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Puggy Potty Alarm


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Ok, so far, so good (excellent actually  ;D)....

Now, how about an auto reset on the trigger? I imagine I could accomplish it myself using an opamp/comparator, but can it be done using fewer parts such as a transistor in combination with another cap and pullup resistor?

Something which'll reset the trigger once the voltage across the two contacts returns to 0v or possibly once the resistance becomes great enough for it not to trigger the alarm.

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Hi Alun,
Your favouite "rail-to-rail" Cmos opamp has some serious limitations with its output current. Like all Cmos ICs, its output current capability reduces as its supply voltage is less.

In this circuit, it is driving the high capacitance (maybe 15nF) of a piezo transducer which is a fairly heavy load (2.7K ohms) at 4KHz. Since its drive is bridged, the 2.7K appears like 1.35K. With only a 9V supply, A CA3130 won't work very well.

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Of, if op-amps aren't good enough try maiking your own inverting complementary pair.

Tr2 2N2905, BC550, BC549 or any similar PNP transistor.
Tr1 & Tr3 2N2222, BC547, BC337 or any similar NPN transistor.

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Hi CRE,
So you had a shorted capacitor and now a fried IC. Maybe they were damaged when you powered the circuit with two batteries in series.

The circuit by Alun provides automatic reset. The 1st 555 is a few seconds timer. It is triggered when the contacts have a current path from wetness. They can remain conducting and the 1st 555 will time-out by itself because the trigger input is capacitor-coupled.

A larger bypass cap can be added in addition to the recommended 0.1uF ceramic disc one to extend the battery's life since the supply voltage won't fluctuate so low during the project's sound output. I always use at least 100uF.

Alun's latest post showing a 3-transistor circuit instead of an opamp will also extend the battery's life, and give the loudest possible output when using only a 9V battery.

A 15V supply made from a 9V battery plus four AAAA, AAA or AA battery cells will make your project much louder.  ;D 

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A 15V supply made from a 9V battery plus four AAAA, AAA or AA battery cells will make your project much louder.  ;D 


I hope you're joking, I don't think this is a good idea because the AAA cells are a different capacity to the 9V battery, and this would create a very large cell inbalence.

So you really want a very loud sound do you?
This will give you the most power from 9V so much so it might destroy the piezo element, I recommend running this of a 6V lantern battery or even 3 1.5V D cells in series to give 4.5V.

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Hi Alun,
A 9V battery has six AAAA cells inside. Additional AAA cells would just last twice as long as the 9V battery. When the sound level begins to drop too low, only replace the 9V battery and replace the AAA cells the next time.

Your idea of using a stepup transformer and only a 4.5V battery would strain the 555 to its limit of output current and therefore it won't produce much output voltage. The 8 ohm to 1K ohm transformer would reflect the 2.7K piezo back to the 555 as a 21.6 ohm resistor. A 555 driving a 21.6 ohm resistor with only a 4.5V supply would have a typical output voltage of only about 1.2Vp-p and supply a peak current of about 55mA. The 1.2V stepped-up by the transformer is only 13.4Vp-p, less than the 15Vp-p if a bridged amp circuit is used with a 9V battery. A Cmos 555 would be worse because it can't deliver much output current at low supply voltages.  ;D 

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The 555 sees 470 plus a 0.7V drop for the transistor, so its output current will only be (4.5-0.7)/470 = 8.1mA.

But you're right about the  output voltage being lower than expected but this is due to the saturation voltage of each transistor not the 555 as they are connected as emmiter followers. Perhaps this circuit will be able to be run of 9V with no problems.

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This might not be quit so good!

The maximum peek voltage for a piezo element is typically 30V.

An 8ohm to 1K audio transformer has a turns ratio of 1:125 so the piezo could be damaged with input voltages as low as 0.25V!

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Hi Alun,
The 555 has a fair amout of voltage loss in its output because it uses cascaded emitter-followers to obtain a high output current rating. I forgot to include your complimentary pair of emitter-followers in my previous calculation, so the total output voltage loss is about 4 X 0.7V = 2.8V. Therefore with a 4.5V supply, the transformer will get only 1.7Vp-p.

The turns ratio of a transformer (and also its voltage step-up ratio) is the square-root of its impedance ratio!
So the 1.7Vp-p is stepped-up to about 19Vp-p.  ;D

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The turns ratio of a transformer (and also its voltage step-up ratio) is the square-root of its impedance ratio!


Oh I remember now from college, I must've been half asleap in that lesson! ;D

I would work well from 9V, if the maximum peak voltage the piezo can stand is 30V then then would be the peak-peak output would be 38V or 19V peak so it would be perfect.
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:-\ whoa..... umm, yeah what that other guy said.  :o


Err, ok, well backing up a few posts, I'm sure the two 9v's in serial didn't fry the cap as I hadn't even used them when that prob occured. It may have fried one of the IC's granted, but as I said, I tested the voltages and they were putting out only about 15v combined, but admittedly I tested them without a load. I think it's more likely that I miplaced a jumper.... I tend to do that.... fortunately enough I only tend to do it on the first circuit and not any to follow.  :P

I'm gonna go through and trace the circuit tomorrow.... hopefully it is just a misplaced lead and not a fried IC or two. I am interested in the complementary pair suggestion though, if the output is higher it should be exactly what I'm looking for.

One of these days I need to either find out how to read and understand the values of transformers that I've found at suppliers'..... or learn how to start wrapping them myself.

At any rate thanks! and I'll be sure to post once I get this initial one debugged.

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