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how can I create a positive logic pulse from a steady high input?

Skidood

Aug 24, 2015
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I kinda forgot about the OR gate..well OK the 2 diodes ...in all the excitement.....the reset input needs to go low and stay low when the alarm condition goes away by itself. Hmmm...aarrrggggg.....

OK just saw your posts Steve..will continue tomorrow..its time to call me Mum and find out what she thought when her grand-daughter told her today that she was a lesbian...lol...
 

Old Steve

Jul 23, 2015
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That diode should do it. Best you calm your Mum down now.....
 

AnalogKid

Jun 10, 2015
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It is a NAND FF, so set and reset are negative true inputs. If any of the three alarm inputs are high, pin 13 is high and the FF is ready to accept a negative pulse on pin 1 (positive pulse out of C1/R5). With the pin 13 node high and current limited by the input pullup resistors, the reset switch can pull it low for a reset. Remember that the set input is low for only a few milliseconds (BTW, the 4093 can respond to sub-microsecond pulses). When all inputs are low, R4 holds the ff in reset mode.

Check your component values. If R4 is 10K instead of 100K, the system will not work. Also, measure the pulse width at pin 4.

Looking at the sch, I should have put C1/R5 between U1B and U1A, with R5 going to Vcc. This gives the pulse former a more consistent drive impedance, but also directly exposes the 4093 to the outside world.

ak
 

AnalogKid

Jun 10, 2015
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Just a followup-
Analog, have you ever noticed that Peart's drumming "riff" during the chorus of Digital Man (while Lee is singing "He's got a force field and a flexible plan" is virtually the same as that during the chorus of My Hero by the Foo Fighters?
Quite a few drummers have taken "inspiration" from NP over the years. I work his fills into all kinds of things. OTOH, some of his work just stands alone. When people asked him which was the hardest song to drum to (next to his solo), for many years his answer was Tom Sawyer. I agree that it takes great stamina, but for me it is Red Barchetta. That puppy is *hard*.

ak
 

Old Steve

Jul 23, 2015
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Of course you're right AK. I'm too distracted to think clearly on this right now, doing too many things at once, so I'll leave it to you.
 

Old Steve

Jul 23, 2015
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This would be my solution:-
In fact, I'd omit the diodes too and use 1 pullup resistor.
scan011.JPG
 

Skidood

Aug 24, 2015
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Haha...

OK I got it...the problem I had with Analogs circuit was that when pressing the silence switch, the alarm would silence but would turn back on again after releaing the switch (hence my saying it wasn't latching)
I put a 100K R in between R4 and Pin 13...and got rid of the pull up resistor on Pin 13 that I thought I should put there earlier.
Solved.

I think the reason is that without the 100K resistor there, when you silence it, C1 instantly discharges through the switch, the FF flops, but then C1 re-charges, causing the Schmitt trigger to fire a pulse, turning the buzzer back on. The 100K kinda isolates C1 from the ground at the silence switch .

At this moment I can't explain why I said earlier "Unfortunately it wont turn off the beeper if the incoming alarm condition clears up on its own.."
That does not seem to be happening anymore...and yes I know trying to follow along with me must have been like chasing a fly around the house...lol...
 

Old Steve

Jul 23, 2015
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Excellent - glad you got it sorted out. And no, you're not at all hard to follow compared to some people.
 

Skidood

Aug 24, 2015
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I think I should use a 2 Meg resistor instead of the 100K...if someone holds the silence switch down for a longer period of time, the problem might come back..will try that...
Thanks very much for your help Old Steve..I can tell you are a kind soul .

Analog...looking forward to your trick LED driver strategy..
The good news out of this exercise was that when I went to the electronics store to buy the 4093, I also treated myself to a new meter, my work meter can now stay out in my vehicle.
 

Old Steve

Jul 23, 2015
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Eliminating the diodes does not let him hang LEDs on the inputs for individual fault indications. But, using three PIC input pins eliminates the diodes and keeps the isolation.

ak
Yeah, right, I forgot about the LEDs. :oops:
 

AnalogKid

Jun 10, 2015
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I think the reason is that without the 100K resistor there, when you silence it, C1 instantly discharges through the switch, the FF flops, but then C1 re-charges, causing the Schmitt trigger to fire a pulse, turning the buzzer back on. The 100K kinda isolates C1 from the ground at the silence switch.

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOPS.

The reason for the connection between pin 13 and R4 is the requirement that the beeper shut off when unattended alarms clear themselves. So technically this is all your fault...

100K for the isolation resistor is more than large enough. If you calculate the change in voltage at the R4/C1 node with the reset switch open and closed, you should get around 0.86 V. That is the step that goes through C1 when the reset switch is opened, and it is not enough to drive pins 5 and 6 through the gate's input transition region (which is up around Vcc/2 - 10%).

BUT - placing the RC network after U1B makes the circuit work with a much wider range of values for the isolation resistor.

ak
 
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Old Steve

Jul 23, 2015
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Here is the full size, updated, corrected, CMOS NAND version, expanded with input LEDs.

ak
View attachment 21759
Looking good. Well done. That shouId do the job nicely. :cool:
I'd personally put a protection diode across R5, though, with the cathode connected to Vcc, just to prevent potential damage to U1a from over-voltage when pin 4 of U1b returns high. The C1 / R5 junction will momentarily rise to 12V above the supply voltage until C1 discharges.
Oh, it disappeared. Last-minute edit, I guess? (Just went to click on 'Like', and was told I didn't have permission.)
Now for that sleep.............
 
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Skidood

Aug 24, 2015
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OK today it looks like I need to do something to my desktop PC, it is being a PITA...plus I need to get a haircut....but as soon as I can, I do intend to give Analog's oscillator a try...love the idea of doing this whole alarm module with just one chip .....
 

Skidood

Aug 24, 2015
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OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOPS.

The reason for the connection between pin 13 and R4 is the requirement that the beeper shut off when unattended alarms clear themselves. So technically this is all your fault...

100K for the isolation resistor is more than large enough. If you calculate the change in voltage at the R4/C1 node with the reset switch open and closed, you should get around 0.86 V. That is the step that goes through C1 when the reset switch is opened, and it is not enough to drive pins 5 and 6 through the gate's input transition region (which is up around Vcc/2 - 10%).

BUT - placing the RC network after U1B makes the circuit work with a much wider range of values for the isolation resistor.

ak

Thanks so much Analog !!!....you are a star for sure.....

In the basement bars,
In the backs of cars,
Conform or be cast out....
 

AnalogKid

Jun 10, 2015
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Here is the full size, updated, corrected, CMOS NAND version, expanded with input LEDs and power-on reset.

ak
PressureAlarm-4a-c.gif
 

Attachments

  • PressureAlarm-4a-c.pdf
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AnalogKid

Jun 10, 2015
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Here is the CMOS version converted back to a single ULN2004 for all logic, pulse forming, and output drive.

Exit the warrior.

ak
PressureAlarm-5-c1.gif
 

Attachments

  • PressureAlarm-5-c.pdf
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