I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek

wrote in news:1152996421.209694.220710@
35g2000cwc.googlegroups.com:

the line voltage is 133 to 138 volts.

That line voltage is WAY too high. The life of lightbulbs and appliances
will be reduced, in addition to the solenoid you mention.

The real solution is to contact the company that furnishes the electricity
and get them to come out and lower the voltage coming into the house.

wrote in message
ups.com...
I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek


You find the value of the resistor by first calculating the current drawn by the
solenoid.
I = P/V
I = 7W/24V = .3A

You can also calculate the impedance of the solenoid's coil.
Z = V/I
Z = 24V/.3A = 80 ohms

Now, calculate the voltage that the resistor needs to drop.
V = Vsupply - Vsolenoid
V = 28 - 24 = 4V

Now, using these two values, calculate the resistor value.
R = V/I
R = 4V/.3A = 13.33 Ohms (a close standard value would be 12 ohms)

Calculate the power that the resistor will dissipate.
P = VI
P = 4V * .3A = 1.2 Watts
Always add a safety factor of at least 2 on the power rating, giving us a
resistor power rating of at least 2.4 Watts. A standard wattage rating for
power resistors is 3W or 5W.

Putting all this together results in a 12 Ohm/5 Watt resistor.

You can deviate from these values a bit if you need... i.e., rework the figures
to further reduce the solenoid voltage.

Cheers!!!
--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

They call it PMS because Mad Cow Disease was already taken.

"Jim Land" wrote in message
. 3.44...
wrote in news:1152996421.209694.220710@
35g2000cwc.googlegroups.com:

the line voltage is 133 to 138 volts.

That line voltage is WAY too high. The life of lightbulbs and appliances
will be reduced, in addition to the solenoid you mention.

The real solution is to contact the company that furnishes the electricity
and get them to come out and lower the voltage coming into the house.

An electric stove or dryer is not going to like 276 V for very long. Also,
check if the line voltage is high at all outlets. If some are high and some
are low, you have an open ground, which the power Co needs to fix ASAP. In
my case, they arrived within the hour.

Tam

DaveM wrote:
wrote in message
ups.com...

I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek



You find the value of the resistor by first calculating the current drawn by the
solenoid.
I = P/V
I = 7W/24V = .3A

You can also calculate the impedance of the solenoid's coil.
Z = V/I
Z = 24V/.3A = 80 ohms

Now, calculate the voltage that the resistor needs to drop.
V = Vsupply - Vsolenoid
V = 28 - 24 = 4V

Now, using these two values, calculate the resistor value.
R = V/I
R = 4V/.3A = 13.33 Ohms (a close standard value would be 12 ohms)

Calculate the power that the resistor will dissipate.
P = VI
P = 4V * .3A = 1.2 Watts
Always add a safety factor of at least 2 on the power rating, giving us a
resistor power rating of at least 2.4 Watts. A standard wattage rating for
power resistors is 3W or 5W.

Putting all this together results in a 12 Ohm/5 Watt resistor.

You can deviate from these values a bit if you need... i.e., rework the figures
to further reduce the solenoid voltage.

Cheers!!!


Excellent reply, Dave. Very instructive and clearly written. You must be a
teacher.

I only want to jump in here to stress that Tam's and Jim's replies should be
considered first as the high line voltage has effects on other household items.

I have a mobile home that survived for about 15 years on about 135 line
volts. We lost a lot of light bulbs and a ventilator fan over that time. We
allowed the high line voltage due to the fact that the home has about 600
feet of #4 AWG between it and the pole. It was thought that the A/C unit
needed the higher voltage to start. About 5 or so years ago we had a
problem, the nature of which I no longer remember, and the power company
came out and lowered the voltage. I was afraid the A/C would not start
properly, but it turned out to not be a problem after all.

In any case, I would just like to say that it is better to cure the problem,
if possible, rather than make modifications. Some day the solenoid may need
replacing and the replacement part may not be the same. There are times,
though, when modifications are called for.

Cheers to all.

John

Thank you for all your replies. The high voltage only to seemed to
happen for a while but we are going to call the electric company,
National Grid, and have them put a recorder on the line.

When I took the reading of 133 volts it was from the hardwired
terminals inside the furnace. I then tried a couple of other outlets
and got similar readings. I then went looking for another meter to
confirm these readings but when I finally got another meter and tried
it the voltage was at 119.

I'm still going to try applying the resistor to the solenoid because
even with the line voltage at 119 and the furnace in an idle state, I
think the only load on the 24 volt transformer is the solenoid and it
is still getting hot.

Thanks again everyone,
Derek

Tam/WB2TT wrote:

"Jim Land" wrote in message
. 3.44...
wrote in news:1152996421.209694.220710@
35g2000cwc.googlegroups.com:

the line voltage is 133 to 138 volts.

That line voltage is WAY too high. The life of lightbulbs and appliances
will be reduced, in addition to the solenoid you mention.

The real solution is to contact the company that furnishes the electricity
and get them to come out and lower the voltage coming into the house.

An electric stove or dryer is not going to like 276 V for very long. Also,
check if the line voltage is high at all outlets. If some are high and some
are low, you have an open ground, which the power Co needs to fix ASAP. In
my case, they arrived within the hour.

Tam


You mean an open neutral.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

wrote:
I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek



Just wire it up without a resistor, 28 volts is fine for a 24 volt
nominal solenoid.

there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

The solenoid coil is not a resistance, it is an impedance composed of
orthogonal resistive and reactive parts, and this impedance is a strong
function of whether the solenoid is in or out, varying over a 2-3 range.
Placing a resistor in series may very well blow the transformer
because the resistor presents too much drop at "inrush" and the solenoid
hangs at a current loading in excess of the transformer VA.
http://www.radiofrigor.com.br/download/coils.pdf
Your line readings are suspect and unreliable. You don't make it clear
about the solenoid being permanently on or not, , thinking it more
important to tell us totally irrelevant things like it is in a house
belonging to your mother's boyfriend. If this is the kind of thing you
focus on, then stay away from the furnace.

"DaveM" wrote in message
. ..
wrote in message
ups.com...
I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek


You find the value of the resistor by first calculating the current drawn
by the
solenoid.
I = P/V
I = 7W/24V = .3A

You can also calculate the impedance of the solenoid's coil.
Z = V/I
Z = 24V/.3A = 80 ohms

Now, calculate the voltage that the resistor needs to drop.
V = Vsupply - Vsolenoid
V = 28 - 24 = 4V

Now, using these two values, calculate the resistor value.
R = V/I
R = 4V/.3A = 13.33 Ohms (a close standard value would be 12 ohms)

Calculate the power that the resistor will dissipate.
P = VI
P = 4V * .3A = 1.2 Watts
Always add a safety factor of at least 2 on the power rating, giving us a
resistor power rating of at least 2.4 Watts. A standard wattage rating
for
power resistors is 3W or 5W.

Putting all this together results in a 12 Ohm/5 Watt resistor.

You can deviate from these values a bit if you need... i.e., rework the
figures
to further reduce the solenoid voltage.


If this were DC, that would be the right way to calculate this, however, he
is dealing with AC. The problem here is that the impedance of the coil is
largely resistive *and* inductive. To make things worse, the inductance will
change, likely a lot when the plunger is in vs. out. Since when the plunger
is out, the magnetic path would have a much larger air gap, and thus making
the resistance of the coil more dominant, and when the plunger is in, the
air path for the magnetic field is much less, and therefore it's inductance
is much higher. The problem comes when the plunger needs to pull in, the
impedance of the coil is at a minimum, causing a higher dissipation in the
resistor, and a much weaker then expected magnetic field in the coil. When
the plunger is in, the impedance increases, dropping the current, and
increasing the voltage. Simply said, adding a series resistor can cause the
plunger to not pull in, and/or keep the power dissipation in the coil high.
I quickly figured out this a long time ago when I wired two contactors with
120V coils in series on a 240V supply. Only 1 contactor would reliably pull
in!

In article . com,
wrote:

I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

Derek-

The solenoid appears to be functioning correctly, from what you've said.
Yes, it will be hot since it is burning up 7 watts of power.

The real problem is the apparently high line voltage. Unless you are
connected to a small town power source that is not connected to the
national grid, it is unlikely that your house is receiving 276 Volts AC.

What is more likely is that the neutral connection between the house and
the power company's transformer has come loose. The two 120 Volt AC
halves do not carry the same current, so voltage divides between them in
relation to the ratio of loads. If this is the case, you may measure
voltage at several places in the house, and find two values: 138 and 102,
for a total of 240 Volts.

The loose connection can be either at the power pole or at your power
meter. The power company should be able to isolate it, but the owner will
have to pay an electrician to fix a problem if it is on the house side of
the power meter.

Fred

wrote:
I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek


If you are going to put a resistor in series,
you need an on delay relay.

24VAC ---o
\ S1
o o--+---[R]---+------+
| | |
+----- | |
N/C ^----+ [Solenoid]
_ | |
| | | |
+--|/|----+ |
| |_|On Delay |
| Relay |
24AC ---------+----------------+

The solenoid gets full current when first switced
on, through the normally closed contact on the relay.
After a delay, the relay transfers, and current to the
solenoid coil is reduce by the resistor.

But it would be a whole lot better to get the real
problem fixed, as others said.

Ed

The solenoid is normally on. I think it is part of a panic circuit that
is left over from the heat pump that we don't using any more. So far,
the best we have been able to figure out, the only time it is off is
when the thermostat is turned off. If the furnace is sitting idle it is
still active. From the wiring diagram, very crude and high level, it
shows that its operation is influenced by a flash detector, temperature
sensor (for over heating) and auxiliary input. Haven't determined if
the the aux is hooked up to anything.

What we have observed since I oroginally posted is this:

If the furnace, the circulator is running, the voltage at the solenoid
is about 24 volts but when it shuts down, it goes up to 28 volts. So
something else is putting a load on the same circuit to cause it to
drop.

Our concern is for the solenoid being hot to the touch. Should we be
trying to change the voltage or the amperage to solenoid?

Derek


The solenoid coil is not a resistance, it is an impedance composed of
orthogonal resistive and reactive parts, and this impedance is a strong
function of whether the solenoid is in or out, varying over a 2-3 range.
Placing a resistor in series may very well blow the transformer
because the resistor presents too much drop at "inrush" and the solenoid
hangs at a current loading in excess of the transformer VA.
http://www.radiofrigor.com.br/download/coils.pdf
Your line readings are suspect and unreliable. You don't make it clear
about the solenoid being permanently on or not, , thinking it more
important to tell us totally irrelevant things like it is in a house
belonging to your mother's boyfriend. If this is the kind of thing you
focus on, then stay away from the furnace.

On 17 Jul 2006 08:42:19 -0700, wrote:

The solenoid is normally on. I think it is part of a panic circuit that
is left over from the heat pump that we don't using any more. So far,
the best we have been able to figure out, the only time it is off is
when the thermostat is turned off. If the furnace is sitting idle it is
still active. From the wiring diagram, very crude and high level, it
shows that its operation is influenced by a flash detector, temperature
sensor (for over heating) and auxiliary input. Haven't determined if
the the aux is hooked up to anything.

What we have observed since I oroginally posted is this:

If the furnace, the circulator is running, the voltage at the solenoid
is about 24 volts but when it shuts down, it goes up to 28 volts. So
something else is putting a load on the same circuit to cause it to
drop.

Our concern is for the solenoid being hot to the touch. Should we be
trying to change the voltage or the amperage to solenoid?

---

From your original post and your latest, you have a circuit that
looks like this: (View in Courier)


ACIN----+-------+---E1
| |
| O--|
| |S1
[SOL] +---O
| |
| |
| [Rl]
| |
ACIN----+-------+---0V

Where S1 is normally closed and E1 is normally 24V, but when S1
opens E1 goes up to 28V and the solenoid runs hot.

From the solenoid's spec's it dissipates 7 watts with 24VRMS across
it, so its resistance is:

E² 24²
R = --- = ----- ~ 82 ohms
P 7W

and the current through it must then be:

E 24V
I = --- = ----- ~ 0.293 amperes
R 82R

When the switch opens and the voltage changes to 28V, the current
rises to:


28V
I = ----- ~ 0.341 amperes
82R

and the power dissipation goes to:


P = IE = 0.341A * 28V ~ 9.55 watts


An easy way to fix the problem would be to keep a constant load on
the transformer, and you can do that by shunting the solenoid with a
pair of 24 Zener diodes in series opposition:


ACIN----+-------+-------+---E1
| | |
| [1N5359B] O--|
| |K |S1
[SOL] | +---O
| | |
| |K |
| [1N5359B] [Rl]
| | |
ACIN----+-------+-------+---0V

That way, with S1 closed and 24V across the diodes they'll be
effectively out of the circuit, but when S1 opens the current which
formerly went to Rl will be diverted through the Zeners and the
voltage on the solenoid clamped to 24V + the Vf of the other diode,
about 0.7V.

As it turns out, the current into the switched load is 48mA and the
Zener test current is 50mA, so that's a pretty good match!


--
John Fields
Professional Circuit Designer

John Fields wrote:
On 17 Jul 2006 08:42:19 -0700, wrote:


The solenoid is normally on. I think it is part of a panic circuit that
is left over from the heat pump that we don't using any more. So far,
the best we have been able to figure out, the only time it is off is
when the thermostat is turned off. If the furnace is sitting idle it is
still active. From the wiring diagram, very crude and high level, it
shows that its operation is influenced by a flash detector, temperature
sensor (for over heating) and auxiliary input. Haven't determined if
the the aux is hooked up to anything.

What we have observed since I oroginally posted is this:

If the furnace, the circulator is running, the voltage at the solenoid
is about 24 volts but when it shuts down, it goes up to 28 volts. So
something else is putting a load on the same circuit to cause it to
drop.

Our concern is for the solenoid being hot to the touch. Should we be
trying to change the voltage or the amperage to solenoid?


---

From your original post and your latest, you have a circuit that
looks like this: (View in Courier)


ACIN----+-------+---E1
| |
| O--|
| |S1
[SOL] +---O
| |
| |
| [Rl]
| |
ACIN----+-------+---0V

Where S1 is normally closed and E1 is normally 24V, but when S1
opens E1 goes up to 28V and the solenoid runs hot.

From the solenoid's spec's it dissipates 7 watts with 24VRMS across
it, so its resistance is:

E² 24²
R = --- = ----- ~ 82 ohms
P 7W

and the current through it must then be:

E 24V
I = --- = ----- ~ 0.293 amperes
R 82R

When the switch opens and the voltage changes to 28V, the current
rises to:


28V
I = ----- ~ 0.341 amperes
82R

and the power dissipation goes to:


P = IE = 0.341A * 28V ~ 9.55 watts


An easy way to fix the problem would be to keep a constant load on
the transformer, and you can do that by shunting the solenoid with a
pair of 24 Zener diodes in series opposition:


ACIN----+-------+-------+---E1
| | |
| [1N5359B] O--|
| |K |S1
[SOL] | +---O
| | |
| |K |
| [1N5359B] [Rl]
| | |
ACIN----+-------+-------+---0V

That way, with S1 closed and 24V across the diodes they'll be
effectively out of the circuit, but when S1 opens the current which
formerly went to Rl will be diverted through the Zeners and the
voltage on the solenoid clamped to 24V + the Vf of the other diode,
about 0.7V.

As it turns out, the current into the switched load is 48mA and the
Zener test current is 50mA, so that's a pretty good match!



I found the DMG coil the OP specified, but it does not show it to be 7 Watts.

http://www.radiofrigor.com.br/download/coils.pdf

wrote:
The solenoid is normally on. I think it is part of a panic circuit that
is left over from the heat pump that we don't using any more. So far,
the best we have been able to figure out, the only time it is off is
when the thermostat is turned off. If the furnace is sitting idle it is
still active. From the wiring diagram, very crude and high level, it
shows that its operation is influenced by a flash detector, temperature
sensor (for over heating) and auxiliary input. Haven't determined if
the the aux is hooked up to anything.

What we have observed since I oroginally posted is this:

If the furnace, the circulator is running,

You're in upstate NY so a hydronic system would be typical.

the voltage at the solenoid
is about 24 volts but when it shuts down, it goes up to 28 volts. So
something else is putting a load on the same circuit to cause it to
drop.

Right- this is normal.


Our concern is for the solenoid being hot to the touch. Should we be
trying to change the voltage or the amperage to solenoid?

Derek


A few volts less is not going to make the solenoid run cool, it will
still be hot to the touch at its nominal rated voltage, it was designed
to run that way and the heat is normal unless the solenoid gets so hot
that it fumes and/or discolors- then it is defective in some way and
should be replaced. You could make an amperage check and verify that the
AC current through the coil is in the range of 0.75 +/-20% amperes just
to be sure. You say this is a dual cooling/heat system, so I assume the
solenoid valve is off when the thermostat is in COOL mode, meaning you
can't just remove it from the system.

John - KD5YI wrote:

I found the DMG coil ....

You're about as useful to this thread as tits on a boar hog, why don't
you take hike.

Fred Bloggs wrote:


John - KD5YI wrote:

I found the DMG coil ....


You're about as useful to this thread as tits on a boar hog, why don't
you take hike.



Sorry to upset you, Fred. I just thought a link to the coil specs might be
helpful.

I guess it depends on what you mean by "hot".

The general rule of thumb is, if you can hold your thumb on it for 30
seconds, it's not too hot.

Most coils these days are wound with wire insulation that can stand at
least 85C if not 105C. Your thumb isnt going to like more than 60C for
a few seconds.

You also have to consider what might happen if you under-drive the
solenoid. It might fail to pull in, then you have a problem. Also
some solenoids used on AC will actually get HOTTER if the plunger
doesnt pull in. The lack of iron will reduce the coil inductance,
which is not a good thing.

John - KD5YI wrote:
Fred Bloggs wrote:



John - KD5YI wrote:

I found the DMG coil ....



You're about as useful to this thread as tits on a boar hog, why don't
you take hike.



Sorry to upset you, Fred. I just thought a link to the coil specs might
be helpful.

I already posted that very same link days ago- thinking it somewhat
relevant to determine the coil characteristics prior to any conjecture
about methods to reduce the self-heating.

Fred Bloggs wrote:


John - KD5YI wrote:

Fred Bloggs wrote:



John - KD5YI wrote:

I found the DMG coil ....




You're about as useful to this thread as tits on a boar hog, why
don't you take hike.



Sorry to upset you, Fred. I just thought a link to the coil specs
might be helpful.


I already posted that very same link days ago- thinking it somewhat
relevant to determine the coil characteristics prior to any conjecture
about methods to reduce the self-heating.

Although I read your post days ago, I forgot that you posted that link.

Thanks for pointing out my failings.

On Tue, 18 Jul 2006 01:17:39 GMT, John - KD5YI
wrote:

John Fields wrote:
On 17 Jul 2006 08:42:19 -0700, wrote:


The solenoid is normally on. I think it is part of a panic circuit that
is left over from the heat pump that we don't using any more. So far,
the best we have been able to figure out, the only time it is off is
when the thermostat is turned off. If the furnace is sitting idle it is
still active. From the wiring diagram, very crude and high level, it
shows that its operation is influenced by a flash detector, temperature
sensor (for over heating) and auxiliary input. Haven't determined if
the the aux is hooked up to anything.

What we have observed since I oroginally posted is this:

If the furnace, the circulator is running, the voltage at the solenoid
is about 24 volts but when it shuts down, it goes up to 28 volts. So
something else is putting a load on the same circuit to cause it to
drop.

Our concern is for the solenoid being hot to the touch. Should we be
trying to change the voltage or the amperage to solenoid?


---

From your original post and your latest, you have a circuit that
looks like this: (View in Courier)


ACIN----+-------+---E1
| |
| O--|
| |S1
[SOL] +---O
| |
| |
| [Rl]
| |
ACIN----+-------+---0V

Where S1 is normally closed and E1 is normally 24V, but when S1
opens E1 goes up to 28V and the solenoid runs hot.

From the solenoid's spec's it dissipates 7 watts with 24VRMS across
it, so its resistance is:

E² 24²
R = --- = ----- ~ 82 ohms
P 7W

and the current through it must then be:

E 24V
I = --- = ----- ~ 0.293 amperes
R 82R

When the switch opens and the voltage changes to 28V, the current
rises to:


28V
I = ----- ~ 0.341 amperes
82R

and the power dissipation goes to:


P = IE = 0.341A * 28V ~ 9.55 watts


An easy way to fix the problem would be to keep a constant load on
the transformer, and you can do that by shunting the solenoid with a
pair of 24 Zener diodes in series opposition:


ACIN----+-------+-------+---E1
| | |
| [1N5359B] O--|
| |K |S1
[SOL] | +---O
| | |
| |K |
| [1N5359B] [Rl]
| | |
ACIN----+-------+-------+---0V

That way, with S1 closed and 24V across the diodes they'll be
effectively out of the circuit, but when S1 opens the current which
formerly went to Rl will be diverted through the Zeners and the
voltage on the solenoid clamped to 24V + the Vf of the other diode,
about 0.7V.

As it turns out, the current into the switched load is 48mA and the
Zener test current is 50mA, so that's a pretty good match!



I found the DMG coil the OP specified, but it does not show it to be 7 Watts.

http://www.radiofrigor.com.br/download/coils.pdf

---
Oh well... The circuit stays the same but the numbers change a
little.


--
John Fields
Professional Circuit Designer