need a simple gadget - an AC current detector audible alarm

[Steve Grubb]

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.

Thanks for any ideas. -Steve

 

[CWatters]

"Steve Grubb" <steve@sgpr.net> wrote in message
news:5edb58d2.0409200517.64bd7563@posting.google.com...

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.

Perhaps you could just wire a mains lamp or a mains powered radio across the
pump motor (tune radio to 24 hour station!)

 

[ThePetPage]

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.

What you are describing is called a "current transformer". It uses a toroid
coil with only one of the ac mains wire going thru it to detect current flow.
The toroid has maybe 30-40 turns of wire wrapped around it which acts like a
"one-turn transformer". That is fed to an op-amp, then a rectifier and filter,
then could feed a sonalert.

But I agree with the person that replied before me. If you don't actually
require that you sense that CURRENT is flowing, (in other words, the pump is
turning), you could just wire a parallel connection so that when the pump has
power, so does a radio or a buzzer (with a transformer), etc.

 

[Watson A.Name - \"Watt Su]

"Steve Grubb" <steve@sgpr.net> wrote in message
news:5edb58d2.0409200517.64bd7563@posting.google.com...

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.

Thanks for any ideas. -Steve

I would use a big, heavy full-wave bridge rectifier, something rated
for more amps than the sump pump will draw. I've seen 25A bridge
rectifiers available for several dollar each. Or I might try a bridge
rectifier out of an old scrapped AT power supply if it's rated at 6A or
more. Depends on the pump.

Then I'd get a pigtail with a plug on it and a socket and insert the
bridge's two "~" AC leads in series with the hot AC wire between the
plug and socket.

Then I'd put a 1 ohm, ten watt resistor across the bridge's + and -
leads. With the pump plugged into it, I'd measure the voltage drop
across the resistor. Be Careful around AC, it can be deadly! The idea
here is to get the voltage drop across the resistor to be enough to make
a buzzer sound. It might take only a 1 ohm resistor if the pump draws a
few amps, giving a few volts drop across the resistor. Or if the pump
is lower current, then it might take 2 or 5 ohms. But the resistor has
to be a power resistor to handle the current. With a 1 ohm resistor, 3
volts drop times 3 amps gives 9 watts of dissipation, so the resistor
will get hot. Of course, a piezo buzzer goes across the resistor, in
the proper polarity.

Make sure everything is well insulated, since the components have the AC
line voltage on them. I would put all of it inside a metal electrical
outlet box, wide enough for both an outlet and a switch. Instead of the
switch, leave the hole for the switch handle empty and mount the buzzer
so that it's below and the sound can come out the hole.

Remember all this has to reliable enough to never prevent the pump from
working. It could be messy if it doesn't do its job.

 

[Watson A.Name - \"Watt Su]

"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:lqC3d.252087$Jo5.13009952@phobos.telenet-ops.be...

"Steve Grubb" <steve@sgpr.net> wrote in message
news:5edb58d2.0409200517.64bd7563@posting.google.com...
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.

Perhaps you could just wire a mains lamp or a mains powered radio
across the
pump motor (tune radio to 24 hour station!)

Problem is that the AC line going to the motor is usually hot all the
time. The soundmaker would have to be wired _after_ the float switch,
which in the sump pumps I've seen, isn't easily accessible because it
has to be insulated from the water.

 

[CWatters]

"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com> wrote
in message > Problem is that

the AC line going to the motor is usually hot all the
time. The soundmaker would have to be wired _after_ the float switch,
which in the sump pumps I've seen, isn't easily accessible because it
has to be insulated from the water.

So you really need to detect the change in current when the motor switches
on rather then the voltage (which is constant).

 

[CWatters]

"Steve Grubb" <steve@sgpr.net> wrote in message
news:5edb58d2.0409200517.64bd7563@posting.google.com...

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.

I saw a gadget advertised in a PC mag recently... It looked like a 4 way
extension block with one special socket into which you plug your PC. It
detected when you switched off your PC (by measuring the current drop) and
then 10 seconds later it automatically switched off power to the other
devices. Perhaps this would do the job.

 

[Bill Bowden]

steve@sgpr.net (Steve Grubb) wrote in message news:<5edb58d2.0409200517.64bd7563@posting.google.com>...

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.

Thanks for any ideas. -Steve

Well, you could use a piezo buzzer and a
few other parts. One resistor, one diode and
one capacitor. The piezo buzzer should be
the type that operates on a low DC voltage
and doesn't require a driver.

Just wire the circuit in parallel with the AC
connections to the pump. You can adjust the 16K
resistor for more or less volume, but don't take
it too low or the thing might fry. The diode needs
to be a 200 volt variety or greater for 120 AC
use. A 1N4003 or 1N4004 should work.


16K Diode
Pump ---\/\/\------->|------+-------+
AC + | |
--- Piezo buzzer
10uF --- |
| |
| |
Pump -----------------------+-------+
AC


-Bill

 

[CWatters]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ad025737.0409212101.32394a31@posting.google.com...

Just wire the circuit in parallel with the AC
connections to the pump.

He said that wasn't possible because the float switch and motor connections
are sealed inside the pump. He only has access to power going in so he has
to measure the current somehow.

 

[John Fields]

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 = .00166ľF ~ 1700ľF

Since the tolerance on the capacitance of cheap electrolytics is
often rated at -20 to +80%, you'd need about 2200ľF to make sure you
got 1700ľF, 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

 

[Tweetldee]

"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:rl_3d.253256$Jo5.13009952@phobos.telenet-ops.be...

"Steve Grubb" <steve@sgpr.net> wrote in message
news:5edb58d2.0409200517.64bd7563@posting.google.com...
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.

I saw a gadget advertised in a PC mag recently... It looked like a 4 way
extension block with one special socket into which you plug your PC. It
detected when you switched off your PC (by measuring the current drop) and
then 10 seconds later it automatically switched off power to the other
devices. Perhaps this would do the job.

Take a look at the circuit at
http://ourworld.compuserve.com/homepages/Bill_Bowden/page8.htm#aclatch.gif,
"AC Line Current Detector". You can build it with common components.
Should do your task nicely.

--
Dave M
MasonDG44 at comcast dot net (Just subsitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!

 

[Watson A.Name - \"Watt Su]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ad025737.0409212101.32394a31@posting.google.com...

steve@sgpr.net (Steve Grubb) wrote in message
news:<5edb58d2.0409200517.64bd7563@posting.google.com>...
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.

Thanks for any ideas. -Steve

Well, you could use a piezo buzzer and a
few other parts. One resistor, one diode and
one capacitor. The piezo buzzer should be
the type that operates on a low DC voltage
and doesn't require a driver.

Just wire the circuit in parallel with the AC
connections to the pump. You can adjust the 16K
resistor for more or less volume, but don't take
it too low or the thing might fry. The diode needs
to be a 200 volt variety or greater for 120 AC
use. A 1N4003 or 1N4004 should work.


16K Diode
Pump ---\/\/\------->|------+-------+
AC + | |
--- Piezo buzzer
10uF --- |
| |
| |
Pump -----------------------+-------+
AC

You omitted one very important specification. The 16k resistor must be
able to handle the power, and IMNSHO, it should be able to handle the
amount of power the whole AC line would put on it. In this case that's
120VAC * 120VAC / 16000, which is 0.9 watts. So call it an even watt.
And it should be flameproof.

> -Bill

 

[petrus bitbyter]

"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 = .00166ľF ~ 1700ľF

Since the tolerance on the capacitance of cheap electrolytics is
often rated at -20 to +80%, you'd need about 2200ľF to make sure you
got 1700ľF, 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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