pulsing ouput to solid

[Chris Moylan]

Hi folks, I was wondering if I could find some help here. I am not
much of a circuit designer, but I have reasonable soldering and repair
experience.

What I need is a circuit that takes an input that pulses at around
1-2Hz cycling from 0V to 12VDC and converts that to a solid 12VDC when
the pulse is happening and 0V when the pulse has stopped. It is ok if
the output returns to 0 after a capacitor finishes disharging, it
needn't be immediate...just relatively soon after the pulse stops.

Now I realize I could use a capacitor sized to charge during the ON
pulse and discharge during the OFF pulse but I do not know how to size
the capacitor or what other components may be needed for loading, etc.

Any help would be greatly appreciated. I realize that this should be
pretty darn easy...

Thanks,

Chris

 

[CFoley1064]

Subject: pulsing ouput to solid
From: crm0922@rocketmail.com (Chris Moylan)
Date: 4/28/2004 11:57 AM Central Standard Time
Message-id: <a0daa417.0404280857.2f6cbd91@posting.google.com

Hi folks, I was wondering if I could find some help here. I am not
much of a circuit designer, but I have reasonable soldering and repair
experience.

What I need is a circuit that takes an input that pulses at around
1-2Hz cycling from 0V to 12VDC and converts that to a solid 12VDC when
the pulse is happening and 0V when the pulse has stopped. It is ok if
the output returns to 0 after a capacitor finishes disharging, it
needn't be immediate...just relatively soon after the pulse stops.

Now I realize I could use a capacitor sized to charge during the ON
pulse and discharge during the OFF pulse but I do not know how to size
the capacitor or what other components may be needed for loading, etc.

Any help would be greatly appreciated. I realize that this should be
pretty darn easy...

Thanks,

Chris

Missing Pulse Detector
+12V
+
|
.---o-------.
| | |
.-. .-. |
| | | |100K |
10K | | | | |
'-' '-' o---.
| | | |
| | | |
.01uF | | .---o---o---.
| | | 8 4 |
|| | | | | Out
o--||----o-------o 2 3 o----o
|| | | |
In (1 to 2 Hz) | | LM555 |
| .-o 7 |
| | | |
o-o-o 6 |
| | 1 5 |
| '---o---o---'
| + | N.C.
10uF ### |
--- |
| |
| |
'-------|
|
===
GND

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

The values of R and C (100K, 10uF) are chosen so that, if there is no
negative-going transition at the input for about 1.1 seconds, the output will
go back low. The only glitch in this circuit is that it will go high for 1.1
seconds on turn-on. All components and the IC are available at Radio Shack or
any hobbyist source.

http://cache.national.com/ds/LM/LM555.pdf

Good luck (and by the way, questions of this type usually get a better response
on sci.electronics.basics)

Chris

 

[John Fields]

On 28 Apr 2004 09:57:04 -0700, crm0922@rocketmail.com (Chris Moylan)
wrote:

Hi folks, I was wondering if I could find some help here. I am not
much of a circuit designer, but I have reasonable soldering and repair
experience.

What I need is a circuit that takes an input that pulses at around
1-2Hz cycling from 0V to 12VDC and converts that to a solid 12VDC when
the pulse is happening and 0V when the pulse has stopped. It is ok if
the output returns to 0 after a capacitor finishes disharging, it
needn't be immediate...just relatively soon after the pulse stops.

Now I realize I could use a capacitor sized to charge during the ON
pulse and discharge during the OFF pulse but I do not know how to size
the capacitor or what other components may be needed for loading, etc.

Any help would be greatly appreciated. I realize that this should be
pretty darn easy...

---
Probably the easiest way is just to use a retriggerable monostable
like a 4538 and set it to time out with a period slightly longer than
the period of your input pulses.

http://www.fairchildsemi.com/ds/CD/CD4538BC.pdf


The next easiest would be to build something like this:

+12v>---------------------------+-------+
| |
| [1000]
| |R3
[10k] +----->OUT
|R2 |
| C
+------B 2N4401
| E Q2
CR1 R1 C |
1Hz>---[1N4148>]---+--[100K]---B 2N4401 |
|+ E Q1 |
[10ľf] | |
| C1 | |
GND>---------------+------------+-------+

With no input pulses, Q1 will be cut off, allowing the base of Q2 to
be pulled up to +12V through R2. This will allow about 1.2mA of
current into the base of Q2, turning it on and pulling its collector
low.

With input pulses occurring, C1 will charge up quickly through CR1
when the pulses are high. When the pulses are low or when they stop
altogether, C1 will only be able to discharge through the base to
emitter junction of Q1 because CR1 will be reverse biased any time its
anode isn't positive with respect to its cathode.

When C1 charges sufficiently to cause Q1's base current to turn on Q1,
Q1's collector will go low, turning off Q2. When this happens, Q2's
collector will be pulled up to 12V through R3 and will remain there
until C1 discharges to the point where it can no longer supply enogh
base cuernt to keep Q1 turned on.

With a 10ľF capacitor and a 100k ohm base resistor it will take about
2-1/2 seconds for the capacitor to discharge to the point where Q1
turns off after the cap has been charged up to 12V by the inputr
pulses.

--
John Fields

 

[Terry Pinnell]

crm0922@rocketmail.com (Chris Moylan) wrote:

Hi folks, I was wondering if I could find some help here. I am not
much of a circuit designer, but I have reasonable soldering and repair
experience.

What I need is a circuit that takes an input that pulses at around
1-2Hz cycling from 0V to 12VDC and converts that to a solid 12VDC when
the pulse is happening and 0V when the pulse has stopped. It is ok if
the output returns to 0 after a capacitor finishes disharging, it
needn't be immediate...just relatively soon after the pulse stops.

Now I realize I could use a capacitor sized to charge during the ON
pulse and discharge during the OFF pulse but I do not know how to size
the capacitor or what other components may be needed for loading, etc.

Any help would be greatly appreciated. I realize that this should be
pretty darn easy...

Thanks,

Chris

There are several things you'd need to specify more precisely, but
here's something to get started:
http://www.terrypin.dial.pipex.com/Images/SmoothingPulse.gif

This uses a simple diode/capacitor arrangement to deliver a more or
less 'smoothed' supply from the very slow pulses you describe. But as
you see from the simulation, how 'solid' it is depends on 3 factors:
1. The source resistance R1
2. The capacitor value C1
3. The load resistance

The comparison shows that lowering R1 and raising C1 by a factor of 10
increases the 'solidity'. Crudely, by a factor of 100 - IOW it takes
100 times as long for the capacitor to lose its charge.

Naturally, you could add a voltage regulator circuit after this as
well. But you're starting with such a low frequency source that I
suspect it would be fruitless.

So, you either need to spell out the three factors above, so that we
can offer better advice, or experiment until you get an adequate
result.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Chris Moylan]

Terry Pinnell <terrypin@dial.pipex.com> wrote in message news:<2i2190tcjtnb26jpuaq7dg7nou0u609ili@4ax.com>...

There are several things you'd need to specify more precisely, but
here's something to get started:
http://www.terrypin.dial.pipex.com/Images/SmoothingPulse.gif

This uses a simple diode/capacitor arrangement to deliver a more or
less 'smoothed' supply from the very slow pulses you describe. But as
you see from the simulation, how 'solid' it is depends on 3 factors:
1. The source resistance R1
2. The capacitor value C1
3. The load resistance

The comparison shows that lowering R1 and raising C1 by a factor of 10
increases the 'solidity'. Crudely, by a factor of 100 - IOW it takes
100 times as long for the capacitor to lose its charge.

Naturally, you could add a voltage regulator circuit after this as
well. But you're starting with such a low frequency source that I
suspect it would be fruitless.

So, you either need to spell out the three factors above, so that we
can offer better advice, or experiment until you get an adequate
result.

Thanks for the excellent advice.

The source supply is going to be a 12V wall wart type supply, probably
500mA which is being powered by 120VAC that is actually pulsing.

For the sake of reference, DCTX-1259 on
http://www.allelectronics.com/matrix/DC_Wall_Transformers.html is
essentially what I am using.

The load could be anything, but a relay was probably what I was going
to put on the solid output end primarily for the electrical separation
the relay would provide and allowing the design to work for various
devices. The relay should probably be rated for 120VAC 5A or so, but
I can use whatever is easiest...

I would bet that the pulse is even less than 2Hz. It is probably. I
know I need to measure it, but this doesn't need to be precise. As
long as the capacitor discharges longer than needed (2-3sec), I don't
care if it hangs over when the pulses stop.

The wrinkle I forgot to mention is that I want the output to shut off
regardless of whether or not the pulse is high or low(off) when the
pulses stop. The pulse could stop at high(12V) output more often than
not.

My overall goal is to convert an alert output from a device that
pulses a 120VAC bulb (while alerting) to other purposes.

Chris

 

[Terry Pinnell]

crm0922@rocketmail.com (Chris Moylan) wrote:

Thanks for the excellent advice.

The source supply is going to be a 12V wall wart type supply, probably
500mA which is being powered by 120VAC that is actually pulsing.

For the sake of reference, DCTX-1259 on
http://www.allelectronics.com/matrix/DC_Wall_Transformers.html is
essentially what I am using.

OK, so that's a low source resistance. (I'd guess a few ohms?)

The load could be anything, but a relay was probably what I was going
to put on the solid output end primarily for the electrical separation
the relay would provide and allowing the design to work for various
devices. The relay should probably be rated for 120VAC 5A or so, but
I can use whatever is easiest...

A typical, robust 12V mains-carrying relay might have a resistance of
anything from say 200 to 1000 ohms. Let's assume 400.

I would bet that the pulse is even less than 2Hz. It is probably. I
know I need to measure it, but this doesn't need to be precise. As
long as the capacitor discharges longer than needed (2-3sec), I don't
care if it hangs over when the pulses stop.

Let's take the worst case of 1 Hz then.

The wrinkle I forgot to mention is that I want the output to shut off
regardless of whether or not the pulse is high or low(off) when the
pulses stop. The pulse could stop at high(12V) output more often than
not.

My overall goal is to convert an alert output from a device that
pulses a 120VAC bulb (while alerting) to other purposes.

Let me play my understanding back and see if I've got it right.

Some mains-powered device is pulsing a mains bulb at 1 Hz. It's on
(lit) for 500 ms and off (unlit) for 500 ms. This is an alert signal.
The mains pulse can also supply a wall transformer rated at 500mA and
suplying 12V DC.

When this is pulsing, you want a relay of say 400 ohm resistance to be
activated permanently (not pulsing), so that you can wire its contacts
for some other purpose. Let's assume that's a subsidiary alarm.

When your source stops pulsing, you want the relay to be deactivated.
One point you haven't mentioned is how fast you want that reaction to
be? Assuming the alert condition ceases right at the end of a 500 ms
pulse, how long is acceptable before the subsidiary alarm ceases?

With a 4700uF electrolytic, a typical relay would stay activated (as
the voltage would drop to only about 8.2V between pulses). But, with
the ultra-simple circuit I suggested, it would then take maybe 2
seconds to deactivate. If you want an 'instant' all clear, then that
won't do.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Chris Moylan]

Terry Pinnell <terrypin@dial.pipex.com> wrote in message > Let me play my understanding back and see if I've got it right.

Some mains-powered device is pulsing a mains bulb at 1 Hz. It's on
(lit) for 500 ms and off (unlit) for 500 ms. This is an alert signal.
The mains pulse can also supply a wall transformer rated at 500mA and
suplying 12V DC.

When this is pulsing, you want a relay of say 400 ohm resistance to be
activated permanently (not pulsing), so that you can wire its contacts
for some other purpose. Let's assume that's a subsidiary alarm.

When your source stops pulsing, you want the relay to be deactivated.
One point you haven't mentioned is how fast you want that reaction to
be? Assuming the alert condition ceases right at the end of a 500 ms
pulse, how long is acceptable before the subsidiary alarm ceases?

With a 4700uF electrolytic, a typical relay would stay activated (as
the voltage would drop to only about 8.2V between pulses). But, with
the ultra-simple circuit I suggested, it would then take maybe 2
seconds to deactivate. If you want an 'instant' all clear, then that
won't do.

Your assessment is correct. The deactivation time is fine anywhere
from 1-4 seconds. The wrinkle is that the bulb will stay lit after
certain alarms and I don't want my subsidiary device to stay activated
in those instances. Is there any way to make the output discharge
even if the input decides to stay high?

Chris

 

[N. Thornton]

hi


Capacitor and relay is no good cos the pulsing may stop in the wall
wart powered on mode. Same problem occurs with a monostable.

The OP needs to use the 2 transistor circuit given in another post,
but mod it by putting the signal thru a biggish coupling cap, adding
diodes to the PS rails to keep Vs within bounds, then rectify and RC
filter that ac and use it for control.


Regards, NT

oh ok then...

+12v>----+----------------------+-------+
| | |
| | [1000]
| | |R3
| [10k] +----->OUT
/_\ Diode |R2 |
| | C
| +------B 2N4401
| | E Q2
cap | D1 R1 C |
1Hz>-||--+---|>|---+--[100K]---B 2N4401 |
| |+ E Q1 |
/_\ diode [10ľf] | |
| | C1 | |
GND>-----+---------+------------+-------+

And oy, who calls diodes CRs? They havent been crystal rectifiers in a
long time


Regards, NT

 

[Terry Pinnell]

crm0922@rocketmail.com (Chris Moylan) wrote:

Your assessment is correct. The deactivation time is fine anywhere
from 1-4 seconds. The wrinkle is that the bulb will stay lit after
certain alarms and I don't want my subsidiary device to stay activated
in those instances. Is there any way to make the output discharge
even if the input decides to stay high?

Ah, understood. I'd been assuming that all mains power would be
removed when alarm stopped, and that the 'wrinkle' was merely the
trivial issue of the last +ve signal being short (e.g shorter than
500mS).

OK, then it's just a matter of adding capacitor isolation. I have to
leave for weekend in a few minutes, but at a glance something along
the lines N Thornton suggested looks fine. Maybe a slightly different
configuration would allow using a single PNP transistor instead of two
NPNs.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Chris Moylan]

Excuse my dumbness, but can you explain how this works in greater
detail? It will solve the problem of the input signal ending in a
high state, yes?

Shall the "condenser" labeled "cap" be of a size similar to the other
circuits presented earlier? ;-)

I also need 12VDC continuous source to power the circuit itself which
allows the transistors to continue to function regardless of the state
of the input pulse....

Boy I wish I understood the operation of transistors and diodes
better. And I took all those courses in college and slid through them
like a bum. Wish I had paid attention now.

Thanks again everyone for the tips. I think I am closing in on a
solution.

Chris




bigcat@meeow.co.uk (N. Thornton) wrote in message news:<a7076635.0404301614.5f81811a@posting.google.com>...

hi


Capacitor and relay is no good cos the pulsing may stop in the wall
wart powered on mode. Same problem occurs with a monostable.

The OP needs to use the 2 transistor circuit given in another post,
but mod it by putting the signal thru a biggish coupling cap, adding
diodes to the PS rails to keep Vs within bounds, then rectify and RC
filter that ac and use it for control.


Regards, NT

oh ok then...

+12v>----+----------------------+-------+
| | |
| | [1000]
| | |R3
| [10k] +----->OUT
/_\ Diode |R2 |
| | C
| +------B 2N4401
| | E Q2
cap | D1 R1 C |
1Hz>-||--+---|>|---+--[100K]---B 2N4401 |
| |+ E Q1 |
/_\ diode [10ľf] | |
| | C1 | |
GND>-----+---------+------------+-------+

And oy, who calls diodes CRs? They havent been crystal rectifiers in a
long time


Regards, NT

 

[Fred Bloggs]

Chris Moylan wrote:

Excuse my dumbness, but can you explain how this works in greater
detail? It will solve the problem of the input signal ending in a
high state, yes?

Shall the "condenser" labeled "cap" be of a size similar to the other
circuits presented earlier? ;-)

I also need 12VDC continuous source to power the circuit itself which
allows the transistors to continue to function regardless of the state
of the input pulse....

Boy I wish I understood the operation of transistors and diodes
better. And I took all those courses in college and slid through them
like a bum. Wish I had paid attention now.

Thanks again everyone for the tips. I think I am closing in on a
solution.

Chris

Pulsing the modular power supply is not a good idea- this will subject
the components to unnecessary repetitive turn-on surges. Because the
alarm output can remain steady state on, you have to include "static"
memory in your circuit, a capacitor will not do, and this usually means
digital logic for the most compact implementation. One way to do this is
like so:
Please view in a fixed-width font such as Courier.

..
..
..
..
.. +----------+ AND
.. ~~~ | | ___
.. >----|ALARM AC |-+--------------------------------| \
.. |DETECTION | | | & >--+
.. +----------+ | +--|__ / |
.. | LATCH | |
.. | +---------+ +------+ | |
.. | | ALARM_AC|------|S Q |-+ |
.. +-->|OFF>100mS| | | |
.. | ? | +--|R | |
.. +---------+ | +------+ |
.. | |
.. +----------------------+
.. |
.. +-------------------------------------------------------+
.. |
.. | +----------------+
.. | | RE-TRIGGERABLE |
.. +-----> MONOSTABLE |--------+
.. | RELAY DRIVER | |
.. +-+ +----------------+ )|| 12VDC
.. | | )|| RELAY
.. | | )|| COIL
.. --+ +-- |
.. |
.. ---
..
..
..

The two fundamental elements are the ALARM AC DETECT and the RELAY
MONOSTABLE DRIVER. The AC DETECT is a simple circuit that detects
whether the alarm unit has placed AC on the line- yes or no, and the
monostable is a simple circuit that keeps steady drive on the relay coil
as long as it is re-triggered before its internal timeout occurs. The
ALARM AC OFF>100mS is a subcircuit that sets the latch whenever ALARM AC
is off for more than 100ms continuously. If the AC is not being pulsed
by the alarm, then both inputs to the AND gate are never true
simultaneously, whether the line is continuos AC on or off, so that the
monostable driver is not triggered and the relay is cut off. When the
alarm begins pulsing the line, the monostable receives triggers on each
AC line off-to-on transition , and the relay is continuously on.
For this type of project, it is simplest to use 74HC logic and make a
small 5V supply from the 12V with a common 5.1V zener diode, so that
only the relay is powered from the 12V. A two-chip solution would look
like so:
Please view in a fixed-width font such as Courier.

..
.. ______
.. AC_DET
..
.. 60Hz WHEN AC PRESENT CONSTANT 5V WHEN AC NOT PRESENT
.. 5V -- ------ ---- 5V --------------------------------
.. | | | |
.. 0V- |_| |_| 0V-
..
..
.. U2:A
.. ______ __
.. AC_DET >----+-----------------------------------------|_ \
.. | | |_o--+
.. | +---|__/ |
.. | 100mS | |
.. | TIMER | |
.. | +--------U1:A | |
.. | | | | |
.. | |__ Q1|-------+ +---------|----------+
.. +--|A1 __| | | | |
.. | Q1| | | | ===
.. | | | | U2:B | 0.1U
.. | | | | __ | |
.. | | | +--|_ \ | 10K |
.. | C1|--+ | | |_o--+---/\/\---+
.. +--|B1 | | | +--|__/ | |
.. | |---- | === | | | |
.. +--|CLR1 |0.47U | | | 22K |
.. | | R1/C1|--+ | +-------------/\/\---+
.. | +-----------+ | | |
.. | 470K | | |
.. 5V>--+------/\/\-------+ | |
.. | |
.. | |
.. +----------+ |
.. 10K | |
.. +----+---/\/\----|----------+--------------+
.. 100P | | | |
.. === | | | RELAY
.. | | | | MONOSTABLE
.. -+- | | | +--------U1:B
.. GND | | | |__ |
.. | | U2 +--|A2 |
.. | U2:C | __ | Q2|-----> TO RELAY
.. | __ +---|_ \ | __| DRIVER
.. +---|_ \ | |_o--+--|B2 Q2|
.. | |_o-------|__/ | | |
.. +---|__/ | | |
.. | RS-LATCH | | C2|--+------+
.. +----------------------+ | | | |
.. |---- | === ===
.. +--|CLR2 |0.47U 0.47U
.. | | R2/C2|--+------+
.. U1 74HC123 | +-----------+ |
.. | 4.7M |
.. U2 74HC132 5V>--+------/\/\-------+
..
..
..
..


The /AC_DET signal is best derived from an OPTOCOUPLER on the line, and
the relay driver is a MOSFET with gate driven by U1:B Q1 output. U2:A-B
form a pulse stretcher and provide the trigger for the U1:B relay driver
monostable with each AC line input pulse, but since these triggers also
CLEAR the RS latch, which inhibits the U1:B monostable B2 input, the
monostable is not retriggered until the AC line voltage is removed for
100ms, U1:A times out, and the RS-latch re-SET , enabling U1:B B-input.
The 10K-100P at the U2:C input is to provide a delay that prevents a
hold time race condition for the U1:B B-input ensuring a solid trigger.
So this takes care of the smarts and all you have left is the
straightforward tidying up: optocoupler to detect 120VAC, a little zener
diode 5.1V shunt regulator, and a MOSFET like the 2N7000, BS-170, or
similar for the relay driver- don't forget to put 1N4001 clamp diode
anti-parallel to relay coil- attached to 12V supply.

 

[John Fields]

On 30 Apr 2004 17:14:56 -0700, bigcat@meeow.co.uk (N. Thornton) wrote:

hi


Capacitor and relay is no good cos the pulsing may stop in the wall
wart powered on mode. Same problem occurs with a monostable.

---
Really?

I suggest you check out the 4538 or the XX123...
---

The OP needs to use the 2 transistor circuit given in another post,
but mod it by putting the signal thru a biggish coupling cap, adding
diodes to the PS rails to keep Vs within bounds, then rectify and RC
filter that ac and use it for control.


Regards, NT

oh ok then...

+12v>----+----------------------+-------+
| | |
| | [1000]
| | |R3
| [10k] +----->OUT
/_\ Diode |R2 |
| | C
| +------B 2N4401
| | E Q2
cap | D1 R1 C |
1Hz>-||--+---|>|---+--[100K]---B 2N4401 |
| |+ E Q1 |
/_\ diode [10ľf] | |
| | C1 | |
GND>-----+---------+------------+-------+

And oy, who calls diodes CRs? They havent been crystal rectifiers in a
long time

---
I do, and I'll continue to. They always were, and still are. What
makes you think they've changed?

BTW, how about some insight as to the value of "cap"?

--
John Fields

 

[N. Thornton]

crm0922@rocketmail.com (Chris Moylan) wrote in message news:<a0daa417.0404302315.2d4f7e4@posting.google.com>...

Excuse my dumbness, but can you explain how this works in greater
detail? It will solve the problem of the input signal ending in a
high state, yes?

yup

Shall the "condenser" labeled "cap" be of a size similar to the other
circuits presented earlier? ;-)

10uF or 22uF should be fine.

I also need 12VDC continuous source to power the circuit itself which
allows the transistors to continue to function regardless of the state
of the input pulse....

not necessarily. If you drive a relay, you can arrange it so the relay
coil is on when the pulsing is gonig on, and have the coil off
otherwise. Now you need no always on 12v supply.


Regards, NT

oh ok then...

+12v>----+----------------------+-------+
| | |
| | [1000]
| | |R3
| [10k] +----->OUT
/_\ Diode |R2 |
| | C
| +------B 2N4401
| | E Q2
cap | D1 R1 C |
1Hz>-||--+---|>|---+--[100K]---B 2N4401 |
| |+ E Q1 |
/_\ diode [10ľf] | |
| | C1 | |
GND>-----+---------+------------+-------+

 

[N. Thornton]

Fred Bloggs <nospam@nospam.com> wrote in message news:<4093B020.6040308@nospam.com>...

Pulsing the modular power supply is not a good idea- this will subject
the components to unnecessary repetitive turn-on surges. Because the
alarm output can remain steady state on, you have to include "static"
memory in your circuit, a capacitor will not do, and this usually means
digital logic for the most compact implementation. One way to do this is
like so:

no need.

Regards, NT

 

[N. Thornton]

John Fields <jfields@austininstruments.com> wrote in message news:<lqb790t5un4t0g4963sq7jef50c5p2kppv@4ax.com>...

On 30 Apr 2004 17:14:56 -0700, bigcat@meeow.co.uk (N. Thornton) wrote:

And oy, who calls diodes CRs? They havent been crystal rectifiers in a
long time

I do, and I'll continue to. They always were, and still are. What
makes you think they've changed?

I know a few folk do, but it always strikes me as odd. Todays diodes
have changed very much from the original crystal rectifiers. But then
I have my odd habits...


Regards, NT

 

[John Fields]

On 1 May 2004 14:48:29 -0700, bigcat@meeow.co.uk (N. Thornton) wrote:

crm0922@rocketmail.com (Chris Moylan) wrote in message news:<a0daa417.0404302315.2d4f7e4@posting.google.com>...
Excuse my dumbness, but can you explain how this works in greater
detail? It will solve the problem of the input signal ending in a
high state, yes?

yup

Shall the "condenser" labeled "cap" be of a size similar to the other
circuits presented earlier? ;-)

10uF or 22uF should be fine.

---
Got some numbers to back that up?
---

I also need 12VDC continuous source to power the circuit itself which
allows the transistors to continue to function regardless of the state
of the input pulse....

not necessarily. If you drive a relay, you can arrange it so the relay
coil is on when the pulsing is gonig on, and have the coil off
otherwise. Now you need no always on 12v supply.

---
For some reason you seem to be assuming that there's going to be
enough energy in the pulsing 12V to, somehow, fill up a reservoir for
the relay coil to draw on while the 12V is off. How would you arrange
for that to happen?

--
John Fields

 

[John Fields]

On 1 May 2004 14:52:31 -0700, bigcat@meeow.co.uk (N. Thornton) wrote:

John Fields <jfields@austininstruments.com> wrote in message news:<lqb790t5un4t0g4963sq7jef50c5p2kppv@4ax.com>...
On 30 Apr 2004 17:14:56 -0700, bigcat@meeow.co.uk (N. Thornton) wrote:

And oy, who calls diodes CRs? They havent been crystal rectifiers in a
long time

I do, and I'll continue to. They always were, and still are. What
makes you think they've changed?

I know a few folk do, but it always strikes me as odd. Todays diodes
have changed very much from the original crystal rectifiers. But then
I have my odd habits...

---
How have they changed? They're still made from P and N doped single
crystal materials (or doped single crystals and an ohmic
contact/junction) and they're still rectifiers...

--
John Fields

 

[Fred Bloggs]

N. Thornton wrote:

Fred Bloggs <nospam@nospam.com> wrote in message news:<4093B020.6040308@nospam.com>...


Pulsing the modular power supply is not a good idea- this will subject
the components to unnecessary repetitive turn-on surges. Because the
alarm output can remain steady state on, you have to include "static"
memory in your circuit, a capacitor will not do, and this usually means
digital logic for the most compact implementation. One way to do this is
like so:


no need.

Regards, NT

You seem to have some hazy idea about ac-coupling the 1Hz output of a
60Hz envelope detector into some kind of half-assed fast charge/slow
discharge excuse for a retriggerable monostable made from discretes. And
you think this is more efficient than a 74HC123- and that the rest of
your junk comes under the quad schmitt NAND 74HC132? I've heard about
cave men like you, but was told you all went extinct 30 years ago...

 

[N. Thornton]

Fred Bloggs <nospam@nospam.com> wrote in message news:<40945108.7070402@nospam.com>...

N. Thornton wrote:

Pulsing the modular power supply is not a good idea- this will subject
the components to unnecessary repetitive turn-on surges. Because the
alarm output can remain steady state on, you have to include "static"
memory in your circuit, a capacitor will not do, and this usually means
digital logic for the most compact implementation. One way to do this is
like so:

no need.

You seem to have some hazy idea about ac-coupling the 1Hz output of a
60Hz envelope detector

quite doable, though I'll admit to a need for a discharge load on the
wall wart to keep its output giving 1Hz.

into some kind of ..... fast charge/slow
discharge ......... monostable made from discretes.

It does the job, I think thats the point here.

And
you think this is more efficient than a 74HC123-

your approach included complications, I thought the simple approach
simpler.

and that the rest of
your .... comes under the quad schmitt NAND 74HC132?

I dont know what you mean.

I've heard about
......... like you, but was told ......

The ad hominem is irrelevant, just stick to the design.


Regards, NT

 

[Chris Moylan]

Easy fellas. ;-)

First of all, the alarm will probably get triggered like once a year,
so I am not worried about the components prematurely wearing because
of a pulsed AC adapter. I just don't want to deal with live AC for
this project, I want to use alarm wire for the run, etc.

Why do I care at all about 60Hz envelope detectors? Will the discrete
design actually work? Because I can build that in short order, and
the digital logic approach would take me considerably longer. I have
no real experience with logic chips so throwing together some caps,
resistors and transistors from RS seems a lot easier.

Thanks again for all the help,

Chris

Fred Bloggs <nospam@nospam.com> wrote in message news:<40945108.7070402@nospam.com>...

N. Thornton wrote:
Fred Bloggs <nospam@nospam.com> wrote in message news:<4093B020.6040308@nospam.com>...


Pulsing the modular power supply is not a good idea- this will subject
the components to unnecessary repetitive turn-on surges. Because the
alarm output can remain steady state on, you have to include "static"
memory in your circuit, a capacitor will not do, and this usually means
digital logic for the most compact implementation. One way to do this is
like so:


no need.

Regards, NT

You seem to have some hazy idea about ac-coupling the 1Hz output of a
60Hz envelope detector into some kind of half-assed fast charge/slow
discharge excuse for a retriggerable monostable made from discretes. And
you think this is more efficient than a 74HC123- and that the rest of
your junk comes under the quad schmitt NAND 74HC132? I've heard about
cave men like you, but was told you all went extinct 30 years ago...