Dumb question

[pmoseley]

I have a landscape lighting unit that can turn on whatever is wired
into it for 2, 4, 6, or 8 hours from when the photo cell trigers the
unit. My problem is that the unit needs a minimum load ot 20 watts to
operate (max 300W). It puts out 12 or 14 volts. I want to use a 14v.
relay to turn on a nouber of light which would far exceed the 300W
max. If I can figure what the relay draws how do I figure what to add
in parallel with the relay to add up to 20 watts to make the unit
work. I guess it would be a resistor of some sort. But how do I
figure the ohm rating and power requirement?

TIA

Peter Moseley

 

[Boris Mohar]

I have a landscape lighting unit that can turn on whatever is wired
into it for 2, 4, 6, or 8 hours from when the photo cell trigers the
unit. My problem is that the unit needs a minimum load ot 20 watts to
operate (max 300W). It puts out 12 or 14 volts. I want to use a 14v.
relay to turn on a nouber of light which would far exceed the 300W
max. If I can figure what the relay draws how do I figure what to add
in parallel with the relay to add up to 20 watts to make the unit
work. I guess it would be a resistor of some sort. But how do I
figure the ohm rating and power requirement?

TIA

Peter Moseley

Light bulb

 

[[email protected]]

Could you just put 20 watts worth of lights in parallel with the relay
to make the photo cell unit happy?

M

 

[Don Lancaster]

I have a landscape lighting unit that can turn on whatever is wired
into it for 2, 4, 6, or 8 hours from when the photo cell trigers the
unit. My problem is that the unit needs a minimum load ot 20 watts to
operate (max 300W). It puts out 12 or 14 volts. I want to use a 14v.
relay to turn on a nouber of light which would far exceed the 300W
max. If I can figure what the relay draws how do I figure what to add
in parallel with the relay to add up to 20 watts to make the unit
work. I guess it would be a resistor of some sort. But how do I
figure the ohm rating and power requirement?

TIA

Peter Moseley

There's been an ugly rumor going around that Ohm's Law might be somewhat
useful here.


--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: [email protected]

Please visit my GURU's LAIR web site at http://www.tinaja.com

 

[pmoseley]

Could you just put 20 watts worth of lights in parallel with the relay
to make the photo cell unit happy?

M




- Show quoted text -

OK the way I understand it is as follows: 20W/V=1.429I. And using
R=V/I (be gentle with me guys. It's been a long time since I've done
this) R=14/1.429=9.8 Ohms. Radio Shack has a 10 Ohm, 10W wire-wound
resistor. So, I guess two of these in parallel would do it. Right?

TIA
Peter Moseley

 

[Fred Bloggs]

I have a landscape lighting unit that can turn on whatever is wired
into it for 2, 4, 6, or 8 hours from when the photo cell trigers the
unit. My problem is that the unit needs a minimum load ot 20 watts
to operate (max 300W). It puts out 12 or 14 volts. I want to use a
14v. relay to turn on a nouber of light which would far exceed the
300W max. If I can figure what the relay draws how do I figure what
to add in parallel with the relay to add up to 20 watts to make the
unit work. I guess it would be a resistor of some sort. But how do
I figure the ohm rating and power requirement?

Sounds like you're using what is usually called an "electronic
transformer" intended for low voltage lighting applications. That would
then be a 12 or 24 volt output selection and not an oddball 14 volts.
The minimum loading requirement of 20W seems to be a standard and is
something they call a "demand circuit design." If your "unit" is in fact
an "electronic transformer" of this type, then it is most likely a high
frequency inverter type, producing a low voltage AC at several tens of
thousands of kilohertz. This would make sense because high frequency
operation makes for much smaller size (volume, "footprint," or what have
you) which is what most people are after when they use the "electronic
transformer." If your limited concentration span has lasted this far
then you need to know that an ordinary low voltage relay will not
operate at this of a high frequency, and this aside from the fact that a
14V relay is rather rare.

 

[David Harmon]

On Mon, 3 Dec 2007 15:42:18 -0800 (PST) in sci.electronics.design,

If I can figure what the relay draws how do I figure what to add
in parallel with the relay to add up to 20 watts to make the unit
work.

Just use 20 watts of your lighting, then put the rest on the relay.
Don't waste the 20 watts.

 

[[email protected]]

On Mon, 3 Dec 2007 15:42:18 -0800 (PST) in sci.electronics.design,


Just use 20 watts of your lighting, then put the rest on the relay.
Don't waste the 20 watts.

Plus, if the original poster uses resistors, depending on how small
(physically) they are, they can get pretty hot, dissipating 20 watts.
That's a lot of heat, especially if placed in an enclosed space -
could possibly start a fire.

Just my 2 cents, and fwiw, I'm not an electronics/electrical expert by
any stretch of the imagination

M

 

[Active8]

OK the way I understand it is as follows: 20W/V=1.429I. And using
R=V/I (be gentle with me guys. It's been a long time since I've done
this) R=14/1.429=9.8 Ohms. Radio Shack has a 10 Ohm, 10W wire-wound
resistor. So, I guess two of these in parallel would do it. Right?

TIA
Peter Moseley

nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20
W per, and goodby resistors. unless you need a space heater also,
forget using resistors, and load it down with some automotive bulbs
which handle 13.8 V or so nicely - if, in fact, that thing really has
a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24
[email protected] auto bulbs are available since, IIRC military
trucks use 2 batts in series for 24 V - can't remember why.

mike

 

[Active8]

OK the way I understand it is as follows: 20W/V=1.429I. And using
R=V/I (be gentle with me guys. It's been a long time since I've done
this) R=14/1.429=9.8 Ohms. Radio Shack has a 10 Ohm, 10W wire-wound
resistor. So, I guess two of these in parallel would do it. Right?

TIA
Peter Moseley

nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20
W per, and goodby resistors. unless you need a space heater also,
forget using resistors, and load it down with some automotive bulbs
which handle 13.8 V or so nicely - if, in fact, that thing really has
a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24
V auto bulbs are available since, IIRC military
trucks use 2 batts in series for 24 V - can't remember why.

mike

 

[pmoseley]

<snip>





nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20
W per, and goodby resistors. unless you need a space heater also,
forget using resistors, and load it down with some automotive bulbs
which handle 13.8 V or so nicely - if, in fact, that thing really has
a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24
V auto bulbs are available since, IIRC military
trucks use 2 batts in series for 24 V - can't remember why.

mike

OK. Thanks Mike. How about a 5 ohm 50 watt resistor with the 12V
output. That would pull 2.4 amps for 28.8 watts. I'm worried that a
12V lamp would burn out. If the resistor doesn't put out too much
heat I can just plug it in and forget it.

peter

 

[Active8]

OK. Thanks Mike. How about a 5 ohm 50 watt resistor with the 12V
output. That would pull 2.4 amps for 28.8 watts. I'm worried that a
12V lamp would burn out. If the resistor doesn't put out too much
heat I can just plug it in and forget it.

peter

Now you get more heat. At one time some people might not care, but
with rising energy costs these days, you might put them in a box for a
bun warmer whatever works. Do it if there's nothing better staring
you in the face.

Mike

 

[ehsjr]

OK. Thanks Mike. How about a 5 ohm 50 watt resistor with the 12V
output. That would pull 2.4 amps for 28.8 watts. I'm worried that a
12V lamp would burn out. If the resistor doesn't put out too much
heat I can just plug it in and forget it.

peter

28.8 watts will make that resistor hot. Use 5 25 ohm
25 watt resistors in parallel instead to spread the
heat out over a much larger area.

Ed