RE: Re: LED lighting for garden

["Keith Doxey" <ukha.diyha@xxxxxxx>]

Coming into this one a bit late but....

Power losses are the same for AC or DC if you use the RMS rating for the AC
which is the average power (0.707 of peak). Peak power losses for Ac will
be
higher because the AC voltage is usually quoted as RMS.

240V mains has a peak of 340V but for quite a period of time is less than
240V and at times Zero.

The reason you can get AC further than DC is that you can use transformers
to step the voltage up and minimise the losses. Power remains constant, it
just gets redistributed. As the voltage is increased the current decreases.

Some simple numbers coming up (coz its late!)...

100v supply send a current of 10A through the circuit. Total power =
1000watts

Cable has a resistance of 1 ohm, power lost in cable is I(sq)R = 10x10x1 =
100W or 10% of total power, load receives 900w or 90% of power.

Step up/step down the voltage by a factor of 10 using transformers.....

1000V sends 1A, still 1000W, still 1 ohm cable,
power lost in cable is still I(sq)R, but now 1x1x1 = 1W or 0.1% loss
Load gets 999 watts or 99.9% of power.

You could of course use a cable 10 times thicker to get the resistance down
but copper is very expensive. the way to minimise losses is to ensure that
the cable is of low enough resistance that the volt drop will be minimised.

Another example using speakers,
a long run of cable has a total resistance of 4 ohms.
fit a 16 ohm speaker, total load is 20 ohms, 20% of power is lost in the
cable
fit a 8 ohm speaker, total load is 12 ohms, 33% of power is lost in cable
fit a 4 ohm speaker, total load is 8 ohms, 505 of power lost in cable.


If you want to use LED's for lighting then one thing you can do with LED's
is to put them in series. Typically an LED requires 2v. you would have to
use a resistor to drop 10v. On a 12v supply, using a 510R resistor would
give the LED a current of about 20mA. If you repeated this 5 times you
would
have 5 LED's lit with a total current of 100mA. 83% of the power would be
wasted in the resistors.

If you put the 5 LED's in series they would require 10V leaving the
resistor
to drop 2V, a 100R resistor would give you the same number of LED's lit at
the same brightness with a total consumption of only 20mA and only 17% of
the power lost in the resistor.

Keith

-----Original Message-----
From: patrickl@xxxxxxx [mailto:patrickl@xxxxxxx]
Sent: 18 April 2001 15:25
To: ukha_d@xxxxxxx
Subject: [ukha_d] Re: LED lighting for garden


> Ian,
>
> Are you sure about that? whilst I know that higher voltages result
in a
> lower loss, (hence the Supergrid 220Kv Distribution pylon I can see
from my
> window)
> I was pretty sure that the losses experienced by simple DC will
always be
> higher than an AC waveform?
>
> Going back to history, isn't that why Tesla's AC distribution was
better
> than Edison's DC plants? because you could get power further away
from the
> generating station?

Phew, even though my memory of A level physics is hazy, I'm glad
someone has the same recollection as me. All to do with P=I^2 R isn't
it?

Patrick





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