How much power saved with power saving mode?

#1
As I understand it there are two levels of standby, one shuts down a few more things to save power. Does anyone know just how much power? The reason I ask is that I would like to put the Humax in the aerial feed rather than on a spur as it is now but to do so I have to start the loop through working in standby which means I can't use power saving mode.

Cheers,
Bob.
 

grahamlthompson

Well-Known Member
#2
If I remember rightly it's about 0.75W low power to 4W with power saving off. Most can simply fit a passive splitter to give the box a discrete feed.
 

Ezra Pound

Well-Known Member
#3
I have measured power in both modes with the following results :-
Power saving on = 0.5Watts
Power saving off = 1.5Watts
The power saving mode was created because new equipment must have a standby mode less than 1 Watt, But as you can see it's only just over that with power saving off and you do get the UHF feed though and Clock feature enabled
 

grahamlthompson

Well-Known Member
#4
I have measured power in both modes with the following results :-
Power saving on = 0.5Watts
Power saving off = 1.5Watts
The power saving mode was created because new equipment must have a standby mode less than 1 Watt, But as you can see it's only just over that with power saving off and you do get the UHF feed though and Clock feature enabled
What did you measure the power with as a point of interest ?. It's very hard without proper kit to measure such low powers without special kit. You could of course use your meter and a stopwatch. Plug in devices which generally only measure current wouldn't cut it. You need accurate voltage, accurate current and the angle between voltage and current.
 

Ezra Pound

Well-Known Member
#5
Most power meters (even the ones with power factor correction) are very unreliable measuring standby current due to waveform distortion, I have two plug-top power meters that give 'power saving off' readings of 7 Watts and 14 Watts. So the answer is to connect it to an old fashioned electric meter (the old spinning wheel sort) and count the revs over a two hour period.
 

grahamlthompson

Well-Known Member
#6
Most power meters (even the ones with power factor correction) are very unreliable measuring standby current due to waveform distortion, I have two plug-top power meters that give 'power saving off' readings of 7 Watts and 14 Watts. So the answer is to connect it to an old fashioned electric meter (the old spinning wheel sort) and count the revs over a two hour period.
Your meter and the stopwatch then (it's basically a shaded pole induction motor), many years ago spent some time calibrating them. - thanks that should be accurate. I admire your dedication :)

You did not need to wait 2 hrs, all meters have a revs/KWHour label. Count say 50 revs and time with a stop watch. Calculating the power is then a trivial calculation.
 

Ezra Pound

Well-Known Member
#7
On the meter I used 1 rev per hour = 5 Watt Hour (200 revs. per KWH), so 0.5 WH = one tenth of a rev. and 1.5WH = three tenths of a rev. 2 hours was needed to get a good enough movement of the wheel to measure. 50 revs. at 0.5WH would have taken 500 Hours
 

Black Hole

May contain traces of nut
#8
The information is interesting, but the units are up the creek. If I may:

On the meter I used 1 rev. = 5 Watt Hours (200 revs. per kWh), so 0.5Wh = one tenth of a rev. and 1.5Wh = three tenths of a rev. 2 hours was needed to get a good enough movement of the wheel to measure. 50 revs. at 0.5W would have taken 500 Hours
Ed.
 

grahamlthompson

Well-Known Member
#9
200 revs/hr at 1000W. 0.2 revs/hr at 1W, 0.1 revs/hr at 0.5W. Doubtfull about the accuracy at such a low loading, friction in the bearings and gear train will cause some undereading of the power. Probally better to have a fixed load of around a kW and impose the extra load of sby on this.
 
#10
I'll give it a go as soon as I get my meter checked - I had to send it away to have the rotational speed accurately calibrated
and there was a delay because the local laser calibration lab said there was a slight speed variation in the speed of light
from their laser so they couldn't be totally accurate until it was checked with intergalactic standards...
 

Ezra Pound

Well-Known Member
#13
Probally better to have a fixed load of around a kW and impose the extra load of sby on this.
It's a fair point, But you would be looking for a diffefrence in the ratio of 2000 : 2001 or to put it another way the difference between 200 revs. and 200.1 revs over 1 Hour. AsI got 0.5 Watts for the power saving = on, my measurement can't be out by more a factor of 2 because this actual figure can't be more that 1Watt
 
#14
Domestic meters were calibrated to be accurate to +- 2% for upf loads within specified current limits typically 20-80A (I imagine the same applies today). For extreme power factors or low loads the error will be significantly higher. Disc inertia means they won't measure very short pulses of current like motor mag inrush or switching on an incandescent bulb very accurately.

As you say in the great scheme of things your results are going to be pretty close to reality.

It would be interesting to know how low you can go before the disc motor simply hasn't enough torque to overcome the friction in the drive train.
 

Ezra Pound

Well-Known Member
#15
grahamlthompson : Domestic meters were calibrated to be accurate to +- 2%
At the end of the day I am being charged for power consumption as measured by the revs. on that wheel, so if it gets the measurement wrong and it's in my favour, that's fine
 

Black Hole

May contain traces of nut
#18
He changed the capitalisation of the units to match SI (i.e. kWh and not KWH)
There's a bit more to it than that - read carefully.

True enough :). If they want to fit a digital meter resist it, it might cost you an extra 50p annually :(
I'm concerned whether the digital meters measure power correctly, or if they measure the RMS voltage and RMS current and just multiply the two (and get AV). It wouldn't be far out for normal domestic loads, but would be hideously wrong for industrial inductive loads.
 
#19
There's a bit more to it than that - read carefully.



I'm concerned whether the digital meters measure power correctly, or if they measure the RMS voltage and RMS current and just multiply the two (and get AV). It wouldn't be far out for normal domestic loads, but would be hideously wrong for industrial inductive loads.
There's no way they could get away with just current and voltage, even domestic loads like induction motors are far from unity power factor. Measuring the time difference between current and voltage zeros ought to be simple using a cpu.
 
#20
Thanks guys for all the info. I've been looking at the cost of low power over a year. Normally, most electric power is very expensive for the first so many units but then much cheaper. The cheap rates seem to be around 12p per unit or 1KWh. There are 8784 hours in a leap year Now 1 watt for that year would cost £1.05 for the whole year. If my maths is correct, the couple of quid saved doesn't seem worth doing without loop through and the clock does it?

Bob.
 
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