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Power in use indicator please ?

Started by Steve Wolf June 6, 2020
I have a water pump in a remote location. Sometimes the pump (120 volts) line springs a leak and then runs all the time with the potential of burning out pump. I want to build a little circuit that will tell me in my house if it is running. I have in the past build and induction coil out of a transformer and then used an LED to light up when the pump is on. I cut out one of the sides of the transformer then run a few winds of the power wire around the transformer that I cut the secondary wires from. It produces enough power to light an led. 

However I would like to try another method. I have two ideas but I need a circuit diagram and instructions to do it.

One person mentioned to me the use of a Shunt. I have read up on shunts, and as far as I can see there is some potential here. Does anyone have a diagram that might be able to use a shunt. To light either an LED, or Neon bulb, or even Incandecent lamp.


Remember the point is not that there is power in the line, but it is to tell me that the pump is actually drawing current.


Regards
On 6/6/2020 6:47 PM, Steve Wolf wrote:
> I have a water pump in a remote location. Sometimes the pump (120 volts) line springs a leak and then runs all the time with the potential of burning out pump. I want to build a little circuit that will tell me in my house if it is running. I have in the past build and induction coil out of a transformer and then used an LED to light up when the pump is on. I cut out one of the sides of the transformer then run a few winds of the power wire around the transformer that I cut the secondary wires from. It produces enough power to light an led. > > However I would like to try another method. I have two ideas but I need a circuit diagram and instructions to do it. > > One person mentioned to me the use of a Shunt. I have read up on shunts, and as far as I can see there is some potential here. Does anyone have a diagram that might be able to use a shunt. To light either an LED, or Neon bulb, or even Incandecent lamp. > > > Remember the point is not that there is power in the line, but it is to tell me that the pump is actually drawing current. > >
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.
> 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
On 6/7/2020 12:23 AM, Steve Wolf wrote:
> >> 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: https://www.dropbox.com/s/ablk0cc0iqfhsta/Load%20presence%20indicator.png?dl=0 Diodes D1-D4 must be able to handle the load, including startup current.
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.
In article <ecfa1684-cdf7-44d8-a667-3dd414da5dfao@googlegroups.com>, 
stevwolf58@gmail.com says...
> > 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. https://www.ebay.com/itm/LED-Digital-Display-Voltmeter-Ammeter-Voltage- Current-Frequency-Tester-Meter/123992634876?_trkparms=aid%3D1110006% 26algo%3DHOMESPLICE.SIM%26ao%3D1%26asc%3D20200520130048%26meid% 3D8dddf9fc793546e39cd00eba869b11b6%26pid%3D100005%26rk%3D1%26rkt%3D12% 26mehot%3Dco%26sd%3D293572031701%26itm%3D123992634876%26pmt%3D1%26noa% 3D0%26pg%3D2047675%26algv%3DSimplAMLv5PairwiseWebWithBBEV2bDemotion% 26brand%3DUnbranded&_trksid=p2047675.c100005.m1851
On Sat, 6 Jun 2020 06:17:13 -0700 (PDT), Steve Wolf
<stevwolf58@gmail.com> wrote:

>I have a water pump in a remote location. Sometimes the pump (120 volts) line springs a leak and then runs all the time with the potential of burning out pump. I want to build a little circuit that will tell me in my house if it is running. I have in the past build and induction coil out of a transformer and then used an LED to light up when the pump is on. I cut out one of the sides of the transformer then run a few winds of the power wire around the transformer that I cut the secondary wires from. It produces enough power to light an led. > >However I would like to try another method. I have two ideas but I need a circuit diagram and instructions to do it. > >One person mentioned to me the use of a Shunt. I have read up on shunts, and as far as I can see there is some potential here. Does anyone have a diagram that might be able to use a shunt. To light either an LED, or Neon bulb, or even Incandecent lamp. > > >Remember the point is not that there is power in the line, but it is to tell me that the pump is actually drawing current. > > >Regards
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.
On Saturday, June 6, 2020 at 6:17:17 AM UTC-7, Steve Wolf wrote:
> I have a water pump in a remote location. Sometimes the pump (120 volts) line springs a leak and then runs all the time...
> Remember the point is not that there is power in the line, but it is to tell me that the pump is actually drawing current.
Get a Kill-a-Watt power monitor, and plug the pump in through it. You'll get LOTS of information. <https://www.amazon.com/dp/B00009MDBU?tag=duckduckgo-d-20&linkCode=ogi&th=1&psc=1> 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.
On 6/7/2020 3:06 AM, Steve Wolf wrote:
> 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. >
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.
On 6/7/2020 12:09 PM, Pimpom wrote:
> On 6/7/2020 3:06 AM, Steve Wolf wrote: >> 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. >> > > 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.