Contactor coil: 50 Hz vs. 60 Hz

[DaveC]

It will be happy. If it were a 60Hz on 50, it would not be as happy.
[David Lesher]

Interesting. Good to know. It's stuff like this I learn here that I wouldn't
otherwise know.

Cheers,
Dave

 

[John Fields]

On Tue, 30 Aug 2011 02:58:19 +0000 (UTC), David Lesher
<wb8foz@panix.com> wrote:

John Fields <jfields@austininstruments.com> writes:


I have an old P&B MR5A here with a 240V 50/60Hz coil.

The coil has a resistance of 4800 ohms, and an open inductance of 14.5
henrys, so it has an impedance of 6616 ohms at 50 Hz, and 7270 ohms at
60 Hz.

What's the inductance while closed?

---
16 henries, but that measurement was made by closing the armature
manually.

If there's any real interest I can measure it energized.

--
JF

 

[John Fields]

On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT <meow2222@care2.com>
wrote:

IME relays pull in at in the region of half rated voltage, and dc
ratings are typically about half the voltage of the ac rating, which
gives an idea of how much current is determined by L and how much by
R.

---
IME, most relays (with either AC or DC coils) are guaranteed to pull
in at about 80% of their rated coil voltage, so I'm at a loss trying
to understand what you meant by: "dc ratings are typically about half
of the ac rating."

Can you elaborate, please?
---

Running your relay on 220v 60Hz it will work fine.

---
Knowing nothing about the contactor, other than that it's specified to
energize when 240V 50Hz is placed across the coil, your imprimatur is
premature.
---

Contact closing speed will be slightly slower. Margin will be reduced,
but its only being reduced from enormous to slightly less enormous,
so its a non-issue except in very unusual situations.

---
It seems you've forgotten that when the armature makes, and the
magnetic circuit is closed, the inductance of the coil will rise.

Such being the case, the current in it will diminish, reducing the
hold on the armature and making the contacts more likely to chatter.
---

The vibration tolerance of
the contacts will be little affected in practice; if your environment
is harsh enough to shake the relay contact open, then you've got
bigger worries than contacts crackling.

If instead you meant you would use it on 110v 60Hz, then dont. But you
could use diodes to get a higher dc voltage and use that.

---
Interesting conjecture.


Something like this?

+-----+
120AC>--|~ +|----+
| | |
| | [COIL]
| | |
120AC>--|~ -|----+
+-----+

Since the coil has an impedance of about 6600 ohms at 50Hz, then the
current through it will be:

E 240V
I = --- = ------- = 0.036A = 36mA
Z 6600R

Then, since the coil has a resistance of 4800 ohms, the DC voltage
across it required to force 36mA through it would be:

E = IR = 0.036A * 4800R ~ 174V.

The peak voltage out of the bridge would be:

E = RMS * sqrt(2) = 120 * 1.414 ~ 170V.

Pretty close, but at 120Hz, the reactance of the coil would increase,
limiting the current to something less than the 36mA needed to close
the armature.

However, the reactance of the coil will smooth the current and the
addition of a capacitor in parallel with the coil will remove some of
the ripple and allow the coil to see more nearly pure DC.

Here's a simulation showing both ways:

Version 4
SHEET 1 880 680
WIRE -144 16 -304 16
WIRE 112 16 -144 16
WIRE 448 16 288 16
WIRE 704 16 448 16
WIRE -304 80 -304 16
WIRE 288 80 288 16
WIRE 448 80 448 64
WIRE 480 80 448 80
WIRE 592 80 560 80
WIRE 704 80 704 64
WIRE 704 80 672 80
WIRE -144 112 -144 80
WIRE -112 112 -144 112
WIRE 0 112 -32 112
WIRE 112 112 112 80
WIRE 112 112 80 112
WIRE -144 160 -144 112
WIRE 112 160 112 112
WIRE 448 160 448 80
WIRE 544 160 448 160
WIRE 704 160 704 80
WIRE 704 160 608 160
WIRE -304 224 -304 160
WIRE -144 224 -304 224
WIRE 112 224 -144 224
WIRE 288 224 288 160
WIRE 448 224 288 224
WIRE 704 224 448 224
WIRE -304 272 -304 224
WIRE 288 272 288 224
FLAG -304 272 0
FLAG 288 272 0
SYMBOL ind -128 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 15
SYMBOL voltage -304 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60)
SYMBOL diode -160 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 128 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 96 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode -128 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL res -16 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 4800
SYMBOL ind 464 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 15
SYMBOL voltage 288 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 170 60)
SYMBOL diode 432 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D5
SYMATTR Value MUR460
SYMBOL diode 720 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D6
SYMATTR Value MUR460
SYMBOL diode 688 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D7
SYMATTR Value MUR460
SYMBOL diode 464 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D8
SYMATTR Value MUR460
SYMBOL res 576 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 4800
SYMBOL cap 608 144 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 10ľ
TEXT -298 246 Left 0 !.tran .05

If the relay is spec'ed as "must make" at 80% of rated current through
the coil (~29mA), then note that with a 10ľF cap in parallel with the
coil the relay will _always_ make using full-wave rectified 120V 60Hz
mains.

--
JF

 

[NT]

On Aug 30, 12:39 pm, John Fields <jfie...@austininstruments.com>
wrote:

On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT <meow2...@care2.com
wrote:

IME relays pull in at in the region of half rated voltage, and dc
ratings are typically about half the voltage of the ac rating, which
gives an idea of how much current is determined by L and how much by
R.

---
IME, most relays (with either AC or DC coils) are guaranteed to pull
in at about 80% of their rated coil voltage, so I'm at a loss trying
to understand what you meant by: "dc ratings are typically about half
of the ac rating."

Can you elaborate, please?

When relays have dual ratings for ac and dc, its normal for the dc
voltage rating to be half the ac voltage rating.

Running your relay on 220v 60Hz it will work fine.

Knowing nothing about the contactor, other than that it's specified to
energize when 240V 50Hz is placed across the coil, your imprimatur is
premature.

I really dont agree. I do know the basics about relays, and one
normally finds that pull-in occurs at around 50% rated voltage. The OP
is welcome to test theirs to see if it behaves the usual way.

Contact closing speed will be slightly slower. Margin will be reduced,
but its only being reduced from enormous to slightly less enormous,
so its a non-issue except in very unusual situations.

It seems you've forgotten that when the armature makes, and the
magnetic circuit is closed, the inductance of the coil will rise.

I dont know why you think I've forgotten it. What's relevant here is
inductance in the closed position.

Such being the case, the current in it will diminish,

true with all relays under all ac conditions. Theyre designed to work
that way.

reducing the
hold on the armature and making the contacts more likely to chatter.

No, its exactly how theyre designed to operate.

The vibration tolerance of
the contacts will be little affected in practice; if your environment
is harsh enough to shake the relay contact open, then you've got
bigger worries than contacts crackling.
If instead you meant you would use it on 110v 60Hz, then dont. But you
could use diodes to get a higher dc voltage and use that.

Interesting conjecture.

Where's the conjecture? I get the feeling you could do with bringing
your skills up to speed on relays.

Something like this?

        +-----+
120AC>--|~   +|----+
        |     |    |  
        |     |  [COIL]
        |     |    |
120AC>--|~   -|----+
        +-----+

That would work.

Since the coil has an impedance of about 6600 ohms at 50Hz, then the
current through it will be:

          E      240V
     I = --- = ------- = 0.036A = 36mA
          Z     6600R

Then, since the coil has a resistance of 4800 ohms, the DC voltage
across it required to force 36mA through it would be:

     E = IR = 0.036A * 4800R ~ 174V.

You're not saying where you got those figures from. Typically dc
rating is half ac rating.

The peak voltage out of the bridge would be:

     E = RMS * sqrt(2) = 120 * 1.414 ~ 170V.

Pretty close, but at 120Hz, the reactance of the coil would increase,
limiting the current to something less than the 36mA needed to close
the armature.

The effect of the relay's inductance, when run off a BR, is simply to
smooth the current flow somewhat. Mean current remains much the same.
So we're looking for 120v rms, which is what the BR would deliver.

However, the reactance of the coil will smooth the current and the
addition of a capacitor in parallel with the coil will remove some of
the ripple and allow the coil to see more nearly pure DC.

and overheat the relay by increasing its rms dc voltage to above 120v.

Here's a simulation showing both ways:

Version 4
SHEET 1 880 680
WIRE -144 16 -304 16
WIRE 112 16 -144 16
WIRE 448 16 288 16
WIRE 704 16 448 16
WIRE -304 80 -304 16
WIRE 288 80 288 16
WIRE 448 80 448 64
WIRE 480 80 448 80
WIRE 592 80 560 80
WIRE 704 80 704 64
WIRE 704 80 672 80
WIRE -144 112 -144 80
WIRE -112 112 -144 112
WIRE 0 112 -32 112
WIRE 112 112 112 80
WIRE 112 112 80 112
WIRE -144 160 -144 112
WIRE 112 160 112 112
WIRE 448 160 448 80
WIRE 544 160 448 160
WIRE 704 160 704 80
WIRE 704 160 608 160
WIRE -304 224 -304 160
WIRE -144 224 -304 224
WIRE 112 224 -144 224
WIRE 288 224 288 160
WIRE 448 224 288 224
WIRE 704 224 448 224
WIRE -304 272 -304 224
WIRE 288 272 288 224
FLAG -304 272 0
FLAG 288 272 0
SYMBOL ind -128 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 15
SYMBOL voltage -304 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60)
SYMBOL diode -160 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 128 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 96 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode -128 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL res -16 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 4800
SYMBOL ind 464 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 15
SYMBOL voltage 288 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 170 60)
SYMBOL diode 432 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D5
SYMATTR Value MUR460
SYMBOL diode 720 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D6
SYMATTR Value MUR460
SYMBOL diode 688 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D7
SYMATTR Value MUR460
SYMBOL diode 464 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D8
SYMATTR Value MUR460
SYMBOL res 576 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 4800
SYMBOL cap 608 144 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 10ľ
TEXT -298 246 Left 0 !.tran .05

If the relay is spec'ed as "must make" at 80% of rated current through
the coil (~29mA), then note that with a 10ľF cap in parallel with the
coil the relay will _always_ make using full-wave rectified 120V 60Hz
mains.  

NT

 

[petrus bitbyter]

"NT" <meow2222@care2.com> schreef in bericht
news:9eae9570-949f-4798-84c2-74062c8df2b1@j1g2000yqn.googlegroups.com...
On Aug 29, 7:03 pm, John Fields <jfie...@austininstruments.com> wrote:

On Mon, 29 Aug 2011 07:28:53 -0700 (PDT), NT <meow2...@care2.com
wrote:



On Aug 29, 3:03 pm, John Fields <jfie...@austininstruments.com> wrote:
On Sun, 28 Aug 2011 23:21:56 -0700, DaveC <inva...@invalid.net> wrote:
I may be able to obtain a very small 2-pole 240 vac contactor I need
rated
for 50 Hz only.

If I install it in N. America, what's the implication? Is the hold-in
magnetism less than if it were 60 Hz? Just noisy?

Please don't ask or suggest other sources. This is a very specific
device and
I've not been able to locate other than this.

Not knowing the specifics about the coil makes predicting what will
happen at 60Hz difficult.

However, assuming that the inductive reactance and resistance of the
coil will remain constant at 50 and 60Hz means that the impedance of
the coil at 60Hz will be 1.2 times (60Hz/50Hz) what it is at 50Hz.

no, only the L component of the impedance will be 1.2 times as high.

---
Well, that's true, so let's just see how far off I was, by using a
real-world example.

I have an old P&B MR5A here with a 240V 50/60Hz coil.

The coil has a resistance of 4800 ohms, and an open inductance of 14.5
henrys, so it has an impedance of 6616 ohms at 50 Hz, and 7270 ohms at
60 Hz.

7270 - 6616 = 1.1, so my error was 1 part in 11, or a little less than
10%

I can live with that.
---



Consequently, the current in the coil at 60Hz will be about 83% of
what it is at 50Hz.

If that turns out to be a problem, a higher drive voltage could be
used in order to increase the current, namely 1.2 times 240V; 288V.

That could easily be accomplished using a transformer to boost the
240V mains to 288V, like this: (View with a fixed pitch font.)

240AC>-----+--+
| |
oP||S
R||E
I||Co
| |
| +-----> \
| > 288AC TO COIL
240AC>-----+--------> /

The transformer would need a 240V primary, a 48V secondary, and a VA
rating greater than or equal to the contactor coil's rating.

these are the kind of 'solutions' that happen when people dont put the
relevant numbers to things.


Well, your tone is certainly insulting, while the solution remains
valid, but since the voltage into the coil will only be 11% low, the
transformer secondary will only have to supply 26V instead of 48.

In reality, 24V will be fine.
|

| The relay has a voltage margin of around 50%, the mains supply wont
| vary more than 10%, so the transformer is of no use.
|
|
|NT
|

"The relay has a voltage margin of around 50%" How do you know? FAIK the op
did not supply this numbers.

petrus bitbyter

 

[Jeffrey Angus]

On 8/30/2011 10:38 AM, David Lesher wrote:

Well anytime you run transformers {Be they two fixed windings,
or one fixed and one rotating aka motor, or one fixed&
one sliding aka solenoid or relay....} on LESS than design
frequency, worry. Lower frequencies need more iron.

Hasn't anybody read the original question?

He wants to know if a 50 Hz relay coil will work at 60 Hz.

Jeff

--
"Everything from Crackers to Coffins"

 

[Rich Grise]

Jeffrey Angus wrote:

On 8/30/2011 10:38 AM, David Lesher wrote:
Well anytime you run transformers {Be they two fixed windings,
or one fixed and one rotating aka motor, or one fixed&
one sliding aka solenoid or relay....} on LESS than design
frequency, worry. Lower frequencies need more iron.

Hasn't anybody read the original question?

Yeah. It was answered some days ago.

Guess you gotta be quick. ;-)

Cheers!
Rich

 

[Rich Grise]

John Fields wrote:

On Tue, 30 Aug 2011 02:58:19 +0000 (UTC), David Lesher
John Fields <jfields@austininstruments.com> writes:

I have an old P&B MR5A here with a 240V 50/60Hz coil.

The coil has a resistance of 4800 ohms, and an open inductance of 14.5
henrys, so it has an impedance of 6616 ohms at 50 Hz, and 7270 ohms at
60 Hz.

What's the inductance while closed?

16 henries, but that measurement was made by closing the armature
manually.

If there's any real interest I can measure it energized.

Well, I guess that depends on what you mean by "real interest" - I'd be

interested in seeing your experimental results, but that's just because
I like seeing experimental results. ;-)

Cheers!
Rich

 

[David Lesher]

DaveC <invalid@invalid.net> writes:

It will be happy. If it were a 60Hz on 50, it would not be as happy.
[David Lesher]

Interesting. Good to know. It's stuff like this I learn here that I wouldn't
otherwise know.

Well anytime you run transformers {Be they two fixed windings,
or one fixed and one rotating aka motor, or one fixed &
one sliding aka solenoid or relay....} on LESS than design
frequency, worry. Lower frequencies need more iron.

I'm not sure the slightly less pull-in power will be relevent but
it's possible.

If you do pursue the "make DC and use that..." approach; don't forget
you need to limit the holdin current. One technique is a cap in parallel
with a resistor.


--
A host is a host from coast to coast.................wb8foz@nrk.com
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

 

[Jeff Liebermann]

On Tue, 30 Aug 2011 10:55:10 -0500, Jeffrey Angus <grendelair@aim.com>
wrote:

Hasn't anybody read the original question?

Nope. With 5 groups in the distribution, I just assumed the original
question to not be worth reading. Besides, some of the groups appear
to be write-only, where nobody (including me) reads the original
question.

He wants to know if a 50 Hz relay coil will work at 60 Hz.

Well, if it's too much of a risk trying it and checking if it will
explode, there are plenty of 50 to/from 60 Hz converters available for
a small fortune:
<http://www.50hz.com/Rotary/rotary.htm>
If sufficiently low power, one could probably just build a converter
from two motors and a suitable gearbox.

Jeff (part of the problem) L.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[David Lesher]

Jeffrey Angus <grendelair@aim.com> writes:

On 8/30/2011 10:38 AM, David Lesher wrote:
Well anytime you run transformers {Be they two fixed windings,
or one fixed and one rotating aka motor, or one fixed&
one sliding aka solenoid or relay....} on LESS than design
frequency, worry. Lower frequencies need more iron.

Hasn't anybody read the original question?

I have.

He wants to know if a 50 Hz relay coil will work at 60 Hz.

And that's my point; if it were the other way, he SHOULD worry.
--
A host is a host from coast to coast.................wb8foz@nrk.com
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

 

[John Fields]

On Tue, 30 Aug 2011 05:11:20 -0700 (PDT), NT <meow2222@care2.com>
wrote:

On Aug 30, 12:39 pm, John Fields <jfie...@austininstruments.com
wrote:
On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT <meow2...@care2.com
wrote:

IME relays pull in at in the region of half rated voltage, and dc
ratings are typically about half the voltage of the ac rating, which
gives an idea of how much current is determined by L and how much by
R.

---
IME, most relays (with either AC or DC coils) are guaranteed to pull
in at about 80% of their rated coil voltage, so I'm at a loss trying
to understand what you meant by: "dc ratings are typically about half
of the ac rating."

Can you elaborate, please?

When relays have dual ratings for ac and dc, its normal for the dc
voltage rating to be half the ac voltage rating.


Running your relay on 220v 60Hz it will work fine.

Knowing nothing about the contactor, other than that it's specified to
energize when 240V 50Hz is placed across the coil, your imprimatur is
premature.

I really dont agree. I do know the basics about relays, and one
normally finds that pull-in occurs at around 50% rated voltage. The OP
is welcome to test theirs to see if it behaves the usual way.


Contact closing speed will be slightly slower. Margin will be reduced,
but its only being reduced from enormous to slightly less enormous,
so its a non-issue except in very unusual situations.

It seems you've forgotten that when the armature makes, and the
magnetic circuit is closed, the inductance of the coil will rise.

I dont know why you think I've forgotten it. What's relevant here is
inductance in the closed position.

---
I disagree.

Since the relay is open when power is applied to the coil, it's the
open inductance (and the resistance, of course) which will determine
how much current will flow through the coil, that current being what
generates the magnetic field to start the armature on its way.

Then when the relay closes, the closed inductance comes into play and
holds the armature in place until the current through the coil is
reduced to a point where the armature's return spring overcomes the
weakened magnetic field, allowing the armature to open.
---

Such being the case, the current in it will diminish,

true with all relays under all ac conditions. Theyre designed to work
that way.

reducing the
hold on the armature and making the contacts more likely to chatter.

No, its exactly how theyre designed to operate.


The vibration tolerance of
the contacts will be little affected in practice; if your environment
is harsh enough to shake the relay contact open, then you've got
bigger worries than contacts crackling.
If instead you meant you would use it on 110v 60Hz, then dont. But you
could use diodes to get a higher dc voltage and use that.

Interesting conjecture.

Where's the conjecture? I get the feeling you could do with bringing
your skills up to speed on relays.

---
Perhaps.
---

Something like this?

        +-----+
120AC>--|~   +|----+
        |     |    |  
        |     |  [COIL]
        |     |    |
120AC>--|~   -|----+
        +-----+

That would work.

---
Not in all cases, certainly.
---

Since the coil has an impedance of about 6600 ohms at 50Hz, then the
current through it will be:

          E      240V
     I = --- = ------- = 0.036A = 36mA
          Z     6600R

Then, since the coil has a resistance of 4800 ohms, the DC voltage
across it required to force 36mA through it would be:

     E = IR = 0.036A * 4800R ~ 174V.

You're not saying where you got those figures from.

---
The P&B MR5A I talked about in an earlier post, which has a 240V
50/60Hz coil, a coil resistance of 4800 ohms, an impedance of ~ 6600
ohms at 50 Hz, an open inductance of 14.5 henrys, and a closed
inductance of 16 henrys
---

Typically dc rating is half ac rating.

---
But I don't think "typical" is what we're after since we want
something that will _always_ work.

Since current is what's doing the work, my real-world example shows
that 240V 50 Hz RMS impressed across a load with an impedance of 6600
ohms will force 36mA RMS of current through the load.

Then, since it's current that's doing the work, 36mA of DC through the
coil should accomplish the same thing.
---

The peak voltage out of the bridge would be:

     E = RMS * sqrt(2) = 120 * 1.414 ~ 170V.

Pretty close, but at 120Hz, the reactance of the coil would increase,
limiting the current to something less than the 36mA needed to close
the armature.

The effect of the relay's inductance, when run off a BR, is simply to
smooth the current flow somewhat.

---
Yeah, I know, said so earlier, and posted a simulation showing the
ripple.
---

Mean current remains much the same.
So we're looking for 120v rms, which is what the BR would deliver.

---
But, what it won't deliver is the worst-case voltage required over the
interval required to guarantee the armature will close.
---

However, the reactance of the coil will smooth the current and the
addition of a capacitor in parallel with the coil will remove some of
the ripple and allow the coil to see more nearly pure DC.

and overheat the relay by increasing its rms dc voltage to above 120v.

---
There's no such thing as "rms dc voltage", and if the relay is
designed to operate on AC with a certain RMS current in its coil, how
can it possibly overheat if that current is DC?
---

Here's a simulation showing both ways:

Version 4
SHEET 1 880 680
WIRE -144 16 -304 16
WIRE 112 16 -144 16
WIRE 448 16 288 16
WIRE 704 16 448 16
WIRE -304 80 -304 16
WIRE 288 80 288 16
WIRE 448 80 448 64
WIRE 480 80 448 80
WIRE 592 80 560 80
WIRE 704 80 704 64
WIRE 704 80 672 80
WIRE -144 112 -144 80
WIRE -112 112 -144 112
WIRE 0 112 -32 112
WIRE 112 112 112 80
WIRE 112 112 80 112
WIRE -144 160 -144 112
WIRE 112 160 112 112
WIRE 448 160 448 80
WIRE 544 160 448 160
WIRE 704 160 704 80
WIRE 704 160 608 160
WIRE -304 224 -304 160
WIRE -144 224 -304 224
WIRE 112 224 -144 224
WIRE 288 224 288 160
WIRE 448 224 288 224
WIRE 704 224 448 224
WIRE -304 272 -304 224
WIRE 288 272 288 224
FLAG -304 272 0
FLAG 288 272 0
SYMBOL ind -128 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 15
SYMBOL voltage -304 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60)
SYMBOL diode -160 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 128 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 96 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode -128 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL res -16 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 4800
SYMBOL ind 464 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 15
SYMBOL voltage 288 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 170 60)
SYMBOL diode 432 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D5
SYMATTR Value MUR460
SYMBOL diode 720 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D6
SYMATTR Value MUR460
SYMBOL diode 688 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D7
SYMATTR Value MUR460
SYMBOL diode 464 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D8
SYMATTR Value MUR460
SYMBOL res 576 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 4800
SYMBOL cap 608 144 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 10ľ
TEXT -298 246 Left 0 !.tran .05

If the relay is spec'ed as "must make" at 80% of rated current through
the coil (~29mA), then note that with a 10ľF cap in parallel with the
coil the relay will _always_ make using full-wave rectified 120V 60Hz
mains.  

--
JF

 

[NT]

On Aug 30, 4:35 pm, "petrus bitbyter" <petrus.bitby...@hotmail.com>
wrote:

"NT" <meow2...@care2.com> schreef in berichtnews:9eae9570-949f-4798-84c2-74062c8df2b1@j1g2000yqn.googlegroups.com...
On Aug 29, 7:03 pm, John Fields <jfie...@austininstruments.com> wrote:



On Mon, 29 Aug 2011 07:28:53 -0700 (PDT), NT <meow2...@care2.com
wrote:

On Aug 29, 3:03 pm, John Fields <jfie...@austininstruments.com> wrote:
On Sun, 28 Aug 2011 23:21:56 -0700, DaveC <inva...@invalid.net> wrote:
I may be able to obtain a very small 2-pole 240 vac contactor I need
rated
for 50 Hz only.

If I install it in N. America, what's the implication? Is the hold-in
magnetism less than if it were 60 Hz? Just noisy?

Please don't ask or suggest other sources. This is a very specific
device and
I've not been able to locate other than this.

Not knowing the specifics about the coil makes predicting what will
happen at 60Hz difficult.

However, assuming that the inductive reactance and resistance of the
coil will remain constant at 50 and 60Hz means that the impedance of
the coil at 60Hz will be 1.2 times (60Hz/50Hz) what it is at 50Hz.

no, only the L component of the impedance will be 1.2 times as high.

---
Well, that's true, so let's just see how far off I was, by using a
real-world example.

I have an old P&B MR5A here with a 240V 50/60Hz coil.

The coil has a resistance of 4800 ohms, and an open inductance of 14.5
henrys, so it has an impedance of 6616 ohms at 50 Hz, and 7270 ohms at
60 Hz.

7270 - 6616 = 1.1, so my error was 1 part in 11, or a little less than
10%

I can live with that.
---

Consequently, the current in the coil at 60Hz will be about 83% of
what it is at 50Hz.

If that turns out to be a problem, a higher drive voltage could be
used in order to increase the current, namely 1.2 times 240V; 288V.

That could easily be accomplished using a transformer to boost the
240V mains to 288V, like this: (View with a fixed pitch font.)

240AC>-----+--+
| |
oP||S
R||E
I||Co
| |
| +-----> \
| > 288AC TO COIL
240AC>-----+--------> /

The transformer would need a 240V primary, a 48V secondary, and a VA
rating greater than or equal to the contactor coil's rating.

these are the kind of 'solutions' that happen when people dont put the
relevant numbers to things.

Well, your tone is certainly insulting, while the solution remains
valid, but since the voltage into the coil will only be 11% low, the
transformer secondary will only have to supply 26V instead of 48.

In reality, 24V will be fine.

|
| The relay has a voltage margin of around 50%, the mains supply wont
| vary more than 10%, so the transformer is of no use.
|
|
|NT
|

"The relay has a voltage margin of around 50%" How do you know? FAIK the op
did not supply this numbers.

petrus bitbyter

experience with various relays


NT

 

[John S]

On 8/31/2011 1:06 PM, The Ghost In The Machine wrote:

YES A-NT-MAN BUT THE RMS REFERS TO THE AC WAVEFORM NOT THE DC OUTPUT.
HENCE THERE IS NO SUCH THING AS RMS DC VOLTAGE.
PATECUM
TGITM

Actually, RMS DC voltage is a redundant expression since DC is RMS.

 

[NT]

On Aug 31, 1:36 am, John Fields <jfie...@austininstruments.com> wrote:

On Tue, 30 Aug 2011 05:11:20 -0700 (PDT), NT <meow2...@care2.com
wrote:
On Aug 30, 12:39 pm, John Fields <jfie...@austininstruments.com
wrote:
On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT <meow2...@care2.com
wrote:

IME relays pull in at in the region of half rated voltage, and dc
ratings are typically about half the voltage of the ac rating, which
gives an idea of how much current is determined by L and how much by
R.

---
IME, most relays (with either AC or DC coils) are guaranteed to pull
in at about 80% of their rated coil voltage, so I'm at a loss trying
to understand what you meant by: "dc ratings are typically about half
of the ac rating."

Can you elaborate, please?

When relays have dual ratings for ac and dc, its normal for the dc
voltage rating to be half the ac voltage rating.

Running your relay on 220v 60Hz it will work fine.

Knowing nothing about the contactor, other than that it's specified to
energize when 240V 50Hz is placed across the coil, your imprimatur is
premature.

I really dont agree. I do know the basics about relays, and one
normally finds that pull-in occurs at around 50% rated voltage. The OP
is welcome to test theirs to see if it behaves the usual way.

Contact closing speed will be slightly slower. Margin will be reduced,
but its only being reduced from enormous to slightly less enormous,
so its a non-issue except in very unusual situations.

It seems you've forgotten that when the armature makes, and the
magnetic circuit is closed, the inductance of the coil will rise.

I dont know why you think I've forgotten it. What's relevant here is
inductance in the closed position.

I disagree.

Since the relay is open when power is applied to the coil, it's the
open inductance (and the resistance, of course) which will determine
how much current will flow through the coil, that current being what
generates the magnetic field to start the armature on its way.

yup

Then when the relay closes, the closed inductance comes into play and
holds the armature in place until the current through the coil is
reduced to a point where the armature's return spring overcomes the
weakened magnetic field, allowing the armature to open.

yes. I guess in theory both matter, one determines closing behaviour,
the other ensures the relay doesnt overheat. In practice though the
margins are very large, and its normal to simply fix holding current
to suit the relay, and not worry about closing current, which will be
so close as to make no real world difference in all but exceptional
circumstances. But yes, we can consider both if need be.

Such being the case, the current in it will diminish,

true with all relays under all ac conditions. Theyre designed to work
that way.

reducing the
hold on the armature and making the contacts more likely to chatter.

No, its exactly how theyre designed to operate.

The vibration tolerance of
the contacts will be little affected in practice; if your environment
is harsh enough to shake the relay contact open, then you've got
bigger worries than contacts crackling.
If instead you meant you would use it on 110v 60Hz, then dont. But you
could use diodes to get a higher dc voltage and use that.

Interesting conjecture.

Where's the conjecture? I get the feeling you could do with bringing
your skills up to speed on relays.

---
Perhaps.
---

Something like this?

        +-----+
120AC>--|~   +|----+
        |     |    |  
        |     |  [COIL]
        |     |    |
120AC>--|~   -|----+
        +-----+

That would work.

Not in all cases, certainly.

I'd like to see you find one single electromechanical relay that wont
work for.

Since the coil has an impedance of about 6600 ohms at 50Hz, then the
current through it will be:

          E      240V
     I = --- = ------- = 0.036A = 36mA
          Z     6600R

Then, since the coil has a resistance of 4800 ohms, the DC voltage
across it required to force 36mA through it would be:

     E = IR = 0.036A * 4800R ~ 174V.

You're not saying where you got those figures from.

---
The P&B MR5A I talked about in an earlier post, which has a 240V
50/60Hz coil, a coil resistance of 4800 ohms, an impedance of ~ 6600
ohms at 50 Hz, an open inductance of 14.5 henrys, and a closed
inductance of 16 henrys

Typically dc rating is half ac rating.

But I don't think "typical" is what we're after since we want
something that will _always_ work.

This 2:1 ratio normally is good for relays, and the OP can check his
to see if it conforms to that. If it does, the thing will always work
when subject to this formula.

FWIW, when ac is applied you get puling force plus vibration. With dc
there is no vibration component when its closed, so less holding
current is needed. How much less I've really no idea.

Some relays are fast movers capable of 100s of Hz, some are slow. Ac
relays can always work on dc, but dc ones often dont work ok on ac.

Since current is what's doing the work, my real-world example shows
that 240V 50 Hz RMS impressed across a load with an impedance of 6600
ohms will force 36mA RMS of current through the load.

Then, since it's current that's doing the work, 36mA of DC through the
coil should accomplish the same thing.
---

The peak voltage out of the bridge would be:

     E = RMS * sqrt(2) = 120 * 1.414 ~ 170V.

Pretty close, but at 120Hz, the reactance of the coil would increase,
limiting the current to something less than the 36mA needed to close
the armature.

The effect of the relay's inductance, when run off a BR, is simply to
smooth the current flow somewhat.

---
Yeah, I know, said so earlier, and posted a simulation showing the
ripple.
---

Mean current remains much the same.
So we're looking for 120v rms, which is what the BR would deliver.

---
But, what it won't deliver is the worst-case voltage required over the
interval required to guarantee the armature will close.

Re ripple: If the relay is designed to run on ac 50 or 60Hz, its
designed and rated to live with the current and force variations that
go along with that, 100-120 times a second. Running it on rectified
mains will only serve to reduce the current variations through the
cycle, it wont cause the relay any issues.

Re rms voltage: With my 2:1 figures, rectified 120v is spot on. With
your 174v figure, 120v is well within the 50% margin. Of course for
some uses that margin would need to be confirmed by testing before
production, and reconfirmed if a new relay type is used. Or as you
say, a cap could be added. Or for off brand consumer goods, in it
goes, relays are good for it.

However, the reactance of the coil will smooth the current and the
addition of a capacitor in parallel with the coil will remove some of
the ripple and allow the coil to see more nearly pure DC.

and overheat the relay by increasing its rms dc voltage to above 120v.

There's no such thing as "rms dc voltage",

RMS can be applied to any and every waveform, dc included. Its very
relevant when working with rectified ac, semismoothed or unsmoothed.

and if the relay is
designed to operate on AC with a certain RMS current in its coil, how
can it possibly overheat if that current is DC?

With the same current it wont, with higher curren ti will. IIRC you
proposed using 174v rms, that would be ok on your specific relay, but
not a universal solution.


NT

Here's a simulation showing both ways:

Version 4
SHEET 1 880 680
WIRE -144 16 -304 16
WIRE 112 16 -144 16
WIRE 448 16 288 16
WIRE 704 16 448 16
WIRE -304 80 -304 16
WIRE 288 80 288 16
WIRE 448 80 448 64
WIRE 480 80 448 80
WIRE 592 80 560 80
WIRE 704 80 704 64
WIRE 704 80 672 80
WIRE -144 112 -144 80
WIRE -112 112 -144 112
WIRE 0 112 -32 112
WIRE 112 112 112 80
WIRE 112 112 80 112
WIRE -144 160 -144 112
WIRE 112 160 112 112
WIRE 448 160 448 80
WIRE 544 160 448 160
WIRE 704 160 704 80
WIRE 704 160 608 160
WIRE -304 224 -304 160
WIRE -144 224 -304 224
WIRE 112 224 -144 224
WIRE 288 224 288 160
WIRE 448 224 288 224
WIRE 704 224 448 224
WIRE -304 272 -304 224
WIRE 288 272 288 224
FLAG -304 272 0
FLAG 288 272 0
SYMBOL ind -128 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 15
SYMBOL voltage -304 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60)
SYMBOL diode -160 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 128 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 96 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode -128 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL res -16 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 4800
SYMBOL ind 464 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 15
SYMBOL voltage 288 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 170 60)
SYMBOL diode 432 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D5
SYMATTR Value MUR460
SYMBOL diode 720 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D6
SYMATTR Value MUR460
SYMBOL diode 688 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D7
SYMATTR Value MUR460
SYMBOL diode 464 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D8
SYMATTR Value MUR460
SYMBOL res 576 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 4800
SYMBOL cap 608 144 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 10ľ
TEXT -298 246 Left 0 !.tran .05

If the relay is spec'ed as "must make" at 80% of rated current through
the coil (~29mA), then note that with a 10ľF cap in parallel with the
coil the relay will _always_ make using full-wave rectified 120V 60Hz
mains.  

--
JF

 

[Jeffrey Angus]

On 8/31/2011 2:33 PM, NT wrote:

I realised it was perhaps not the best phrasing. But... would the dc
component be the average V or the rms?

DC would be the RMS value. Because, RMS means "This is what the DC
value would be."

Jeff


--
"Everything from Crackers to Coffins"

 

[The Ghost In The Machine]

On Aug 31, 12:38 pm, NT <meow2...@care2.com> wrote:

On Aug 31, 1:36 am, John Fields <jfie...@austininstruments.com> wrote:





On Tue, 30 Aug 2011 05:11:20 -0700 (PDT), NT <meow2...@care2.com
wrote:
On Aug 30, 12:39 pm, John Fields <jfie...@austininstruments.com
wrote:
On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT <meow2...@care2.com
wrote:

IME relays pull in at in the region of half rated voltage, and dc
ratings are typically about half the voltage of the ac rating, which
gives an idea of how much current is determined by L and how much by
R.

---
IME, most relays (with either AC or DC coils) are guaranteed to pull
in at about 80% of their rated coil voltage, so I'm at a loss trying
to understand what you meant by: "dc ratings are typically about half
of the ac rating."

Can you elaborate, please?

When relays have dual ratings for ac and dc, its normal for the dc
voltage rating to be half the ac voltage rating.

Running your relay on 220v 60Hz it will work fine.

Knowing nothing about the contactor, other than that it's specified to
energize when 240V 50Hz is placed across the coil, your imprimatur is
premature.

I really dont agree. I do know the basics about relays, and one
normally finds that pull-in occurs at around 50% rated voltage. The OP
is welcome to test theirs to see if it behaves the usual way.

Contact closing speed will be slightly slower. Margin will be reduced,
but its only being reduced from enormous to slightly less enormous,
so its a non-issue except in very unusual situations.

It seems you've forgotten that when the armature makes, and the
magnetic circuit is closed, the inductance of the coil will rise.

I dont know why you think I've forgotten it. What's relevant here is
inductance in the closed position.

I disagree.

Since the relay is open when power is applied to the coil, it's the
open inductance (and the resistance, of course) which will determine
how much current will flow through the coil, that current being what
generates the magnetic field to start the armature on its way.

yup

Then when the relay closes, the closed inductance comes into play and
holds the armature in place until the current through the coil is
reduced to a point where the armature's return spring overcomes the
weakened magnetic field, allowing the armature to open.

yes. I guess in theory both matter, one determines closing behaviour,
the other ensures the relay doesnt overheat. In practice though the
margins are very large, and its normal to simply fix holding current
to suit the relay, and not worry about closing current, which will be
so close as to make no real world difference in all but exceptional
circumstances. But yes, we can consider both if need be.





Such being the case, the current in it will diminish,

true with all relays under all ac conditions. Theyre designed to work
that way.

reducing the
hold on the armature and making the contacts more likely to chatter.

No, its exactly how theyre designed to operate.

The vibration tolerance of
the contacts will be little affected in practice; if your environment
is harsh enough to shake the relay contact open, then you've got
bigger worries than contacts crackling.
If instead you meant you would use it on 110v 60Hz, then dont. But you
could use diodes to get a higher dc voltage and use that.

Interesting conjecture.

Where's the conjecture? I get the feeling you could do with bringing
your skills up to speed on relays.

---
Perhaps.
---

Something like this?

        +-----+
120AC>--|~   +|----+
        |     |    |  
        |     |  [COIL]
        |     |    |
120AC>--|~   -|----+
        +-----+

That would work.

Not in all cases, certainly.

I'd like to see you find one single electromechanical relay that wont
work for.





Since the coil has an impedance of about 6600 ohms at 50Hz, then the
current through it will be:

          E      240V
     I = --- = ------- = 0.036A = 36mA
          Z     6600R

Then, since the coil has a resistance of 4800 ohms, the DC voltage
across it required to force 36mA through it would be:

     E = IR = 0.036A * 4800R ~ 174V.

You're not saying where you got those figures from.

---
The P&B MR5A I talked about in an earlier post, which has a 240V
50/60Hz coil, a coil resistance of 4800 ohms, an impedance of ~ 6600
ohms at 50 Hz, an open inductance of 14.5 henrys, and a closed
inductance of 16 henrys
Typically dc rating is half ac rating.

But I don't think "typical" is what we're after since we want
something that will _always_ work.

This 2:1 ratio normally is good for relays, and the OP can check his
to see if it conforms to that. If it does, the thing will always work
when subject to this formula.

FWIW, when ac is applied you get puling force plus vibration. With dc
there is no vibration component when its closed, so less holding
current is needed. How much less I've really no idea.

Some relays are fast movers capable of 100s of Hz, some are slow. Ac
relays can always work on dc, but dc ones often dont work ok on ac.





Since current is what's doing the work, my real-world example shows
that 240V 50 Hz RMS impressed across a load with an impedance of 6600
ohms will force 36mA RMS of current through the load.

Then, since it's current that's doing the work, 36mA of DC through the
coil should accomplish the same thing.
---

The peak voltage out of the bridge would be:

     E = RMS * sqrt(2) = 120 * 1.414 ~ 170V.

Pretty close, but at 120Hz, the reactance of the coil would increase,
limiting the current to something less than the 36mA needed to close
the armature.

The effect of the relay's inductance, when run off a BR, is simply to
smooth the current flow somewhat.

---
Yeah, I know, said so earlier, and posted a simulation showing the
ripple.
---

Mean current remains much the same.
So we're looking for 120v rms, which is what the BR would deliver.

---
But, what it won't deliver is the worst-case voltage required over the
interval required to guarantee the armature will close.

Re ripple: If the relay is designed to run on ac 50 or 60Hz, its
designed and rated to live with the current and force variations that
go along with that, 100-120 times a second. Running it on rectified
mains will only serve to reduce the current variations through the
cycle, it wont cause the relay any issues.

Re rms voltage: With my 2:1 figures, rectified 120v is spot on. With
your 174v figure, 120v is well within the 50% margin. Of course for
some uses that margin would need to be confirmed by testing before
production, and reconfirmed if a new relay type is used. Or as you
say, a cap could be added. Or for off brand consumer goods, in it
goes, relays are good for it.

However, the reactance of the coil will smooth the current and the
addition of a capacitor in parallel with the coil will remove some of
the ripple and allow the coil to see more nearly pure DC.

and overheat the relay by increasing its rms dc voltage to above 120v.

There's no such thing as "rms dc voltage",

RMS can be applied to any and every waveform, dc included. Its very
relevant when working with rectified ac, semismoothed or unsmoothed.

and if the relay is
designed to operate on AC with a certain RMS current in its coil, how
can it possibly overheat if that current is DC?

With the same current it wont, with higher curren ti will. IIRC you
proposed using 174v rms, that would be ok on your specific relay, but
not a universal solution.

NT



Here's a simulation showing both ways:

Version 4
SHEET1 880 680
WIRE -144 16 -304 16
WIRE112 16 -144 16
WIRE448 16 288 16
WIRE704 16 448 16
WIRE -304 80 -304 16
WIRE288 80 288 16
WIRE448 80 448 64
WIRE480 80 448 80
WIRE592 80 560 80
WIRE704 80 704 64
WIRE704 80 672 80
WIRE -144 112 -144 80
WIRE -112 112 -144 112
WIRE0 112 -32 112
WIRE112 112 112 80
WIRE112 112 80 112
WIRE -144 160 -144 112
WIRE112 160 112 112
WIRE448 160 448 80
WIRE544 160 448 160
WIRE704 160 704 80
WIRE704 160 608 160
WIRE -304 224 -304 160
WIRE -144 224 -304 224
WIRE112 224 -144 224
WIRE288 224 288 160
WIRE448 224 288 224
WIRE704 224 448 224
WIRE -304 272 -304 224
WIRE288 272 288 224
FLAG -304 272 0
FLAG288 272 0
SYMBOL ind -128 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 15
SYMBOL voltage -304 64 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 170 60)
SYMBOL diode -160 16 R0
WINDOW 0 4 -52 Left 0
WINDOW 3 -28 -24 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 128 80 R180
WINDOW 0 1 119 Left 0
WINDOW 3 -33 84 Left 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 96 160 R0
WINDOW 0 3 89 Left 0
WINDOW 3 -27 118 Left 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode -128 224 R180
WINDOW 0 2 -28 Left 0
WINDOW 3 -31 -57 Left 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL res -16 128 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 4800
SYMBOL ind 464 96 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 15
SYMBOL voltage 288 64 R0
WINDOW 3 24 104 Invisible 0...

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YES A-NT-MAN BUT THE RMS REFERS TO THE AC WAVEFORM NOT THE DC OUTPUT.
HENCE THERE IS NO SUCH THING AS RMS DC VOLTAGE.
PATECUM
TGITM

 

[Ian Field]

"Jeffrey Angus" <grendelair@aim.com> wrote in message
news:j3m408$fqf$1@dont-email.me...

On 8/31/2011 2:33 PM, NT wrote:
I realised it was perhaps not the best phrasing. But... would the dc
component be the average V or the rms?

DC would be the RMS value. Because, RMS means "This is what the DC
value would be."

Years ago when I needed more power from a soldering iron I used to feed the
240VRMS through a rectifier/reservoir to get aproximately 320VDC.

 

[NT]

On Aug 31, 7:11 pm, John S <soph...@invalid.org> wrote:

On 8/31/2011 1:06 PM, The Ghost In The Machine wrote:



YES A-NT-MAN BUT THE RMS REFERS TO THE AC WAVEFORM NOT THE DC OUTPUT.
HENCE THERE IS NO SUCH THING AS RMS DC VOLTAGE.
PATECUM
TGITM

Root Mean Square does not imply an ac waveform, its jsut most commonly
used for ac waveforms. Every stable waveform has an rms value, even
perfect dc.

Actually, RMS DC voltage is a redundant expression since DC is RMS.

I realised it was perhaps not the best phrasing. But... would the dc
component be the average V or the rms?


NT

 

[Spehro Pefhany]

On Wed, 31 Aug 2011 12:33:06 -0700 (PDT), NT <meow2222@care2.com>
wrote:

On Aug 31, 7:11 pm, John S <soph...@invalid.org> wrote:
On 8/31/2011 1:06 PM, The Ghost In The Machine wrote:



YES A-NT-MAN BUT THE RMS REFERS TO THE AC WAVEFORM NOT THE DC OUTPUT.
HENCE THERE IS NO SUCH THING AS RMS DC VOLTAGE.
PATECUM
TGITM

Root Mean Square does not imply an ac waveform, its jsut most commonly
used for ac waveforms. Every stable waveform has an rms value, even
perfect dc.


Actually, RMS DC voltage is a redundant expression since DC is RMS.

I realised it was perhaps not the best phrasing. But... would the dc
component be the average V or the rms?


NT

Generally, in electronics, "DC component" is defined as the average
value (say, over a period of a periodic waveform). So a 1V peak sine
wave sitting on top of 1VDC would have DC component of 1.0V.
A 1V peak sine wave has a DC component of 0.

The RMS value is the heating value- a 1 ohm resistor with 1VDC across
it will dissipate 1W. A 1 ohm resistor with 1.414V peak sine wave
across it (1 V RMS) will dissipate 1W.

A 1 ohm resistor powered with a 1V peak sine wave sitting on top of
1VDC will dissipate a bit more than 1 watt (RMS value is sqrt(3/2) if
you want to get analytical about it, so about 1.22W).