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I'll try explain everything as clearly as I can as it's slightly complicated. I really need help as I'm quite new to this kind of experiment and work so any help would be extremely appreciated.

I did some LIV tests with some LED lights and gathered some results. However I'm not an expert and this is just lab work so I'm not sure I understand what all the data that the machine retrieved means.

My final goal was to plot the External Quantum Efficiency (EQE) against the injected current into the device, where I gradually increase the injected current. Therefore I needed to find the Light output power (LOP) and divide it with the input power to give the EQE, as EQE = Output Power / Input Power.

Look at the first file which consist of all the data retrieved for the first LED: SPREADSHEET 1 [http://imgur.com/bpoadSP][1]

NOTE: Spreadsheets can be zoomed in by clicking on the picture!

I'm not sure what all this means but I know that the column under "PS Amps" is the supplied current, the one under "PS Volts" is the supplied voltage, and I assumed under "Watts" is the output power. Therefore I found the input power to be simply the supplied current x supplied voltage. This resulting value is higher than the output power column so I was quite confident that the the column under "Watts" in that spreadsheet is the output power.

So simply I divided each element of that row of output power under watts with each element of the total input power. This gave me the corresponding EQE for each row as the injected current increases.

However I'm still confused if I'm doing this right as there is a problem - in the second spreadsheet here: SPREADSHEET 2 [http://imgur.com/yBMJwAO][2]

NOTE: Spreadsheets can be zoomed in by clicking on the picture!

As you can see in this second spreadsheet I'm using the same example of test data for the same LED, after 0.001amps in the current row its the exact same data as spread sheet 1. Here I simply listed the supplied voltage, current, and the light output power together for easier access.

However the data above 0.001amps where the LOP is all listed wrongly as "1" (as a place holder as remember I don't know how to find output power with this new data), was retrieved as I was asked to take measurements without using the LIV test system as the machine couldn't take readings with such low current. These sets of reading I only got the supplied current and voltage like shown in the diagram, and I've reassured with a lab partner that is all we were supposed to take.

How am I supposed to find the total Light output power like I have with the data in spreadsheet 1 then? In spread sheet 1 all the data is given in such detail including the watts column which as I have mentioned I think is the light output power as its lower than the supplied I x supplied V. Is there a way to calculate the light output power so that I can find the EQE for the recordings above 0.001amps from the second spreadsheet?

To summarise in short:

1) Need to retrieve data of LEDs to plot their EQE (external quantum efficiency) against the injected current.

1) With the data in Spreadsheet 1 I retrieved readings of the output power while increasing injected current. Supplied voltage is also listed, so I could do injected current * supplied voltage = input power, then assume the colum under 'watts' is the output power as it was a reading from the test system, and therefore do output power / input power to find the EQE as I have learned.

3) However more readings of injected current under 0.001amps had to be taken with another machine, and the data is shown in Spreadsheet 2 but does not include LOP or a "Watts" column so I don't know how to find output power. I still have input power as I have both supplied current and supplied voltage there.

Is there something I don't know about the data in spreadsheet 1 that can help me find the light output power in spreadsheet 2 with the extra readings? Or am I calculating everything wrong and the light output power is given in another way? Please help as I'm not sure what the data all means in the first spreadsheet and I'm struggling to find the way to find LOP in the second.

[1]: Spreadsheet 1 - DATA of LED [2]: Spreadsheet 2 - DATA of LED with Extra Readings Under 0.001amps

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  • \$\begingroup\$ It's unclear what you're asking. It seems that your spreadsheet formula is incorrectly multiplying two numbers (the given result is 1, where it shouldn't be.) For example, 0.00001 x 2.1283 should not be 1, therefore there is something wrong with your formula. Just fix that! \$\endgroup\$ – CL22 Apr 28 '15 at 9:00
  • \$\begingroup\$ This is beacuse the the "1's" are placeholders as I don't know how to calculate the LOP or light out put power. Data below the 1's in spreadsheet 2 are the LOP because in spreadsheet 1 i get these values from the "watts" column from the machine. However in spread sheet 2 I had to take extra data in which I only got supplied current / voltage, and can calculate input power form that but I'm stuck on how to find LOP. I also have calibration data for each lamp sample is that needed? \$\endgroup\$ – Beanbear Apr 28 '15 at 17:03
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I needed to find the Light output power (LOP) and divide it with the input power to give the EQE, as EQE = Output Power / Input Power.

This is not the usual definition of quantum efficiency. Quantum efficiency should be a ratio of numbers of quanta. For an LED this would be $$\frac{\mathrm{Photons\ emitted}}{\mathrm{charge\ carriers\ injected}}$$

If you have the input current and output power you can calculate this as

$$\frac{P_o/h\nu}{I/e}$$

where \$P_o\$ is optical output power, \$h\nu\$ is the photon energy at the emission frequency \$\nu\$, \$I\$ is the injection current, and \$e\$ is the electron charge.

You do not need the diode forward voltage to calculate the quantum efficiency. You might need it in the future if you are asked to calculate the power efficiency or wall-plug efficiency.

How am I supposed to find the total Light output power like I have with the data in spreadsheet 1 then?

The ususal way to get the total light output of an LED is to put the LED at the input port of an integrating sphere, and measure the optical power at the output port with a large-area photodiode. Then scale the measured power by the loss factor of the integrating sphere.

Or am I calculating everything wrong and the light output power is given in another way?

It looks to me like you did not collect enough data when you were doing the experiment. You need to re-do the experiment and measure the light output for all of the input current levels you are interested in.

Realistically, there's no reason the quantum efficiency (my definition, not yours) should change dramatically at current levels below 1 mA. Maybe your next experiment will be with a laser instead of an LED, then you will see some interesting behavior changes at low currents.

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  • \$\begingroup\$ Thank you! This answer was helpful, I'm 100% sure its measurements im missing now will take this up with lab supervisor. \$\endgroup\$ – Beanbear Apr 28 '15 at 22:25
  • \$\begingroup\$ Why do you say there should be no reason for QE to decrease below 1mA? (That certainly doesn't fit with my limited measurements.) \$\endgroup\$ – George Herold Apr 29 '15 at 13:04
  • \$\begingroup\$ @GeorgeHerold, not sure what you mean. The plot in your answer shows near-constant QE from 100 nA to 1 mA. Usually the behavior is nearer to ideal at low currents, and QE decreases at high currents (like in your answer). \$\endgroup\$ – The Photon Apr 29 '15 at 15:42
  • \$\begingroup\$ Hi Photon. I'm being a bit of a grouch, but the QE is hardly constant at low currents. If you just take it as the ratio of the two currents. It's 10^-7/10^-11 ~10^-4 at 100nA bias and 10^-7/10^-4 ~10^-3 at 0.1 mA bias. Above 1mA the slope is 1 and QE is constant. (roughly) \$\endgroup\$ – George Herold Apr 29 '15 at 18:14
  • \$\begingroup\$ @GeorgeHerold, OK, fair point. The fact the slope of your log-log fitting line is 2/3 means the QE is not constant. \$\endgroup\$ – The Photon Apr 29 '15 at 18:30
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Wow there are a lot of questions there.
1.) know your instruments.

I measured photo current vs bias current for a few LEDs
Here's a log/ log plot. The line is added by hand and has a slope of 2/3.
The photocurrent was linear with bias current between ~ 1mA -10mA. (Higher current caused heating which reduced the light.)

(Photocurrent measured with a photodiode.)

I enter image description here

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  • \$\begingroup\$ Thanks for your answer, but I'm looking for a more specific answer in how to find light output power for the extra data in spreadsheet 2. Look at it, the "1's" are placeholders as I don't know how to calculate the LOP or light out put power. 1's in spreadsheet 2 are the LOP because in spreadsheet 1 i get these values from the "watts" column from the machine. However in spread sheet 2 I had to take extra data (the data with LOP = 1) in which I only got supplied current / voltage, and can calculate input power from that but I'm stuck on how to find LOP as its not given like the data in sheet 1. \$\endgroup\$ – Beanbear Apr 28 '15 at 17:06
  • \$\begingroup\$ You could extrapolate based on the trend above, as long as you make it clear that is what you are doing. There is no way to know the actual light output if you did not measure it previously. I suggest you graph input current (not power) vs light output power for all the points where you have such data. Perhaps the nature of the curve will give you confidence to extrapolate your own data. In my experience, which does not include low bias levels, the light output is fairly linear with input current. \$\endgroup\$ – mkeith Apr 28 '15 at 19:26
  • \$\begingroup\$ @mkeith, are you talking about my data? The linear region is ~1 order of magnitude. and goes as the 2/3 power for at least 4 orders of magnitude. \$\endgroup\$ – George Herold Apr 30 '15 at 1:19
  • \$\begingroup\$ @GeorgeHerold, when I said "trend above" I was referring to your data. The data you present is sufficient to allow the OP to extrapolate over a wide range, although that is probably not consistent with the goals of the class. My experience involves measuring luminance of LED backlit displays over a range of less than 1 order of magnitude. In that range, the nits/mA relationship is very linear. I was urging the OP to compare his/her graph with yours. \$\endgroup\$ – mkeith Apr 30 '15 at 3:07

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