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Power in use indicator please ?...

S

Steve Wolf

Guest
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
 
S

Steve Wolf

Guest
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.
 
S

Steve Wolf

Guest
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.
 
S

Steve Wolf

Guest
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.
 
J

Jasen Betts

Guest
On 2020-06-06, 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.
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.

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.
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: https://www.amazon.com/G/dp/B078HXZG3H


--
Jasen.
 
J

Jasen Betts

Guest
On 2020-06-06, 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.
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.

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.
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: https://www.amazon.com/G/dp/B078HXZG3H


--
Jasen.
 
R

Ralph Mowery

Guest
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
 
R

Ralph Mowery

Guest
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
 
J

Jasen Betts

Guest
On 2020-06-06, 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.
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.

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.
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: https://www.amazon.com/G/dp/B078HXZG3H


--
Jasen.
 
D

default

Guest
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.
 
D

default

Guest
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.
 
W

whit3rd

Guest
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.
 
W

whit3rd

Guest
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.
 
P

Pimpom

Guest
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.
 
P

Pimpom

Guest
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.
 
P

Pimpom

Guest
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.
 
P

Pimpom

Guest
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.
 
A

amdx

Guest
On 6/6/2020 10:51 PM, whit3rd wrote:
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.
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
 
A

amdx

Guest
On 6/6/2020 10:51 PM, whit3rd wrote:
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.
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
 
B

Bob Engelhardt

Guest
On 6/6/2020 3:35 PM, Pimpom wrote:
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.
I like it. KISS. A small point: only need one of the D3/D4 pair.
 
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