Simple plan - will it work?

B

Bdyne

Guest
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!
 
"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!
Click to expand...
?

Generally speaking the idea will work. But I can imagine some things you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly connected
to the 5V power you will have no problem but when it is switched somewhere,
the switch (a transistor for instance) may not be able to provide the extra
current for the relay.

When you connected the fan and the relay, check the circuit, particularly
the relay when it switches the fan off. If there's a heavy sparking at the
contacts you may suppress them by placing a capacitor parallel to the
contacts. I guess a 1nF/250VAC will do.

petrus


---
Outgoing mail is certified Virus Free.
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Version: 6.0.532 / Virus Database: 326 - Release Date: 27-10-2003
 
In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...
"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly connected
to the 5V power you will have no problem but when it is switched somewhere,
the switch (a transistor for instance) may not be able to provide the extra
current for the relay.
Click to expand...
Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).
When you connected the fan and the relay, check the circuit, particularly
the relay when it switches the fan off. If there's a heavy sparking at the
contacts you may suppress them by placing a capacitor parallel to the
contacts. I guess a 1nF/250VAC will do.
Click to expand...
--
Keith
 
In article <MPG.1a1aa7e09be09538989adf@enews.newsguy.com>,
Keith R. Williams <krw@attglobal.net> wrote:

In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...

"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly connected
to the 5V power you will have no problem but when it is switched somewhere,
the switch (a transistor for instance) may not be able to provide the extra
current for the relay.

Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).
Click to expand...
Ummm... Why?

Perhaps I'm daft, but isn't a freewheel diode only needed in cases where
a permanent magnet type (which almost always means DC-powered) motor is
being controlled? Unless I've misunderstood things really badly, your
typical AC "fan motor" type doesn't act as a generator/alternator when
freewheeling down after power-off, since there's no juice on the stator,
therefore there's no magnetic field to be cutting/cut by the windings of
the rotor, which renders it functionally nothing more or less than a
flywheel. (Albeit it fairly poor one, since they're usually pretty light)

Conversely, cutting the juice to a PM-type motor DOESN'T "turn off" the
magnets that make up the "stator", so the magnetic lines of force are
cutting/being cut by the windings of the rotor, producing juice as long
as the rotor continues to spin after power-off.

--
Don Bruder - dakidd@sonic.net <--- Preferred Email - SpamAssassinated.
Hate SPAM? See <http://www.spamassassin.org> for some seriously great info.
I will choose a path that's clear: I will choose Free Will! - N. Peart
Fly trap info pages: <http://www.sonic.net/~dakidd/Horses/FlyTrap/index.html>
 
In article <z37sb.5951$Wy2.76893@typhoon.sonic.net>, dakidd@sonic.net
says...
In article <MPG.1a1aa7e09be09538989adf@enews.newsguy.com>,
Keith R. Williams <krw@attglobal.net> wrote:

In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...

"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly connected
to the 5V power you will have no problem but when it is switched somewhere,
the switch (a transistor for instance) may not be able to provide the extra
current for the relay.

Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).

Ummm... Why?

Perhaps I'm daft, but isn't a freewheel diode only needed in cases where
a permanent magnet type (which almost always means DC-powered) motor is
being controlled? Unless I've misunderstood things really badly, your
typical AC "fan motor" type doesn't act as a generator/alternator when
freewheeling down after power-off, since there's no juice on the stator,
therefore there's no magnetic field to be cutting/cut by the windings of
the rotor, which renders it functionally nothing more or less than a
flywheel. (Albeit it fairly poor one, since they're usually pretty light)
Click to expand...
It's not the motor that's in question.

Conversely, cutting the juice to a PM-type motor DOESN'T "turn off" the
magnets that make up the "stator", so the magnetic lines of force are
cutting/being cut by the windings of the rotor, producing juice as long
as the rotor continues to spin after power-off.
Click to expand...
Unless I've missed the whole point here, he wants to pick a 5VDC relay
coil to control a 115V fan. Since the relay coil (and the wire between
the the driver and the coil) has significant inductance, a free
wheeling diode is needed to clip the transient when the driver turns
off. ...otherwise the driver goes to driver heaven and he's out a
motherboard.

--
Keith
 
"Don Bruder" <dakidd@sonic.net> wrote in message
news:z37sb.5951$Wy2.76893@typhoon.sonic.net...
In article <MPG.1a1aa7e09be09538989adf@enews.newsguy.com>,
Keith R. Williams <krw@attglobal.net> wrote:

In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...

"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will
be
mounted in an enclosure feeding 3" diameter flex hose to the
computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things
you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay
on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly
connected
to the 5V power you will have no problem but when it is switched
somewhere,
the switch (a transistor for instance) may not be able to provide the
extra
current for the relay.

Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).

Ummm... Why?

Perhaps I'm daft, but isn't a freewheel diode only needed in cases where
a permanent magnet type (which almost always means DC-powered) motor is
being controlled? Unless I've misunderstood things really badly, your
typical AC "fan motor" type doesn't act as a generator/alternator when
freewheeling down after power-off, since there's no juice on the stator,
therefore there's no magnetic field to be cutting/cut by the windings of
the rotor, which renders it functionally nothing more or less than a
flywheel. (Albeit it fairly poor one, since they're usually pretty light)

Conversely, cutting the juice to a PM-type motor DOESN'T "turn off" the
magnets that make up the "stator", so the magnetic lines of force are
cutting/being cut by the windings of the rotor, producing juice as long
as the rotor continues to spin after power-off.

--
Don Bruder - dakidd@sonic.net <--- Preferred Email - SpamAssassinated.
Click to expand...
The poster said to put the diode across the RELAY COIL not the motor.
When the relay de-energizes, you will create a reverse EMF. The diode
will shunt that so as not to possibly damage the driver transistor.
 
In article <JP7sb.20$Op4.16797743@newssvr11.news.prodigy.com>,
"Lord Garth" <LGarth@Tantalus.net> wrote:


The poster said to put the diode across the RELAY COIL not the motor.
When the relay de-energizes, you will create a reverse EMF. The diode
will shunt that so as not to possibly damage the driver transistor.
Click to expand...
D'oh! Musta missed that part. Somehow, for some reason, I was thinking
that the poster was saying put the diode across the MOTOR side of
things, not the relay coil side. In that case, I withdraw the question.

--
Don Bruder - dakidd@sonic.net <--- Preferred Email - SpamAssassinated.
Hate SPAM? See <http://www.spamassassin.org> for some seriously great info.
I will choose a path that's clear: I will choose Free Will! - N. Peart
Fly trap info pages: <http://www.sonic.net/~dakidd/Horses/FlyTrap/index.html>
 
In article <MPG.1a1ace4eb4993406989ae3@enews.newsguy.com>,
Keith R. Williams <krw@attglobal.net> wrote:


Unless I've missed the whole point here, he wants to pick a 5VDC relay
coil to control a 115V fan. Since the relay coil (and the wire between
the the driver and the coil) has significant inductance, a free
wheeling diode is needed to clip the transient when the driver turns
off. ...otherwise the driver goes to driver heaven and he's out a
motherboard.
Click to expand...
Now that it's been pointed out to me that you were saying "put it across
the relay coil", you're 100% correct. I must have missed that when I
read the message I replied to. Had I seen/understood that as your
intent, I would have been completely in agreement, and wouldn't have
bothered to post my response. Chalk it up to a brain-fart, I guess.

--
Don Bruder - dakidd@sonic.net <--- Preferred Email - SpamAssassinated.
Hate SPAM? See <http://www.spamassassin.org> for some seriously great info.
I will choose a path that's clear: I will choose Free Will! - N. Peart
Fly trap info pages: <http://www.sonic.net/~dakidd/Horses/FlyTrap/index.html>
 
Don Bruder <dakidd@sonic.net> wrote in message news:<z37sb.5951$Wy2.76893@typhoon.sonic.net>...
In article <MPG.1a1aa7e09be09538989adf@enews.newsguy.com>,
Keith R. Williams <krw@attglobal.net> wrote:

In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...

"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly connected
to the 5V power you will have no problem but when it is switched somewhere,
the switch (a transistor for instance) may not be able to provide the extra
current for the relay.

Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).

Ummm... Why?

Perhaps I'm daft, but isn't a freewheel diode only needed in cases where
a permanent magnet type (which almost always means DC-powered) motor is
being controlled? Unless I've misunderstood things really badly, your
typical AC "fan motor" type doesn't act as a generator/alternator when
freewheeling down after power-off, since there's no juice on the stator,
therefore there's no magnetic field to be cutting/cut by the windings of
the rotor, which renders it functionally nothing more or less than a
flywheel. (Albeit it fairly poor one, since they're usually pretty light)

Conversely, cutting the juice to a PM-type motor DOESN'T "turn off" the
magnets that make up the "stator", so the magnetic lines of force are
cutting/being cut by the windings of the rotor, producing juice as long
as the rotor continues to spin after power-off.
Click to expand...
The diode goes across the relay coil. The coil is a inductor
and produces back emf of L*di/dt which is a bad thing without
the diode.

-Bill
 
"Keith R. Williams" <krw@attglobal.net> schreef in bericht
news:MPG.1a1aa7e09be09538989adf@enews.newsguy.com...
In article <CT3sb.78229$di.20439976@amsnews02.chello.com>,
p.kralt@hccnet.nl says...

"Bdyne" <well@Ido.say> schreef in bericht
news:aog0rvcd09m49icf90fkvpb4l6kgiv6fp9@4ax.com...
I am installing a 115VAC cooling fan for my computer. The fan will be
mounted in an enclosure feeding 3" diameter flex hose to the computer
case. I will be using a 5 VDC signal from a fan connection on the
motherboard to close a small radio shack relay (5VDC - 120VAC) and
start the fan. Is it this simple or am I forgetting something? I
don't want to burn the house down!


?

Generally speaking the idea will work. But I can imagine some things
you'll
have to check.
- Check the current rating of the contacts of the relay. (Just to stay
on
the safe side.)
- Check the 5V control voltage you want to use. If it is directly
connected
to the 5V power you will have no problem but when it is switched
somewhere,
the switch (a transistor for instance) may not be able to provide the
extra
current for the relay.

Don't forget the free-wheeling diode across the relay coil (at the
driver end of the cable).
Click to expand...
You're right, I should have mentioned it. The 5V power will not be harmed,
but if the control voltage is switched by a transistor you can easily blow
it (the transistor) if you don't use a free-wheeling-diode.

petrus

When you connected the fan and the relay, check the circuit,
particularly
the relay when it switches the fan off. If there's a heavy sparking at
the
contacts you may suppress them by placing a capacitor parallel to the
contacts. I guess a 1nF/250VAC will do.

--
Keith
Click to expand...

---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.532 / Virus Database: 326 - Release Date: 27-10-2003
 
You guys kind of lost me (noob)- but I'm sure glad I asked here before
cooking my new mobo! Thank you all very much.

I understand what a diode is - like a check valve. If I look at the
bottom of the relay I have 2 contacts for the DC and 3 for the AC.
Physically where in the circuit are you putting the diode? And is the
diode one of those resistor-looking ones or the three wire ones (I
would guess the 2 wire).

Again thanks for helping me NOT fry my new board.
 
In article <57a3rvkthkv05j0t3iir2ma415ep628acu@4ax.com>, well@Ido.say
says...
You guys kind of lost me (noob)- but I'm sure glad I asked here before
cooking my new mobo! Thank you all very much.

I understand what a diode is - like a check valve. If I look at the
bottom of the relay I have 2 contacts for the DC and 3 for the AC.
Physically where in the circuit are you putting the diode? And is the
diode one of those resistor-looking ones or the three wire ones (I
would guess the 2 wire).
Click to expand...
Since the current in an inductor (the relay coil and long wire) cannot
change instantaneously, when the driver transistor turns off the
current will try to go somewhere. Without the freewheeling diode there
is no other path so the voltage rises until it can go somewhere. Zap,
your off transistor is no longer off (if it's anywhere to be found. ;-)
With the freewheeling diode, the current flows through it into the 5V
supply, effectively shorting out the inductor. Of course when the
driver is ON the diode is backwards biased, so it's not in the picture.

+5V
| |
C| o /
relay C| /
C| ./
| o
| |
|
+5V |
| |
- Freewheeling diode |
^ |
| |
+-------------------------+
| long wire
|/
-| driver
|>
|

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

Again thanks for helping me NOT fry my new board.

You'd better check the current rating of the driver too. ...and please
Click to expand...
be careful with the 120V circuit! It'll do some damage to your board
too (and you).

--
Keith
 
Just a nitpick, but it should be:

+5V
|
+----------+
| | |
| C| o /
Freewheeling diode - relay C| /
^ C| ./
| | o
+----------+ |
|
|
|
|
|
|
+----------------------+
| long wire
|/
-| driver
|>
|

Cheers!
Rich

"Keith R. Williams" wrote:
In article <57a3rvkthkv05j0t3iir2ma415ep628acu@4ax.com>, well@Ido.say
says...
You guys kind of lost me (noob)- but I'm sure glad I asked here before
cooking my new mobo! Thank you all very much.

I understand what a diode is - like a check valve. If I look at the
bottom of the relay I have 2 contacts for the DC and 3 for the AC.
Physically where in the circuit are you putting the diode? And is the
diode one of those resistor-looking ones or the three wire ones (I
would guess the 2 wire).

Since the current in an inductor (the relay coil and long wire) cannot
change instantaneously, when the driver transistor turns off the
current will try to go somewhere. Without the freewheeling diode there
is no other path so the voltage rises until it can go somewhere. Zap,
your off transistor is no longer off (if it's anywhere to be found. ;-)
With the freewheeling diode, the current flows through it into the 5V
supply, effectively shorting out the inductor. Of course when the
driver is ON the diode is backwards biased, so it's not in the picture.

+5V
| |
C| o /
relay C| /
C| ./
| o
| |
|
+5V |
| |
- Freewheeling diode |
^ |
| |
+-------------------------+
| long wire
|/
-| driver
|
|

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de


Again thanks for helping me NOT fry my new board.

You'd better check the current rating of the driver too. ...and please
be careful with the 120V circuit! It'll do some damage to your board
too (and you).

--
Keith
Click to expand...
 
Rich Grise wrote:
Just a nitpick, but it should be:

+5V
|
+----------+
| | |
| C| o /
Freewheeling diode - relay C| /
^ C| ./
| | o
+----------+ |
|
|
|
|
|
|
+----------------------+
| long wire
|/
-| driver
|
|
Click to expand...
Since we are picking nits, you should keep in mind that wiring has
inductance, too. The cleanest way to use the diode is at the switch
as follows:

+5V
|
+--------+
| | |
| C| o /
| relay C| /
| C| ./
| two | o
+5v | long | |
| | wires |
+-------------+ |
| |
- |
^ |
| |
+----------------------+
|
|/
-| driver
|>
|
0v
--
John Popelish
 
Yeah - you've got all that inductance in the wire in series with the
inductance in the relay coil, so you get an even bigger spike! (and
more concomitant radiated noise)

:)
Cheers!
Rich

John Popelish wrote:
Rich Grise wrote:

Just a nitpick, but it should be:

+5V
|
+----------+
| | |
| C| o /
Freewheeling diode - relay C| /
^ C| ./
| | o
+----------+ |
|
|
|
|
|
|
+----------------------+
| long wire
|/
-| driver
|
|

Since we are picking nits, you should keep in mind that wiring has
inductance, too. The cleanest way to use the diode is at the switch
as follows:

+5V
|
+--------+
| | |
| C| o /
| relay C| /
| C| ./
| two | o
+5v | long | |
| | wires |
+-------------+ |
| |
- |
^ |
| |
+----------------------+
|
|/
-| driver
|
|
0v
--
John Popelish
Click to expand...
 
In article <3FC5E61A.EE2BD1F9@spam.org>, spam@spam.org says...
Yeah - you've got all that inductance in the wire in series with the
inductance in the relay coil, so you get an even bigger spike! (and
more concomitant radiated noise)
Click to expand...
It's better than frying the driver! We had several IBM 3033s
catch fire because of exactly this problem (inrush limiters stuck
on). Customers weren't impressed and management was *not* amused
at this little slight. Especially so, since the drivers had the
diodes built=in, but the designer didn't want to wire 48V on the
card and instead mounted diodes on the contactors. No, the free-
wheeling diode goes on the driver. After all, it's not there to
protect the relay coil. ;-)

--
Keith
 
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