Power in use indicator please ?

What are the nameplate ratings of the pump motor? Remember that working with mains voltages is very dangerous and not to be encouraged unless you know exactly what you're doing.

Unfortunately I'm not local to it. But it is a Jet pump for pumping household water from lake to house. Its 1/2 horse 120vac. Based on other similar pumps draws about 4-6amps.

Reasonably common enough that you can get at any hardware. Regards

This should work:

Diodes D1-D4 must be able to handle the load, including startup current.

Your home-built current transformer (above) seems like the ideal solution. just connect the output directo to a bi-directional LED or to whatever amplifier you want.

A shunt produces only a small signal. it needs some sort of amplifier to get enough to light a LED, so that means that you need to build an amplifier and find a power source to run it, and it's all mains-live so dangerous to mess with.

You can get power strips with a current detecting circuit built in that's capabile of turning on any second appliance (like a lamp)

eg:

That's great . Can you explain it a bit. I'd like to understand it somewhat . You have created what appears to me to be a rectifier but on one line and then bypassed that with led.

Regards.

Each diode will have a voltage drop of about .6 volts. So you have a voltage of around 1.2 volts. The LED needs about that voltage to light up. Depending on the LED it may take 3 diodes in each direction.

The 100 ohm resistor limits the current to the LED. LEDs are mostly current devices and you have to limit the current through them.

The diodes have to be in each direction for an AC motor so the AC will pass in both directions.

I had though of a similar circuit but using a very low value resistor to give a voltage drop of about 2 or 3 volts. Then it would take a regular low current diode and resistor to limit the current and change the AC to DC so the LED would work unless you get a special LED that works in both directions.. One minor problem might be the motor starting up and drawing too much current for the resistor/LED combination.

Other than the above diode solution, a current meter may be the best way to go. Ebay has some from China for $ 5 to $ 15 shipped. Hardly worth trying to roll your own. You can see how much current is being drawn incase the motor just slightly overloads at some point.

I asked pretty much the same question awhile ago regarding DC power.

If the shut provides enough voltage drop to light an LED directly, it will use power itself. If the pump draws 10 amps, and you need 2.5V to light a red led, that's 25 watts of heat in the shunt, and 2.5 volts the pump isn't getting.

That would be inefficient and may be costly.

Your home-made current transformer is a better option IMO. I'm using that on my water heater to know when it has stopped heating.

Alternatively you can take an AC relay and turn it into a current relay. Cut away the coil. First it would be a nice idea to take the relay specs, and see how many turns of wire it takes at say, it's rated 12VAC coil (for illustration only) then calculate backwards to see how many ampere turns that is then wind a new coil using much thicker wire. You don't need to count the turns, you can come close enough by knowing how many turns of whatever size wire will fit on the bobbin.

My application was for DC current and as someone suggested wind a few turns of wire around a reed switch and use that to switch a LED. Works like a champ... I'm using it in an electronic circuit breaker to switch a larger relay and kill power when the current is exceeded, and in another application to sense when some cooling fans are running.

Get a Kill-a-Watt power monitor, and plug the pump in through it. You'll get LOTS of information.

Actually adding something at the power panel, though, is tricky; nominally, your current transformer solutions could go inside the breaker box, if they had the right safety stamps, but otherwise your insurance won't like the idea.

Ralph Mowery's explanation is correct, except that when there's a considerable load on a silicon diode, the voltage drop is more than the oft-quoted 0.6V. It's more like 1V and two of them is about right to light up a red or amber LED. It won't work well with green or white LEDs.

The 1V or so drop is not constant. It varies with the AC wave, so will the instantaneous brightness at any given moment. What your eyes see is the average.

The average current, and therefore the apparent brightness, will also vary with the load current - from a fraction of a milliamp to some mAs.

BTW, you can add several more LEDs in parallel with the first for better visibility if you like, preferably each with its own series resistor. You can also add them in the opposite direction.

I'd like to see someone get fancy and develop a circuit that starts a beeper after the pump has been running an adjustable amount of time. Even put a reset button on it, so if you know it's running for a reason, you can reset it for a second or third time period. Beep, beep, beep, beep. (556)

Mikek

I like it. KISS. A small point: only need one of the D3/D4 pair.

If I wanted to find out the amperage of a Diode for 120 volts, would the following be true.

I happen to have a package of Silicon Rectifier Diodes. 1N5404TR. It says on it 400v 3A.

Would this be 1200 Watts.

then at 120 Volts it would be 10 Amps.

Regards.

That would produce a small amount of DC in the pump voltage. Shaded pole motors are very sensitive to that - with some DC you can easily use them as brake. Better "Keep It Symmetrical [and] Safer".

Arie

A simple RC network charging a cap that switches a mosfet connected to a piezo buzzer is all that takes. When the charge on the cap reaches the threshold voltage of the fet ~2-4V buzzer sounds, reset by putting a pushbutton across the cap to discharge it fast.

No.

Yes, the drop depends on current drawn. I was just taking the simple short cut explination.

I think I mentioned it may take another diode or two in each direction to get enough voltage for some LEDs.

The diodes may need to be 20 or 30 amp diodes in this case as the pump may draw a lot more than the 6 amps that the motor may draw in normal usage as it starts up. Diodes will handle a lot more current than the rating for a few cycles of the AC wave, but I don't know if they will take it long enough for the motor to reach speed.

Then the diodes will need some heat sinking to handle the load.

I still like may 5 to 10 buck China solution.

There was a question about some 3 amp 400 volt diodes and 1200 watts. That is not the way a diode rates. The 3 amps is the maximum current (not counting the ability to handle a lot more for a few AC cycles) at any voltage, and 400 volts is the maximum voltage at any current .

So at 120 volts you still only have a max of 3 amps for the load. Even if the load was 6 volts ,you could only draw 3 amps through the diode before it is destroied.

That's because the AC voltage in one direction would be a diode drop different from the other direction? So the dc voltage would be equal to the diode drop, or is it more complicated than that?

That's more or less it. Divide that voltage drop by the total DC resistance in the circuit (assume the mains is zero ohms), and you know the DC current flowing. Simulate that with a DC supply connected to the motor, rotate the motor manually, and check how much breaking action occurs. I've been amazed at it (and used it as a controllable motor torque load).

Arie