Motion detector

[AZ Nomad]

On Thu, 24 Dec 2009 14:17:12 +1100, Trevor Wilson <trevor@SPAMBLOCKrageaudio.com.au> wrote:

"AZ Nomad" <aznomad.3@PremoveOBthisOX.COM> wrote in message
news:slrnhj5hhp.mom.aznomad.3@ip70-176-155-130.ph.ph.cox.net...
On Thu, 24 Dec 2009 12:13:39 +1100, Trevor Wilson
trevor@SPAMBLOCKrageaudio.com.au> wrote:

"Dave Platt" <dplatt@radagast.org> wrote in message
news:22ra07-mn9.ln1@radagast.org...
In article <7pfpjsFrh0U1@mid.individual.net>,
Phil Allison <phil_a@tpg.com.au> wrote:

One advantage of a relay is that it really does cut the load current
to
zero, near as makes no difference. I haven't seen a solidstate device
that
does that, and at 240V, even a few milliamps adds up to a watt or so,
and
is running 24/7.


** A triac cuts off load current as near as make no difference.

Leakage is in the uA range - not mA.

You IMBECILE !!

In the overview of triac-based solid state relays that I find at
Omega.com (they make these), they write:

The output-circuit ratings of the more common isolated
SSR's, most of which are designed to control ac load
circuits, are very similar to those described above,
except that OFF-state leakage is usually higher---on the
order of 5 mA at 140 V for a 5-ampere device---still only
about one-thousandth of the load current rating.

The data sheet for the AQ-R 10-to-40-ampere solid state relays
specifies a maximum off-state leakage current of 2.5 mA at 100 volts
AC, and 5 mA at 200 volts AC.
...

**We can be absolutely certain that a motion detector light switch (unless
it is in a top secret, government funded military situation) will not be
using a SSR.

and I think you can be pretty sure it won't switch 40A, although
you never know when you might want to have 2400watts of lighting.

**No problems over here. A small (TO-220), heat sunk TRIAC will easily
switch 2.4kVA of resistive or, (more importantly) inductive loads. 2.4KW
lighting tends to be very inductive. Relays die very quickly under such
loads.

I just don't see the point of using a relay in such a circuit. I'm
can't do AC analog worth a damn, but I've never had the slightest
difficulty switching lights. Since the 70's, I've just used two
parts, a optoisolated trigger diac and a triac. Can't imagine
anywhere in such a circuit where a relay would be the slightest benefit.

 

[Phil Allison]

"Trevor Wilson"

"Phil Allison"

**Complete bollocks. A TRIAC has miniscule leakage. Less leakage
current, in fact, than the suppression capacitor across the relay
contacts. MUCH less.


** The leakage current flowing in such a suppression cap produces no
heat and consumes no power.

Cos the PF is zero.

**Indeed it is, unless it the relay contacts happens to be in series
with a resistive load (like an incadescent lamp).


** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....


**Misunderstanding and poor wording by me. I do not dispute the PF. I was
making the point that the current through a typical suppression cap
exceeds the leakage current through a typical TRIAC.

** Like fucking hell you were.

You drunk ?


..... Phil

 

[Baron]

Van Chocstraw Inscribed thus:

Baron wrote:
Van Chocstraw Inscribed thus:

My yard light won't go off any more. I block the detector wait 60
seconds and I hear the relay click off but the light now stays on
when it used to go off. It clicks again when I block the detector,
it times out and clicks again but the light never goes out. What's
the deal?

Welded relay contacts... New relay required !

Yea. welded by ice.

Outdoors, thats always a possibility. I thought IP35 was supposed to be
splash proof !

--
Best Regards:
Baron.

 

[Robert Macy]

On Dec 23, 4:29 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Trevor Wilson"

" Stupidest Bitch on Earth Sylvia Else"



One advantage of a relay is that it really does cut the load current to
zero, near as makes no difference. I haven't seen a solidstate device
that does that, and at 240V, even a few milliamps adds up to a watt or
so, and is running 24/7.

**Complete bollocks. A TRIAC has miniscule leakage. Less leakage current,
in fact, than the suppression capacitor across the relay contacts. MUCH
less.

** The leakage current flowing in such a suppression cap produces no heat
and consumes no power.

  Cos the PF is zero.

.....  Phil

Actually, it produces no 'billable' power consumption, but it eats
power throughout the distribution system in cable and transformer
losses. Thus, reactive losses, especially inductive losses, are
discouraged by the utilities companies.

 

[Sylvia Else]

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less leakage
current, in fact, than the suppression capacitor across the relay
contacts. MUCH less.
** The leakage current flowing in such a suppression cap produces no
heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in series
with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the PF. I was
making the point that the current through a typical suppression cap
exceeds the leakage current through a typical TRIAC. It was for Sylvia's
benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point related
to power consumption.

**It is very relevant. The leakage current through a typical suppression cap
across a relay is in the order of milliamps. The leakage across a tyical
TRIAC is in the order of microamps. The use of a relay in such an
application does not guarantee zero power consumption when non-operational.

As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.

Sylvia.

 

[Phil Allison]

"Robert Macy"
"Phil Allison" > "Trevor Wilson"

" Stupidest Bitch on Earth Sylvia Else"


One advantage of a relay is that it really does cut the load current to
zero, near as makes no difference. I haven't seen a solidstate device
that does that, and at 240V, even a few milliamps adds up to a watt or
so, and is running 24/7.

**Complete bollocks. A TRIAC has miniscule leakage. Less leakage
current,
in fact, than the suppression capacitor across the relay contacts. MUCH
less.

** The leakage current flowing in such a suppression cap produces no heat
and consumes no power.

Cos the PF is zero.

Actually, it produces no 'billable' power consumption, but it eats
power throughout the distribution system in cable and transformer
losses.

** Really ??

The largest " I squared R " copper loss will be where the largest resistance
exists in the AC supply.

Typical domestic AC outlets have internal resistances of about 0.2 to 1
m - mostly due to the wiring installed in the premises. Resistances
values further back in the supply system are very much lower.

A suppression cap draws only about 3.5mA from a 240 volt AC supply.

So the loss in watts is between 2 and 12 uW - that is MICRO watts !!!

Absolutely SFA.

Piss off - you pedantic fool.



..... Phil

 

[Trevor Wilson]

Sylvia Else wrote:

Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.
** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.

**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.

As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.

**As was I. The POWER consumed by the device using a relay will probaby be
significantly higher than if it uses a TRIAC. The reason is that the current
flowing through the load will be much higher with a relay.


--
Trevor Wilson
www.rageaudio.com.au

 

[Sylvia Else]

Trevor Wilson wrote:

Sylvia Else wrote:
Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.
** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.
**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.
As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.

**As was I. The POWER consumed by the device using a relay will probaby be
significantly higher than if it uses a TRIAC. The reason is that the current
flowing through the load will be much higher with a relay.

It might be, but it's not the power consumed by the load that is the
main concern. It is the power consumed by the switching element, since
it is where most of the voltage drop occurs. The TRIAC is essentially
resistive, so it is consuming real power. The relay, with a capacitor
across its contacts, is only consuming reactive power, for which there
is no charge to a retail customer.

Sylvia.

 

[Trevor Wilson]

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00b04b29$0$12969$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:
Sylvia Else wrote:
Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.
** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.
**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.
As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.

**As was I. The POWER consumed by the device using a relay will probaby
be significantly higher than if it uses a TRIAC. The reason is that the
current flowing through the load will be much higher with a relay.



It might be, but it's not the power consumed by the load that is the main
concern.

**Yes, it is.

It is the power consumed by the switching element, since

it is where most of the voltage drop occurs.

**Change of discussion, duly noted. There will be a Voltage drop across any
semiconductor, whilst there will be essentially zero drop across relay
contacts. THAT, however, is not your original contention.

The TRIAC is essentially

resistive, so it is consuming real power.

**Wrong. The TRIAC is essentially a solid state switch. It does, however,
dissipate some power, due to the valtage drop across the device. A relay
(usually) exhibits a far lower Voltage drop.

The relay, with a capacitor

across its contacts, is only consuming reactive power, for which there is
no charge to a retail customer.

**WRONG! The capacitor across the contacts allows a small current to flow
into the load at all times. This power will certainly be charged to the
consumer. A TRIAC, OTOH, will usually allow a much smaller current to flow
into the load. Thus, the consumer will pay less.


--
Trevor Wilson
www.rageaudio.com.au

 

[Sylvia Else]

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00b04b29$0$12969$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
Sylvia Else wrote:
Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.
** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.
**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.
As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.
**As was I. The POWER consumed by the device using a relay will probaby
be significantly higher than if it uses a TRIAC. The reason is that the
current flowing through the load will be much higher with a relay.


It might be, but it's not the power consumed by the load that is the main
concern.

**Yes, it is.

It is the power consumed by the switching element, since
it is where most of the voltage drop occurs.

**Change of discussion, duly noted. There will be a Voltage drop across any
semiconductor, whilst there will be essentially zero drop across relay
contacts. THAT, however, is not your original contention.

The TRIAC is essentially
resistive, so it is consuming real power.

**Wrong. The TRIAC is essentially a solid state switch. It does, however,
dissipate some power, due to the valtage drop across the device. A relay
(usually) exhibits a far lower Voltage drop.

Remember, we're talking about the off state. The voltage drop across the
device is the input voltage, near enough.

The relay, with a capacitor
across its contacts, is only consuming reactive power, for which there is
no charge to a retail customer.

**WRONG! The capacitor across the contacts allows a small current to flow
into the load at all times. This power will certainly be charged to the
consumer. A TRIAC, OTOH, will usually allow a much smaller current to flow
into the load. Thus, the consumer will pay less.

Your ignoring the power dissipated in the TRIAC. That's most of the
power consumption of the combined system in the off state.

Sylvia.

 

[Trevor Wilson]

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00ade816$0$1485$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00b04b29$0$12969$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
Sylvia Else wrote:
Trevor Wilson wrote:
"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...
Trevor Wilson wrote:
"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...
"Trevor Wilson"
"Phil Allison"
**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.
** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.
**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....
**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.
**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.
As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.
**As was I. The POWER consumed by the device using a relay will probaby
be significantly higher than if it uses a TRIAC. The reason is that the
current flowing through the load will be much higher with a relay.


It might be, but it's not the power consumed by the load that is the
main concern.

**Yes, it is.

It is the power consumed by the switching element, since
it is where most of the voltage drop occurs.

**Change of discussion, duly noted. There will be a Voltage drop across
any semiconductor, whilst there will be essentially zero drop across
relay contacts. THAT, however, is not your original contention.

The TRIAC is essentially
resistive, so it is consuming real power.

**Wrong. The TRIAC is essentially a solid state switch. It does, however,
dissipate some power, due to the valtage drop across the device. A relay
(usually) exhibits a far lower Voltage drop.

Remember, we're talking about the off state. The voltage drop across the
device is the input voltage, near enough.

**Correct. Since the off state of a TRIAC exhibits a leakage current in the
order of a few microAmps, it is several orders of magnitude less than the
current flow through a typical suppression cap across relay contacts.

The relay, with a capacitor
across its contacts, is only consuming reactive power, for which there
is no charge to a retail customer.

**WRONG! The capacitor across the contacts allows a small current to flow
into the load at all times. This power will certainly be charged to the
consumer. A TRIAC, OTOH, will usually allow a much smaller current to
flow into the load. Thus, the consumer will pay less.



Your ignoring the power dissipated in the TRIAC.

**Well, no, I'm not. The Pdiss is miniscule in the off state.

That's most of the

power consumption of the combined system in the off state.

**Yes, it is. It is also far less than the Pdiss of the load, when using a
relay. The TOTAL power consumption of a TRIAC device (when connected to the
load) is generally lower than a relay device.


--
Trevor Wilson
www.rageaudio.com.au

 

[Sylvia Else]

Jamie wrote:

Sylvia Else wrote:

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00b04b29$0$12969$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:

Sylvia Else wrote:

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:

"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...

"Trevor Wilson"

"Phil Allison"

**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.

** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.

**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....

**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.

**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.

As Phil has pointed out, and you have conceded, the PFs are
different,
and I was clearly focussing on power, not current.

**As was I. The POWER consumed by the device using a relay will
probaby be significantly higher than if it uses a TRIAC. The reason
is that the current flowing through the load will be much higher
with a relay.


It might be, but it's not the power consumed by the load that is the
main concern.


**Yes, it is.

It is the power consumed by the switching element, since

it is where most of the voltage drop occurs.


**Change of discussion, duly noted. There will be a Voltage drop
across any semiconductor, whilst there will be essentially zero drop
across relay contacts. THAT, however, is not your original contention.

The TRIAC is essentially

resistive, so it is consuming real power.


**Wrong. The TRIAC is essentially a solid state switch. It does,
however, dissipate some power, due to the valtage drop across the
device. A relay (usually) exhibits a far lower Voltage drop.


Remember, we're talking about the off state. The voltage drop across
the device is the input voltage, near enough.


The relay, with a capacitor

across its contacts, is only consuming reactive power, for which
there is no charge to a retail customer.


**WRONG! The capacitor across the contacts allows a small current to
flow into the load at all times. This power will certainly be charged
to the consumer. A TRIAC, OTOH, will usually allow a much smaller
current to flow into the load. Thus, the consumer will pay less.



Your ignoring the power dissipated in the TRIAC. That's most of the
power consumption of the combined system in the off state.

Sylvia.
Maybe I am miss reading your statement how ever, if I understand you
correctly, I think you have a misconception with the operation of
thyristors.

To enable a thyristor in the on state, it still requires less power
with the added components than what a relay would require considering
that a relay needs to power it's coil(constantly). The operation of a
coil will use more energy than maintaining a thyristor in the on state.

And for it being off, leakage is far less than you think. Energy being
lost in the thyristor is minuscule. You'll have more loss from heat of a
relay coil.

Leakage figures vary.

The issue I was raising was the power consumption in the off state. You
may be right in asserting the relays require more power than TRIACS in
the one state, but that means that one would need to consider the duty
cycle to reach a conclusion. It seems that a motion detector system
would spend most of its time in the off state.

Sylvia.

 

[Jamie]

Sylvia Else wrote:

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:00b04b29$0$12969$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:

Sylvia Else wrote:

Trevor Wilson wrote:

"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:0011d0d7$0$2124$c3e8da3@news.astraweb.com...

Trevor Wilson wrote:

"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7pfv32FkduU1@mid.individual.net...

"Trevor Wilson"

"Phil Allison"

**Complete bollocks. A TRIAC has miniscule leakage. Less
leakage current, in fact, than the suppression capacitor
across the relay contacts. MUCH less.

** The leakage current flowing in such a suppression cap
produces no heat and consumes no power.

Cos the PF is zero.

**Indeed it is, unless it the relay contacts happens to be in
series with a resistive load (like an incadescent lamp).

** Huh ????

Cold lamps have typical resistances under 100ohms

The AC current from a 47nF cap at 240 volts is 3.5 mA

Do the math.

I get a PF of 0.0014

Over to you.....

**Misunderstanding and poor wording by me. I do not dispute the
PF. I was making the point that the current through a typical
suppression cap exceeds the leakage current through a typical
TRIAC. It was for Sylvia's benefit, since you already knew that.


It didn't, and doesn't, seem relevant though, given that my point
related to power consumption.

**It is very relevant. The leakage current through a typical
suppression cap across a relay is in the order of milliamps. The
leakage across a tyical TRIAC is in the order of microamps. The use
of a relay in such an application does not guarantee zero power
consumption when non-operational.

As Phil has pointed out, and you have conceded, the PFs are different,
and I was clearly focussing on power, not current.

**As was I. The POWER consumed by the device using a relay will
probaby be significantly higher than if it uses a TRIAC. The reason
is that the current flowing through the load will be much higher
with a relay.


It might be, but it's not the power consumed by the load that is the
main concern.


**Yes, it is.

It is the power consumed by the switching element, since

it is where most of the voltage drop occurs.


**Change of discussion, duly noted. There will be a Voltage drop
across any semiconductor, whilst there will be essentially zero drop
across relay contacts. THAT, however, is not your original contention.

The TRIAC is essentially

resistive, so it is consuming real power.


**Wrong. The TRIAC is essentially a solid state switch. It does,
however, dissipate some power, due to the valtage drop across the
device. A relay (usually) exhibits a far lower Voltage drop.


Remember, we're talking about the off state. The voltage drop across the
device is the input voltage, near enough.


The relay, with a capacitor

across its contacts, is only consuming reactive power, for which
there is no charge to a retail customer.


**WRONG! The capacitor across the contacts allows a small current to
flow into the load at all times. This power will certainly be charged
to the consumer. A TRIAC, OTOH, will usually allow a much smaller
current to flow into the load. Thus, the consumer will pay less.



Your ignoring the power dissipated in the TRIAC. That's most of the
power consumption of the combined system in the off state.

Sylvia.
Maybe I am miss reading your statement how ever, if I understand you

correctly, I think you have a misconception with the operation of
thyristors.

To enable a thyristor in the on state, it still requires less power
with the added components than what a relay would require considering
that a relay needs to power it's coil(constantly). The operation of a
coil will use more energy than maintaining a thyristor in the on state.

And for it being off, leakage is far less than you think. Energy being
lost in the thyristor is minuscule. You'll have more loss from heat of a
relay coil.

 

[Robert Macy]

On Dec 24, 6:07 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

 "Phil Allison" > "Trevor Wilson"
....snip...
**  Really ??

The largest " I squared R " copper loss will be where the largest resistance
exists in the AC supply.

Typical domestic AC outlets have internal resistances of about 0.2 to 1
m   -   mostly due to the wiring installed in the premises.  Resistances
values further back in the supply system are very much lower.

A suppression cap draws only about 3.5mA from a 240 volt AC supply.

So the loss in watts is between 2 and 12 uW -  that is MICRO watts !!!

Absolutely SFA.

Piss off  - you pedantic fool.

....  Phil

My favorite film!, but the quote is, "...pedantic old fool..."

Hmm, 1500W room heater drops the voltage from 125 to 119, implying
resistance of around 0.48 ohms. Assuming wiring is either 12 Awg at
1.6 ohms/kft, or 14 Awg at 2.5 ohms/kft that means there was 150 [300
total out and back], or 81 ft of run to that 'low resistance' feed.
Expecting 0.2 to 1 ohm seems reasonable.

I remember 'green standby is less than 1 W', or 10 mA, so 3.5 mA is
less, and not power. The loss inside residences would be on the order
of 6 uW. If everybody had 5 of these and 10million people did this
[not including businesses], that would only amount to 300 watts
TOTAL. There is MUCH more than that lost along the transmission paths
delivering primary power, interestingly low.