Is it possible to operate a 12 VDC relay from a 24 VAC suppl

[John Larkin]

On Wed, 20 Feb 2008 07:52:51 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 18 Feb 2008 17:20:45 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


Did you actually test a bunch of relays just to prove me wrong?

---
Of course. With your proclivity for argumentative generalization
and the ol' bob and weave I sometimes find it necessary to use
experimentally derived real-world data in order to pin you down.
---

I'm flattered.

---
Sorry, that wasn't my intention.
---

Better would be to put the resistor *before* the bridge, so
the diodes clamp the coil voltage close to zero.

---
You've lost me there.

What are you talking about?


The L/R thing. Look it up.

---
Better yet, look at this:

Version 4
SHEET 1 960 932
WIRE -1168 -48 -1248 -48
WIRE -1024 -48 -1088 -48
WIRE -704 -48 -1024 -48
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SYMATTR InstName V1
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WINDOW 3 -34 59 VTop 0
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SYMATTR Value 50
SYMBOL diode 192 304 R180
WINDOW 0 47 34 Left 0
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SYMATTR InstName D2
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SYMBOL diode -224 48 R180
WINDOW 0 -44 32 Left 0
WINDOW 3 -107 64 Left 0
SYMATTR InstName D3
SYMATTR Value MURS120
SYMBOL diode -256 240 R0
WINDOW 0 39 34 Left 0
SYMATTR InstName D4
SYMATTR Value MURS120
SYMBOL res 160 112 R90
WINDOW 0 -38 58 VBottom 0
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SYMATTR Value 50
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WINDOW 0 32 56 VTop 0
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SYMATTR Value MURS120
SYMBOL res -864 112 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R5
SYMATTR Value 50
SYMBOL diode -688 304 R180
WINDOW 0 47 34 Left 0
WINDOW 3 23 -2 Left 0
SYMATTR InstName D7
SYMATTR Value MURS120
SYMBOL diode -1008 48 R180
WINDOW 0 -44 32 Left 0
WINDOW 3 -107 64 Left 0
SYMATTR InstName D8
SYMATTR Value MURS120
SYMBOL diode -1040 240 R0
WINDOW 0 39 34 Left 0
SYMATTR InstName D9
SYMATTR Value MURS120
SYMBOL res -1072 -64 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R7
SYMATTR Value 50
TEXT -1248 400 Left 0 !.tran .05 uic

Note that either way (with the 50 ohm resistor internal or external
to the bridge) clamps the coil to a couple of diode drops _below_
ground, so I still don't know what you're talking about.

The difference is in the ripple current in the coil. The current
difference is modest for the L/R values in your model, especially with
two diode drops. One interesting thing about relays is that the
inductance increases, often by a huge factor, between the de-energized
and pulled-in state [1]. Which L did you use in your sim? [2]

However, there _does_ seem to be an advantage to using the external
resistor.

Can you tell what it is?

Dunno. Tell us.

John


[1] which gives power AC relays a huge advantage.

[2] It would be awful to do a fully time-accurate model of a relay
pulling in, what with L being a function of the varying magnetic loop
reluctance. Ugh.

 

[John Fields]

On Mon, 18 Feb 2008 17:20:45 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Did you actually test a bunch of relays just to prove me wrong?

---
Of course. With your proclivity for argumentative generalization
and the ol' bob and weave I sometimes find it necessary to use
experimentally derived real-world data in order to pin you down.
---

I'm flattered.

---
Sorry, that wasn't my intention.
---

Better would be to put the resistor *before* the bridge, so
the diodes clamp the coil voltage close to zero.

---
You've lost me there.

What are you talking about?


The L/R thing. Look it up.

---
Better yet, look at this:

Version 4
SHEET 1 960 932
WIRE -1168 -48 -1248 -48
WIRE -1024 -48 -1088 -48
WIRE -704 -48 -1024 -48
WIRE -240 -48 -464 -48
WIRE 176 -48 -240 -48
WIRE -1024 -16 -1024 -48
WIRE -704 -16 -704 -48
WIRE -240 -16 -240 -48
WIRE 176 -16 176 -48
WIRE -1024 128 -1024 48
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WIRE -240 368 -464 368
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WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
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WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 34 60)
SYMATTR InstName V1
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WINDOW 0 -45 31 Left 0
WINDOW 3 -109 -7 Left 0
SYMATTR InstName D1
SYMATTR Value MURS120
SYMBOL res -96 112 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R1
SYMATTR Value 50
SYMBOL diode 192 304 R180
WINDOW 0 47 34 Left 0
WINDOW 3 23 -2 Left 0
SYMATTR InstName D2
SYMATTR Value MURS120
SYMBOL diode -224 48 R180
WINDOW 0 -44 32 Left 0
WINDOW 3 -107 64 Left 0
SYMATTR InstName D3
SYMATTR Value MURS120
SYMBOL diode -256 240 R0
WINDOW 0 39 34 Left 0
SYMATTR InstName D4
SYMATTR Value MURS120
SYMBOL res 160 112 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R4
SYMATTR Value 50
SYMBOL ind -864 144 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L3
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SYMBOL voltage -1248 224 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
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SYMBOL diode -720 -16 R0
WINDOW 0 -45 31 Left 0
WINDOW 3 -109 -7 Left 0
SYMATTR InstName D5
SYMATTR Value MURS120
SYMBOL res -864 112 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R5
SYMATTR Value 50
SYMBOL diode -688 304 R180
WINDOW 0 47 34 Left 0
WINDOW 3 23 -2 Left 0
SYMATTR InstName D7
SYMATTR Value MURS120
SYMBOL diode -1008 48 R180
WINDOW 0 -44 32 Left 0
WINDOW 3 -107 64 Left 0
SYMATTR InstName D8
SYMATTR Value MURS120
SYMBOL diode -1040 240 R0
WINDOW 0 39 34 Left 0
SYMATTR InstName D9
SYMATTR Value MURS120
SYMBOL res -1072 -64 R90
WINDOW 0 -38 58 VBottom 0
WINDOW 3 -34 59 VTop 0
SYMATTR InstName R7
SYMATTR Value 50
TEXT -1248 400 Left 0 !.tran .05 uic

Note that either way (with the 50 ohm resistor internal or external
to the bridge) clamps the coil to a couple of diode drops _below_
ground, so I still don't know what you're talking about.

However, there _does_ seem to be an advantage to using the external
resistor.

Can you tell what it is?


--
JF

 

[whit3rd]

On Feb 15, 8:19 pm, HC <hboo...@gte.net> wrote:

I want to be able to control 120VAC devices some distance away from a
controller, say, up to 100 feet or so.

To do it safely, there are some concerns about the '12V' relays; if
they are
intended for auto use, for instance, the application of 120VAC may not
be wise.

Why not use the canned solution, an X-10 appliance module and suitable
remote controllers? It has the advantage of UL approval. In case
of mishap your insurance provider might find that interesting.

 

[John Larkin]

On Sat, 16 Feb 2008 13:47:28 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.

It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

John

 

[John Larkin]

On Mon, 18 Feb 2008 18:31:04 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 17 Feb 2008 09:39:58 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 17 Feb 2008 10:12:44 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 15:06:17 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 16 Feb 2008 13:47:28 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.


It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

---

"It probably won't"?

For certain values of probably. A fast relay is going to need a
capacitor somewhere, since no power is available at zero crossing, no
matter how many diodes you use.

---
That's certainly true for your circuit, but not necessarily for mine
since, without a capacitor, your circuit is much more likely to
drive the coil discontinuously during the long wait for the rise of
a new edge after the fall of the current one.

John Popelish pointed that out by noting that the time constant of
the coils is too short to keep the contacts engaged between cycles
with one half-cycle missing, and that was also borne out by my
experiment, where your method failed 4 out of five times using a
random selection of relays.

Did you actually test a bunch of relays just to prove me wrong? I'm
flattered.

---

I've found the opposite to be true, empirically.

Here:

COIL R COIL L
MFG PART NO OHMS HENRY REF
--------|------------|-------|--------|-----
AROMAT JT1a-DC12V 150 0.2 A

AROMAT HC3-P-DC12V 160 0.27 B

ORIGINAL SRUT..12VDC 430 0.41 C

AROMAT JW2EN-EDC12V 275 0.52 D

ECI 8501-3023-12 50 0.148 E


For your circuit:


ACIN>--[D1>]--+-----+
|K |
[D2] [COIL]
| |
ACIN>---------+-----+

I got:

PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 24.3 24.3

B >30 N/A

C >30 N/A

D >30 N/A

E 14.8 9.3


For my circuit:

+-----+
ACIN>--|~ +|--[Rs]--+
| | |
| | [COIL]
| | |
ACIN>--|~ -|--------+
+-----+

where Rs is a fixed resistor equal to the DC resistance of the relay
coil, I got:


Your circuit adds the Rs resistance to the coil circuit, decreasing
L/R by about a factor of 2:1, which negates the improvement of going
full-wave.

---
Not true.

By increasing the time constant of the coil circuit, what it does
(in addition to keeping the relay coil from dissipating more power
than it's rated for) is to change the phase of the current in the
coil so that the current never goes to zero, thereby making sure the
armature isn't tempted to abandon the pole piece.
---

Better would be to put the resistor *before* the bridge, so
the diodes clamp the coil voltage close to zero.

---
You've lost me there.

What are you talking about?

The L/R thing. Look it up.

John

 

[John Fields]

On Sun, 17 Feb 2008 09:39:58 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 17 Feb 2008 10:12:44 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 15:06:17 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 16 Feb 2008 13:47:28 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.


It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

---

"It probably won't"?

For certain values of probably. A fast relay is going to need a
capacitor somewhere, since no power is available at zero crossing, no
matter how many diodes you use.

---
That's certainly true for your circuit, but not necessarily for mine
since, without a capacitor, your circuit is much more likely to
drive the coil discontinuously during the long wait for the rise of
a new edge after the fall of the current one.

John Popelish pointed that out by noting that the time constant of
the coils is too short to keep the contacts engaged between cycles
with one half-cycle missing, and that was also borne out by my
experiment, where your method failed 4 out of five times using a
random selection of relays.
---

I've found the opposite to be true, empirically.

Here:

COIL R COIL L
MFG PART NO OHMS HENRY REF
--------|------------|-------|--------|-----
AROMAT JT1a-DC12V 150 0.2 A

AROMAT HC3-P-DC12V 160 0.27 B

ORIGINAL SRUT..12VDC 430 0.41 C

AROMAT JW2EN-EDC12V 275 0.52 D

ECI 8501-3023-12 50 0.148 E


For your circuit:


ACIN>--[D1>]--+-----+
|K |
[D2] [COIL]
| |
ACIN>---------+-----+

I got:

PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 24.3 24.3

B >30 N/A

C >30 N/A

D >30 N/A

E 14.8 9.3


For my circuit:

+-----+
ACIN>--|~ +|--[Rs]--+
| | |
| | [COIL]
| | |
ACIN>--|~ -|--------+
+-----+

where Rs is a fixed resistor equal to the DC resistance of the relay
coil, I got:


Your circuit adds the Rs resistance to the coil circuit, decreasing
L/R by about a factor of 2:1, which negates the improvement of going
full-wave.

---
Not true.

By increasing the time constant of the coil circuit, what it does
(in addition to keeping the relay coil from dissipating more power
than it's rated for) is to change the phase of the current in the
coil so that the current never goes to zero, thereby making sure the
armature isn't tempted to abandon the pole piece.
---

Better would be to put the resistor *before* the bridge, so
the diodes clamp the coil voltage close to zero.

---
You've lost me there.

What are you talking about?

--
JF

 

[John Popelish]

HC wrote:

Hey, John, I've been thinking about what you've said here and I think
there is a concept I had not realized before: the voltage drop across
the relay coil will determine its ability to work; no enough voltage
and it won't activate or won't activate fully. The wire before and
after the relay coil (what comes from and returns to the power supply)
will act as series resistors with the load (the coil). So, assuming
that 90% of the rated coil capacity is sufficient to activate the coil
fully (and I can do some testing to be sure) then the coil needs to
"see" 10.8VDC minimum.

So, if I'm right: 10.8VDC across a 308 Ohm coil (as measured on my
relay) gives me a current of 0.035A. Knowing that I can drop 0.6
volts across each leg of wire going to this remote coil for a total of
1.2 volts dropped I can calculate the resistance allowable in the
wire: 0.6V / 0.035A = roughly 17 Ohms. So, I could have a resistance
in series with the relay coil of a total of 34 Ohms. It should not
then matter if this resistance is actually one or more "discrete" (if
I'm using that term correctly) resistors or just the resistance of the
wire that the circuit is comprised of. I could then use that
information to calculate the maximum wire length for ANY relay as long
as I know it's coil resistance, minimum activating voltage, and the
wire resistance.

Is that correct?

Yes. now you are designing. You could also use this
analysis to determine what DC supply voltage you need to
have 12 volts left at the coil after the wiring drops its
voltage.

I will test my relay to find out what the minimum activating voltage
is.

Then derate it by making sure you give it at least 5% more
than that to make it operate even when it is hot (which
raises its coil resistance).

--
Regards,

John Popelish

 

[John Popelish]

HC wrote:

On Feb 16, 11:29 am, John Larkin wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC? I'll try this a little later today and
see what happens. John P has given me information that shows that I
could run these relays remotely with 12VDC without a problem (I had
originally feared drastic line-loss) but I'm still going to try to
find a way to use the 24VAC for no other reason than that I hate to
try something and fail. So, I'll probably wind up using the 12VDC
route, but I'm going to spend some more time trying to use 24VAC "just
because". Ed suggested using a Zener and a transistor and I've come
up with an idea that might work so I'm learning some stuff in the
process.

That circuit uses the inductance of the relay as an
averaging mechanism for the half wave rectified voltage.
For half a cycle, the current increases as the half sine
wave pushes current through the coil. Then, for a half
cycle, the inductance tries to reverse voltage as the
current falls, but the right diode turns on and shorts it
out, so that the current falls more slowly than if the coil
was left open circuit. The open question is whether the
current ripple is smooth enough that the relay will not buzz
and burn the contacts. And that depends on the L/R time
constant of the coil inductance (L) and the coil resistance
(R). If that time constant is much larger than the line
cycle's period, then the current will be fairly steady.

--
Regards,

John Popelish

 

[Ross Herbert]

On Sat, 16 Feb 2008 09:16:08 -0800 (PST), HC <hboothe@gte.net> wrote:

:On Feb 15, 10:41 pm, Ross Herbert <rherb...@bigpond.net.au> wrote:
:> On Fri, 15 Feb 2008 20:19:36 -0800 (PST), HC <hboo...@gte.net> wrote:
:>
:> :Hey, all, I'm not sure if this can be done but here's what I am trying
:> :to do and how I've tried to go about doing it.
:> :
:> :I want to be able to control 120VAC devices some distance away from a
:> :controller, say, up to 100 feet or so. I would like to run the 120VAC
:> :to the device through a switch (relay) at the device with no other
:> :switches or control devices in line from the breaker. Then I would
:> :like to have the controller turn that relay on and off to control the
:> :device. The idea is that I could run the thicker, high-voltage lines
:> :directly to the device and then use smaller wire to operate a relay at
:> :the device to turn it on and off instead of running the high-voltage
:> :wire to each switch I would like to use.
:> :
:> :Since I have a boat-load of low-cost 12VDC relays that can switch up
:> :to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
:> :be a lot more expensive and harder to come by (they seem to be, in my
:> :searching, related to HVAC and other "industrial" uses; they're not
:> :like the overly-abundant 12VDC relays we have for our cars and such) I
:> :would like to use a 12VDC relay at the device. However, I'm afraid
:> :that if I attempt to use 12VDC to control these relays over a distance
:> :like I mention of up to 100 feet that the line-loss will be
:> :significant (on 12VDC). I was thinking that using 24VAC would be much
:> :better (it's higher voltage and it's AC, so line-loss should be quite
:> :a bit less than 12VDC).
:> :
:> :I tried this: I took the 24VAC and rectified it with a single diode
:> half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
:> :that buffered output through a LM78L12 (with input and output caps as
:> :detailed in the datasheet I was reading) and that output to the 12VDC
:> :relay I wanted to operate. When I apply 24VAC to the circuit the
:> :relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
:> :the relay turns off like a marshmallow; slow and makes some light
:> :clicking noises. So, the input cap (2,200 uF) is still powering the
:> :relay coil and is letting it down slowly; at least, that's my hack-boy
:> :assessment: I'm not an expert at any of this stuff. I tried whatever
:> :I could including "pull-down" resistors (if I'm using that term
:> :correctly); I put a 10k resistor from the relay input to ground. I'm
:> :afraid the slow turn-off is going to cause arcing and fry the relay
:> :contacts.
:> :
:> :I don't have a lot of caps that are rated at 35 volts or higher (which
:> :could handle the 24VAC, rectified) so experimentation was limited in
:> :various cap sizes (like, could a 1,000 uF input cap allow the relay to
:> :turn off quickly?); somewhere I got the idea that if you put more
:> :voltage across an electrolytic cap than it can handle it can "explode"
:> r "pop" or "blow-up" or whatever so I'm reluctant to use 16V
:> :electrolytics on what should be 24V or higher.
:> :
:> :Anyway, I hope I've done a good job of explaining what I'm trying to
:> :do and what I've tried to do to achieve it. Is there an effective way
:> :to run a 12VDC relay from a supply circuit of 24VAC?
:> :
:> :Thanks in advance.
:> :
:> :--HC
:>
:> The first thing to sort out is whether your "boat load of 12V relays" is
:> suitable for controlling 120Vac devices. Current rating sounds reasonable but
:> the critical detail is "isolation voltage rating" between the 12Vdc
side(coil)
:> and the 120Vac side (contacts). Any relay which is to be used for this
purpose
:> must meet certain standards and automotive types are not suitable. Many
:> "industrial" types will be suitable.
:>
:> Your approch to producing the dc coil operating voltage is ok and your
reasoning
:> for the long release time is also correct. The release time problem is easy
to
:> overcome. When you want the relay to release, you don't switch off the 24V
:> supply at the ac wall switch, you simply disconnect the relay coil from the
dc
:> output voltage of your power supply.
:
:Hey, Ross, thanks for your reply. You make a very good point about
:the isolation rating/capabilities of the automotive-style relays. I
:do, also, have some 12VDC relays that are rated for 250VAC and 10A and
:a good supplier of those, so I will still have some 12VDC relays I can
:use if the auto style won't work. I'll check to see what the
:isolation is on those auto relays; I bet you're right that it won't be
:enough.
:
:The problem that I see with disconnecting the relay coil from its
ower is that switch would need to be a relay, as well; all of this is
:supposed to be remote. On the bench I did what you suggested
breaking the connection from the coil to the PS) and it works great;
:the problem is how to switch that off remotely without using a relay
r similar device. What John Popelish has suggested is going to work
:for me, I think; the loss across the wire won't be as significant as I
riginally feared.
:
:--HC

Are you referring to the operating method suggested by John Larkin with the
relay powered from 24Vac?

If so, that is a resonable and simple method of using a higher voltage to
activate the 12V coil, but as John Popelish hinted there is a sort of balancing
act to be considered in order to avoid relay buzz.

I don't see a problem with using another relay to switch the DC supply. In fact
that would be an ideal use for one of your cheap 12V automotive relays. You
could use the suggested circuit from John Larkin to power that relay. Even small
gauge wire would not produce too much potential drop in the operating loop of
the power switching relay.

 

[John Larkin]

On Sun, 17 Feb 2008 10:12:44 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 15:06:17 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 16 Feb 2008 13:47:28 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.


It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

---

"It probably won't"?

For certain values of probably. A fast relay is going to need a
capacitor somewhere, since no power is available at zero crossing, no
matter how many diodes you use.

I've found the opposite to be true, empirically.

Here:

COIL R COIL L
MFG PART NO OHMS HENRY REF
--------|------------|-------|--------|-----
AROMAT JT1a-DC12V 150 0.2 A

AROMAT HC3-P-DC12V 160 0.27 B

ORIGINAL SRUT..12VDC 430 0.41 C

AROMAT JW2EN-EDC12V 275 0.52 D

ECI 8501-3023-12 50 0.148 E


For your circuit:


ACIN>--[D1>]--+-----+
|K |
[D2] [COIL]
| |
ACIN>---------+-----+

I got:

PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 24.3 24.3

B >30 N/A

C >30 N/A

D >30 N/A

E 14.8 9.3


For my circuit:

+-----+
ACIN>--|~ +|--[Rs]--+
| | |
| | [COIL]
| | |
ACIN>--|~ -|--------+
+-----+

where Rs is a fixed resistor equal to the DC resistance of the relay
coil, I got:

Your circuit adds the Rs resistance to the coil circuit, decreasing
L/R by about a factor of 2:1, which negates the improvement of going
full-wave. Better would be to put the resistor *before* the bridge, so
the diodes clamp the coil voltage close to zero.

John

 

[John Popelish]

John Fields wrote:

On Sat, 16 Feb 2008 15:06:17 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

---

"It probably won't"?

I have added a third column to your relay data that is L/R,
the time constant of the coil. I think this needs to be
larger than 1/F (1/60Hz=0.0167s) for the two diode circuit
to work well (little vibration).

I've found the opposite to be true, empirically.

Here:

COIL R COIL L
MFG PART NO OHMS HENRY REF L/R
--------|------------|-------|--------|-----|---
AROMAT JT1a-DC12V 150 0.2 A 0.0013s

AROMAT HC3-P-DC12V 160 0.27 B 0.0017s

ORIGINAL SRUT..12VDC 430 0.41 C 0.00095s

AROMAT JW2EN-EDC12V 275 0.52 D 0.0019s

ECI 8501-3023-12 50 0.148 E 0.003s

The last one has the longest L/R time constant, but it still
is too short for me to trust to not vibrate the contacts
with the two diode circuit, unless the supply frequency was
400 Hz (0.0025s/cycle). I am a little surprised it worked
as well as it did on 6 Hz. It must have a rather massive
armature whose mechanical time constant helped smooth the bumps.

For your circuit:


ACIN>--[D1>]--+-----+
|K |
[D2] [COIL]
| |
ACIN>---------+-----+

I got:

PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 24.3 24.3

B >30 N/A

C >30 N/A

D >30 N/A

E 14.8 9.3


For my circuit:

+-----+
ACIN>--|~ +|--[Rs]--+
| | |
| | [COIL]
| | |
ACIN>--|~ -|--------+
+-----+

where Rs is a fixed resistor equal to the DC resistance of the relay
coil, I got:


PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 15.1 9.8

B 15.5 9.3

C 14.6 8.8

D 14.5 10.9

E 18.5 7.5

I just picked the first five 12VDC relays I could find here and
tested them using a VARIAC to vary the input voltage, an AC
voltmeter to measure it and an ohmmeter to determine contact closure
and opening.

So, it looks like my circuit got all five relays to work, while
yours only got one to work, the ECI unit, which is a 12PDT T-BAR
relay. Not something you come across every day.

--
Regards,

John Popelish

 

[John Fields]

On Sat, 16 Feb 2008 15:06:17 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 16 Feb 2008 13:47:28 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.


It probably won't. The second diode keeps current circulating in the
coil inductance between half-cycles, and most DC relays have a pretty
high pullin:dropout ratio.

---

"It probably won't"?

I've found the opposite to be true, empirically.

Here:

COIL R COIL L
MFG PART NO OHMS HENRY REF
--------|------------|-------|--------|-----
AROMAT JT1a-DC12V 150 0.2 A

AROMAT HC3-P-DC12V 160 0.27 B

ORIGINAL SRUT..12VDC 430 0.41 C

AROMAT JW2EN-EDC12V 275 0.52 D

ECI 8501-3023-12 50 0.148 E


For your circuit:


ACIN>--[D1>]--+-----+
|K |
[D2] [COIL]
| |
ACIN>---------+-----+

I got:

PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 24.3 24.3

B >30 N/A

C >30 N/A

D >30 N/A

E 14.8 9.3


For my circuit:

+-----+
ACIN>--|~ +|--[Rs]--+
| | |
| | [COIL]
| | |
ACIN>--|~ -|--------+
+-----+

where Rs is a fixed resistor equal to the DC resistance of the relay
coil, I got:


PULLIN DROPOUT
REF VRMS VRMS
-----+--------+---------
A 15.1 9.8

B 15.5 9.3

C 14.6 8.8

D 14.5 10.9

E 18.5 7.5

I just picked the first five 12VDC relays I could find here and
tested them using a VARIAC to vary the input voltage, an AC
voltmeter to measure it and an ohmmeter to determine contact closure
and opening.

So, it looks like my circuit got all five relays to work, while
yours only got one to work, the ECI unit, which is a 12PDT T-BAR
relay. Not something you come across every day.

--
JF

 

[ehsjr]

HC wrote:

On Feb 15, 11:28 pm, ehsjr <eh...@bellatlantic.net> wrote:

HC wrote:

Hey, all, I'm not sure if this can be done but here's what I am trying
to do and how I've tried to go about doing it.

I want to be able to control 120VAC devices some distance away from a
controller, say, up to 100 feet or so. I would like to run the 120VAC
to the device through a switch (relay) at the device with no other
switches or control devices in line from the breaker. Then I would
like to have the controller turn that relay on and off to control the
device. The idea is that I could run the thicker, high-voltage lines
directly to the device and then use smaller wire to operate a relay at
the device to turn it on and off instead of running the high-voltage
wire to each switch I would like to use.

Since I have a boat-load of low-cost 12VDC relays that can switch up
to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
be a lot more expensive and harder to come by (they seem to be, in my
searching, related to HVAC and other "industrial" uses; they're not
like the overly-abundant 12VDC relays we have for our cars and such) I
would like to use a 12VDC relay at the device. However, I'm afraid
that if I attempt to use 12VDC to control these relays over a distance
like I mention of up to 100 feet that the line-loss will be
significant (on 12VDC). I was thinking that using 24VAC would be much
better (it's higher voltage and it's AC, so line-loss should be quite
a bit less than 12VDC).

I tried this: I took the 24VAC and rectified it with a single diode
(half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
that buffered output through a LM78L12 (with input and output caps as
detailed in the datasheet I was reading) and that output to the 12VDC
relay I wanted to operate. When I apply 24VAC to the circuit the
relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
the relay turns off like a marshmallow; slow and makes some light
clicking noises. So, the input cap (2,200 uF) is still powering the
relay coil and is letting it down slowly; at least, that's my hack-boy
assessment: I'm not an expert at any of this stuff. I tried whatever
I could including "pull-down" resistors (if I'm using that term
correctly); I put a 10k resistor from the relay input to ground. I'm
afraid the slow turn-off is going to cause arcing and fry the relay
contacts.

I don't have a lot of caps that are rated at 35 volts or higher (which
could handle the 24VAC, rectified) so experimentation was limited in
various cap sizes (like, could a 1,000 uF input cap allow the relay to
turn off quickly?); somewhere I got the idea that if you put more
voltage across an electrolytic cap than it can handle it can "explode"
or "pop" or "blow-up" or whatever so I'm reluctant to use 16V
electrolytics on what should be 24V or higher.

Anyway, I hope I've done a good job of explaining what I'm trying to
do and what I've tried to do to achieve it. Is there an effective way
to run a 12VDC relay from a supply circuit of 24VAC?

Thanks in advance.

--HC

John Popelish gave you the answer - you won't
have a problem driving those 12 volt relays
with 100 feet of wire.

But as a learning/experimental/fun thing, you could
try using a zener & transistor circuit with the 24 volt
approach you mentioned.

Ed


Hey, Ed, I'm game. I did some thinking along the lines you suggested
and the best I can come up with is this: rectify and buffer (diode and
capacitor) the 24VAC which gives me a little over 30DCV. Run that
through my voltage regulator (LM78L12) to get 12VDC. That can power
my 12VDC relay coil. I need a way to break that connection crisply at
the relay coil so I could use a Zener diode from the (roughly) 33VDC
supply (rectified and buffered) to the base of a transistor. I would
select a Zener that could pass, say, over 30 volts (1N4751 might work
according to the table I found here: http://sound.westhost.com/appnotes/an008.htm)
and, with proper resistors to bias the transistor, it would turn on
when power was supplied, switching on the relay coil. When power was
cut the voltage/current from the capacitor would start to drop
relatively slowly. There would be enough for the LM78L12 to continue
to provide 12VDC to the relay coil but would quickly fall below the
Zener voltage and turn the transistor off sharply in turn shutting off
the relay coil sharply.

Is that right? I had not thought about something like that. I'm not
sure I got that right but if it is right that is pretty cool.

Thanks for the suggestion. I don't have any Zeners in my collection
of stuff and my supply store is about 100 miles away so testing is
going to be impossible until I make the trip. I dunno, I might get a
wild hair and go later today.

--HC

You understand it correctly. There's a little more to it,
however. You would want to drop the relay when the supply
is voltage has dropped considerably lower than 30 volts.

If you want to experiment with it, you don't need a zener
in the neighborhood of 30 V - use a 6V zener, and add
a two resistor voltage divider. Here's a diagram:

+ ---+---+---7812---+------+------+
| | | | | | c
| [C1] | [C2] [Relay] [D2] 1N914
R1[1K] | | | | | a
| +----+-----+ +------+
| | 6v /
+------------[Zd]---| NPN
| | \
R2[1K] | |
| | |
Gnd -+--------+------------+

C1 & C2 are the standard caps used with a 7812, typically
..33 uF for C1 and .1 uF for C2. The 1N914 (can be a 1N4148,
1N4001 etc) diode is installed "backwards" (that is, with the
striped end toward +) across the relay coil to prevent the
inductive spike that occurs when the relay is suddenly
de-energized from doing damage.

R1 and R2 form a voltage divider, so the zener "sees"
1/2 the supply voltage. That also limits the current
into the base and reduces the dissipation in the zener.
As long as the supply voltage has not dropped below
about 12 volts, the relay stays energized, but when the
supply drops below 12 volts the relay drops out.
A general purpose NPN transistor, like a 2N2222,
will work well with your relay.

By the way, the R1/R2 voltage divider is not mandatory
if you change the zener diode. You could eliminate
R2 and use a zener rated around 9 to 12V. It is also
possible to design the circuit with the zener on the
output side of the 7812. There's a lot of room for
experimentation.


Ed

 

[John Fields]

On Sat, 16 Feb 2008 10:05:21 -0800 (PST), HC <hboothe@gte.net>
wrote:

On Feb 16, 11:29 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC?

---
No, it'll be 1/2 wave rectified AC, and John's saying that by using
only the positive portion of the sine wave the relay will dissipate
the same power it would if it were hooked up to 12VDC. That's a
trick often used to good advantage with devices like heaters, but I
don't think it's a good idea for your application because the period
of the signal will be about 17ms, which may make the relay chatter.

A better way, IMO, would be to full-wave rectify the AC to get the
period of the DC pulses down to 8ms and then use a 300 ohm 1 watt
resistor in series with the coil in order to get the DC down to 12V.

Either that or (since you've got a lot of them lying around) use
another relay coil in series with the first one to drop the voltage.
That way you'd also have twice the current carrying capacity of the
contacts if you wired them in parallel.

24AWG wire has a resistance of 2.57 ohms per 100', so your 200'
(100' out and 100' back) will have a resistance of about 5 ohms,
which will have an insignificant effect on the operation of the
relay(s).

--
JF

 

[HC]

On Feb 16, 11:29 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 15 Feb 2008 20:19:36 -0800 (PST), HC <hboo...@gte.net> wrote:
Hey, all, I'm not sure if this can be done but here's what I am trying
to do and how I've tried to go about doing it.

I want to be able to control 120VAC devices some distance away from a
controller, say, up to 100 feet or so. I would like to run the 120VAC
to the device through a switch (relay) at the device with no other
switches or control devices in line from the breaker. Then I would
like to have the controller turn that relay on and off to control the
device. The idea is that I could run the thicker, high-voltage lines
directly to the device and then use smaller wire to operate a relay at
the device to turn it on and off instead of running the high-voltage
wire to each switch I would like to use.

Since I have a boat-load of low-cost 12VDC relays that can switch up
to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
be a lot more expensive and harder to come by (they seem to be, in my
searching, related to HVAC and other "industrial" uses; they're not
like the overly-abundant 12VDC relays we have for our cars and such) I
would like to use a 12VDC relay at the device. However, I'm afraid
that if I attempt to use 12VDC to control these relays over a distance
like I mention of up to 100 feet that the line-loss will be
significant (on 12VDC). I was thinking that using 24VAC would be much
better (it's higher voltage and it's AC, so line-loss should be quite
a bit less than 12VDC).

I tried this: I took the 24VAC and rectified it with a single diode
(half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
that buffered output through a LM78L12 (with input and output caps as
detailed in the datasheet I was reading) and that output to the 12VDC
relay I wanted to operate. When I apply 24VAC to the circuit the
relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
the relay turns off like a marshmallow; slow and makes some light
clicking noises. So, the input cap (2,200 uF) is still powering the
relay coil and is letting it down slowly; at least, that's my hack-boy
assessment: I'm not an expert at any of this stuff. I tried whatever
I could including "pull-down" resistors (if I'm using that term
correctly); I put a 10k resistor from the relay input to ground. I'm
afraid the slow turn-off is going to cause arcing and fry the relay
contacts.

I don't have a lot of caps that are rated at 35 volts or higher (which
could handle the 24VAC, rectified) so experimentation was limited in
various cap sizes (like, could a 1,000 uF input cap allow the relay to
turn off quickly?); somewhere I got the idea that if you put more
voltage across an electrolytic cap than it can handle it can "explode"
or "pop" or "blow-up" or whatever so I'm reluctant to use 16V
electrolytics on what should be 24V or higher.

Anyway, I hope I've done a good job of explaining what I'm trying to
do and what I've tried to do to achieve it. Is there an effective way
to run a 12VDC relay from a supply circuit of 24VAC?

Thanks in advance.

--HC

Assuming the relay contacts can handle the load, do this:

|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+

where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

John, thank you for the reply. I put that into Notepad (to see it in
a fixed-width font) and that's pretty simple. But won't the voltage
across the relay still be AC? I'll try this a little later today and
see what happens. John P has given me information that shows that I
could run these relays remotely with 12VDC without a problem (I had
originally feared drastic line-loss) but I'm still going to try to
find a way to use the 24VAC for no other reason than that I hate to
try something and fail. So, I'll probably wind up using the 12VDC
route, but I'm going to spend some more time trying to use 24VAC "just
because". Ed suggested using a Zener and a transistor and I've come
up with an idea that might work so I'm learning some stuff in the
process.

--HC

 

[HC]

On Feb 16, 12:44 am, John Popelish <jpopel...@rica.net> wrote:

HC wrote:
Hey, John, thank you for your reply. That wire link is cool and I've
bookmarked that.

What I've been intending to use as the "messenger" wire for the relay
is some 24 gauge wire I have already (from where and when I don't
recall; it's followed me for about 10 years now).

If I understand you correctly, the resistance of the wire for this
distance and for this load (the relay coil) is not significant enough
to worry about, within some parameters. I went and measured my
relay's coil resistance and found it to be 308 ohms. I then applied
11.89 volts (according to my DVM) from my proto-board and measured
37.7 mA (using the same DVM) across the relay coil with that voltage.
I guess all I needed to measure was the 308 ohms, but since I was
there it seemed useful to measure the rest. I ran it through the
equations I have for power (Watts) and Ohm's Law, just to see if my
measurements were right and got numbers that seem good.

1/20th of the 308 would be 15.4 ohms, so, you're saying I could add up
to 15.4 ohms of round-trip "messenger" wire to the relay, on a 12VDC
supply, and it would be okay? So, looking at the chart you linked to,
24 gauge has 38.958 feet per ohm so I could run, round-trip, roughly
600 feet?

Sounds about right. 300 feet out and back, with about a 5%
voltage loss. Most DC relays will pull in with less than
90% of their rated coil voltage.

I really am an amateur at this stuff and I've seen your posts all over
the place helping people so I ask the following with no ill-will or
attitude towards you; you said that a 2W coil would be about a 288 ohm
coil. I tried to get that (because I want to be able to calculate
this information for other relays and to better understand this stuff)
assuming a 12 volt DC supply and I did a calculation for current with
P = I E so I = 2W / 12V and got I = 0.1667A. I put that into E = I R
solving for R = 12V / 0.1667A, R = 72 ohms. What have I done wrong?

You caught me making a math error. I was using P=V^2/R, or
2W=12V*12V/R, but I hit a wrong key, somewhere. Its getting
late, here. Sorry.

By the way, this means that your 308 ohm coil will consume
only about 12*12/308= 0.47 watts. That makes the wiring
resistance a lot less of a problem.

Thank you very much again for your reply. It helped a lot. I will
set up a test rig to verify that the numbers I work out on paper are
accurate but I'm confident this will work. This is going to simplify
my project quite a bit (not having to use a 24 VAC supply to try to
run 12 VDC relays).

You should also test your relay with a slowly rising coil
voltage and make sure it pulls in at well below 12 volts.

--HC

--
Regards,

John Popelish

Hey, John, I've been thinking about what you've said here and I think
there is a concept I had not realized before: the voltage drop across
the relay coil will determine its ability to work; no enough voltage
and it won't activate or won't activate fully. The wire before and
after the relay coil (what comes from and returns to the power supply)
will act as series resistors with the load (the coil). So, assuming
that 90% of the rated coil capacity is sufficient to activate the coil
fully (and I can do some testing to be sure) then the coil needs to
"see" 10.8VDC minimum.

So, if I'm right: 10.8VDC across a 308 Ohm coil (as measured on my
relay) gives me a current of 0.035A. Knowing that I can drop 0.6
volts across each leg of wire going to this remote coil for a total of
1.2 volts dropped I can calculate the resistance allowable in the
wire: 0.6V / 0.035A = roughly 17 Ohms. So, I could have a resistance
in series with the relay coil of a total of 34 Ohms. It should not
then matter if this resistance is actually one or more "discrete" (if
I'm using that term correctly) resistors or just the resistance of the
wire that the circuit is comprised of. I could then use that
information to calculate the maximum wire length for ANY relay as long
as I know it's coil resistance, minimum activating voltage, and the
wire resistance.

Is that correct?

I will test my relay to find out what the minimum activating voltage
is.

Thank you again.

--HC

 

[HC]

On Feb 15, 11:28 pm, ehsjr <eh...@bellatlantic.net> wrote:

HC wrote:
Hey, all, I'm not sure if this can be done but here's what I am trying
to do and how I've tried to go about doing it.

I want to be able to control 120VAC devices some distance away from a
controller, say, up to 100 feet or so. I would like to run the 120VAC
to the device through a switch (relay) at the device with no other
switches or control devices in line from the breaker. Then I would
like to have the controller turn that relay on and off to control the
device. The idea is that I could run the thicker, high-voltage lines
directly to the device and then use smaller wire to operate a relay at
the device to turn it on and off instead of running the high-voltage
wire to each switch I would like to use.

Since I have a boat-load of low-cost 12VDC relays that can switch up
to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
be a lot more expensive and harder to come by (they seem to be, in my
searching, related to HVAC and other "industrial" uses; they're not
like the overly-abundant 12VDC relays we have for our cars and such) I
would like to use a 12VDC relay at the device. However, I'm afraid
that if I attempt to use 12VDC to control these relays over a distance
like I mention of up to 100 feet that the line-loss will be
significant (on 12VDC). I was thinking that using 24VAC would be much
better (it's higher voltage and it's AC, so line-loss should be quite
a bit less than 12VDC).

I tried this: I took the 24VAC and rectified it with a single diode
(half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
that buffered output through a LM78L12 (with input and output caps as
detailed in the datasheet I was reading) and that output to the 12VDC
relay I wanted to operate. When I apply 24VAC to the circuit the
relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
the relay turns off like a marshmallow; slow and makes some light
clicking noises. So, the input cap (2,200 uF) is still powering the
relay coil and is letting it down slowly; at least, that's my hack-boy
assessment: I'm not an expert at any of this stuff. I tried whatever
I could including "pull-down" resistors (if I'm using that term
correctly); I put a 10k resistor from the relay input to ground. I'm
afraid the slow turn-off is going to cause arcing and fry the relay
contacts.

I don't have a lot of caps that are rated at 35 volts or higher (which
could handle the 24VAC, rectified) so experimentation was limited in
various cap sizes (like, could a 1,000 uF input cap allow the relay to
turn off quickly?); somewhere I got the idea that if you put more
voltage across an electrolytic cap than it can handle it can "explode"
or "pop" or "blow-up" or whatever so I'm reluctant to use 16V
electrolytics on what should be 24V or higher.

Anyway, I hope I've done a good job of explaining what I'm trying to
do and what I've tried to do to achieve it. Is there an effective way
to run a 12VDC relay from a supply circuit of 24VAC?

Thanks in advance.

--HC

John Popelish gave you the answer - you won't
have a problem driving those 12 volt relays
with 100 feet of wire.

But as a learning/experimental/fun thing, you could
try using a zener & transistor circuit with the 24 volt
approach you mentioned.

Ed

Hey, Ed, I'm game. I did some thinking along the lines you suggested
and the best I can come up with is this: rectify and buffer (diode and
capacitor) the 24VAC which gives me a little over 30DCV. Run that
through my voltage regulator (LM78L12) to get 12VDC. That can power
my 12VDC relay coil. I need a way to break that connection crisply at
the relay coil so I could use a Zener diode from the (roughly) 33VDC
supply (rectified and buffered) to the base of a transistor. I would
select a Zener that could pass, say, over 30 volts (1N4751 might work
according to the table I found here: http://sound.westhost.com/appnotes/an008.htm)
and, with proper resistors to bias the transistor, it would turn on
when power was supplied, switching on the relay coil. When power was
cut the voltage/current from the capacitor would start to drop
relatively slowly. There would be enough for the LM78L12 to continue
to provide 12VDC to the relay coil but would quickly fall below the
Zener voltage and turn the transistor off sharply in turn shutting off
the relay coil sharply.

Is that right? I had not thought about something like that. I'm not
sure I got that right but if it is right that is pretty cool.

Thanks for the suggestion. I don't have any Zeners in my collection
of stuff and my supply store is about 100 miles away so testing is
going to be impossible until I make the trip. I dunno, I might get a
wild hair and go later today.

--HC

 

[HC]

On Feb 15, 10:41 pm, Ross Herbert <rherb...@bigpond.net.au> wrote:

On Fri, 15 Feb 2008 20:19:36 -0800 (PST), HC <hboo...@gte.net> wrote:

:Hey, all, I'm not sure if this can be done but here's what I am trying
:to do and how I've tried to go about doing it.
:
:I want to be able to control 120VAC devices some distance away from a
:controller, say, up to 100 feet or so. I would like to run the 120VAC
:to the device through a switch (relay) at the device with no other
:switches or control devices in line from the breaker. Then I would
:like to have the controller turn that relay on and off to control the
:device. The idea is that I could run the thicker, high-voltage lines
:directly to the device and then use smaller wire to operate a relay at
:the device to turn it on and off instead of running the high-voltage
:wire to each switch I would like to use.
:
:Since I have a boat-load of low-cost 12VDC relays that can switch up
:to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
:be a lot more expensive and harder to come by (they seem to be, in my
:searching, related to HVAC and other "industrial" uses; they're not
:like the overly-abundant 12VDC relays we have for our cars and such) I
:would like to use a 12VDC relay at the device. However, I'm afraid
:that if I attempt to use 12VDC to control these relays over a distance
:like I mention of up to 100 feet that the line-loss will be
:significant (on 12VDC). I was thinking that using 24VAC would be much
:better (it's higher voltage and it's AC, so line-loss should be quite
:a bit less than 12VDC).
:
:I tried this: I took the 24VAC and rectified it with a single diode
half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
:that buffered output through a LM78L12 (with input and output caps as
:detailed in the datasheet I was reading) and that output to the 12VDC
:relay I wanted to operate. When I apply 24VAC to the circuit the
:relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
:the relay turns off like a marshmallow; slow and makes some light
:clicking noises. So, the input cap (2,200 uF) is still powering the
:relay coil and is letting it down slowly; at least, that's my hack-boy
:assessment: I'm not an expert at any of this stuff. I tried whatever
:I could including "pull-down" resistors (if I'm using that term
:correctly); I put a 10k resistor from the relay input to ground. I'm
:afraid the slow turn-off is going to cause arcing and fry the relay
:contacts.
:
:I don't have a lot of caps that are rated at 35 volts or higher (which
:could handle the 24VAC, rectified) so experimentation was limited in
:various cap sizes (like, could a 1,000 uF input cap allow the relay to
:turn off quickly?); somewhere I got the idea that if you put more
:voltage across an electrolytic cap than it can handle it can "explode"
r "pop" or "blow-up" or whatever so I'm reluctant to use 16V
:electrolytics on what should be 24V or higher.
:
:Anyway, I hope I've done a good job of explaining what I'm trying to
:do and what I've tried to do to achieve it. Is there an effective way
:to run a 12VDC relay from a supply circuit of 24VAC?
:
:Thanks in advance.
:
:--HC

The first thing to sort out is whether your "boat load of 12V relays" is
suitable for controlling 120Vac devices. Current rating sounds reasonable but
the critical detail is "isolation voltage rating" between the 12Vdc side(coil)
and the 120Vac side (contacts). Any relay which is to be used for this purpose
must meet certain standards and automotive types are not suitable. Many
"industrial" types will be suitable.

Your approch to producing the dc coil operating voltage is ok and your reasoning
for the long release time is also correct. The release time problem is easy to
overcome. When you want the relay to release, you don't switch off the 24V
supply at the ac wall switch, you simply disconnect the relay coil from the dc
output voltage of your power supply.

Hey, Ross, thanks for your reply. You make a very good point about
the isolation rating/capabilities of the automotive-style relays. I
do, also, have some 12VDC relays that are rated for 250VAC and 10A and
a good supplier of those, so I will still have some 12VDC relays I can
use if the auto style won't work. I'll check to see what the
isolation is on those auto relays; I bet you're right that it won't be
enough.

The problem that I see with disconnecting the relay coil from its
power is that switch would need to be a relay, as well; all of this is
supposed to be remote. On the bench I did what you suggested
(breaking the connection from the coil to the PS) and it works great;
the problem is how to switch that off remotely without using a relay
or similar device. What John Popelish has suggested is going to work
for me, I think; the loss across the wire won't be as significant as I
originally feared.

--HC

 

[John Larkin]

On Fri, 15 Feb 2008 20:19:36 -0800 (PST), HC <hboothe@gte.net> wrote:

Hey, all, I'm not sure if this can be done but here's what I am trying
to do and how I've tried to go about doing it.

I want to be able to control 120VAC devices some distance away from a
controller, say, up to 100 feet or so. I would like to run the 120VAC
to the device through a switch (relay) at the device with no other
switches or control devices in line from the breaker. Then I would
like to have the controller turn that relay on and off to control the
device. The idea is that I could run the thicker, high-voltage lines
directly to the device and then use smaller wire to operate a relay at
the device to turn it on and off instead of running the high-voltage
wire to each switch I would like to use.

Since I have a boat-load of low-cost 12VDC relays that can switch up
to 250VAC and 15 amps and since 24VAC relays seem, in my searching, to
be a lot more expensive and harder to come by (they seem to be, in my
searching, related to HVAC and other "industrial" uses; they're not
like the overly-abundant 12VDC relays we have for our cars and such) I
would like to use a 12VDC relay at the device. However, I'm afraid
that if I attempt to use 12VDC to control these relays over a distance
like I mention of up to 100 feet that the line-loss will be
significant (on 12VDC). I was thinking that using 24VAC would be much
better (it's higher voltage and it's AC, so line-loss should be quite
a bit less than 12VDC).

I tried this: I took the 24VAC and rectified it with a single diode
(half-wave) then buffered it to "ground" with a 2,200 uF cap. I ran
that buffered output through a LM78L12 (with input and output caps as
detailed in the datasheet I was reading) and that output to the 12VDC
relay I wanted to operate. When I apply 24VAC to the circuit the
relay turns on like a mousetrap: SNAP! But, when I remove the 24VAC
the relay turns off like a marshmallow; slow and makes some light
clicking noises. So, the input cap (2,200 uF) is still powering the
relay coil and is letting it down slowly; at least, that's my hack-boy
assessment: I'm not an expert at any of this stuff. I tried whatever
I could including "pull-down" resistors (if I'm using that term
correctly); I put a 10k resistor from the relay input to ground. I'm
afraid the slow turn-off is going to cause arcing and fry the relay
contacts.

I don't have a lot of caps that are rated at 35 volts or higher (which
could handle the 24VAC, rectified) so experimentation was limited in
various cap sizes (like, could a 1,000 uF input cap allow the relay to
turn off quickly?); somewhere I got the idea that if you put more
voltage across an electrolytic cap than it can handle it can "explode"
or "pop" or "blow-up" or whatever so I'm reluctant to use 16V
electrolytics on what should be 24V or higher.

Anyway, I hope I've done a good job of explaining what I'm trying to
do and what I've tried to do to achieve it. Is there an effective way
to run a 12VDC relay from a supply circuit of 24VAC?

Thanks in advance.

--HC

Assuming the relay contacts can handle the load, do this:


|\|
---------------| |----------------+---------------+
|/| | |
| |
24 ac | |
in relay -----
coil / \
| ---
| |
| |
| |
----------------------------------+---------------+


where the goofy looking things are any old power diodes, 1N4001 or
whatever.

The relay coil will see pretty close to the equivalent of 12 volts.

John

 

[John Popelish]

HC wrote:

Hey, John, thank you for your reply. That wire link is cool and I've
bookmarked that.

What I've been intending to use as the "messenger" wire for the relay
is some 24 gauge wire I have already (from where and when I don't
recall; it's followed me for about 10 years now).

If I understand you correctly, the resistance of the wire for this
distance and for this load (the relay coil) is not significant enough
to worry about, within some parameters. I went and measured my
relay's coil resistance and found it to be 308 ohms. I then applied
11.89 volts (according to my DVM) from my proto-board and measured
37.7 mA (using the same DVM) across the relay coil with that voltage.
I guess all I needed to measure was the 308 ohms, but since I was
there it seemed useful to measure the rest. I ran it through the
equations I have for power (Watts) and Ohm's Law, just to see if my
measurements were right and got numbers that seem good.

1/20th of the 308 would be 15.4 ohms, so, you're saying I could add up
to 15.4 ohms of round-trip "messenger" wire to the relay, on a 12VDC
supply, and it would be okay? So, looking at the chart you linked to,
24 gauge has 38.958 feet per ohm so I could run, round-trip, roughly
600 feet?

Sounds about right. 300 feet out and back, with about a 5%
voltage loss. Most DC relays will pull in with less than
90% of their rated coil voltage.

I really am an amateur at this stuff and I've seen your posts all over
the place helping people so I ask the following with no ill-will or
attitude towards you; you said that a 2W coil would be about a 288 ohm
coil. I tried to get that (because I want to be able to calculate
this information for other relays and to better understand this stuff)
assuming a 12 volt DC supply and I did a calculation for current with
P = I E so I = 2W / 12V and got I = 0.1667A. I put that into E = I R
solving for R = 12V / 0.1667A, R = 72 ohms. What have I done wrong?

You caught me making a math error. I was using P=V^2/R, or
2W=12V*12V/R, but I hit a wrong key, somewhere. Its getting
late, here. Sorry.

By the way, this means that your 308 ohm coil will consume
only about 12*12/308= 0.47 watts. That makes the wiring
resistance a lot less of a problem.

Thank you very much again for your reply. It helped a lot. I will
set up a test rig to verify that the numbers I work out on paper are
accurate but I'm confident this will work. This is going to simplify
my project quite a bit (not having to use a 24 VAC supply to try to
run 12 VDC relays).

You should also test your relay with a slowly rising coil
voltage and make sure it pulls in at well below 12 volts.

--HC

--
Regards,

John Popelish