Homemade LED anchor light circuit

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A few numbers of some possible relevance:

The requirement is for a light to be visible at 2 nautical miles, so at 2 x 1852 = 3704 metres.

The luminosity which equates to 'visible' in night time conditions is 2 x 10-7 lumens per sq meter according to the IALA recommendations E-200-2, December 2008.

A bright white LED driven at ~20 mA is of the order of 9 lumens, into a +/- 22.5° solid angle (one should check the spec of what one's using).

Thus this LED, if spread all round 360 degrees of azimuth but still with elevation of +/- 22.5°(achievable for instance by pointing it vertically and reflecting off a silvered cone, as is common in 'solar' garden lights) will be 9 / 360 * 45 = 1.125 lumens.

Applying Allard's law for how it attenuates with distance but assuming no atmospheric absorption gives 1.125 / 37042 = 8 x 10-8 lumens per sq metre, ie a bit less than half the luminosity required.

Thus one has three choices:

1. Drive the LED harder / get a more powerful one, order 25 lumens is easily available.
2. Arrange several LEDs in a 'crown', nominally 8 of them spaced by 45° will be 8x brighter
3. Use an array of, say, 9 LEDs pointing at the silvered cone.

Any of these would easily satisfy the rules.
 
jdc said:
2. Arrange several LEDs in a 'crown', nominally 8 of them spaced by 45° will be 8x brighter
Click to expand...

My Bebi light uses two rows of LEDs, either 12 or 16 per row (I can't quite remember). The two rows are offset slightly to give a more even pattern, and are powered via two independent regulators for improved reliability - even if one of them fails, the light will still exceed the mandatory minimum brightness in all directions.

Pete
 
jdc said:
A few numbers of some possible relevance:

The requirement is for a light to be visible at 2 nautical miles, so at 2 x 1852 = 3704 metres.

The luminosity which equates to 'visible' in night time conditions is 2 x 10-7 lumens per sq meter according to the IALA recommendations E-200-2, December 2008.

A bright white LED driven at ~20 mA is of the order of 9 lumens, into a +/- 22.5° solid angle (one should check the spec of what one's using).

Thus this LED, if spread all round 360 degrees of azimuth but still with elevation of +/- 22.5°(achievable for instance by pointing it vertically and reflecting off a silvered cone, as is common in 'solar' garden lights) will be 9 / 360 * 45 = 1.125 lumens.

Applying Allard's law for how it attenuates with distance but assuming no atmospheric absorption gives 1.125 / 37042 = 8 x 10-8 lumens per sq metre, ie a bit less than half the luminosity required.

Thus one has three choices:

1. Drive the LED harder / get a more powerful one, order 25 lumens is easily available.
2. Arrange several LEDs in a 'crown', nominally 8 of them spaced by 45° will be 8x brighter
3. Use an array of, say, 9 LEDs pointing at the silvered cone.

Any of these would easily satisfy the rules.
Click to expand...
Interesting, ta :cool:

Though isn't it the colregs we want to go by?

They spec the following...

  1. Tthe minimum luminous intensity of lights shall be calculated by using :
    I = 3.43 x 106 x T x D2 x K-D
    where I is luminous intensity in candelas under service conditions,
    T is threshold factor 2 x 10-7 lux,
    D is range of visibility (luminous range) of the light in nautical miles,
    K is atmospheric transmissivity.
    For prescribed lights the value of K shall be 0.8, corresponding to a meteorological visibility of approximately 13 nautical miles.
  2. A selection of figures derived from the formula is given in the following table:

    [TABLE="width: 1"]
    [TR]
    [TH]Range of visibility (luminous range) of light in miles[/TH]
    [TH]Luminous intensity of light in candelas for K=0.8[/TH]
    [/TR]
    [TR]
    [TD]D[/TD]
    [TD]I[/TD]
    [/TR]
    [TR]
    [TD]1[/TD]
    [TD]0.9[/TD]
    [/TR]
    [TR]
    [TD]2[/TD]
    [TD]4.3[/TD]
    [/TR]
    [TR]
    [TD]3[/TD]
    [TD]12[/TD]
    [/TR]
    [TR]
    [TD]4[/TD]
    [TD]27[/TD]
    [/TR]
    [TR]
    [TD]5[/TD]
    [TD]52[/TD]
    [/TR]
    [TR]
    [TD]6[/TD]
    [TD]94[/TD]
    [/TR]
    [/TABLE]

The cree leds mentioned above are 40,000mcd at 30mA, driving them at 20mA is approx 1/1.25 so 32 candelas, not 2 miles but close.

If those numbers are right, this is all new to me :)

And does adding more leds actually make the range further? I've heard explained like a garden Hose. If there's only so much pressure in the hose it makes no difference how many hoses you have next to each other, it won't squirt any further.
 
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alahol2 said:
I told you I know nothing about electronics... I've been investigating (as far as my limited knowledge will allow). The BFY51 is apparently an NPN transistor so have I just got the position of the arrowhead in the wrong position (and the letters in the correct positions) or have I got the arrow in the right position but the letters in the wrong position?
Click to expand...

It's not an 'elegant' circuit, when the LDR is at switching point, the transistor will be partly on, so passing a fair current and dropping some volts, so will get hot.
But if it works, don't worry!
I think the arrows are wrong, should be on the E leg pointing out if you see what I mean, assuming it is intended to be a Darlington connection and the LDR resistance is higher in the dark?
That would work will B, C and E as placed.
 
GHA said:
Interesting, ta :cool:

Though isn't it the colregs we want to go by?

They spec the following...

  1. Tthe minimum luminous intensity of lights shall be calculated by using :
    I = 3.43 x 106 x T x D2 x K-D
    where I is luminous intensity in candelas under service conditions,
    T is threshold factor 2 x 10-7 lux,
    D is range of visibility (luminous range) of the light in nautical miles,
    K is atmospheric transmissivity.
    For prescribed lights the value of K shall be 0.8, corresponding to a meteorological visibility of approximately 13 nautical miles.
  2. A selection of figures derived from the formula is given in the following table:

    [TABLE="width: 1"]
    [TR]
    [TH]Range of visibility (luminous range) of light in miles[/TH]
    [TH]Luminous intensity of light in candelas for K=0.8[/TH]
    [/TR]
    [TR]
    [TD]D[/TD]
    [TD]I[/TD]
    [/TR]
    [TR]
    [TD]1[/TD]
    [TD]0.9[/TD]
    [/TR]
    [TR]
    [TD]2[/TD]
    [TD]4.3[/TD]
    [/TR]
    [TR]
    [TD]3[/TD]
    [TD]12[/TD]
    [/TR]
    [TR]
    [TD]4[/TD]
    [TD]27[/TD]
    [/TR]
    [TR]
    [TD]5[/TD]
    [TD]52[/TD]
    [/TR]
    [TR]
    [TD]6[/TD]
    [TD]94[/TD]
    [/TR]
    [/TABLE]

The cree leds mentioned above are 40,000mcd at 30mA, driving them at 20mA is approx 1/1.25 so 32 candelas, not 2 miles but close.

If those numbers are right, this is all new to me :)

And does adding more leds actually make the range further? I've heard explained like a garden Hose. If there's only so much pressure in the hose it makes no difference how many hoses you have next to each other, it won't squirt any further.
Click to expand...

I think it amounts to the same thing, although you found the reference visibility which I hadn't found, thank you, which defines the additional factor of 0.64 for the transmission through 2 miles of atmosphere. The problem with the col regs approach however is that it assumes I think that the emitter is an incandescent filament source(1) rather than an LED, which emits into a more restricted solid angle. It's reassuring that the threshold, of 2e-7 lumens per square metre, is the same in both the colregs formula and the lighthouse authority's one.

In answer to the question, does adding more LEDs make it go further, the answer is yes in my example because I was contrasting spreading the light from one LED over a whole 360 degrees of azimuth, which clearly diminishes its perceived brightness, with a crown - or staggered double crown as prv describes - wherein each LED emits only in a smaller solid angle - maybe 0.5 steradians - and so is brighter.

(1) I now realise that I was wrong about this: The two formulae are exactly the same, just rearranged and one converted to miles from metres. There is still the complication with LED specs in that the beam pattern needs to be taken into account, hence I prefer to work in watts, spreading them around the optical system's 'footprint' at 2 miles away to work out power per sq m. This seems to me the conceptually easiest thing to do, but ymmv.
 
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lw395 said:
It's not an 'elegant' circuit, when the LDR is at switching point, the transistor will be partly on, so passing a fair current and dropping some volts, so will get hot.
But if it works, don't worry!
I think the arrows are wrong, should be on the E leg pointing out if you see what I mean, assuming it is intended to be a Darlington connection and the LDR resistance is higher in the dark?
That would work will B, C and E as placed.
Click to expand...

Thanks, I suspect that the original diagram just had the B, C and E marked without the arrows in the diagram. When I re-drew it I chose the wrong diagram for a transistor. You are correct, the LDR gets a higher resistance in the dark and I can say it definitely works.
I'll now have to try to correct the diagram so it doesn't confuse anyone else whilst keeping all the postings relevant...

EDIT: Diagram has now been amended. Thanks to LW395.
 
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GHA said:
.....
And does adding more leds actually make the range further? I've heard explained like a garden Hose. If there's only so much pressure in the hose it makes no difference how many hoses you have next to each other, it won't squirt any further.
Click to expand...

Light is not like that, thankfully.
Think of the light as shining out towards the surface of a sphere.
The amount of light which reaches a square metre of the sphere will double if you put two candles at the centre instead of 1.
If you double the radius of the sphere, the area of the sphere is 4 times, so you need fork handles.

In 'real life' often there is background light so the theoretical light you can see won't compete.
 
lw395 said:
Light is not like that, thankfully.
Think of the light as shining out towards the surface of a sphere.
The amount of light which reaches a square metre of the sphere will double if you put two candles at the centre instead of 1.
If you double the radius of the sphere, the area of the sphere is 4 times, so you need fork handles.

In 'real life' often there is background light so the theoretical light you can see won't compete.
Click to expand...
Not completely convinced by that but unknown territory..
As the light gets attenuated by distance couldn't the brightness be the deciding factor, rather than the amount.
If you can just see one candle from a mile away could you see fork handles in a row from further?
The irpcs figures are given in candelas, not lumins.
Hmmm.... :)
 
GHA said:
Not completely convinced by that but unknown territory..
As the light gets attenuated by distance couldn't the brightness be the deciding factor, rather than the amount.
If you can just see one candle from a mile away could you see fork handles in a row from further?
The irpcs figures are given in candelas, not lumins.
Hmmm.... :)
Click to expand...
To see fork handles in a row, they'd have to be far enough apart that you discerned them as separate light sources, in which case, each one would need to be bright enough to see on its own.
 
Possibly useful to remind ourselves what a candela is.
It is the amount of light emitted into a 'solid angle'.
It's equal to 1 lumen per 'steraradian'.
A whole sphere is 4.pi steraradians.

So if you have 1 lumen, the more you focus it into a smaller solid angle, the more candelas you get.

As you move away from the source, a candela is spread over a greater area, so becomes less 'lumens per sq m' or 'lux'.
Also some of the light may be absorbed.
 
lw395 said:
To see fork handles in a row, they'd have to be far enough apart that you discerned them as separate light sources, in which case, each one would need to be bright enough to see on its own.
Click to expand...

I don't think the question was whether you'd see them as distinct lights, but whether you could see the four-candle light (as one target) where you couldn't see the single-candle one.

Pete
 
prv said:
I don't think the question was whether you'd see them as distinct lights, but whether you could see the four-candle light (as one target) where you couldn't see the single-candle one.

Pete
Click to expand...
Four candles so close together they looked like one light would be visible from twice as far away, as they are putting four times the energy into each direction. The energy your eye picks up reduces with (distance squared), assuming no absorption. So the distance can double before the light drops below the eye's sensitivity.
 
lw395 said:
Four candles so close together they looked like one light would be visible from twice as far away, as they are putting four times the energy into each direction. The energy your eye picks up reduces with (distance squared), assuming no absorption. So the distance can double before the light drops below the eye's sensitivity.
Click to expand...


I agree - in fact it couldn't be otherwise.

The confusion is perhaps because the intensity in watts per square metre (or more like per sq millimetre) at the emitting source is not increased - that would be to increase the thermodynamic temperature which would disobey the 2nd law of thermodynamics. But instead what one does do is increase the emitting area four-fold, which increases the total power radiated four-fold. It's just a bigger flame, which common sense and experience says will be detectable from further away.

It will be perceived as brighter provided that the angle between the multiple candles, as seen from 2 miles away, is less than the angular resolution of the eye, which it will be - thus more photons will fall on a particular rod or cone in the eye.

The digression into candelas vs lumens is, in my opinion, not relevant. In physical terms which are more intuitive than photometric units (for me anyway), lumens are just power, related to Watts by a dimensionless constant, one lumen being 1/683 Watts for green light. So if we radiate the power over a wider beam there's less power per sq metre, or if we focus it down there's more power per sq metre.
 
jdc said:
I agree - in fact it couldn't be otherwise.

The confusion is perhaps because the intensity in watts per square metre (or more like per sq millimetre) at the emitting source is not increased - that would be to increase the thermodynamic temperature which would disobey the 2nd law of thermodynamics. But instead what one does do is increase the emitting area four-fold, which increases the total power radiated four-fold. It's just a bigger flame, which common sense and experience says will be detectable from further away.

It will be perceived as brighter provided that the angle between the multiple candles, as seen from 2 miles away, is less than the angular resolution of the eye, which it will be - thus more photons will fall on a particular rod or cone in the eye.

The digression into candelas vs lumens is, in my opinion, not relevant. In physical terms which are more intuitive than photometric units (for me anyway), lumens are just power, related to Watts by a dimensionless constant, one lumen being 1/683 Watts for green light. So if we radiate the power over a wider beam there's less power per sq metre, or if we focus it down there's more power per sq metre.
Click to expand...

Not quite - the candles will interfere with each other, so that not all 4 candles will be visible in every direction. The flame of a candle is opaque, so if one is behind another, the latter will block light from the former.

There's another affect (for candles!), which is that the increased heat output of four candles close together would cause increased convection, perhaps that would increase the brightness of the candles.
 
Ta the little side line, that makes more sense now, I was getting too interested in transmissivity, and ignoring the extra area which is involved. Same as gravity, a bit. It's there in the formula all along, D squared. :)

I = 3.43 x 106 x T x D2 x K-D
where I is luminous intensity in candelas under service conditions,
T is threshold factor 2 x 10-7 lux,
D is range of visibility (luminous range) of the light in nautical miles,
K is atmospheric transmissivity.
For prescribed lights the value of K shall be 0.8, corresponding to a meteorological visibility of approximately 13 nautical miles.
 
Almost all of the lumen ratings you will see are for LEDs are for the the bare LED unit in perfect lab test conditions, not as it will be when fitted to a base and badly driven, usually giving a loss of around 30-40% of the base spec.

There are purpose made simple IC's for driving LEDs, a common one is the PT4115, http://www.datasheetdir.com/PT4115+LED-Drivers which appears quite often in the cheap Ebay/Chinese driver boards and doesn't require many other components if you fancy building your own driver circuit, and supports dimming if you need it.
 
DinghyMan said:
Almost all of the lumen ratings you will see are for LEDs are for the the bare LED unit in perfect lab test conditions, not as it will be when fitted to a base and badly driven, usually giving a loss of around 30-40% of the base spec.

There are purpose made simple IC's for driving LEDs, a common one is the PT4115, http://www.datasheetdir.com/PT4115+LED-Drivers which appears quite often in the cheap Ebay/Chinese driver boards and doesn't require many other components if you fancy building your own driver circuit, and supports dimming if you need it.
Click to expand...
Switch mode circuits are an option, but you might want some filtering added to that circuit.
There are a lot of other switching circuits for LEDs which step up the volts, so you can many LEDs in series, so they all get the same current.
If you encase the bases of the LEDs in gle or resin leaving the tops exposed, you don't need to have any losses from enclosures.
The one I made is still going after about 8 years (?), might make another one as the LEDs available have got brighter I think.
Many ways of skinning a cat....
 
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