Ask an Expert: Are LEDs the Future? Sensor help.
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 Posts: 3
 Joined: Fri Oct 06, 2017 10:33 am
 Occupation: Student
Are LEDs the Future? Sensor help.
Hey forum,
So I've been thinking of doing the a project from ScienceBuddies, but I want to make sure it will all work out before I commit to it.
The project is called "Are LEDs the Future? Energy Savings with LED Lighting." In short, it has you calculate and compare the relative efficiency of LED vs Incandescent light bulbs. It has you use a Light to Voltage Converter, and you measure it with a multimeter.
To calculate the efficiency, you use a bunch of complicated math equations, that I don't really understand. Could someone explain them to me?
1. Optical output power of LED (watts) ∝ Voltage drop across resistor (volts)
2. Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
3. Input Electrical Power (watts) = Current going through the device (amperes) × Voltage across the device (volts)
Pin =Iin × Vin
4. Relative Wall Plug Efficiency (dimensionless) = Optical Output Power/Input Electrical Power
Thanks in advance!
So I've been thinking of doing the a project from ScienceBuddies, but I want to make sure it will all work out before I commit to it.
The project is called "Are LEDs the Future? Energy Savings with LED Lighting." In short, it has you calculate and compare the relative efficiency of LED vs Incandescent light bulbs. It has you use a Light to Voltage Converter, and you measure it with a multimeter.
To calculate the efficiency, you use a bunch of complicated math equations, that I don't really understand. Could someone explain them to me?
1. Optical output power of LED (watts) ∝ Voltage drop across resistor (volts)
2. Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
3. Input Electrical Power (watts) = Current going through the device (amperes) × Voltage across the device (volts)
Pin =Iin × Vin
4. Relative Wall Plug Efficiency (dimensionless) = Optical Output Power/Input Electrical Power
Thanks in advance!

 Expert
 Posts: 350
 Joined: Mon Jan 12, 2009 11:15 pm
 Occupation: Electrical Engineer
Re: Are LEDs the Future? Sensor help.
Hi bellabichon,
Sure, I will give this a shot.
For Equation 1, the symbol means that the variables are directly proportional. Or put into words. The optical output power of the LED is directly proportional to the voltage drop across the resistor.
Equation 2 expands on this directly proportional concept. Pout is the Optical output power and Vres is the voltage drop across the resistor. so Pout = "linear factor" mulitiplied by Vres. Linear factor means that this is constant slope. (Please think of this linear factor as the "m term" in y = mx + b equation that you studied in algebra). Once you have completed this part, you have the Power Output.
Equation 3 is a equation that helps you find the power that you put in to get the optical power output that you just calculated with Equation #2. . Input Power = Voltage times Current (V*I = P)
Finally, Equation #4 helps you calculate the efficiency by taking the answer from Equation #2 and dividing it by Equation #3, i.e. Power_out / Power_In. If something is 100% efficient, i.e. you get out everything that you put in, Power_out = Power_In and you would get a efficiency of 100%. In real world system, there will always be some losses so you should expect something less than 100%.
Ok, I hope that helps. Please post again if we can help with any additional clarification.
Thanks and Good Luck with your experiment!
Willey
Sure, I will give this a shot.
For Equation 1, the symbol means that the variables are directly proportional. Or put into words. The optical output power of the LED is directly proportional to the voltage drop across the resistor.
Equation 2 expands on this directly proportional concept. Pout is the Optical output power and Vres is the voltage drop across the resistor. so Pout = "linear factor" mulitiplied by Vres. Linear factor means that this is constant slope. (Please think of this linear factor as the "m term" in y = mx + b equation that you studied in algebra). Once you have completed this part, you have the Power Output.
Equation 3 is a equation that helps you find the power that you put in to get the optical power output that you just calculated with Equation #2. . Input Power = Voltage times Current (V*I = P)
Finally, Equation #4 helps you calculate the efficiency by taking the answer from Equation #2 and dividing it by Equation #3, i.e. Power_out / Power_In. If something is 100% efficient, i.e. you get out everything that you put in, Power_out = Power_In and you would get a efficiency of 100%. In real world system, there will always be some losses so you should expect something less than 100%.
Ok, I hope that helps. Please post again if we can help with any additional clarification.
Thanks and Good Luck with your experiment!
Willey
Re: Are LEDs the Future? Sensor help.
Hi bellabichon,
Replying here instead of email so other users can see it in the future. I wanted to add somethign to LeungWilley's response. There's a somewhat subtle and possibly confusing point in the procedure that I think might be causing your problem  step 4 in "Testing and Data Collection":
This project assumes that optical output power is the power collected by the detector. Using this assumption, N is the same for both the LED and the incandescent light sources, and the output power can be written in units of N in the data table (for example, write 0.75 N in the table). N cannot be determined easily, as it depends on the light emission vs. angle for each source.
In other words, what that means is that in the equation Optical output power of LED (watts) = N_linear factor x Voltage drop across resistor (volts), you don't actually know N. To use a rough analogy  imagine shining a flashlight at someone's face (that's not nice). Not 100% of the light from the flashlight hits the person's eyes  some of it hits their skin, and some of it goes past their head. Only some percentage of the light hits the person's eyes  and that exact percentage will vary depending on whether the flashlight has a wide or focused beam. Same idea here  the light from your LED or incandescent bulb is spreading out across the whole room, and only a small amount of it is hitting your light sensor. You don't know exactly how much of it, however  which is why you don't know N. You do still know that the amount of light is proportional to the voltage drop (that's what that weird squiggly symbol means in Equation 1). Since you don't have a value for N in Equation 2, you can write the optical output power "in units of N" as described in step 4. For example, if you measured a voltage drop of 2V, your optical output power would be 2N watts.
Ultimately, even though you don't know what N is, you can assume it is the same for both the LED and the incandescent bulb  so when you do the very last step in the project, the N's will cancel out, and you can still directly compare the efficiency of the LED and the incandescent.
Hope that helps  depending on how much algebra you've had, it might seem a little weird to leave a variable in there like that when you don't have a value for it.
One more thing: this goes above and beyond what is written in the project, but I believe there is a way to calculate the actual optical output power in watts (i.e., getting rid of the N). Take a look at the datasheet for the TSL252:
https://www.mouser.com/datasheet/2/588/ ... 149939.pdf
On page 6 there is a variable called "Ne" which represents "Irradiance responsivity". It has units of mV/(uW/cm^2), or millivolts per microwatt per square centimeter. That's the value that determines how the amount of light the sensor detects (in microwatts per square centimeter) is converted to volts. So ultimately, given that you have a measurement in volts, you should be able to use that information in the datasheet to calculate a value that's in watts. If you want to try that, let me know and I can explain in more detail; if not, don't worry about it.
Hope that helps!
Replying here instead of email so other users can see it in the future. I wanted to add somethign to LeungWilley's response. There's a somewhat subtle and possibly confusing point in the procedure that I think might be causing your problem  step 4 in "Testing and Data Collection":
This project assumes that optical output power is the power collected by the detector. Using this assumption, N is the same for both the LED and the incandescent light sources, and the output power can be written in units of N in the data table (for example, write 0.75 N in the table). N cannot be determined easily, as it depends on the light emission vs. angle for each source.
In other words, what that means is that in the equation Optical output power of LED (watts) = N_linear factor x Voltage drop across resistor (volts), you don't actually know N. To use a rough analogy  imagine shining a flashlight at someone's face (that's not nice). Not 100% of the light from the flashlight hits the person's eyes  some of it hits their skin, and some of it goes past their head. Only some percentage of the light hits the person's eyes  and that exact percentage will vary depending on whether the flashlight has a wide or focused beam. Same idea here  the light from your LED or incandescent bulb is spreading out across the whole room, and only a small amount of it is hitting your light sensor. You don't know exactly how much of it, however  which is why you don't know N. You do still know that the amount of light is proportional to the voltage drop (that's what that weird squiggly symbol means in Equation 1). Since you don't have a value for N in Equation 2, you can write the optical output power "in units of N" as described in step 4. For example, if you measured a voltage drop of 2V, your optical output power would be 2N watts.
Ultimately, even though you don't know what N is, you can assume it is the same for both the LED and the incandescent bulb  so when you do the very last step in the project, the N's will cancel out, and you can still directly compare the efficiency of the LED and the incandescent.
Hope that helps  depending on how much algebra you've had, it might seem a little weird to leave a variable in there like that when you don't have a value for it.
One more thing: this goes above and beyond what is written in the project, but I believe there is a way to calculate the actual optical output power in watts (i.e., getting rid of the N). Take a look at the datasheet for the TSL252:
https://www.mouser.com/datasheet/2/588/ ... 149939.pdf
On page 6 there is a variable called "Ne" which represents "Irradiance responsivity". It has units of mV/(uW/cm^2), or millivolts per microwatt per square centimeter. That's the value that determines how the amount of light the sensor detects (in microwatts per square centimeter) is converted to volts. So ultimately, given that you have a measurement in volts, you should be able to use that information in the datasheet to calculate a value that's in watts. If you want to try that, let me know and I can explain in more detail; if not, don't worry about it.
Hope that helps!

 Posts: 3
 Joined: Fri Oct 06, 2017 10:33 am
 Occupation: Student
Re: Are LEDs the Future? Sensor help.
Hello,
I'll try to walk myself through your forum post (thank you for that), and you can try to tell me where I'm making a mistake.
Not all of the light given off by our LED or incandescent will reach the detector, as some of it goes to the sides. This means we'll never get a perfect efficiency reading across our resistor. We don't know exactly how much light is actually hitting our detector, so we'll never actually know N.
This is the part that confuses me. If we don't know N, do we just call the Optical Output Power (voltage)N? I've tried doing this, and according to the efficiency equation that is given to us, (efficiency = output/input x 100%) it just doesn't work with the numbers we have.
Distance LED is from LighttoVoltage Converter
Voltage Reading Across Resistor (volts)
Output Optical Power (watts)
Voltage across LED (volts)
Current Going Through LED (amperes)
Input Electrical Power (watts)
Relative Wall Plug Efficiency
LED #1
25cm
1.733
1.733N
3.15
0.359
1.13085
LED #2
25cm
1.745
1.745N
3.17
0.353
1.11901
LED #3
25cm
1.725
1.725N
3.21
0.353
1.13313
Average LED wall plug efficiency =
Using the efficiency equation, and the data we have, it should be 1.733N/1.13085 x 100% (we calculated the Input Electrical Power by multiplying the Voltage across the LED by the Current going through the LED.) But, that doesn't work. We get more than 100% efficiency, which doesn't work, right?
The same appears true for our incandescent bulb.
We calculated the Input Electrical Power the same way as with the LED. Voltage across the Incandescent x Current going through the Incandescent. If we assume that the Output Optical Power is N, then we have 1.742N/0.96064. Again, this doesn't work, because it's more than 100 efficient.
What do you suggest I do here to complete the equations? After I finish this data, we will be able to complete the rest of our science fair project fairly easily. This is all we need to know how to do at this point.
Thank you so much for your help so far. Get back to me whenever you can.
I'll try to walk myself through your forum post (thank you for that), and you can try to tell me where I'm making a mistake.
Not all of the light given off by our LED or incandescent will reach the detector, as some of it goes to the sides. This means we'll never get a perfect efficiency reading across our resistor. We don't know exactly how much light is actually hitting our detector, so we'll never actually know N.
This is the part that confuses me. If we don't know N, do we just call the Optical Output Power (voltage)N? I've tried doing this, and according to the efficiency equation that is given to us, (efficiency = output/input x 100%) it just doesn't work with the numbers we have.
Distance LED is from LighttoVoltage Converter
Voltage Reading Across Resistor (volts)
Output Optical Power (watts)
Voltage across LED (volts)
Current Going Through LED (amperes)
Input Electrical Power (watts)
Relative Wall Plug Efficiency
LED #1
25cm
1.733
1.733N
3.15
0.359
1.13085
LED #2
25cm
1.745
1.745N
3.17
0.353
1.11901
LED #3
25cm
1.725
1.725N
3.21
0.353
1.13313
Average LED wall plug efficiency =
Using the efficiency equation, and the data we have, it should be 1.733N/1.13085 x 100% (we calculated the Input Electrical Power by multiplying the Voltage across the LED by the Current going through the LED.) But, that doesn't work. We get more than 100% efficiency, which doesn't work, right?
The same appears true for our incandescent bulb.
We calculated the Input Electrical Power the same way as with the LED. Voltage across the Incandescent x Current going through the Incandescent. If we assume that the Output Optical Power is N, then we have 1.742N/0.96064. Again, this doesn't work, because it's more than 100 efficient.
What do you suggest I do here to complete the equations? After I finish this data, we will be able to complete the rest of our science fair project fairly easily. This is all we need to know how to do at this point.
Thank you so much for your help so far. Get back to me whenever you can.
Re: Are LEDs the Future? Sensor help.
Hi,
Important point of confusion: you aren't actually getting an efficiency value over 100%. From your message:
"then we have 1.742N/0.96064. Again, this doesn't work, because it's more than 100 efficient."
That's not true because there is still an "N" in that equation. IF there was no N, then you'd be correct that you get a value over 100%. But you can't ignore the N. For example, imagine that N = 0.1. You would get an efficiency less than 100%. However, since you don't know the actual value of N, you have to leave the efficiency in terms of N. So in that case you would divide 1.742 by 0.96064 and get an efficiency of 1.813N. The N is still there.
Since you don't know N, you can't calculate the absolute efficiency for the LED or incandescent or LED individually, but you can still compare them. If you divide the efficiency of the LED by the efficiency of the incandescent, the N's will cancel out.
Here's another algebra example in case it helps. Say that you ate 10 apples and I ate 5 apples. We know exactly how many apples each one of us ate, and we can say that you at twice as many apples as I did (10/5 = 2). That's easy.
Now say that you ate 10N apples and I ate 5N apples, and we don't know what N is. We don't know exactly how many apples either one of us ate individually, BUT we can still say that you ate twice as many apples as I did, because 10N/5N = 2 (the N's cancel out).
Ultimately, that is what this project is having you do. It doesn't allow you to calculate a number for the efficiency of the LED or incandescent individually, but it lets you calculate how much more efficient the LED is than the incandescent.
Let me know if that makes sense  it might be confusing depending on how much algebra you've had.
Important point of confusion: you aren't actually getting an efficiency value over 100%. From your message:
"then we have 1.742N/0.96064. Again, this doesn't work, because it's more than 100 efficient."
That's not true because there is still an "N" in that equation. IF there was no N, then you'd be correct that you get a value over 100%. But you can't ignore the N. For example, imagine that N = 0.1. You would get an efficiency less than 100%. However, since you don't know the actual value of N, you have to leave the efficiency in terms of N. So in that case you would divide 1.742 by 0.96064 and get an efficiency of 1.813N. The N is still there.
Since you don't know N, you can't calculate the absolute efficiency for the LED or incandescent or LED individually, but you can still compare them. If you divide the efficiency of the LED by the efficiency of the incandescent, the N's will cancel out.
Here's another algebra example in case it helps. Say that you ate 10 apples and I ate 5 apples. We know exactly how many apples each one of us ate, and we can say that you at twice as many apples as I did (10/5 = 2). That's easy.
Now say that you ate 10N apples and I ate 5N apples, and we don't know what N is. We don't know exactly how many apples either one of us ate individually, BUT we can still say that you ate twice as many apples as I did, because 10N/5N = 2 (the N's cancel out).
Ultimately, that is what this project is having you do. It doesn't allow you to calculate a number for the efficiency of the LED or incandescent individually, but it lets you calculate how much more efficient the LED is than the incandescent.
Let me know if that makes sense  it might be confusing depending on how much algebra you've had.

 Posts: 3
 Joined: Fri Oct 06, 2017 10:33 am
 Occupation: Student
Re: Are LEDs the Future? Sensor help.
Hey again,
I followed your advice for the calculations, and I think I have my final numbers. Now, if you have some time, I would like to make sure that I represent them correctly in our report and when we write everything up for our board. In the end, the numbers averaged out to 1.538073923039399N for the LED and 1.784504645246501N for the Incandescent. I followed your advice, and divided the LED by the Incandescent from that, I got 0.861905250365396. How exactly should I choose to interpret this? Should I simply round it to 86.19% and say that LEDs are 86.19% more efficient than Incandescent bulbs, or is there a bit more math to do here?
Thanks for all your help.
I followed your advice for the calculations, and I think I have my final numbers. Now, if you have some time, I would like to make sure that I represent them correctly in our report and when we write everything up for our board. In the end, the numbers averaged out to 1.538073923039399N for the LED and 1.784504645246501N for the Incandescent. I followed your advice, and divided the LED by the Incandescent from that, I got 0.861905250365396. How exactly should I choose to interpret this? Should I simply round it to 86.19% and say that LEDs are 86.19% more efficient than Incandescent bulbs, or is there a bit more math to do here?
Thanks for all your help.
Re: Are LEDs the Future? Sensor help.
Two things:
1) Just making up data with round numbers for an example: say you got an efficiency of 3N for the LED and 1.5N for the incandescent. Dividing those gives you 3N/1.5N = 2, meaning the LED is twice as efficient as the incandescent. So with your data, you are getting a number less than 1  which means the LED is less efficient than the incandescent (about 86% as efficient, or you could say it's 14% less efficient). That is certainly not expected but that doesn't mean it's "wrong"  it could be the correct number according to the data you collected (although obviously it would be good to double check your math). The important thing for a science fair if you got unexpected results is that you can still explain to a judge WHY you think you got those results, and what you could do differently in the future to prove or disprove those results (if you had more time and money to do another experiment).
2) Have you learned about significant digits in science class at all? The very short explanation is that there's a limit to how many decimal places you can use when reporting results, based on how accurate your measurements were. For example, if you use a metric ruler with millimeter marks on it, that gives you an accuracy of 0.001 meters. You couldn't have something in your results with a value of 0.53426236234512 meters because your measurements aren't that accurate. There are rules you need to follow when you do math and add/subtract/multiply/divide different measurements that might have different units or levels of accuracy (like you do in this project  for example in the input electrical power equation, you are multiplying amps by volts, which you can probably only measure to one or two decimal places of accuracy on your multimeter). I don't remember the rules off the top of my head, so you should ask your science teacher about significant digits, or do some research about them online.
1) Just making up data with round numbers for an example: say you got an efficiency of 3N for the LED and 1.5N for the incandescent. Dividing those gives you 3N/1.5N = 2, meaning the LED is twice as efficient as the incandescent. So with your data, you are getting a number less than 1  which means the LED is less efficient than the incandescent (about 86% as efficient, or you could say it's 14% less efficient). That is certainly not expected but that doesn't mean it's "wrong"  it could be the correct number according to the data you collected (although obviously it would be good to double check your math). The important thing for a science fair if you got unexpected results is that you can still explain to a judge WHY you think you got those results, and what you could do differently in the future to prove or disprove those results (if you had more time and money to do another experiment).
2) Have you learned about significant digits in science class at all? The very short explanation is that there's a limit to how many decimal places you can use when reporting results, based on how accurate your measurements were. For example, if you use a metric ruler with millimeter marks on it, that gives you an accuracy of 0.001 meters. You couldn't have something in your results with a value of 0.53426236234512 meters because your measurements aren't that accurate. There are rules you need to follow when you do math and add/subtract/multiply/divide different measurements that might have different units or levels of accuracy (like you do in this project  for example in the input electrical power equation, you are multiplying amps by volts, which you can probably only measure to one or two decimal places of accuracy on your multimeter). I don't remember the rules off the top of my head, so you should ask your science teacher about significant digits, or do some research about them online.