555 relay current reducer

Jan 02, 2016 153 Replies

Then how would it have arrived in that state? Unless it's a simple connection of course.

joe

Sorry, I'm an engineer with a (shut up Sloman!) scientific education. :p

On the other hand, if you can't measure an array's latching current, I first would check my multimeter if I were you. :)

joe

In a relay that would be dPout/dPin = 0 until the distance between the contacts is so small (depending on voltage) that ionisation starts to occur. From there on, until the pressure on the contacts are sufficiently strong so there is no more current increase (if this condition really exists in a partially latched array) upon further latching, I would prefer to introduce the difference quotient DeltaPout/DeltaPin, because it's totally unclear what the condition of the contact surface is and how it affects the build up of the current with increasing 'latchiness'. Once latched, dPout/dPin = 0 again.

Not undefined, depending on the circuit in which it is incorporated.

No, just funny.

One day someone here filed a proposal to study the influence of low and high tide on the behaviour of the gay sea mussel, with an emphasis on its night life. I thought that was funny to. (Or was that politically incorrect?)

joe

While after U_EB > U_EB,max the gain will be undetermined...

joe

And some people want to relax from time to time, playing with words that are meaningful in their profession or education.

joe

Ionization only occurs when you have quite a bit of voltage across the contacts. At low voltage the first thing that happens is capacitive displacement current, followed by tunnelling, followed by contact and possibly bouncing.

Sure, that's the point. Differential gain is a perfectly good and useful concept that doesn't apply to latching relays in any sensible way. For a latching relay, time-dependent average gain, i.e.

P_load(t)/P_coil(t)

makes more sense for thinking abstractly about it, and

integral (0->t(dead battery)) P_load(t') dt'

------------------------------------------- integral (0->t(dead battery)) P_coil(t') dt'

for thinking about battery life.

The time-dependent average gain of a latching relay is indefinitely large when the load is powered and the coil isn't. Given the thermocouple offsets, the coil could even be generating a small amount of power due to its previous heating, so its average gain could even change sign. Doesn't matter.

Zero over zero is mathematically undefined. That is, contacts open, coil not energized. Average gain can also be finite in some instances, i.e. nonzero load power, coil energized.

You have an odd sense of humour.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net

OK, so you don't understand limits, either.

We all understand that, and we like to have fun too.

Here it's not a controversy about accuracy or design, not a philosophical brain-teaser just for fun, it's about silly, petty stuff instead.

Cheers, James Arthur

If we're going to be strict about everything, any statement of alleged power 'gain' /must/ henceforth include the power WASTED mining, refining, transporting, and otherwise making the 'gain' element, its supporting circuitry, siblings, close relatives, and second cousins.

After all, isn't the term 'power gain' misleading, unless all the power inputs are included? ;-)

Cheers, James Arthur

It should be obvious, to any EE, that a latching relay can't have infinite gain, so anyone who makes the claim that a latching relay _does_ have infinite gain is wrong. The problem here seems to be that Larkin, who originally proposed that latching relays have infinite gain, would rather mount innumerable subterfuges than simply admit that he was wrong. John Fields

Why have you been brooding over this for years? Don't you have anything useful to do?

John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com

Close. :) P(out) Since G = ------- P(in) Where G is gain, . P(out) is power into the load, in watts, . and P(in) is the power, in watts, required to latch the relay, Regardless of the power controlled by the numerator, if the power in the denominator is more than zero, the gain can never be infinite. John Fields

But just look at the power to be grabbed by outlawing all that stuff. ;)

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net

That's like saying this 120 ohm resistor is extremely unlikely to be 120.0000000 ohms therefore calling it 120 ohm is wrong. It's a fool's argument.

The problem is your failure to grasp what EEs can reasonably assume from other EEs, namely that there are tolerances, and that such a claim is sufficiently close for some purposes but not necessarily for all. The rest of us got that no problem.

Not much point continuing this any more.

NT

power requred to maintain the latched state is zero, so P(in) is usually 0

mostly infinite then.

\_(?)_

I've got a resistor with only black bands in my junk box, so that should be 0 ohms +/- 20%, but I get the feeling it isnt.

\_(?)_

Ionization occurs when the value of the electrical field (Volt/meter) is high enough. At a low voltage ionization could occur if the contacts are close enough, and even more if they have some 'pointy' irregularities that would increase the local field strength, from previous operations.

However I wouldn't know what comes first, tunnelling or ionization. :)

tunnelling... I like that one :)

You have some funny things to bring up, but I agree.

joe

P.S.: Oh? Well, I'm happy with it.

I also have a feeling that the 2nd law of thermodynamics tries to warns us that there is no such thing as power gain.

joe

Nope. Ionization requires a voltage exceeding the activation energy for the ionization process, which means that there's a voltage (a few volts) below which ionization doesn't occur no matter how slowly the contacts close.

Cheers

Phil Hobbs

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