Fuse for an AC motor

[Uriah]

You CAN protect them but a regular fuse is unlikely to help. What sort of
HP/amps are they?

Here is what I found out. When I wire them up the draw .37 amps.
This is with them not being installed in the machine. When I stop the
motor with a pair of vise-grips it draws .49 amps. So I have about .
10 of an amp to deal with. But if I put them in the machine the are
probably going to drag it down a little. So that might cut my margin
to less then .10 amps. Maybe .07 Amps. I think that this is too
little of a difference to be able to catch with a fuse? I would have
to find a really acurate fuse that blows at exactly .45 amps. I think
it will have to be a Rockwell CB fuse but will it work with something
with this small of a margin? As far as the thermal I guess I will
have to let the motor run in the machine check the temp and then add 5
degrees? Is that the right idea?

 

[Homer J Simpson]

Here is what I found out. When I wire them up the draw .37 amps.
This is with them not being installed in the machine. When I stop the
motor with a pair of vise-grips it draws .49 amps. So I have about .
10 of an amp to deal with. But if I put them in the machine the are
probably going to drag it down a little. So that might cut my margin
to less then .10 amps. Maybe .07 Amps. I think that this is too
little of a difference to be able to catch with a fuse?

Yes, that's why a fuse is no good. You really need a proper thermal overload
for the motor. It's possible to wire up a three phase motor starter to run a
single phase motor but that's expensive. You should be OK with something
like this http://www.sensata.com/products/controls/motor-phenolics.htm if
you can enclose it safely or
http://www.geindustrial.com/cwc/products?pnlid=3&famid=35&catid=110&id=s-fvman&lang=en_US
for a packaged product.

 

[ehsjr]

Here is what I found out. When I wire them up the draw .37 amps.
This is with them not being installed in the machine. When I stop the
motor with a pair of vise-grips it draws .49 amps. So I have about .
10 of an amp to deal with. But if I put them in the machine the are
probably going to drag it down a little. So that might cut my margin
to less then .10 amps. Maybe .07 Amps. I think that this is too
little of a difference to be able to catch with a fuse? I would have
to find a really acurate fuse that blows at exactly .45 amps. I think
it will have to be a Rockwell CB fuse but will it work with something
with this small of a margin? As far as the thermal I guess I will
have to let the motor run in the machine check the temp and then add 5
degrees? Is that the right idea?

You can use a 6 ohm 5 watt resistor in series with the
motor, to create a 2.22 volt drop under the normal
condition, and a 2.94 volt drop when jammed. Rectify,
filter and feed that into a 4N35 opto coupler circuit
that disconnects power to the DPDT motor relay until
manually reset. Something like this for the opto sensing
portion:

|> | D1
AC ---o o---[Motor]---+--->|---+---+
RLY1-2 | | | Pot = 500 ohms
| | P
[6R] [47uF] O<---[LED]---+
| | T |
| D2 | | |
AC --------------------+---|<---+---+---[33R]----+

The LED is the input side of the opto coupler.
Adjust the 500 ohm pot so the relay drops out when the
motor slows way down but is not yet stalled.

For the relay portion:

+ 12 ----+--------+-----+----+
| | | |
[1K] [10K] [RLY] [D5]
| | | |a
| | +----+
| | |
| D3 | /c
+---|<---+----| Q1 NPN
| \e
| |
| +-----+
/c | |
opto o-- o
\e RLY1-1 o-^ |= PB
| | o
| D4 | |
Gnd ----+-----|<--------+-----+

When you press the pushbutton, current flows
through Q1, which is biased on by the 10K resistor.
This energizes the relay and transfers the contacts,
which starts the motor through RLY1-2. The RLY1-1
contact maintains the Q1 emitter path to ground,
keeping the relay energized when you release the
button.

When the motor stalls, sufficient voltage is
developed across the 6 ohm resistor to charge
the capacitor to ~2.5 volts. That lights the
LED portion of the optocoupler, which causes
the opto transistor to conduct through the 1K
resistor. That biases Q1 off, and the relay
drops out, stopping the motor. Since that opens
the emitter path to Q1, it cannot energize the
relay again until the pushbutton is pressed.

With the low current requirement, most any
DPDT (Double Pole Double Throw) relay rated
for a 12 volt coil and 120 VAC contacts will
work. You can use 1N4004 diodes and any NPN
transistor. The 6 ohm resistor is made with
a 1 ohm and a 5 ohm resistor in series.
6 ohms resistance will disspiate about 1.5
watts, so use 2 watt or higher resistors.

Ed

 

[Chris]

Here is what I found out. When I wire them up the draw .37 amps.
This is with them not being installed in the machine. When I stop the
motor with a pair of vise-grips it draws .49 amps. So I have about .
10 of an amp to deal with. But if I put them in the machine the are
probably going to drag it down a little. So that might cut my margin
to less then .10 amps. Maybe .07 Amps. I think that this is too
little of a difference to be able to catch with a fuse? I would have
to find a really acurate fuse that blows at exactly .45 amps. I think
it will have to be a Rockwell CB fuse but will it work with something
with this small of a margin? As far as the thermal I guess I will
have to let the motor run in the machine check the temp and then add 5
degrees? Is that the right idea?

Hi, Uriah. It does sound tough. For very small AC motors, this can
be a difficult problem to solve with either a fuse or a circuit
breaker. Have you considered a mechanical aid like a breakaway
clutch? In the event of excessive torque, the cluch disengages from
the load, saving the motor.

Cheers
Chris

 

[Homer J Simpson]

You can use a 6 ohm 5 watt resistor in series with the
motor, to create a 2.22 volt drop under the normal
condition, and a 2.94 volt drop when jammed. Rectify,
filter and feed that into a 4N35 opto coupler circuit
that disconnects power to the DPDT motor relay until
manually reset.

Bad idea. Have you analysed this circuit for each component failure and each
combination of component failures?

http://www.geindustrial.com/cwc/products?pnlid=3&famid=35&catid=110&id=s-fvman&lang=en_US

is a better idea - they are a well known and well understood component and
any electrician can maintain them.

 

[ehsjr]

Bad idea. Have you analysed this circuit for each component failure and each
combination of component failures?

Sure. If the resistor opens the cap, pot and opto
die. So he's out $1.86, Mouser prices. If you count the
resistors, they are 39 cents and 35 cents, bringing
the total replacement cost to less than 3 dollars.
And what is the probability either resistors opening?
They are running at less than 35% of rating for the
5 ohm resistor, and less than 25% of rating for the
1 ohm resistor. Failures in the other parts would
render the jam protection ineffective, or the motor
inoperable if the relay failed or the circuit
failed to energize it.

So suppose you tell us why it is a bad idea.

http://www.geindustrial.com/cwc/products?pnlid=3&famid=35&catid=110&id=s-fvman&lang=en_US

is a better idea - they are a well known and well understood component and
any electrician can maintain them.

But will it continue running the motor at .37 to .45
amps, and disconnect it when current rises above
that? Nope. The CR101 is rated at 16 amps. This motor
draws .37 amps, nominal. And what happens if it fails?
The same scenario as above, either the motor won't run,
or jam detection won't work. Except in your device, jam
detection won't work in the first place. So it's a
useless expenditure of money - and the expense will be
relatively huge compared to proposed circuit. Parts
for the circuit I proposed cost about $12.00, not
including the junction box, receptacle, miscellaneous
hardware. What does the device you propose cost?

Then you can throw in the cost of an electrician, which
you imply is a good idea.

Ed

 

[ehsjr]

Sure. If the resistor opens the cap, pot and opto
die. So he's out $1.86, Mouser prices. If you count the
resistors, they are 39 cents and 35 cents, bringing
the total replacement cost to less than 3 dollars.
And what is the probability either resistors opening?
They are running at less than 35% of rating for the
5 ohm resistor, and less than 25% of rating for the
1 ohm resistor. Failures in the other parts would
render the jam protection ineffective, or the motor
inoperable if the relay failed or the circuit
failed to energize it.

So suppose you tell us why it is a bad idea.




But will it continue running the motor at .37 to .45
amps, and disconnect it when current rises above
that? Nope. The CR101 is rated at 16 amps. This motor
draws .37 amps, nominal.

That creates the wrong impression. The contacts are
rated 16 amps, but the tripping mechanism can be rated
as low as .48 full load amps. Still won't work - you'd
need to draw 125% of that to trip it, and the max this
motor draws ia .49 amps, locked rotor.

Ed

 

[Homer J Simpson]

Sure. If the resistor opens the cap, pot and opto ....
failed to energize it.

So suppose you tell us why it is a bad idea.

What happens if that motor fails? What are the safety implications? Are
there fire or injury risks?

But will it continue running the motor at .37 to .45
amps, and disconnect it when current rises above
that? Nope. The CR101 is rated at 16 amps. This motor
draws .37 amps, nominal. And what happens if it fails?

You fail to understand this device. As part of the installation, a small,
carefully chosen, heater element is installed which automatically opens the
switch if that motor has a sustained overload.

16 Amps is the maximum rating only.

The same scenario as above, either the motor won't run,
or jam detection won't work. Except in your device, jam
detection won't work in the first place. So it's a
useless expenditure of money - and the expense will be
relatively huge compared to proposed circuit. Parts
for the circuit I proposed cost about $12.00, not
including the junction box, receptacle, miscellaneous
hardware.

Plus $400 or more for development and testing. Plus $1,000 to get it passed
by the inspectors (assuming it ever is). Plus $200 or more for repair on
each failure since no electrician will touch it.

What does the device you propose cost?

List Price: $38.00 UL/CSA listed.

 

[ehsjr]

What happens if that motor fails? What are the safety implications? Are
there fire or injury risks?

It's in the first post. If the motor draws
more current than the set point, the overload
circuit I proposed disconnects power to it, and
must be manually reset before the motor can be
started again.

Stating it for the second time: tell us why the
circuit is a bad idea. Support your contention -
maybe you have something worth discussing.

You fail to understand this device. As part of the installation, a small,
carefully chosen, heater element is installed which automatically opens the
switch if that motor has a sustained overload.

It was spelled out in my post at 9:36 this morning.
I'll repeat it: The CR101 will *not* protect the motor.
the smallest CR101 heating element is rated at .48
amps, and requires a 125% draw to trip. The locked
rotor amps on this motor is .49 amps. The CR101 will
do nothing - it can't trip unless current drawn equals
or exceed .60 amps, and the motor draws only .49 amps
when it is jammed.
That information appears in the GE catalog on page 164 at
http://www.gecatalogs.com/content/offline/control/01_CC.pdf

16 Amps is the maximum rating only.




Plus $400 or more for development and testing. Plus $1,000 to get it passed
by the inspectors (assuming it ever is). Plus $200 or more for repair on
each failure since no electrician will touch it.

Why are you attempting to twist the DIY project I proposed
into a commercial product? Is it because you cannot support
your contention that it is a bad idea, or are you saying
that a DIY project (not the circuit) is a bad idea for the
op when a commercial product is available?

List Price: $38.00 UL/CSA listed.

That's 38 dollars, plus the electrician's fee, down
the toilet, because it cannot protect his motor.
Then you can add the cost of the replacement motor
and installation. Plus there's whatever safety
implications and fire and injury risks you had in
mind in your question, and any associated costs.
I call *that* a bad idea, and support my contention
with the information on page 164 of the url:
http://www.gecatalogs.com/content/offline/control/01_CC.pdf


The good news is that the (considerably more expensive
at $99, plus $12 for the required enclosure, plus $27 for
the heater because you have to order at least 3, plus
installation cost) CR1062R at the url you posted will
most likely do what he wants. You can get a heater for
it rated at .37 amps, so the trip current would be .4625.
That should work, except the CR1062 is designed for
infrequent starting. That may preclude it - I can't
say.

So, all that remains if for you to tell us what is bad
about the circuit I proposed. Or is your point that
he'd be better off to use the CR1062, which makes
using a DIY circuit bad in your view?

Ed

 

[Homer J Simpson]

So, all that remains if for you to tell us what is bad
about the circuit I proposed. Or is your point that
he'd be better off to use the CR1062, which makes
using a DIY circuit bad in your view?

Simple. You cannot install electrical items in any factory which do not
comply with code and home made protection circuits are right up there.

I am a registered electrician AND an experienced electronics technician and
I wouldn't put my ticket on the line for some home made gadget like that.

What happens when there is a fire? You really think they won't point fingers
at the thing they understand least of all?

 

[ehsjr]

Simple. You cannot install electrical items in any factory which do not
comply with code and home made protection circuits are right up there.

Thank you! Whether it's a factory or not,
now I understand what you were thinking.

Ed