On Tue, 30 Aug 2011 05:11:20 -0700 (PDT), NT
wrote:

On Aug 30, 12:39*pm, John Fields
wrote:
On Mon, 29 Aug 2011 07:25:23 -0700 (PDT), NT
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 10µ
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 10µF cap in parallel with the
coil the relay will _always_ make using full-wave rectified 120V 60Hz
mains. *

--
JF