# Why is my measured LED activation voltage lower than theoretical?

I have done an experiment to measure Planck's constant. I measured the activation voltage of the LED for each wavelength used. I calculated theoretical activation voltage results using the energy of photon emitted (E=hf) and V=E/e. The experimental activation voltage results were 50% lower than they should have been. This is not what I was expected because I thought that I would be seeing more light than one photon and therefore the activation voltage would be higher. I assume this has something to do with an energy gain in the system which was unaccounted for. Does this have something to do with an energy gain in the system to do with the depletion zone? Or something to do with the electrons increasing in energy as they move from the p-type material back to the electrical wire? I used a breaboard to wire the circuit

I wired the LED, a resistor and potentiometer to a breadboard which I then connected to a 2V battery. To measure the activation voltage I used a digital multimeter across the two ends of the LED. The experiment was conducted in a dark room. The potentiometer was turned until light was seen from the LED. At this point the voltage drop across the LED was recorded. I am trying to prove the value of Planck's constant. So far for red visible light, the activation voltage is 0.770V compared to calculated theoretical of 1.86V, orange of 0.895V compared to 2.02V, green of 1.111V compared to 2.3V, blue of 1.165V compared to 2.6V.

• Your post would be better with a description of experimental setup and results so far, as well as of the theory that you are trying to prove. Jun 2, 2019 at 8:46
• I wired the LED, a resistor and potentiometer to a breadboard which I then connected to a 2V battery. To measure the activation voltage I used a digital multimeter across the two ends of the LED. The experiment was conducted in a dark room. The potentiometer was turned until light was seen from the LED. At this point the voltage drop across the LED was recorded. I am trying to prove the value of Planck's constant. So far for red visible light, the activation voltage is 0.770V compared to calculated theoretical of 1.86V, orange of 0.895V compared to 2.02V, green of 1.111V compared to 2.3V. Jun 2, 2019 at 8:52

What you are measuring is roughly the turn-on voltage of the diode. Photons aren't produced from somehow directly sapping the voltage drop across the diode. Diodes have a $$p$$-doped region and an $$n$$-doped region. Applying some amount of forward bias allows electrons from the $$n$$ region to flow over to the $$p$$ region. When this happens, electrons can recombine with holes that are already separated by $$hf$$ and emit a photon.

• So I measured the voltage required for electrons from the n region to flow to the p region, not actually the voltage required for photons to be produced? Thank you Jun 2, 2019 at 9:27
• @Eloise Correct. Or more specifically, the current is exponential in the applied voltage. You found the "knee" of the curve where the current begins to really take off. Jun 2, 2019 at 9:29
• Is it possible to measure the band gap experimentally to find an experimental value of hf? Jun 2, 2019 at 11:33
• @Eloise, the most straightforward way to measure the band gap would be to use the diode as a photodiode and see what wavelengths it absorbs and what wavelengths it doesn't. Jun 2, 2019 at 15:32
• Do you think this could be because my measurements are not quite the activation voltage? Perhaps I am measuring the voltage before the activation voltage where small amounts of current are still moving through the diode Jun 3, 2019 at 9:04

You've described a very simply experiment. Elegant, really.

Now let's talk about experimental realities. There are at least three issues which could be confounding your measurements.

1. The diode's behavior may be temperature dependent If you ran all the steps of the experiment in sequence the diode may have been warmer for the late runs than the early ones.

2. Visual response curve You are counting on your eye as a part of the apparatus, and your eye has different sensitivities to different wavelengths. Sensitivity is highest near the middle of the visible spectrum (in the green) and drops off to both sides.

3. Dark adaptation Again, you're using your eye as a detector, and your eye actually has two detection regimes. Color detection dominated by the code cells and monochrome detection dominated by the rod cells. The chemical reaction that trigger detection in the rods is saturated in bright light (even ordinary room levels) meaning they are effectively turned off. They turn on slowly over the course of about 30 minutes in the dark which means that your sensitivity threshold for determining "hey, there's light coming from it" gets lower and lower over the course of about 30 minutes. Unless the diode comes on bright enough to wash out the rod pigments again which would re-start the clock.

The fun part come in your trying to figure out how each of these might have (or not have) messed up your analysis and then trying to work around them.

• Okay thank you, I see how this would affect the reliability of the results however I think most of these would only mean that when I measured the voltage drop across the LED it was higher than it should have been -- because I was seeing more than just one photon of light. Do you agree with @HiddenBabel in that the experimental design is flawed and isn't actually measuring the band gap? Jun 3, 2019 at 7:05