On 09/05/2011 13:11, wrote:
On May 8, 9:13 pm, Dave wrote:
I've just watched "Richard Hammond's Engineering Connections" on BBC2
about a new skyscraper in Dubai. Some expert told him that capacitors
and inductors were installed to smooth out the chopped up current from
the lighting dimmers, to stop the wiring overheating. We were treated
to a ridiculous demonstration in which hot wires ignited cotton wool
soaked in inflammable liquid, producing a spectacular destruction of a
wooden shed in a field.

My whole working life has been spent in electronics design, but I have
never heard of such a heating effect, and I still cannot understand
what it was all about. Can anybody provide a convincing technical
explanation?

Dave W

It's called Power Factor Correction, dimmers and many other
appliances, computers are bad for this too, draw a current that is out
of phase with the voltage, resulting in excessive 'wattless current'
flowing in the wiring, the supply mains, and the transformers, if
capacitors are not fitted near the load to absorb the 'reactive
power'.

Correct.

The supply boards used to charge extra for energy supplied to commercial and
industrial custonmers if the supply was "seeing" a highly inductive or highly
capacitive load. PFC (by placing a massively cpacative or massively inductive
load across the supply) was a way of getting he power demand back more nearly
in phase (and thereby lowering the PF surcharge).

JNugent wrote:
On 09/05/2011 13:11, wrote:
On May 8, 9:13 pm, Dave wrote:
I've just watched "Richard Hammond's Engineering Connections" on BBC2
about a new skyscraper in Dubai. Some expert told him that capacitors
and inductors were installed to smooth out the chopped up current from
the lighting dimmers, to stop the wiring overheating. We were treated
to a ridiculous demonstration in which hot wires ignited cotton wool
soaked in inflammable liquid, producing a spectacular destruction of a
wooden shed in a field.

My whole working life has been spent in electronics design, but I have
never heard of such a heating effect, and I still cannot understand
what it was all about. Can anybody provide a convincing technical
explanation?

Dave W

It's called Power Factor Correction, dimmers and many other
appliances, computers are bad for this too, draw a current that is out
of phase with the voltage, resulting in excessive 'wattless current'
flowing in the wiring, the supply mains, and the transformers, if
capacitors are not fitted near the load to absorb the 'reactive
power'.

Correct.

The supply boards used to charge extra for energy supplied to commercial
and industrial custonmers if the supply was "seeing" a highly inductive
or highly capacitive load. PFC (by placing a massively cpacative or
massively inductive load across the supply) was a way of getting he
power demand back more nearly in phase (and thereby lowering the PF
surcharge).

Its quite correct. The model aircraft boys have found that about 80-90%
throttle is the absolute worst case for controller heating.

I tried to analyse it but there were too many imponderables with an
inductive load with a variable back emf. The bulb case is much simpler
to analyse.

The bulb will cool, and therefore peak currents will be higher. How MUCH
of that peak is NOT smoothed by the interference choke is the issue, but
there's bound to be most of it.

Heating is I^2 * R *TF where R is the on resistance of the controller
and TF is the on time duty cycle.

as I gets bigger with falling output a lot faster than the time interval
its on gets smaller, there is definitely a peak at less than full power.

With motors, is that fact you are sending full voltage in bursts to a
motor that is not generating enough back EMF cos its too slow..instead
of the bulb not being hot enough, that causes the peak currents to rise.

On 09/05/2011 19:13, The Natural Philosopher wrote:

[...]
as I gets bigger with falling output a lot faster than the time interval
its on gets smaller, there is definitely a peak at less than full power.

That's not the case for tungsten filament lamps it seems.

Quick experiment (dangerous lash-ups-R-us):

- 100 W GLS bulb (from secret hoard)
- bog standard triac phase control dimmer (Hamilton)
- true RMS ammeter (Fluke 87)
- power and PF readout from Maplin 'energy monitor'
- supply voltage ~235 V

Result:

- as the dimmer is turned down from max. the RMS current falls
monotonically. There is no current peak at less than full power due to
the filament resistance falling with temperature.

Some readings:

Irms Power PF
------ ----- ----
Max. 412 mA 95 W 0.99
Mid. 324 mA 50 W 0.65
Min. 280 mA 33 W 0.51

No setting of the dimmer gave a current higher than 412 mA.

There's reasonable agreement between 235 V * Irms * PF and the measured
power for each setting.

--
Andy

Andy Wade wrote:
On 09/05/2011 19:13, The Natural Philosopher wrote:

[...]
as I gets bigger with falling output a lot faster than the time interval
its on gets smaller, there is definitely a peak at less than full power.

That's not the case for tungsten filament lamps it seems.

Quick experiment (dangerous lash-ups-R-us):

- 100 W GLS bulb (from secret hoard)
- bog standard triac phase control dimmer (Hamilton)
- true RMS ammeter (Fluke 87)
- power and PF readout from Maplin 'energy monitor'
- supply voltage ~235 V

Result:

- as the dimmer is turned down from max. the RMS current falls
monotonically.
So waht? we are not talking RMS current we are talking power in the wire.


Assuming you even have a meter that reads true RMS current in a heavily
clipped non sinusoidal wave.

What you dont do of course, was to measure how much power was actually
being dissipated in the WIRES.

It must vary in relation to what is being dissipated in the bulb becuase
the hot bulb takes several orders less current than the cold one, as we
know that cold bulbs have big switch on surges, yet you are assuming
that the proportion of power lost to the wire, is the same whether its n
series with a hot bulb or a cod one?

Go back and rethink ohms law dude.


Hint. The voltage drop across the wire is not a constant between the
full on and part on state,.

Hint: you wont see the current peaks when you are measuring RMS.

Hint: how sure are you that you are ersuin RMS anyay.


Hint: Consider a 50% square wave of 1A peak through a resistor of one
ohm the power dissipated is 0.5W. (0.5 * 1^2 * 1)

The average current drawn is 0.5A So surely the power dissipated by
that current is (0.5 * 0.5 * 1)= 0.25W.


Reconcile the difference.

On 10/05/2011 03:12, The Natural Philosopher wrote:
Andy Wade wrote:
- as the dimmer is turned down from max. the RMS current falls
monotonically.
So waht? we are not talking RMS current we are talking power in the wire.

You've really lost it now. The dissipation in the wires is just I^2 *
R, where I is the RMS current and R is the resistance. This assumes
that R is constant over the integration period (10 ms in this case),
which is it is.

Assuming you even have a meter that reads true RMS current in a heavily
clipped non sinusoidal wave.

Yes I do. True-RMS reading multi-meters have been widely available for
a long time now. You do have to watch the crest factor on very peaky
waveforms, to ensure that clipping doesn't occur in the meter, but
that's certainly not a concern in this case.

What you dont do of course, was to measure how much power was actually
being dissipated in the WIRES.

That was exactly the purpose of the experiment. The RMS current in the
circuit and the resistance are all you need to know.

It must vary in relation to what is being dissipated in the bulb becuase
the hot bulb takes several orders less current than the cold one, as we
know that cold bulbs have big switch on surges, yet you are assuming
that the proportion of power lost to the wire, is the same whether its n
series with a hot bulb or a cod one?

If you have a valid measurement of the RMS current in the wires, and
their resistance is constant you don't need to know anything about the
load. Oh and BTW the hot/cold resistance ratio for a GLS lamp is about
15:1 - hardly "several orders"

Go back and rethink ohms law dude.

Pah, you need to go and re-sit electrical basics 101.

Hint. The voltage drop across the wire is not a constant between the
full on and part on state,.

Agreed, so what.

Hint: you wont see the current peaks when you are measuring RMS.

So long as the meter sees them you're OK. See comments above re crest
factor. Anyway the chopped dimmer waveform is an easy one to handle
from this point of view.

Hint: how sure are you that you are ersuin RMS anyay.

Very sure. Very very sure. Better than 1% accuracy.

Hint: Consider a 50% square wave of 1A peak through a resistor of one
ohm the power dissipated is 0.5W. (0.5 * 1^2 * 1)

Agreed (assuming you mean a square wave between 0 and 1 A).

The average current drawn is 0.5A So surely the power dissipated by that
current is (0.5 * 0.5 * 1)= 0.25W.

No, because you've used the mean current, not the RMS value. In this
example the mean-squared current is 0.5 A, so the RMS current is 0.707 A
and 0.707^2 * 1 ohm gives 0.5 W. QED.

Reconcile the difference.

Done. Trivial.

--
Andy

In message , Andy Wade
writes
On 10/05/2011 03:12, The Natural Philosopher wrote:
Andy Wade wrote:
- as the dimmer is turned down from max. the RMS current falls
monotonically.
So waht? we are not talking RMS current we are talking power in the wire.

You've really lost it now. The dissipation in the wires is just I^2 *
R, where I is the RMS current and R is the resistance. This assumes
that R is constant over the integration period (10 ms in this case),
which is it is.

Assuming you even have a meter that reads true RMS current in a heavily
clipped non sinusoidal wave.

Yes I do. True-RMS reading multi-meters have been widely available for
a long time now. You do have to watch the crest factor on very peaky
waveforms, to ensure that clipping doesn't occur in the meter, but
that's certainly not a concern in this case.

What you dont do of course, was to measure how much power was actually
being dissipated in the WIRES.

That was exactly the purpose of the experiment. The RMS current in the
circuit and the resistance are all you need to know.

It must vary in relation to what is being dissipated in the bulb becuase
the hot bulb takes several orders less current than the cold one, as we
know that cold bulbs have big switch on surges, yet you are assuming
that the proportion of power lost to the wire, is the same whether its n
series with a hot bulb or a cod one?

If you have a valid measurement of the RMS current in the wires, and
their resistance is constant you don't need to know anything about the
load. Oh and BTW the hot/cold resistance ratio for a GLS lamp is about
15:1 - hardly "several orders"

Go back and rethink ohms law dude.

Pah, you need to go and re-sit electrical basics 101.

Hint. The voltage drop across the wire is not a constant between the
full on and part on state,.

Agreed, so what.

Hint: you wont see the current peaks when you are measuring RMS.

So long as the meter sees them you're OK. See comments above re crest
factor. Anyway the chopped dimmer waveform is an easy one to handle
from this point of view.

Hint: how sure are you that you are ersuin RMS anyay.

Very sure. Very very sure. Better than 1% accuracy.

Hint: Consider a 50% square wave of 1A peak through a resistor of one
ohm the power dissipated is 0.5W. (0.5 * 1^2 * 1)

Agreed (assuming you mean a square wave between 0 and 1 A).

The average current drawn is 0.5A So surely the power dissipated by that
current is (0.5 * 0.5 * 1)= 0.25W.

No, because you've used the mean current, not the RMS value. In this
example the mean-squared current is 0.5 A, so the RMS current is 0.707
A and 0.707^2 * 1 ohm gives 0.5 W. QED.

Reconcile the difference.

Done. Trivial.

A squarewave is the easiest example to get your head around (and without
having to resort to 'hard sums'). One simplistic way of thinking about
it is that (for a 50/50 squarewave), the power is 1W for 50% of the
time, and zero for the other 50%. Over the cycle, the average is
therefore 0.5W. The same logic can be applied to other mark-space
ratios.
--
Ian

On May 10, 3:12*am, The Natural Philosopher
wrote:
Andy Wade wrote:
On 09/05/2011 19:13, The Natural Philosopher wrote:

[...]
as I gets bigger with falling output a lot faster than the time interval
its on gets smaller, there is definitely a peak at less than full power.

On May 10, 9:12*am, Ian Jackson
wrote:
In message , Andy Wade
writes







On 10/05/2011 03:12, The Natural Philosopher wrote:
Andy Wade wrote:
- as the dimmer is turned down from max. the RMS current falls
monotonically.
So waht? we are not talking RMS current we are talking power in the wire.

You've really lost it now. *The dissipation in the wires is just I^2 *
R, where I is the RMS current and R is the resistance. *This assumes
that R is constant over the integration period (10 ms in this case),
which is it is.

Assuming you even have a meter that reads true RMS current in a heavily
clipped non sinusoidal wave.

Yes I do. *True-RMS reading multi-meters have been widely available for
a long time now. *You do have to watch the crest factor on very peaky
waveforms, to ensure that clipping doesn't occur in the meter, but
that's certainly not a concern in this case.

What you dont do of course, was to measure how much power was actually
being dissipated in the WIRES.

That was exactly the purpose of the experiment. *The RMS current in the
circuit and the resistance are all you need to know.

It must vary in relation to what is being dissipated in the bulb becuase
the hot bulb takes several orders less current than the cold one, as we
know that cold bulbs have big switch on surges, yet you are assuming
that the proportion of power lost to the wire, is the same whether its n
series with a hot bulb or a cod one?

If you have a valid measurement of the RMS current in the wires, and
their resistance is constant you don't need to know anything about the
load. *Oh and BTW the hot/cold resistance ratio for a GLS lamp is about
15:1 - hardly "several orders"

Go back and rethink ohms law dude.

Pah, you need to go and re-sit electrical basics 101.

Hint. The voltage drop across the wire is not a constant between the
full on and part on state,.

Agreed, so what.

Hint: you wont see the current peaks when you are measuring RMS.

So long as the meter sees them you're OK. *See comments above re crest
factor. *Anyway the chopped dimmer waveform is an easy one to handle
from this point of view.

Hint: how sure are you that you are ersuin RMS anyay.

Very sure. *Very very sure. *Better than 1% accuracy.

Hint: Consider a 50% square wave of 1A peak through a resistor of one
ohm the power dissipated is 0.5W. (0.5 * 1^2 * 1)

Agreed (assuming you mean a square wave between 0 and 1 A).

The average current drawn is 0.5A So surely the power dissipated by that
current is (0.5 * 0.5 * 1)= 0.25W.

No, because you've used the mean current, not the RMS value. *In this
example the mean-squared current is 0.5 A, so the RMS current is 0.707
A and 0.707^2 * 1 ohm gives 0.5 W. *QED.

Reconcile the difference.

Done. *Trivial.

A squarewave is the easiest example to get your head around (and without
having to resort to 'hard sums'). One simplistic way of thinking about
it is that (for a 50/50 squarewave), the power is 1W for 50% of the
time, and zero for the other 50%. Over the cycle, the average is
therefore 0.5W. The same logic can be applied to other mark-space
ratios.

If it's AC with no DC offset then the power is 1W 100% of the time,
regardless of the mark space ratio.

Square it, calculate the mean, take the square root. The RMS voltage
of an AC square wave with no DC offset is just the peak voltage.

MBQ

On May 9, 7:13*pm, The Natural Philosopher
wrote:
JNugent wrote:
On 09/05/2011 13:11, wrote:
On May 8, 9:13 pm, Dave *wrote:
I've just watched "Richard Hammond's Engineering Connections" on BBC2
about a new skyscraper in Dubai. Some expert told him that capacitors
and inductors were installed to smooth out the chopped up current from
the lighting dimmers, to stop the wiring overheating. We were treated
to a ridiculous demonstration in which hot wires ignited cotton wool
soaked in inflammable liquid, producing a spectacular destruction of a
wooden shed in a field.

My whole working life has been spent in electronics design, but I have
never heard of such a heating effect, and I still cannot understand
what it was all about. Can anybody provide a convincing technical
explanation?

Dave W

It's called Power Factor Correction, dimmers and many other
appliances, computers are bad for this too, draw a current that is out
of phase with the voltage, resulting in excessive 'wattless current'
flowing in the wiring, the supply mains, and the transformers, if
capacitors are not fitted near the load to absorb the 'reactive
power'.

Correct.

The supply boards used to charge extra for energy supplied to commercial
and industrial custonmers if the supply was "seeing" a highly inductive
or highly capacitive load. PFC (by placing a massively cpacative or
massively inductive load across the supply) was a way of getting he
power demand back more nearly in phase (and thereby lowering the PF
surcharge).

Its quite correct. The model aircraft boys have found that about 80-90%
throttle is the absolute worst case for controller heating.

Do they use switching or linear controllers? Bipolar or MOSFET? It
could well be more to do with poor controller design than power
factor.

MBQ

On 10/05/2011 10:55, Man at B&Q wrote:

Square it, calculate the mean, take the square root. The RMS voltage
of an AC square wave with no DC offset is just the peak voltage.

Quite correct, but from the context it was clear that both TNP and Ian
were both talking about an asymmetrical square wave with a DC component
equal to half the peak value. Here the RMS value is 1/sqrt(2) times the
peak value.

--
Andy

On May 10, 10:57*am, "Man at B&Q" wrote:
On May 9, 7:13*pm, The Natural Philosopher
wrote:





JNugent wrote:
On 09/05/2011 13:11, wrote:
On May 8, 9:13 pm, Dave *wrote:
I've just watched "Richard Hammond's Engineering Connections" on BBC2
about a new skyscraper in Dubai. Some expert told him that capacitors
and inductors were installed to smooth out the chopped up current from
the lighting dimmers, to stop the wiring overheating. We were treated
to a ridiculous demonstration in which hot wires ignited cotton wool
soaked in inflammable liquid, producing a spectacular destruction of a
wooden shed in a field.

My whole working life has been spent in electronics design, but I have
never heard of such a heating effect, and I still cannot understand
what it was all about. Can anybody provide a convincing technical
explanation?

Dave W

It's called Power Factor Correction, dimmers and many other
appliances, computers are bad for this too, draw a current that is out
of phase with the voltage, resulting in excessive 'wattless current'
flowing in the wiring, the supply mains, and the transformers, if
capacitors are not fitted near the load to absorb the 'reactive
power'.

Correct.

The supply boards used to charge extra for energy supplied to commercial
and industrial custonmers if the supply was "seeing" a highly inductive
or highly capacitive load. PFC (by placing a massively cpacative or
massively inductive load across the supply) was a way of getting he
power demand back more nearly in phase (and thereby lowering the PF
surcharge).

Its quite correct. The model aircraft boys have found that about 80-90%
throttle is the absolute worst case for controller heating.

Do they use switching or linear controllers? Bipolar or MOSFET? It
could well be more to do with poor controller design than power
factor.

MBQ- Hide quoted text -

- Show quoted text -

Phase shift (ie bad power factor/non unity can be caused by electronic
devices that reduce voltage by "chopping" the wave rather then
"lopping".
This has the effect of shifting the centre of the wave.

In message , Andy Wade
writes
On 10/05/2011 10:55, Man at B&Q wrote:

Square it, calculate the mean, take the square root. The RMS voltage
of an AC square wave with no DC offset is just the peak voltage.

Quite correct, but from the context it was clear that both TNP and Ian
were both talking about an asymmetrical square wave with a DC component
equal to half the peak value. Here the RMS value is 1/sqrt(2) times
the peak value.

Industry takes power factor correction very seriously.
The vast increase in the use of switchmode power supplies, it's much
more complicated these days.
Have a read:
(http://www.copperinfo.co.uk/power-qu...33-active-harm
onic-conditioners.pdf
http://www.copperinfo.co.uk/power-qu...51-neutral-siz
ing-in-harmonic-rich-installations.pdf

--
Ian

On 10/05/2011 21:56, John Rumm wrote:

Power dissipated in the wire will be directly proportional to the RMS
current.

That's not what you meant to write. Insert "the square of" :~)

I^2 R and all that...

Quite.

--
Andy

On May 10, 4:23*pm, harry wrote:
On May 10, 10:57*am, "Man at B&Q" wrote:









On May 9, 7:13*pm, The Natural Philosopher
wrote:

JNugent wrote:
On 09/05/2011 13:11, wrote:
On May 8, 9:13 pm, Dave *wrote:
I've just watched "Richard Hammond's Engineering Connections" on BBC2
about a new skyscraper in Dubai. Some expert told him that capacitors
and inductors were installed to smooth out the chopped up current from
the lighting dimmers, to stop the wiring overheating. We were treated
to a ridiculous demonstration in which hot wires ignited cotton wool
soaked in inflammable liquid, producing a spectacular destruction of a
wooden shed in a field.

My whole working life has been spent in electronics design, but I have
never heard of such a heating effect, and I still cannot understand
what it was all about. Can anybody provide a convincing technical
explanation?

Dave W

It's called Power Factor Correction, dimmers and many other
appliances, computers are bad for this too, draw a current that is out
of phase with the voltage, resulting in excessive 'wattless current'
flowing in the wiring, the supply mains, and the transformers, if
capacitors are not fitted near the load to absorb the 'reactive
power'.

Correct.

The supply boards used to charge extra for energy supplied to commercial
and industrial custonmers if the supply was "seeing" a highly inductive
or highly capacitive load. PFC (by placing a massively cpacative or
massively inductive load across the supply) was a way of getting he
power demand back more nearly in phase (and thereby lowering the PF
surcharge).

Its quite correct. The model aircraft boys have found that about 80-90%
throttle is the absolute worst case for controller heating.

Do they use switching or linear controllers? Bipolar or MOSFET? It
could well be more to do with poor controller design than power
factor.

MBQ- Hide quoted text -

- Show quoted text -

Phase shift (ie bad power factor/non unity can be caused by electronic
devices that reduce voltage by "chopping" the wave rather then
"lopping".
This has the effect of shifting the centre of the wave.

Eh?

"Man at B&Q" wrote in message
...

Phase shift (ie bad power factor/non unity can be caused by electronic
devices that reduce voltage by "chopping" the wave rather then
"lopping".
This has the effect of shifting the centre of the wave.

Eh?

I think he means if its chopped asymmetrically it introduces a DC bias.

On 11/05/2011 11:16, dennis@home wrote:


"Man at B&Q" wrote in message
...

Phase shift (ie bad power factor/non unity can be caused by electronic
devices that reduce voltage by "chopping" the wave rather then
"lopping".
This has the effect of shifting the centre of the wave.

Eh?

I think he means if its chopped asymmetrically it introduces a DC bias.

It doesn't have to be DC. As an example, taking the part of each half
wave moving away from zero volts would move the current wave relative to
the voltage.

Andy