Re: need a simple gadget - an AC current detector audible alarm
From: petrus bitbyter (p.kralt_at_reducespamforchello.nl)
Date: 09/24/04
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Date: Fri, 24 Sep 2004 11:04:01 GMT
"John Fields" <jfields@austininstruments.com> schreef in bericht
news:8lp2l0pd4gb9ful29altkn0lkunko9dg6d@4ax.com...
> On 20 Sep 2004 06:17:32 -0700, steve@sgpr.net (Steve Grubb) wrote:
>
>>Hi, I'm looking for something with AC male on one end, and AC female
>>on the other, that can plug an AC device (sump pump) into, that will
>>emit an audible tone when the device is running and consuming current.
>> Kind of like an audible pilot light, or an electric meter that
>>sings.. I want this so I can be aware upstairs when the sump pump runs
>>in the cellar, since occasionally it sticks on. This would also alert
>>us to a plumbing problem. It's a dirt cellar.
>>
>>It seems like would be a simple device, but I don't know what it would
>>be called.
>
>
> ---
> FWB
> +----+
> +-------------|~ +|------+-----+
> | +-----|~ -|--+ |+ |
> | | +----+ | [C] [BUZZER]
> MAINS>---+--[R]--+----+ | | |
> | +---+-----+
> [PUMP]
> |
> MAINS>----------------+
>
> Select the lowest voltage buzzer (beeper, Sonalert, etc.) you can
> find, and choose R so that the current drawn by the pump motor will
> drop the voltage needed by the buzzer plus 1.4V for the diodes in the
> bridge. That is:
>
>
> R = (Vbuzzer + 1.4V) / I motor
>
>
> The capacitor, C, will be:
>
>
> C = IdT/dV
>
> where C = the capacitance in Farads
> I = the steady-state current ion the buzzer
> dt = the period of the rectified waveform
> dv = the permissible ripple voltage across the buzzer
>
> For example, lets say you have a pump which draws 1A and you've chosen
> a 6 volt Sonalert which needs 100mA.
>
> For R,
>
> R = (Vb + 1.4V) / 1A = 7.4/1 = 7.4 ohms.
>
> 7.5 ohms is a standard value, and then, since the resistor will drop
> 7.5V at 1A, it will dissipate 7.5 watts, so you'll need to get
> something like a 15 or 20 watt resistor.
> A 20 watter won't get as hot as a 10 watter, so that would be the way
> to go, IMO. A good idea would be to use a wirewound metal-cased 25
> watt resistor which you could bolt onto the inside of a small metal
> minibox which you could use to house the rest of the components.
> Restricting all of the electrical connections to the inside of the
> enclosure and making sure there were no electrical connections made to
> the [metal] housing itself would go a long way to reducing the danger
> of shock, since you're using the unisolated mains to run the buzzer.
>
>
> For C, assuming you can tolerate 1/2 a volt of ripple:
>
>
> C = Idt / dV = 0.1A*0.0083s/0.5V = .00166µF ~ 1700µF
>
> Since the tolerance on the capacitance of cheap electrolytics is
> often rated at -20 to +80%, you'd need about 2200µF to make sure you
> got 1700µF, and the cap would need to be rated for at least 6V. 12V
> wouldn't hurt.
>
> Pretty much any old bridge would work, or you could make up something
> out of 1N4000s, so your circuit would wind up looking like this:
>
> FWB
> +----+
> +----------------|~ +|-------+-------+
> | +-----|~ -|--+ |+ |
> | | +----+ | [2200] [BUZZER]
> MAINS>---+--[7.5R]--+----+ | | |
> | +----+-------+
> [PUMP]
> |
> MAINS>-------------------+
>
>
> If you wanted to cut down on the dissipation you could substitute a
> current transformer for the 7.5 ohm resistor, but that's another
> story.
>
> Whatever you decide to do be careful, since mains voltage can kill.
>
> --
> John Fields
This solution is simple and cheap. The only drawback is heat. The heat
production can be lowered by using schottky- or germanium diodes in the
rectifier bridge and finding a buzzer that requires a low voltage. The lower
the better.
A low voltage AC buzzer should be even better as you don't need a rectifier
and a capacitor to drive it.
Another possibility is replacing the DC buzzer by a solid state relay (SSR).
The common ones can be driven by a voltage as low as 3V. The switch can
drive a mains powered buzzer. A simple electrical doorbel for instance
(through its transformer of course.) A drawback may be the cost. A simple
SSR will do about $10,--.
As for producing the less heat possible, a current transformer is the best
solution I can imagine. But even if you can find one it will be expensive.
You can make one yourself by disassembling a mains to low voltage
transformer. An old wallwart will contain one. (Not the newer ones. They are
SMPSs.) Remove the mains coil and replace it by some turns of thick wire,
thick enough to handle the current to your pump. The series resistor has to
be replaced by that coil. To find out the number of turns required you will
have to do some experimenting. Start with let's say ten turns and increase
this number until the secundary voltage is high enough to drive the buzzer.
(The voltage/turn will be more or less constant. So if ten turns produce 2V
you will require twentyfive turns to produce 5V.)
The most important and may be most expensive part is the enclosure. To stay
on the safe side you need a solid one. Make sure that any metal part that
can be touched from the outside has been firmly tied to the protective
ground. Otherwise you may have build a killing gadget.
petrus bitbyter
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