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Voltage versus Lumens


Mattyboy

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There are a number of factors at work.

 

The Voltage drives current through a resistance (the lamp).

A Short thick lamp filament it has a small resistance, a long (coiled) thin one has a large resistance.

Thus you can have the same Power of lamp powered at 230V or 12V (for reasonably small powers, no-one would try to make a 12V 2000W lamp as the current is unwieldy.

The Voltage times the Resistance is the Power consumed by the lamp. [As Mikepoll points out below I have made an error here].

But not all Lamps use this Power equally efficiently.

Standard incandescent Lamps turn 90% of this power into heat and only 10% into light

Fluorescents turn about 50% to heat and 50% to light.

 

Then there is the quality of the light. The human eye does not perceive all frequencies of light as being equally bright.

 

PS: This thread probably belongs in "Lighting" rather than "General tech Chat"

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A lot has to do with the fillaments material and the innert gas arround it. The designing of bubbles is quite complex. For example - headlights. They are 12v bubbles, but they are extreamly bright and a relatively high wattage. However - you can buy bubbles that are the same wattage, but brighter due to a number of design factors, as mentioned, fillament and innert gas, then there is the design of the fillament, shape of the bubble, shape of the refelector, type of glass, thickness of glass, and the list goes on. There are probably thousands of factors that in some way increase light output.

 

12V or 240V, it matters little, really only changing the number of number of amps the device draws to give the same amount of power. If I had a 1200W fixture, I could power it with 240v (5A) or 12v (100A). What you need to look at is power (as mentioned above) which is measured in Watts (W). As a general rule, the higher the wattage the greater the light output. HOWEVER, that applies only within the species of bubble (ie incandecent or florescent).

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The Voltage times the Resistance is the Power consumed by the lamp.

 

Sorry to correct you Ekij just the relationship is actually Power equals Voltage squared(^2) divided by Resistance.

 

But all the rest of what you said was perfectly valid. :unsure:

 

Mike

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The Voltage times the Resistance is the Power consumed by the lamp.

Sorry to correct you Ekij just the relationship is actually Power equals Voltage squared(^2) divided by Resistance.

But all the rest of what you said was perfectly valid. :mods:

Mike

I can't believe I got that wrong :unsure:

I must have been concentrating too much on getting my bold bits right!

What I intended to say was

The Voltage times the Current is the Power consumed by the lamp.

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Thankyou all for your fantastic input. On a final note what about with Arc lamps? How does that differ

 

Dear Matt

How does what differ?

One of the key factors here is whether it is a short or long arc. Xenon lamps, for example, have a very short arc and are thus suitable for compact and accurate optical systems - film projectors being a case in point. (Obviously, there are other issues with Xenon sources......internal pressure being one.)

 

Fitt & Thornley are good on this http://www.amazon.co.uk/exec/obidos/ASIN/0...3108549-4247650

 

Or I have some lecture notes on the topic that I'm happy to send.

 

Cheers

 

Ken

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