B
Bob Engelhardt
Guest
On 6/6/2020 3:35 PM, Pimpom wrote:
I like it. KISS. A small point: only need one of the D3/D4 pair.
I like it. KISS. A small point: only need one of the D3/D4 pair.
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S
Steve Wolf
Guest
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.
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.
S
Steve Wolf
Guest
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.
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.
D
default
Guest
On Sat, 6 Jun 2020 06:17:13 -0700 (PDT), Steve Wolf
<stevwolf58@gmail.com> wrote:
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.
<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.
A
Arie de Muynck
Guest
On 2020-06-07 14:07, Bob Engelhardt wrote:
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
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.
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
Arie de Muynck
Guest
On 2020-06-07 14:07, Bob Engelhardt wrote:
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
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.
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
Arie de Muynck
Guest
On 2020-06-07 14:07, Bob Engelhardt wrote:
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
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.
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
D
default
Guest
On Sun, 7 Jun 2020 06:59:21 -0500, amdx <nojunk@knology.net> wrote:
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.
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 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.
D
default
Guest
On Sun, 7 Jun 2020 06:59:21 -0500, amdx <nojunk@knology.net> wrote:
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.
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 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.
D
default
Guest
On Sun, 7 Jun 2020 06:59:21 -0500, amdx <nojunk@knology.net> wrote:
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.
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 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.
D
default
Guest
On Sun, 7 Jun 2020 06:18:45 -0700 (PDT), Steve Wolf
<stevwolf58@gmail.com> wrote:
No.
3 amps is the rating for the diodes, 10 would fry it
<stevwolf58@gmail.com> wrote:
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.
No.
3 amps is the rating for the diodes, 10 would fry it
D
default
Guest
On Sun, 7 Jun 2020 06:18:45 -0700 (PDT), Steve Wolf
<stevwolf58@gmail.com> wrote:
No.
3 amps is the rating for the diodes, 10 would fry it
<stevwolf58@gmail.com> wrote:
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.
No.
3 amps is the rating for the diodes, 10 would fry it
D
default
Guest
On Sun, 7 Jun 2020 06:18:45 -0700 (PDT), Steve Wolf
<stevwolf58@gmail.com> wrote:
No.
3 amps is the rating for the diodes, 10 would fry it
<stevwolf58@gmail.com> wrote:
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.
No.
3 amps is the rating for the diodes, 10 would fry it
R
Ralph Mowery
Guest
In article <0Z%CG.115934$vu3.48753@fx21.ams1>, nobody@nowhere.com
says...
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.
says...
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.
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.
R
Ralph Mowery
Guest
In article <0Z%CG.115934$vu3.48753@fx21.ams1>, nobody@nowhere.com
says...
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.
says...
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.
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.
R
Ralph Mowery
Guest
In article <0Z%CG.115934$vu3.48753@fx21.ams1>, nobody@nowhere.com
says...
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.
says...
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.
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.
B
Bob Engelhardt
Guest
On 6/7/2020 9:23 AM, Arie de Muynck wrote:
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?
On 2020-06-07 14:07, Bob Engelhardt wrote:
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.
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
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?
B
Bob Engelhardt
Guest
On 6/7/2020 9:23 AM, Arie de Muynck wrote:
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?
On 2020-06-07 14:07, Bob Engelhardt wrote:
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.
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
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?
B
Bob Engelhardt
Guest
On 6/7/2020 9:23 AM, Arie de Muynck wrote:
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?
On 2020-06-07 14:07, Bob Engelhardt wrote:
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.
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
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?
A
Arie de Muynck
Guest
On 2020-06-07 18:02, Bob Engelhardt wrote:
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
On 6/7/2020 9:23 AM, Arie de Muynck wrote:
On 2020-06-07 14:07, Bob Engelhardt wrote:
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.
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
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