Question about electricity

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On Sun, 24 Aug 2008 12:02:21 -0500, AZ Nomad
[snip]

A transformer is an AC-only device. Maybe you're (incorrectly) referring to a power supply which contains a transformer and a rectifier. The transformer is on the AC side.

There might be some conversion table around, something like use xVAC for a yVDC relay. Still, I'd rather not do that.

Most of the relays I use are made for 12VDC and work with small DC wall-warts.
A half-wave rectifier has an even greater need for filtering. The capacitor needs to store enough electricity to fill in the gaps. With a half-wave circuit the gaps are half as frequent but much longer.
BTW, most wall-warts contain a full-wave rectifier that uses only 2 diodes. This is possible with a transformer with double the secondary voltage and a center tap.
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Halfwave rectification also causes a DC component to flow through the transformer's secondary winding. That may cause the core to saturate when the primary's "magnetizing current" is in the same direction.

I have seen fullwave ones. The 2-diode-center-tap scheme has higher ratio of RMS to average current in its secondary windings since the secondary windings are used only half the time. Copper has gotten really expensive over the past 3 years and diodes are cheap.
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On Sun, 24 Aug 2008 11:17:45 -0400, "TWayne"

Peak voltage is the height of a peak above the centerline (0V). RMS is a common, but complicated measurement which is equal to .707 of this (that is, peak is 1.414 of RMS). Average (mean) is equal to .637 of peak.
Peak-to-peak is the (unsigned) sum of the heights of both positive and negative peaks. That is, it's twice peak. For one example:
12V RMS = 17V peak = 10.8V average = 34V peak-to peak.
BTW, 1.414 is also one of the square roots of 2 (the other being -1.414 of course). For something completely OT here, how about the square roots of -2?
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On Sun, 24 Aug 2008 18:06:34 -0500, Mark Lloyd
<snip>

Average???? Most of what you said is correct, but the average of an AC voltage is zero, not .637 of peak. That's why they came up with RMS. Using RMS voltages & currents is essentially a way to enable the use of Ohm's law on AC circuits as though they were DC. A simple average won't work because the average of the voltage and current is always zero, but RMS works because the "S" part squares the voltage to make both half cycles positive. Of course, then the "R" part (square "R"oot) is used to restore the voltage back to the correct value after having squared it. The M is mean. So RMS is the square root of the mean of the squared value. You are right... it *does* sound complicated ;-)
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greenpjs wrote:

I don't think that was the motivation. RMS is equivalent heating value.
Using RMS voltages & currents is essentially a way to enable the

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wrote:

I don't necessarily remember everything from college, but average voltage (.637 of peak) is a real thing. I thing it assumes ideal full wave rectification.
For one thing, analog meters respond to average. The fact they seem to read RMS is because of the calibration, and is valid for sine waves only.

Yes. 12VRMS creates the same amount of heating as 12VDC.

Voltage and current are different things. It doesn't make sense to combine them that way, and I think you didn't mean to.

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On Sun, 24 Aug 2008 20:04:05 -0500, Mark Lloyd
<snip>

Thanks for giving me the benefit of the doubt. I didn't mean to combine them in any way other than to say the average of each (separately) is zero. I should have chosen my words more carefully.
And, yes, as you and others have said, the whole discussion assumes a sine wave.
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wrote:

OK. I know I've made mistakes like that before. It's too easy to do.

Interestingly, when I was writing my earlier post (mentioning RMS and average), I originally put in something about this material applying only to sine waves. I took that out in consideration for those who don't know and aren't able to understand.
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On Sun, 24 Aug 2008 20:04:05 -0500, Mark Lloyd

He just means the average of AC voltage is always zero, and the average of AC current is always zero. He combined the sentences, not the voltage and current, to get what he wrote.
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wrote:

That's what I thought.
Of course, there's still the term "average current", that describes what an analog meter responds to.
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wrote:

Presuming that the initial wave form of the ac is a sine wave.
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On Mon, 25 Aug 2008 00:16:14 GMT, "Dave + Gloria"

[stale sig snipped]

Right. Assuming sine waves is a very common simplification.
Also, considering the differences between AC and DC relays.
I seem to have forgotten something even bigger. In AC, each cycle has the OPPOSITE polarity to the preceding one. This means that the magnetic field generated by that current is opposing the residual magnetic field from the previous half cycle. This would make the AC relay less efficient.
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Mark Lloyd wrote:

Don't think so. The magnetic field is smoothly reversed by the sine wave current.
The energy stored in the magnetic field comes back out as current returned to the source. (This produces the 90 degree phase shift between applied voltage and current through an ideal inductance.)
The magnetic pull of the coil pulses at 2x the applied frequency which can be a problem. A "shaded pole" (heavy shorted turn) is often used to smooth out the magnetic pull.
Magnetic materials have some 'magnetic resistance' to magnetic field changes which produces "hysteresis" losses.
But in a DC coil (and AC coil) there are losses in the resistance.
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Irrelevent. The current consumed by the coil is fairly insignifigant, typically less than 200ma for a 12VAC relay with 15A contacts.
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wrote:

If a DC relay is used on AC it will chatter or vibrate. If a AC relay is used on DC the coil will draw way too much current and burn up. It may be possible to use a AC relay on DC if the voltage is cut way down to limit the current.
The current in a DC relay is limited by the resistance of the wire in the coil. The current in an AC relay is limited to a small part by the resistance of the coil but mostly by the reactance of the coil due to being operated on AC.
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On Sat, 23 Aug 2008 22:38:20 -0500, "Ralph Mowery"

Aha!
Thnnks. I've used a few relays, that I got out of scrap**, and never paid attention. :( I guess everything was always DC, so I was lucky.

**I have about 90 relays in my scrap box now, 30 high quality 8PDT from an old AT&T phone system, 20 from office photocopier from around 1975 (those might be AC I guess), and maybe 40 miscellaneous.
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wrote:

To follow up on my own post, two of those relays from the Royal photocopier were latching relays. If the machine jambed, and I turned it off or unplugged it, later on it wouldn't work. Even though I'd bought the manual, it was not a narrative and I missed the bit about latching relays. So they were in the wrong position, and when the machine was plugged in again, it wouldn't work. There was even a window in the top of the relay to show it was closed, and a button to reset it, each one of them, but it took me a long time to figure out.
By the time I got that fixed, something else broke.
But I paid a dollar plust the fixer element and had it working for about 5 years, and made a few hundred copies.

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For most applications the "last known" latched position after a power failure is the desireable one. Like if you are using a latching relay for a light switch, you get a power failure, when power is restored the light is still in its last known state. This is the same as if it were a regular wall switch. But If the power restored position of a latching relay must always be a forced-certain state, then you will need some logic that tests the position of the relay after a power up and sets it right if its not in the proper state.
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Yes it has resistance (and impedance). If you are using a transformer, make sure you have an AC relay. A battery, being DC, would require a DC relay.
HTH,
Paul
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Many brick "transformers" incorporate diode/s and a cap to produce DC. You have to check the label.
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