Re: Relay

[Raymond <Reb.barnett@xxxxxxx>]

Cheers Keith! You've convinced me (and explained what mystical zero volt
switching means). However, to re-muddy the waters, another post in the
Comfort group suggests bad things concerning the (small) voltage drop on
the mains side of the device:

Solid state relays have no moving parts, and are tend to be more
reliable.
However, dropping voltage on the mains lines is not a good thing if you
are
driving a large load like central heat (wastes some energy).  I would go
with a
"real" mechanical relay myself.  I have had bad experience with
AC
motors and
compressors burning out because of not getting proper main voltage.

I'm not worried about the 'wastes some energy' point, especially since
you're logic suggests that quite the reverse is true. However, I could
do without my central heating pump burning out because unless I come up
with a clever way of hinging the kitchen worktop, it won't be very
accessible when we put the kitchen back together again...

Any thoughts as to whether 1.6v is worth worrying about. My theory is
that 1.6v is within the parameters for the mains supply anyway, so
probably isn't significant...

Ray.

> -----Original Message-----
> From: keith.doxey@xxxxxxx [mailto:keith.doxey@xxxxxxx]
> Sent: 30 September 1999 22:21
> To: REB.Barnett@xxxxxxx
> Subject: [ukha_d] Re: Relay
>
>
> Hi Raymond,
>
> Go for the SSR Zero Volt Switching.
> No moving parts and no RF Interference.
> SSR's (instant or zerovoltage) all switch off at mains zero unlike a
> conventional relay which could release at peak mains with
> some arcing at the
> contacts.
> Normal relays require a diode in inverse parallel with the
> coil to protect
> the driver circuitry from the back EMF when the relay is
> de-energised. (a
> 12V relay can easily give a back EMF or 300V or more)
> No clunks from the cupboard when the boiler gets turned on.
>
> Some decidedly dodgy maths coming up....
>
> ...looking at the figures in the Maplin catalogue the SSR
> drops 1.6V @ 5A
> which works out to be 8W.
> BUT....to draw 5A at 240V would require a resistive load of
> 240/5 = 48 ohms.
> If the SSR was dropping the 1.6V then the supply to the load
> would only be
> 238.4V
> Small difference but....238.4/48 = 4.97A
> Power =Current(squared)*Resistance = 4.97 * 4.97 * 48 = 1186W
>
> Power used by the load plus the power lost in the SSR
> 1186 + 8 = 1194W
>
> THATS 6 WATTS LESS THAN CONNECTING IT DIRECTLY TO THE MAINS!!!
> You will SAVE money :-))
>
> It wont work out EXACTLY like that as the load will not be
> purely resistive
> but the point is that any power lost in control circuitry
> will reduce the
> available voltage to the load by a small percentage thus
> lessening its power
> consumption slightly.
>
> Dont worry about the power consumption, its so low as to be
> non existant.
> Make sure you put a Hi-speed ceramic fuse (of the lowest
> possible rating) in
> series with the SSR to protect it in the event of a fault.
>
> As to your point about mixing mains with low voltage...no
> problem. The SSR
> is designed to withstand a minimum of 2500V without the
> isolation breaking
> down. Use heatshrink sleeving over the connections if you are
> soldering
> wiring direct to the SSR. If you are making a printed circuit
> board to mount
> them make sure you have maximim separation between the tracks
> by keeping all
> the mains stuff to one side and the low voltage to the other.
> Also put an
> earthed track between the low and high voltage sides of the
> PCB keeping it
> away from the mains side to avoid any possibility of a short circuit.
>
> Dont forget to take a few pictures of the finished project
> for us all to
> see. :-))
>
> Keith
>
> Keith Doxey
> http://www.btinternet.com/~krazy.keith
> Krazy Keith's World of DIY HomeAutomation
>
>
> > -----Original Message-----
> > From: Raymond [mailto:Reb.barnett@xxxxxxx]
> > Sent: 30 September 1999 17:13
> > To: 'ukha_d@xxxxxxx'
> > Subject: [ukha_d] Re: Relay
> >
> >
> > Hi Keith,
> >
> > Thanks for the reply: I was trying to work out which was
> best solution
> > out of a solid state relay versus a 'real' relay.
> >
> > I was using the Maplin catalogue specs for their solid
> state relays to
> > get the 5W power loss, but I may well have got my maths
> wrong, and will
> > certainly have calculated it for the worst case if the
> heating system
> > drew 5A from the relay - I don't know how much current will
> be drawn so
> > that's a project for tonight with the multimeter... But anyway,
the
> > thing is, if the solid state version wastes 5W, that's a
> point to the
> > 'real' relay.
> >
> > However, if the 'real' relay requires an auxilliary 12V
> power supply for
> > it to work without overloading Comfort, then a second
> transformer will
> > probably waster 5W, so we're back to evens.
> >
> > The solid state relay has no moving parts, so presumably will
last
> > longer than a real one; the solid state relay moves ahead.
> >
> > I understand how a real relay works, and am a touch
> concerned about a
> > chip that has mains voltage on one pin and 12v on another -
> I don't want
> > to start a fire or blow up Comfort. So the real relay is
> back on evens.
> > The Maplin catalogue details two types of solid state relay
> (zero volts
> > and something different - I don't have the catalogue at work) and
I
> > haven't a clue what the difference is and whether it matters...
> >
> > So which one should I use? I think that if you reckon a solid
state
> > relay will work, I'll order a couple of these and play with
> them. In the
> > mean time, I'll check the current drawn by the boiler and check
what
> > spare capacity I've got on Comfort's 12V supply - I've added more
> > detectors since last time I checked.
> >
> > Ray.
> >
>
>
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