Re: relays

On Sun, 23 Nov 2008 13:52:35 -0600, John Fields
<jfields@xxxxxxxxxxxxxxxxxxxxx> wrote:

On Sun, 23 Nov 2008 07:45:00 -0800, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

On Sun, 23 Nov 2008 04:33:22 -0600, John Fields
<jfields@xxxxxxxxxxxxxxxxxxxxx> wrote:


Go ahead, try a real example, with real units. Do the math. Look up a
real relay data*** and compute its power gain. Make and state any
appropriate assumptions.

Try this one:

http://www.cel.com/pdf/datasheets/ps72141.pdf

---
Strangely, its "contacts" aren't rated in VA...

Show us some math. This one's really easy.

---
You came up with the example, so it's incumbent on _you_ to do the math.

Nothing is incumbent on me but breakfast (coffee and cherry cobbler.)
If neither you nor JKK can read a data*** and work some real
numbers, well, nobody will be really shocked. Engineering is not a
high-school debating team, much less a gaggle of little girls calling
names. Only the numbers matter.


But that wasn't the point; what was was that you claimed, earlier on,
that relay contacts are rated in VA.


---
from:

news:h0kdi4lck2gjis8rhb36qcp54on0a0p1v9@xxxxxxx

You wrote: "VA has the engineering units of watts, and the VA rating of
a relay contact (which is what they put on datasheets)"
.
.
.

But now you write:
---

Some are in fact rated in VA.
But many are rated for maximum working voltage, and maximum current,

---
So you've changed your tune and are now admitting that what's on most
datasheets is current and voltage, just like I said.


My God, you don't know how to multiply.



Do a little
research and I'm sure you'll have trouble finding very few with VA rated
contacts, the reason being that there are an infinite number of voltage
and current combinations which will satisfy the VA product, while the
relay was designed with a specific maximum contact voltage and current
in mind.
---

and there is a mathematical operation that will convert those into VA.
See if you can discover that.

---
Pitiful attempt at sarcasm, Larkin, especially since I'm the one who did
all the math which proved you wrong, while all you did was slobber all
over yourself with that "infinite gain" garbage.

Ah, but that was a day or two ago, so I guess you've forgotten...
---

This one's output seems to be rated in amperes.

Well, I suppose that forbids you from even thinking in terms of VA.

---
Certainly not, I just prefer to consider the relay's limits in terms of
current and voltage and then relate that to the load's requirements.
---

You are NOT ALLOWED TO MULTIPLY.

---
Tsk, tsk, tsk... Better watch that old blood pressure, you insufferable
ass.

Well, let's see..

It's kind of a rainy day and I've got nothing better to do, so let's
play your little game for a while.

If I recall correctly, you said to multiply the maximum contact voltage
by the maximum contact current and that would give the VA capability of
the relay, yes?

OK, from the spec ***'s absolute maximum values, we have 100V for the
breakdown voltage, Vl, and 400 mA for the load current, Il.

For a DC load, that would be:

Pl = Vl * Il = 100V * 0.4A = 40 VA = 40 watts.

and the load resistance would be:

Vl 100V
Rl = ---- = ------ = 250 ohms
Il 0.4A

But there's a fly in the ointment already, in that the relay has a 1.2
ohm worst-case on-state resistance and the MOSFETS in it can only
dissipate 300mW.

That means that in order to keep the part safe, the current through it
must be lowered to:


Pd 0.3W
Il = sqrt ----- = sqrt ------ = 0.25A
Ron 1.2R



Why would NEC rate a part for 400 mA continuous, when it can only
handle 250?

The answer is obvious: your math is once again nonsense.

(0.4)^2 * 1.2 = 0.192, comfortably below 0.3 watts.

WHY can't you check your work? (You forgot to actually take the root.)



So, the rest is all wrong too...



It also means that the drop across the MOSFETS won't be available to the
load, so instead of 100V the load will see:

Vl = Vs - Von = 100V - 1.2V = 98.8 volts

and it'll dissipate:

Pl = Vl * Il = 98.8V * 0.25A = 24.7 watts

Quite a bit different from that obtained by following your simplistic
"maximum voltage multiplied by maximum current = VA", huh?

Let's see if we can make it better, though...

Since we now know that we can only get 98.8 volts and 250mA out of the
relay, the lowest resistance it can safely drive will be:

El 98.8V
Rl = ---- = ------- = 395.2 ohms
Il 0.25A

Quite a bit different from 250 ohms, huh?

The power gain, BTW, would be


Pload
Pg = -----------------
Pled + Pmosfet


El Il
= --------------------
(Vf If) + (Von Il)



98.8V * 0.25A
= --------------------------------
(1.4V * 0.01A) + (1.2V * 0.25A)


~ 78.7



More nonsense! The fet doesn't drop 1.2 volts with a 250 mA load. At
1.2 ohms on resistance, the drop is 300 millivolts. Wrong by 4:1.

Some people really shouldn't do math in public.




And that's with the maximum legal DC load on it.


NEC specs 400 mA, in multiple places on their data***. You have
chosed to ignore their data*** and re-define what is "legal" for
their part. This disconnect didn't suggest that your math ought to be
checked? Wouldn't a usenet dispute over mathematical accuracy also
suggest checking?



For AC it would depend on the reactance of the load.

Wanna see how to work it out?

JF


Based on what I've seen so far, I think not.


John


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