2 pulses from 1

Terry Pinnell · Feb 28, 2005

I'm sure there must be a smarter way to do this than my first design.
I want to produce separate +ve going 12V pulses at the leading and
trailing edges of a low impedance 12V input.

My first draft looks too complex:

- Use R/C/diode edge detector to produce OUT 1.
- Also take input to half a 4013 to produce FF
- Logic NOR the input with the inverted FF
- Use R/C/diode edge detector to produce OUT 2

Here's the timing chart:
http://www.terrypin.dial.pipex.com/Images/2From1.gif

I think those crude output pulses should be adequate without further
cleaning up, as they go to the respective Set and Reset inputs of a
simple 4001 bistable. Of course, squared up outputs would be a bonus;
I might do that anyway if I have any spare 4001 gates.

For the curious, it's an add-on for my curtain controller, to open and
close them either at dawn/dusk or at specific times of day.

Any simpler way to achieve this please?

--
Terry Pinnell
Hobbyist, West Sussex, UK

Ken Moffett · Feb 28, 2005

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote in
news:f71621tpfvur92en3q138qdlkjk12ioca3@4ax.com:

http://www.terrypin.dial.pipex.com/Images/2From1.gif

Terry,
Someplace in the past I remember using an edge detecter circuit for the
count output of a rotory encoder. I think it was from one of the "Design
Ideas" in Electronic Design(?). It used an XOR gate with an RC delay on
one of the two inputs. The other input was direct. The output pulse
width was a function of the RC time constant.

-- --- - -
| | | | || || || ||
--- --- XOR -- --- --- -- ---
__
---+----------------| |
| ___ | |-------
+---|___|---+----|__|
|
---
---
|
---------------+---------------

Edge Detector
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Fred Bloggs · Feb 28, 2005

Terry Pinnell wrote:

I'm sure there must be a smarter way to do this than my first design.
I want to produce separate +ve going 12V pulses at the leading and
trailing edges of a low impedance 12V input.

My first draft looks too complex:

- Use R/C/diode edge detector to produce OUT 1.
- Also take input to half a 4013 to produce FF
- Logic NOR the input with the inverted FF
- Use R/C/diode edge detector to produce OUT 2

Here's the timing chart:
http://www.terrypin.dial.pipex.com/Images/2From1.gif

I think those crude output pulses should be adequate without further
cleaning up, as they go to the respective Set and Reset inputs of a
simple 4001 bistable. Of course, squared up outputs would be a bonus;
I might do that anyway if I have any spare 4001 gates.

For the curious, it's an add-on for my curtain controller, to open and
close them either at dawn/dusk or at specific times of day.

Any simpler way to achieve this please?

View in a fixed-width font such as Courier.

4070_
+----------------\ \ \
| | | >--+------>OUTN
| |\ +-/ /__/ | |\
IN>---+-[R]-+--| o --+ +-| o-->OUTP
| |/ |/
[C] 40106 40106
|
|
---

------------
| |
| |
IN ------- -----------

------- --------- -------
| | | |
OUTN | | | |
-- --

~0.7RC
-- --
->| |<- | |
OUTP | | | |
------- --------- ------

Fred Bartoli · Feb 28, 2005

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> a écrit dans le message
de news:f71621tpfvur92en3q138qdlkjk12ioca3@4ax.com...

I'm sure there must be a smarter way to do this than my first design.
I want to produce separate +ve going 12V pulses at the leading and
trailing edges of a low impedance 12V input.

My first draft looks too complex:

- Use R/C/diode edge detector to produce OUT 1.
- Also take input to half a 4013 to produce FF
- Logic NOR the input with the inverted FF
- Use R/C/diode edge detector to produce OUT 2

Here's the timing chart:
http://www.terrypin.dial.pipex.com/Images/2From1.gif

I think those crude output pulses should be adequate without further
cleaning up, as they go to the respective Set and Reset inputs of a
simple 4001 bistable. Of course, squared up outputs would be a bonus;
I might do that anyway if I have any spare 4001 gates.

For the curious, it's an add-on for my curtain controller, to open and
close them either at dawn/dusk or at specific times of day.

Any simpler way to achieve this please?

Terry,

You want separate positive pulses for your NOR RS bistable.

If your 12 pulses are clean, you can do this:

10K
___
.-|___|-+------+-----VDD
| | |
1K 1n | | |
12V ___ || | | |<

-|___|--||--+-------|----|
pulse || | | |\

| | |

| | |
|-----' |

/| |
| |
Set <-----+ +-----> Reset
| |
.-. .-.
| | | |
| | | |
'-' '-'
| |
| |
=== ===
GND GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

Adjust the 1K 1n time constant to your taste.

--
Thanks,
Fred.

Fred Bloggs · Feb 28, 2005

Terry Pinnell wrote:

I'm sure there must be a smarter way to do this than my first design.
I want to produce separate +ve going 12V pulses at the leading and
trailing edges of a low impedance 12V input.

My first draft looks too complex:

- Use R/C/diode edge detector to produce OUT 1.
- Also take input to half a 4013 to produce FF
- Logic NOR the input with the inverted FF
- Use R/C/diode edge detector to produce OUT 2

Here's the timing chart:
http://www.terrypin.dial.pipex.com/Images/2From1.gif

I think those crude output pulses should be adequate without further
cleaning up, as they go to the respective Set and Reset inputs of a
simple 4001 bistable. Of course, squared up outputs would be a bonus;
I might do that anyway if I have any spare 4001 gates.

For the curious, it's an add-on for my curtain controller, to open and
close them either at dawn/dusk or at specific times of day.

Any simpler way to achieve this please?

This should be more in line with what you want:
View in a fixed-width font such as Courier.

4070_
+----------------\ \ \
| | | >--+------>OUTN
| |\ +-/ /__/ | |\
IN>---+-[R]-+--| o --+ +-| o-->OUTP
| |/ |/
[C] 40106 40106
|
|
---

------------
| |
| |
IN ------- -----------

------- --------- -------
| | | |
OUTN | | | |
-- --

~0.7RC
-- --
->| |<- | |
OUTP | | | |
------- --------- ------

+--------------------------+ 4001
| _ +-\ \
+----------------\ \ \ | o-->OUTNN
| | | >-+---/__/
| |\ +-/ /__/ |
IN>---+-[R]-+--| o --+ 4070 |
| | |/ |
| [C] 40106 |
| | |
| | |
| --- |
| | 4011
| | __
+----------------------------| \ |\
| | o-| o --->OUTPP
+---|__/ |/
40106

------------
| |
| |
IN ------- -----------

~0.7RC
--
->| |<-
OUTNN | |
-------------------- -------

~0.7RC
--
->| |<-
OUTPP | |
------- -------------------

Fred Bloggs · Feb 28, 2005

Fred Bloggs wrote:
scratch that- do this:
View in a fixed-width font such as Courier.

4070_
+----------------\ \ \
| | | >--+------>OUTN
| |\ +-/ /__/ | |\
IN>---+-[R]-+--| o --+ +-| o-->OUTP
| |/ |/
[C] 40106 40106
|
|
---

------------
| |
| |
IN ------- -----------

------- --------- -------
| | | |
OUTN | | | |
-- --

~0.7RC
-- --
->| |<- | |
OUTP | | | |
------- --------- ------

+--------------------------+ 4001
| _ +-\ \
+----------------\ \ \ | o-->OUTNN
| | | >-+---/__/
| |\ +-/ /__/ |
IN>---+-[R]-+--| o --+ 4070 |
| | |/ |
| [C] 40106 |
| | |
| | |
| --- |
| |
| |\ | 4001
+------------| o ------------\ \
|/ | | o-->OUTPP
40106 +---/__/

------------
| |
| |
IN ------- -----------

~0.7RC
--
->| |<-
OUTNN | |
-------------------- -------

~0.7RC
--
->| |<-
OUTPP | |
------- -------------------

Terry Pinnell · Feb 28, 2005

Tim Hubberstey <bogus@bogusname.com> wrote:

I'm almost reluctant to say this since it seems to be the answer to so
many simple projects: use an 8-pin PIC or other similar microcontroller
(uC).

Extreme overkill IMO, and that's even *if* I had:
- the parts to hand
- the programming skills necessary
- the time to remind myself how to use the programmer

There are other considerations that are not addressed by a simple edge
detector. I'm assuming your input comes from a photo-detector of some
kind so your edges are probably very slow and subject to "chatter".

Further, you probably don't want to trigger if a shadow from a bird
passes over the sensor during the day, or a headlight beam hits it at
night. This means you need to debounce the input before edge detection.
This all adds up to extra components if you do it in the analog world.

The dawn/dusk detector is based on an LDR, but via a Schmitt with
hysteresis, so those problems don't arise. The second input option I
mentioned will come from a plain mains-based programmable timer, and
I'll ensure it delivers clean 12V pulses to the pulse-generating
circuit.

Using a uC with an internal clock, you need only the uC and perhaps an
input clamp. Plus you can easily vary the width of your output pulse or
delay one edge if you want (if you find you'd prefer an hour after dawn,
for instance).

Versus a couple of chips and a few passives - come on! OK, it
certainly offers more versatility (not particularly important in this
application), but I think every other factor weighs against a PIC
approach here. Even my ponderous design takes only 6 passives plus 2
dirt cheap ICs. And some of the other recommendations, for which
detailed schematics have been posted up-thread, are much simpler.

I appreciate your good intentions, but, like quite a few posters,
you've tacitly assumed everyone here knows PICs and works with them
every day. Although I still have that as a goal, unhappily I haven't
achieved it yet! <g>

--
Terry Pinnell
Hobbyist, West Sussex, UK

Rich Grise · Feb 28, 2005

On Mon, 28 Feb 2005 21:16:27 +0000, Fred Bloggs wrote:

John Fields wrote:
What's wrong with the old standard way of doing it?

IN>-----+-----------A
| 4070 Y---->OUT
+--[R]--+---B
|
[C]
|
GND>------------+

_______________
IN _____| |__________

_ _
OUT_____| |_____________| |________

Nothing- he wants those pulses separated out- the IN + edge pulse to one
output, and IN - edge pulse to another channel.

Then just differentiate the square, and pick off the negative-going
polarity with an inverter. Clamp them both, of course.

Cheers!
Rich

Keith Williams · Mar 1, 2005

In article <d02h0n$ktm$1@sparta.btinternet.com>,
g4fgq.regp@ZZZbtinternet.com says...

Just differentiate the square pulse with a CR circuit. Use both the leading
and trailing edges to trigger something.

I'm not particularly fond of differentiators (noise). An integrator

with a couple of gates works. For a single output an XOR gate can be
used as a discriminator. For separate rising/falling outputs use a
couple of NAND/NOR gates.

--
Keith

Reg Edwards · Mar 1, 2005

Just differentiate the square pulse with a CR circuit. Use both the leading
and trailing edges to trigger something.

Reg Edwards · Mar 1, 2005

See Blumlein, early television, circa 1935.

Unfortunately Blumlein was killed in an air-crash around 1943/4 while
testing yet another of his circuits in a war-time, top-secret, centimetric,
airborne Radar navigation equipment intended for guidance of Lancaster
bombers, the ultimate killing machines, over German towns and cities for the
purpose incinerating their populations in fire storms.

We'll never know the circuit details but a monument has been erected to his
memory.

Terry Pinnell · Mar 1, 2005

"Reg Edwards" <g4fgq.regp@ZZZbtinternet.com> wrote:

Just differentiate the square pulse with a CR circuit. Use both the leading
and trailing edges to trigger something.

Read the requirement again. Look at the timing chart...

--
Terry Pinnell
Hobbyist, West Sussex, UK

Mar 2, 2005

Is the contest still open? Here's my entry:

.. +12v
.. _
.. |
.. R1 |
.. +----/\/\/\------*----------+
.. C1 | | |
.. | Q1 / |
.. | | | PNP |e |
.. >-----+ +---*-------------|-------\ |
.. | | |\ e| |
.. | |---+
.. | /|
.. | | Q2
.. | | PNP
.. OUT- <-------+ +-------> OUT+
.. | |
.. R2 R3
.. | |
.. | |
.. GND GND

James Arthur

Mar 2, 2005

Oops -- just saw this. With apologies to Fred, please ignore my
earlier post...

James

Terry Pinnell · Mar 9, 2005

This morning I breadboarded both Fred Bloggs' final circuit
news:4224845A.6080401@nospam.com and Fred Bartoli's. Both work fine
thanks.

The latter is especially attractive because of its delightfully low
component count, as well as its novelty. My test input was a clean
square wave from a function generator, so of course the circuit might
need a front-end clean-up in practice. However, the outputs would
probably be fine without further squaring up, as their leading edges
were fast.

But how does it work please?! What function does that left hand PNP
serve, with its gate permanently at Vdd?

As it turns out, I haven't needed to build either of them after all.
Instead, I improvised from my existing LDR-based dawn/dusk circuit, as
shown here, and this appears to work well.
http://www.terrypin.dial.pipex.com/Images/CurtainTimerAddOn.gif

The dawn/dusk switch wasn't working reliably, mainly due to
unpredictable variations in light conditions, so I've abandoned it now
in favour of this 'Timer Add-on', which is set to open and close
curtains at fixed times.

--
Terry Pinnell
Hobbyist, West Sussex, UK

Fred Bartoli · Mar 9, 2005

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> a écrit dans le message
de news:ivnt219o2uqv0aui047q2h3p9vkcu6d111@4ax.com...

This morning I breadboarded both Fred Bloggs' final circuit
news:4224845A.6080401@nospam.com and Fred Bartoli's. Both work fine
thanks.

The latter is especially attractive because of its delightfully low
component count, as well as its novelty. My test input was a clean
square wave from a function generator, so of course the circuit might
need a front-end clean-up in practice. However, the outputs would
probably be fine without further squaring up, as their leading edges
were fast.

But how does it work please?! What function does that left hand PNP
serve, with its gate permanently at Vdd?

Hi Terry,
don't forget its emitter

There are 3 basic circuits to build around a BJT : common emitter, common
base and common collector (emitter follower).
Google for those and you'll find all the needed information.

As you've seen, there are 2 transistors, the right one in common emitter
configuration, the left one in common base.

10K
___
.-|___|-+------+-----VDD
| | |
1K 1n | | |
12V ___ || | A | |<

-|___|--||--+-------|----| Q2
pulse || | | |\

| | |

| | |
Q1 |-----' |

/| |
| |
Set <-----+ +-----> Reset
| |
.-. .-.
| | | |
R1 | | | | R2
'-' '-'
| |
| |
=== ===
GND GND

Due to the base emitter junctions, the A point is clamped to +/- 0.6V around
VDD.
On rising edge, Q2 is reverse biased and the capacitor current is injected
into Q1 emitter => positive pulse on the set output.
On falling edge Q1 is reverse biased and the capacitor current turns Q2 on
=> positive pulse on the reset output.

The 1K resistor obviously limits injected current.
R1 needs to be > 1K*VDD/(12-VDD-0.6).

BTW, since those transistors drive a regenerative circuit (your NOR SR
latch) you don't need squaring after the transistors: the hysteresis is
built into the SR latch.

--
Thanks,
Fred.

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