell me again how this is gonna work with a 1Meg resistor.

mike

That's 1Meg resistor, over the long term, over temperature, etc.

I'm not saying that it can't be done -- just that you can probably do it
much easier and more precisely with a cheap uP than with an electrolytic.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[John Popelish]

John wrote:

I think you need both a timing function and a gate function (to disable
any inputs to the timer when it is timing).

Do you also want the output to remain low after the time interval, if the
input is low at that point? Should the output go high immediately, after
that if the input goes high?

Yes, youve got it. If input goes high during the time period then the output
should go high after the end of the time period. If input is low after the
time interval then output should also be low. After the time period if the
input goes high the output should immediatley go high.

The output should be the same as the input - with the exception that once
input goes low it remain low for 5 mins regardless of what the input does.
After the time period it follows the input again. i.e stays low untill input
goes high (then output goes high), or goes immedialy high is input is high.
When input falls low then time starts/cycle starts again.

The 5 mins isnt critical if it varys it doesnt matter as long as it is say
within 1min either way. (i.e 20% tollerance)

Go to alt.binaries,schematics.electronic to see a schematic that I
think will do what you want. I couldn't easily figure out a way to
eliminate the odd NAND gate, but it could be replaced with a diode
transistor combination to eliminate that pack.

--
John Popelish

 

[John Fields]

On Thu, 22 Apr 2004 14:21:06 -0700, mike <spamme0@juno.com> wrote:

John Fields wrote:
On Thu, 22 Apr 2004 12:57:59 -0700, mike <spamme0@juno.com> wrote:


A 555 definitely won't do
5 minutes with an electrolytic cap...at least not one you can afford.


---
Sure it can!

Even a bipolar 555 with 250nA of threshold current will let you put a
1M timing resistor in there and, depending on the cap's leakage
current, about a 300ľF cap to get the 300 seconds, and 300ľF 6V
aluminum electrolytics are cheap.

A CMOS 555 (7555) with about 100pA of threshold current is 2500 times
better and will let you use a smaller capacitor. Again, depending on
the capacitor's own leakage current. Something reasonable might be 10
megohms and 30ľF which will get you a _cheap_ aluminum electrolytic
or, if you need lower leakage, a tantalum electrolytic.
---


I'll admit I didn't search very hard, but just digging thru the
Panasonic caps in the digikey catalog, a typical electrolytic
leakage spec is i<.01CV or 3uA whichever is greater. at 20 degrees C.
300uF * 6V = 1800 *.01 = 18uA.
VCC can't be over 9V with a 6V cap. So, you've got 3uA of current to
supply the 555 input current plus 18uA of cap leakage.

So, tell me again how this is gonna work with a 1Meg resistor.

---

http://www.mouser.com/catalog/specsheets/llrl.pdf0

For Vcc = 9V, Vth = 6V,

So, if Il = 0.002 CV, Il = 6V * 330ľF * 0.002 = 3.96ľA.

Add to that 250nA for Ith, and the total will be 4.21ľA, so since I =
ER = 9V/1MR = 9ľA, it'll start off OK, but once the cap starts
charging up its leakage current will increase while the current the
resistor will be able to source will decrease because the voltage
across it will drop. With 6V across the cap there will only be 3V
across the R, so it will only be able to source 3ľA, while to charge
the cap to 6V and supply Ith will reguire 4.1ľA. So, worst case, it
won't charge up enough. Close though, just a little less leakage.
The search goes on...


--
John Fields

 

[Terry Pinnell]

Ben Bradley <ben_nospam_bradley@mindspring.example.com> wrote:

In sci.electronics.design, Tim Wescott <tim@wescottnospamdesign.com
wrote:

For a five minute time interval you really ought to be looking at
something with a counter.

Agreed - no one has discussed this. An earlier post's link shows a
470uF cap and 1 meg resistor, which may be correct values for the
proper R-C time constant, but leakage current in the cap is greater
than the 1 meg can supply.

The earlier post was mine. Whatever the theory, in practice I've built
maybe half a dozen circuits over many years with RC time constants
giving me 2 to 3 hour monostable periods. Needless to say, experiment
is needed to establish R values needed at higher end of scale. I
wouldn't for example have expected the 2 hour setting of my 12W rotary
on a kitchen timer to switch in a resistor precisely 8 times larger
than the 15 minute setting.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Wayne]

"John" <John.Mason1212@hotmail.com> wrote in message
news:yIShc.4$Rc.3@nurse.blueyonder.net...

I think you need both a timing function and a gate function (to disable
any inputs to the timer when it is timing).

Do you also want the output to remain low after the time interval, if
the
input is low at that point? Should the output go high immediately, after
that if the input goes high?

Yes, youve got it. If input goes high during the time period then the
output
should go high after the end of the time period. If input is low after the
time interval then output should also be low. After the time period if the
input goes high the output should immediatley go high.

The output should be the same as the input - with the exception that once
input goes low it remain low for 5 mins regardless of what the input does.
After the time period it follows the input again. i.e stays low untill
input
goes high (then output goes high), or goes immedialy high is input is
high.
When input falls low then time starts/cycle starts again.

The 5 mins isnt critical if it varys it doesnt matter as long as it is say
within 1min either way. (i.e 20% tollerance)

It would be good to know how short the input pulse may be. Are we talking
nanoseconds, millisecs or just secs.?

Wayne

 

[John]

It would be good to know how short the input pulse may be. Are we talking
nanoseconds, millisecs or just secs.?

Havent had chance to study the suggestions given so far yet, but to answer
your query the input would be be at least 20 seconds - not just a short
trigger.

 

[John Fields]

On Thu, 22 Apr 2004 19:27:44 -0500, John Fields
<jfields@austininstruments.com> wrote:

http://www.mouser.com/catalog/specsheets/llrl.pdf0

For Vcc = 9V, Vth = 6V,

So, if Il = 0.002 CV, Il = 6V * 330ľF * 0.002 = 3.96ľA.

Add to that 250nA for Ith, and the total will be 4.21ľA, so since I =
ER = 9V/1MR = 9ľA, it'll start off OK, but once the cap starts
charging up its leakage current will increase while the current the
resistor will be able to source will decrease because the voltage
across it will drop. With 6V across the cap there will only be 3V
across the R, so it will only be able to source 3ľA, while to charge
the cap to 6V and supply Ith will reguire 4.1ľA. So, worst case, it
won't charge up enough. Close though, just a little less leakage.
The search goes on...

---
The problem: Getting the cap to charge up to Vth in 5 minutes with a 1
megohm timing resistor and a 330ľF timing capacitor with a leakage
current of 3.96ľA.

The solution: Lower Vth by lowering the 555's Vcc, or increase the the
voltage into the timing network.

Using 9V for the charging voltage and 5V for Vcc will lower Vth to
3.33V, so that will leave 5.67V across the resistor when the cap
charges up to Vth, which will allow the R to supply 5.42ľA of current
into the cap if Ith is 250nA. With a leakage current of 0.002 * 330 *
5.42 = 2.64ľA, that leaves 3.58ľA of headroom, so there ya go!

Now, a 555 solution for the OP:

+-[78L05]-+----------------------------------------+
| | | |
9V>-+----|----|-----------------------------+ [R]
__ | | | |
IN>------|--- |----+------------------------|--[<CR]---+-->OUT
| | | | |
| | [0.1ľF] +-------+ [1M] |
| | | |_ | | C
| | +-----+-----O|T OUT|---|---[R]---B NPN
| | | |A |_ __| | E
| | [10K] [CR] +-O|R TH|O--+ |
| | | | | | _| | |
| +----+--+--+---+-O|V+ D|O--+ |
| | | V- | |+ |
| *[0.1ľF] +---+---+ [330ľF] |
| | (7)555 | | |
GND>-----+------------+-------------+-------+----------+-->GND

* Not needed for 7555.

For 555 place directly across V+ and V-

---
But... there's a little fix that needs to be made, and that's that
with a 9V charging voltage and a Vth of 3.3V, 1.1RC will no longer get
5 minutes, it'll get something less than that.

I just got some paying work in so I don't have time to do it now, but
if anybody needs to know how, post a request and I'll show you how to
do it later. Or, anybody else wanna do it?


--
John Fields

 

[Walter Harley]

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:215g80d278ggdgtto3t880lka58qliuagb@4ax.com...

I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

I think that with input high, the multivibrator part of that is astable
(oscillates), so the output will oscillate? I think this circuit doesn't do
what you want, or else I'm not analyzing it properly.

 

[Peter H Anderson]

John Popelish <jpopelish@rica.net> wrote in message news:<4087EB8A.AC17D56B@rica.net>...

John wrote:

I have a output that provides a voltage of 5v when high/0v when low.

This output can go repeatedly high & low. I want to smooth the output, so
when in goes low it cannot go high again for at least 5 mins.

The timer needs to start when the input goes low, and not start the timer
again if the input goes high/low during the "off" time.

If the input is high but the output is low (off time), I want the output to
go high again at the end of the time period.

I have tried to do this myself, I could not get a 555 timer to take the
input going low as a trigger as it held off the timer completely which the
input was high.

My second attempt used two AND gates and two capacitors and a transisitor,
which sort of works except that at the end of the time period if the input
is still high the output only goes up to 2.5v (see schmatic in abse)

Any idea how do do this easily?

I don't know who the original poster is.

Long time delays with a 555, even the CMOS version, are difficult.

The PICAXE -08 from http://www.picaxe.co/uk is a Basic Stamp like
device, but costs between $3.00 and $4.00 and can be programmed again
and again with a PC serial port and two resistors. The development
software is free from the PICAXE website.

A sample program;

TOP:
If Pin0 = 0 then Delay_15_Mins
' else
High 1 ' make outout a high
Goto TOP

Delay_15_Mins:
Low 1
For Mins = 1 to 15
Pause 60000 ' about one second
Next
GoTo TOP

Note that with this simple implementation, there is only the 8-pin
DIP. No timing caps and resistors, etc.

I have more complex examples using the PICAXE-18X at
http://www.phanderson.com/picaxe/picaxe.html

Peter H Anderson, http://www.phanderson.com

 

[Terry Pinnell]

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:215g80d278ggdgtto3t880lka58qliuagb@4ax.com...
I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

I think that with input high, the multivibrator part of that is astable
(oscillates), so the output will oscillate? I think this circuit doesn't do
what you want, or else I'm not analyzing it properly.

Thanks, that was carelessly drawn. I can see at leat two mistakes. For

a start, R1 should go to Vcc. And a 4001-based mono needs a +ve
trigger, so John's input should go direct to pin 6, not get inverted
first. I'll redraw when I get a minute - hopefully more carefully this
time!

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

Terry Pinnell <terrypin@dial.pipex.com> wrote:

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:215g80d278ggdgtto3t880lka58qliuagb@4ax.com...
I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

I think that with input high, the multivibrator part of that is astable
(oscillates), so the output will oscillate? I think this circuit doesn't do
what you want, or else I'm not analyzing it properly.

Thanks, that was carelessly drawn. I can see at leat two mistakes. For
a start, R1 should go to Vcc. And a 4001-based mono needs a +ve
trigger, so John's input should go direct to pin 6, not get inverted
first. I'll redraw when I get a minute - hopefully more carefully this
time!

Oops. Maybe I'll get this right sometime! I see John specified a -ve
going trigger, so I was right first time on that point.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:215g80d278ggdgtto3t880lka58qliuagb@4ax.com...
I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

I think that with input high, the multivibrator part of that is astable
(oscillates), so the output will oscillate? I think this circuit doesn't do
what you want, or else I'm not analyzing it properly.

Here's the actual result on breadboard (with the correction to R1 as

per my subsequent post). I've naturally reduced timescale for testing.
Seems to meet stated requirement.

http://www.terrypin.dial.pipex.com/Images/John.gif

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Robert C Monsen]

"Mr Anon" <anon@anon.co.uk> wrote in message
newsdDhc.35556$Y%6.4635002@wards.force9.net...

If anyone has any ideas on how to make this work, or an alternative
circuit,
I'd really appreciate your help.

If I start the circuit [see attached image] with the switch open, the
components blow because of the current [about 4A at the fuse]. And if I
start the circuit with the switch closed, the relay closes, but does not
respond to the switch opening or closing.

As a reminder, what I'm trying to do is drive a relay to open or close
according to the presence or absence of a graphite pencil line drawn on
paper. Drawing a line would drive the relay to close, erasing the line
would
break the circuit and open the relay [which switches a lamp on or off]. In
the circuit diagram, VR1 represents the resistance of the pencil line
which
can be anything from 100k that you measured, up to 18M that I measured.
SW1
simulates a joined or a broken pencil line. Is it possible to make a
driver
that can work with such a range of resistance, without blowing when the
resistance is low [for example, if the contacts accidentally touch].

If the resistance through the switch is zero, you have a diode from 12V to
ground
through the NPN transistor. The transistor will just fry if you do that.

You could add in another 100k resistor in series with the line. Then, the
total
resistance at the base of the NPN transistor is 100k + Rl, where Rl is the
resistance
of the graphite line.

If your transistor has a beta of 100, and your Rl is 18Meg, you will have a
base
current of

11.6/18e6 = 0.65uA.

That will allow about 65uA, which will cause the gate voltage at the mosfet
to go to 0.2V, allowing to turn on. Now, make sure your relay is a 12V
relay,
otherwise you'll need a resistor to limit current through it.

If Rl is more like 100k, then the base current is

11.6/200e3 = 58uA

This will again allow the transistor to saturate, allowing the mosfet to
turn on.

So, just put a 100k resistor in series with the connection from 12V to the
transistor base.

I could just use a normal switch to control the lamps, but hey, I'm just
being picky!

"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:jiB_b.377519$I06.4134130@attbi_s01...

"Mr Anon" <anon@anon.co.uk> wrote in message
news:IXv_b.9718$h44.1089599@stones.force9.net...
Dimitrij,

Thanks so much for your very kind help - the lesson I've learnt is
that
electronics from a text book is far too ideal! I've revised the
circuit
with
your suggestions (see the attachment).

THe circuit appears to work well at VR1 of up to 12M, but my
multimeter
shows that the resistance of a thick line of graphite from a pencil
drawn
on
paper is actually up to 15-18M. Am I asking too much to have a
simple
circuit that can cope with such a variance in resistance?
Are the values for C1 and C2 enough to attenuate any AC?

I just measured a thick 2cm graphite line to be about 100k ohms. Maybe
it
depends on how thick the line is?

With the circuit you have (the original one, not the 'improved' one)
the
current through the relay is max about 10,000 times the current into
the
base. If you have 18M across 12-1.4V=10.6V, then the current will be
just
shy of a microamp. Thus, the current through the relay will be 10,000
times
that, or 10mA, which might not be enough to close the relay.

Instead of using a darlington connected pair, you could use the
following
circuit:

(view with courier font)

12V
-+-------------------+
| | |
| .-. |
| | | |
| | |1MEG |
| '-' |
o | |
+-------G||-+S
o | ||-> P-MOSFET
| | ||-+D
| |c |
'----b| NPN |
|e Relay Coil Here
| |
| |
GND------+-----------+

When there is no connection between the 'o's, the transistor is off,
and
so
the P-MOSFET base is at 12V, turning it off.

When there is a connection through a resistance of 100MEG or less, the
NPN
will turn on fully, causing the gate of the P-MOSFET to go near
ground.
This
turns it on, thus closing the relay.

Any P-MOSFET where Vgs(th) is < 8V should work. You don't need to
parallel
a
diode, because the mosfet has a substrate diode which should prevent
any
damage when the coil is turned off. If the relay isn't a 12V relay
(say,
its
a 5V relay), you should add in a resistor between the P-MOSFET and the
coil.

Oh, and you may need a 0.2uF cap _before_ your 7812, between in and
gnd.

Regards,
Bob Monsen

 

[Terry Pinnell]

Terry Pinnell <terrypin@dial.pipex.com> wrote:

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypin@dial.pipex.com> wrote in message
news:215g80d278ggdgtto3t880lka58qliuagb@4ax.com...
I think this might do it, using just one cheap IC.
http://www.terrypin.dial.pipex.com/Images/JohnProblem.gif

I think that with input high, the multivibrator part of that is astable
(oscillates), so the output will oscillate? I think this circuit doesn't do
what you want, or else I'm not analyzing it properly.

Here's the actual result on breadboard (with the correction to R1 as
per my subsequent post). I've naturally reduced timescale for testing.
Seems to meet stated requirement.

http://www.terrypin.dial.pipex.com/Images/John.gif

Ignore the above, as it's the result of my misunderstanding John's
spec.

For a version that I believe actually *does* meet the spec, see my
latest in alt.binaries.schematics.electronic. This uses a 555 and a
couple of the 4 gates from a quad 4001 NOR. Breadboarded today. Actual
input and output waveforms now (finally!) appear to be correct...

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Fred Bloggs]

Terry Pinnell wrote:

For a version that I believe actually *does* meet the spec, see my
latest in alt.binaries.schematics.electronic. This uses a 555 and a
couple of the 4 gates from a quad 4001 NOR. Breadboarded today. Actual
input and output waveforms now (finally!) appear to be correct...

Actually there is a way to do this, and for much longer timing than 5
minutes, with the LMC555 and a more stable capacitor technology that is
intermediate to your bulk timer circuit and a crystal timebase counter.
Taking an off-the-shelf Radio Shack 50V 0.22uF "polyester", 5% carbon
film resistors, and some other readily available parts, I was able to
construct a delay that ran 5:10 at 25oC, 5:40 at 70oC, and now looking
at 6:05 at -20oC -having left it in the freezer all night. I am
impressed with this unadjusted performance with no special provisions
for temperature compensation. Try running an actual circuit through that
one.