What is efficiency and how is it measured?

[sygyzy]

People often talk about how efficient a flashlight (driver) is. What exactly does this mean and how do you measure it?

 

[drmaxx]

Genereally, it means how much percent of the energy that is taken from the energy source (battery) ends up at the bulb. The rest is converted to heat by the driver.
Calculation: [(Current through the bulb)*(Voltage across the bulb)] / [(Current from the battery)*(Voltage across the battery)]

There are other definition of efficiency - but from your question I take that you mean this one.

 

[Oddjob]

People often talk about how efficient a flashlight (driver) is. What exactly does this mean and how do you measure it?

How does it translate to the non-technically minded? A more efficient emitter will give you longer runtimes than a less efficient emitter at the same brightness.

 

[mudman cj]

Efficiency (power out/power in) is not to be confused with luminous efficacy, which is lumens per watt. To measure power into and out of a driver, you need to measure the voltages and currents similar to what drmaxx posted, just substitute "LED" for "bulb" as needed.

 

[old4570]

If you measure current from the tail ... Lets say 1amp ..

You measure current to the LED from the driver ... lets say its 800mA

200mA is lost to the driver , so in theory the driver would be 80% efficient .

Hope this helps :

 

[Russel]

If you measure current from the tail ... Lets say 1amp ..

You measure current to the LED from the driver ... lets say its 800mA

200mA is lost to the driver , so in theory the driver would be 80% efficient .

Hope this helps :

Forgive me if I am wrong, but isn't that assuming the voltage from the battery to the driver is the same as voltage from the driver to the LED?

 

[old4570]

Forgive me if I am wrong, but isn't that assuming the voltage from the battery to the driver is the same as voltage from the driver to the LED?

A lot of drivers regulate the voltage to the LED , so it depends a lot on battery set up etc , so you may be better off measuring Watts ..

So if your using a 18650 batt @ 4.2v voltage to LED may be 4.2v as well or possibly 3.7v , but its not going to be a huge difference ..

I have a R2 that draws up to 1.7A at the tail with a 18650 , so if only 1A is going to the LED from the driver , im not really going to worry about a few % here or there . What im going to worry about is where 700mA are going .

But then if your running 2 x CR123A 6v to 8.4v then watts may be the way to go ..

In which case you would need to measure Volts + Amps from driver and the same from the tail , and compare Watts in and out of the driver :

Better ?

 

[drmaxx]

How does it translate to the non-technically minded? A more efficient emitter will give you longer runtimes than a less efficient emitter at the same brightness.

Or: A more efficient driver consumes less energy and therefore increases runtime of a flashlight.

This can be quite significant, especially in a step up driver for a LED that converts the 1.2 V of a NiMH into a 3.x V for the LED. If I am remembering correctly, for this setup the driver of Nightcore D10 consumes ~25% energy.

 

[Justin Case]

If you measure current from the tail ... Lets say 1amp ..

You measure current to the LED from the driver ... lets say its 800mA

200mA is lost to the driver , so in theory the driver would be 80% efficient .

Hope this helps :

This is completely wrong.

The driver efficiency is as described by drmaxx and mudman cj.

Just because you have a tailcap current of Iin and an LED current of Iout doesn't mean that the efficiency is 1- (Iin-Iout)/Iin = 1 - (1 - Iout/Iin) = Iout/Iin. Let's suppose you measure 1.6A at the tailcap for a driver board that is rated to deliver 1000mA (1A) to the LED. Is the driver efficiency = 1.0/1.6 = 62.5%? No! You are probably running a boost driver. Assuming that you are powering the driver with 2xNiMH at a nominal voltage of 2.4V, then

% driver efficiency * Vbatt * Ibatt = Vf * If

Thus,

% driver efficiency = (Vf*If)/(Vbatt*Ibatt)

For a typical Vf of 3.5V, we get

% driver efficiency = (3.5V*1A)/(2.4V*1.6A) = 91%

Similarly, suppose you were using a buck driver rated to deliver 1000mA to the LED and ran the driver with 2xRCR123A, with nominal voltage of 7.4V. Let's say you measure 0.55A at the tailcap. Using your approach, driver efficiency is greater than 100%, i.e., 1.0/0.55 = 182%. I don't think so. Using the same equation as above, we get

% driver efficiency = (3.5V*1A)/(7.4V*0.55A) = 86%

 

[old4570]

Eeeer , I did repost saying a Watts calculation might be better , considering all the variables , such as driver / batt combination / voltage etc .

That simple formula is a rough guesstimate if in and out voltage is close , IE voltage to Driver is 4.2 voltage to LED is 4.2 . or even 3.7 , it will change things , how critical it is ? , just depends . [ My bad ]

So one more Time ! Shall we ...

Voltage in as well as Amps in [ to driver ]
Voltage to LED and Amps to LED , work out the watts , and see the difference . :tired:

 

[VidPro]

Eeeer , I did repost saying a Watts calculation might be better , considering all the variables , such as driver / batt combination / voltage etc .

that fixes it, I figured you were speaking loosly.


-------------------------------------------------------------------------------------
driver
Watts in - Watts Wasted = Watts actually delivered to led
led
Watts delivered - watts wasted + short run phosphor output - light loss internally +- a endless array of factors = Light quantity output - color

with some of these PWM things, it can be hard to read the actual power Exactally on normal meters. Meters can be tricked, dont be fooled by things that cant even read the exact totals, that is where REAL actual runtime and output can verify reality. or a scope of course. I mention that because you can see 2 different analisis/reviews for the same driver that show way more or less efficency, sometimes its just the instrumentation used to measure.

 

[Justin Case]

Eeeer , I did repost saying a Watts calculation might be better , considering all the variables , such as driver / batt combination / voltage etc .

That simple formula is a rough guesstimate if in and out voltage is close , IE voltage to Driver is 4.2 voltage to LED is 4.2 . or even 3.7 , it will change things , how critical it is ? , just depends . [ My bad ]

So one more Time ! Shall we ...

Voltage in as well as Amps in [ to driver ]
Voltage to LED and Amps to LED , work out the watts , and see the difference . :tired:

Using watts "might be better"? It is better, not might be, because that is the correct approach.

The formula you presented is just wrong. Why bother with even using it as a "rough guesstimate"? It's wrong. It will give you greater than 100% efficiency for any buck driver configuration you encounter, regardless of whether or not the "voltage is close".