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Old 07-27-2011, 12:27 PM #1
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Default Calculating Efficiency Questions.

For a 660nm laser, Assuming a silicon diode is 0.7V and you need 4 of them for 660nm Lasers, 4x 0.7 = 2.8V.

So if I apply 3.35Vs (2.8 + 0.550) to the diode I would get 550mA of current.

So power going into the LD would = 3.35V x 0.550A = 1.8425Ws

Does that look right?

Now my Optical Output for this laser is 384mWs.

Now to calculate the efficiency of the energy conversion.

Output/Input x 100 = Efficiency

0.384 / 1.8425 x 100 = 20.84%

Some feedback would be nice

Cheers!


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Old 07-27-2011, 01:44 PM #2
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Default Re: Calculating Efficiency Questions.

That looks fine

Edit: I think the 826's vf doesn't match the 815's... Not too sure though

Oh yeh, did you receive my last pm?
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Old 07-27-2011, 07:58 PM #3
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Default Re: Calculating Efficiency Questions.

Measure the voltage - don't guess or calculate. A diode's voltage isn't exactly 0.7V and it varies with current just like with a laser diode. It also varies between diodes and it varies with temperature.
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Old 07-27-2011, 09:51 PM #4
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Default Re: Calculating Efficiency Questions.

sweet yeah i know this, just wanted to confirm the theory behind it was right
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Old 07-27-2011, 11:54 PM #5
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Default Re: Calculating Efficiency Questions.

Everything makes sense except the 0.550 added to the voltage. Where is that coming from? If it is the driver, than your calc is describing the overall efficiency of the (driver+4 diodes in series), not the efficiency of strictly the diode system. To measure the conversion efficiency of the diodes it would be (optical power out / power dissipated by diodes)

= 100*(.384)/(.550*2.80) = 25.9% efficiency.
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Old 07-28-2011, 02:08 AM #6
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Default Re: Calculating Efficiency Questions.

Quote:
Originally Posted by Fiddy View Post
Now to calculate the efficiency of the energy conversion.

Output/Input x 100 = Efficiency

0.384 / 1.8425 x 100 = 20.84%
In addition to what has already been said, I have a slight concern with your theory / terminology here. You mentioned calculating the efficiency of the energy conversion.

If you're measuring the optical output the way we generally do on the forum (with the laser ~12 inches or more from the sensor), then you're clearly going to be using a lens. Lenses account for some additional loss.

While lenses are part of the overall system, they are not part of the energy conversion. They interfere, so to speak, after the energy conversion.

So to measure the energy conversion efficiency, you really need to be using the raw optical output of the diode, with no lens. Depending on your testing setup, this may introduce a slight additional complication in that doing so requires positioning the diode fairly close to the sensor, and this can in some cases introduce radiant heat into the measurement of optical output.
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Old 07-28-2011, 06:24 AM #7
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Default Re: Calculating Efficiency Questions.

Quote:
Originally Posted by Jubathoph View Post
Everything makes sense except the 0.550 added to the voltage. Where is that coming from? If it is the driver, than your calc is describing the overall efficiency of the (driver+4 diodes in series), not the efficiency of strictly the diode system. To measure the conversion efficiency of the diodes it would be (optical power out / power dissipated by diodes)

= 100*(.384)/(.550*2.80) = 25.9% efficiency.
In my interpretation if i apply 2.8V to test load with 4 diodes & 1 ohm resistor (for a 660nm) wouldn't the 4 diodes drop out all of the Input Voltage (4x 0.7 = 2.8V)?

and because its a 1 ohm circuit after that, 1V =~ 1A so to apply 550mA you need 550mV across that 1 ohm resistor.

so you need 2.8V's to overcome the Vdrop of the 4x diodes and 550mV to get the current you require across the 1ohm resistor?


Quote:
Originally Posted by rhd View Post
In addition to what has already been said, I have a slight concern with your theory / terminology here. You mentioned calculating the efficiency of the energy conversion.

If you're measuring the optical output the way we generally do on the forum (with the laser ~12 inches or more from the sensor), then you're clearly going to be using a lens. Lenses account for some additional loss.

While lenses are part of the overall system, they are not part of the energy conversion. They interfere, so to speak, after the energy conversion.

So to measure the energy conversion efficiency, you really need to be using the raw optical output of the diode, with no lens. Depending on your testing setup, this may introduce a slight additional complication in that doing so requires positioning the diode fairly close to the sensor, and this can in some cases introduce radiant heat into the measurement of optical output.
oh ok, im aiming for what i deem 'useable' output which means using a lens.

So then as you said it wouldn't be a accurate conversion calculation cause the lens has a power drop, perhaps i should change the thread tittle?

So comparing the LD Input Power to the LD Output Power with lens is what im aiming for because i have little use for Raw Outputs atm.
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Last edited by Fiddy; 07-28-2011 at 06:28 AM.
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Old 07-29-2011, 06:16 PM #8
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Default Re: Calculating Efficiency Questions.

Just a note: For measuring the raw output you'd need an integrating sphere. Thermal laser power meters pick up the heat from the diode and a photodiode or any glas window to block that will reflect depending on the angle of incidence which varies a lot with the raw output. An integrating sphere can use a photodiode to prevent measuring the heat of the diode but is insensitive to the output beam shape and angles.
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Old 07-29-2011, 07:32 PM #9
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Default Re: Calculating Efficiency Questions.

Quote:
Originally Posted by Bluefan View Post
Just a note: For measuring the raw output you'd need an integrating sphere.
Or what's much cheaper, an optical meter
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Old 07-29-2011, 10:39 PM #10
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Default Re: Calculating Efficiency Questions.

Care to explain? an "optical meter" doesn't say much...
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Old 07-29-2011, 10:57 PM #11
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Default Re: Calculating Efficiency Questions.

A semiconductor-based laser power meter.
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Old 07-31-2011, 10:55 PM #12
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Default Re: Calculating Efficiency Questions.

Quote:
Originally Posted by Bluefan View Post
...a photodiode ... will reflect depending on the angle of incidence which varies a lot with the raw output.
As I've said, photodiode based laser power meter don't work properly because photodiodes have a reflecting surface. The reflection at large angles can be large, and the raw output of an LD has harge angles.
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Old 08-01-2011, 12:25 AM #13
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Default Re: Calculating Efficiency Questions.

How about we test it? (I mean someone with an LPM so not me ). Take a laser, de focus till it emits as latge of an angle possible, and measure the output. Compare that with its collimated output.

Another way to measure would simply measure the outputs at different angles of incidence and plot a graph, I.e. plot output from normal to +45 degrees, at 5 degree increments. You would need a stable laser to do this.

One other factor to consider is that reflectance is a function of wavelength. So results at one wavelength will not necessarily transfer to all wavelengths.
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