0

Consider a setup of an experiment to measure the photoelectric effect. We have a photodiode which is linked to an ammeter and a source of voltage which can apply voltage in either direction. Let the light shone onto the photodiode be such that the energy of photons exceeds the work function of the metal and a current is observed to flow when the voltage is set to zero.

I understand that when a reverse voltage is applied - one that opposes the flow of electrons through the diode - eventually a stage is reached when the kinetic energy of even the most energetic electrons cannot overcome the potential energy attracting the electron back in the direction it came, and no current flows because no electrons reach the other side of the diode.

However, what I don't understand is that when a voltage is applied in the same direction as the electrons, the current does not increase. However, it looks to me as if the current is dependent on the kinetic energies of the electrons, because when the kinetic energy of the electrons is restricted by a reverse voltage, the current drops.

I like to compare this to an analogy of an object being thrown off a high place. The gravitational force at the top of the cliff depends on the height, and a change in height could be compared to a change in voltage - they both lead to a change in the potential energy of the object/electron. The initial velocity of the object when it is thrown downwards is analogous to the initial kinetic energy on the electron created by the photoelectric effect. It remains constant, no matter from what height you throw the object.

With this analogy in mind, it looks to me as if it does not matter what height you throw the object down from - it will always have the same kinetic energy when it hits the bottom. This is obviously incorrect in mechanics, but seems to be the case in the photoelectric effect.

What is going on here?

Cormac Stopes
  • 1

1 Answers1

0

Diode (not necessarily a photodiode for a moment) is a pn-junction, which allows the current to flow in one direction and does not permit it to flow when reverse biased, so that the respective charge carriers on the two sides of the junction (i.e. the electrons and the holes) are sucked out of the junction region and cannot recombine.

It is this letter regime that is used in a photodiode, where the new electron-hole pairs are immediately separated by the applied voltage, thus creating a current. Note, that these electron-hole pairs do not need to have high kinetic energy (they may be even created with zero kinetic energy) - they are accelerated by the electric field.

When the field is applied in the forward direction, the excess electrons of the n-region move towards the excess holes of the p-region, recombine and thus create the current. Exciting a few more electron-hole pairs changes this current a little bit, which is not good enough for practical purposes. In other words: the reverse biased regime has a much higher signal-to-noise ratio.

Roger V.
  • 68,984