# Optoelectronics FAQ

**Q. Why doesn't the photoelectric effect generate more than one electron that is emitted from the surface?**

A. In order to emit more than one electron we firstly need a photon with at least twice the work function energy for the photocathode - try to put some numbers in and work out what wavelength photon we would need to have twice the minimum work function energy (work func. >1.25eV). Secondly we need 3 'particles' to interact in order to emit 2 electrons (if the photon has enough energy). Statistically this 3 'particle' interaction is far less likely than the 2 'particle' interaction where one electron is emitted.

**Q. What causes the charge carriers generated in the depletion region to leave the depletion region in a photodiode absorbing light?**

A. The depletion region in a photodiode is set-up when charge carriers from each side of the p-n junction diffuse across the 'interface' of the p-n junction. Thus consider the n-type material - some of the electrons have diffused out of the material and some holes have diffused into the material. This generates a net +ve charge in the n-type material and vice-versa for the p-type material. This then sets up an electric field across each side of the junction extending into the material. An equilibrium position is reached where the diffusion of charge carriers is balanced against the electric field that diffusion sets up which tries to 'pull' the native charge carriers back into their 'own' host material. When light of high enough energy is absorbed in the depletion region an electron-hole pair is generated where the electron has sufficient energy to be promoted to the conduction band. The equilibrium setup by diffusion and electric field no longer holds and the 'excess' charge carriers are transported from the depletion region by the electric field - linking each end of the diode to form a circuit allows this current to flow around the circuit.

**Q. In second harmonic generation does the intensity of the second harmonic just get bigger the further it travels through the crystal?**

A. Basically yes provided that we are phase-matched, and through each part of the crystal we are effectively constructively interfering the second harmonic component. There will come a point of course where you start to "run out" of intensity for the fundamental component as the energy is transfered to the frequency doubled or second harmonic component.

**Q. The equation for the beam width of the confocal cavity - where has it come from?**

A. The cross-sectional beam profile in a cavity is actually a 2D Gaussian function and the quoted width is the value to which the intensity falls off to 1/e of its maximum value. There is quite a bit of maths in the derivation and I felt deriving the equation would not add to the understanding of the physics - it would just be a maths exercise. So just try to understand what the terms are and what they mean - don't worry too much as you will not be asked to derive it, or get marks for remembering it. HOWEVER THERE ARE SOME IMPORTANT FUNDAMENTAL EQUATIONS THAT YOU WILL NEED TO REMEMBER THAT WE DID NOT DERIVE - YOU WERE TOLD ABOUT THE IMPORTANCE OF THESE DURING THE COURSE.

**Q. Where are the answers to the course questions?**

A. Email me a request for the course question answers and I will email them to you - or buy them from the departmental office.

**Q. I' currently having some trouble with question 4 on the June 1999 paper for your course. The particular part is to show that: (Gopt)^m = (4KThf) / (m-2)PRe^2n I have looked in the Wilson and Hawkes book but can't find the answer in the form required.**

A. Firstly you will not find the exact method in the text book - some parts of exam questions have to be "unseen".

I can't go through it step by step here but I will try to explain.

- Get the term for the noise - it should look like equn. 7.38 in Wilson and Hawkes except we are told that their function {F(M)} for the excess noise factor is G^x our case, and x is between 1 and 0.
- The optimum gain is where we get the best signal to noise - write a term for the signal to noise ratio from the noise current / the signal current.
- Next to keep the algebra clear and easy write anything that is not a function of the gain G as some constant.
- Now differentiate the signal:noise expression wrt the gain G - its zero at a maximum (or minimum).

From the above we get a value for G that corresponds to the best signal:noise ratio. What takes people by surprise at first may be the fact that the best gain for the device is not as high as possible - its usually around 10 - 100.

**Q. The logical arguement for the population inversion not being higher than Nth when lasing is confusing me a little - is there another way to view it? Also, how does the laser output increase if population inversion is maxed out at Nth?**

A. Ok - try this - imagine the laser is operating under steady state conditions (same as the start of the argument in the course text). This means pumping rate is steady, laser output is steady and all the transitions are steady. The key thing though is to think about the pumping rate being steady and the output being steady. Now, if we are just at threshold then population inversion has reached Nth and we are just about to get laser light output. The gain of the laser medium is proportional to the population inversion and we also showed how it depends on loss factors and energy lost at mirror reflections through the round trip gain. When we just reach threshold gain=1 in the cavity. If the pumping rate is steady and the gain increases above 1 then the energy would indefinately rise in the cavity - imagine how many times a second light bounes back and forth in the cavity - it wouldn't take long to get ridiculous levels of energy in the cavity. Where would this energy come from? We know the only source of energy we have is pumping but that is constant. Thus, the only way that we can satisfy the conservation of energy or maintain a steady state is that the gain never rises above one or population inversion never rises above Nth. REMEMBER THIS ONLY APPLIES WHEN LASING AT A STEADY STATE - A CW LASER. THERE ARE CASES WHERE POPULATION INVERSION RISES ABOVE Nth BUT IN THOSE CASES THE SYSTEM IS NOT LASING AT THE WAVELENGTH WHERE GAIN>1 OR POPULATION INVERSION IS >Nth. EXAMPLES ARE SEEN IN Q_SWITCHING BEFORE LASING STARTS AND IN HOLE BURNING IN GAIN CURVES.

Now for the second part of the question - whilst population inversion is locked to Nth when lasing, one can increase the pumping rate to a higher amount. This will increase the rate of stimulated emissions and thus the energy density in the cavity to a higher but fixed and stable level. The argument for population inversion being locked at Nth still holds of course, so the way we are increasing the output power is by increasing the pumping rate. Remember we drew a couple of graphs to illustrate how population inversion and output power changed with pumping rate.