Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In this post Why not use our own light production to produce new energy instead of wasting it?, I naively asked if it was possible to also recycle our own lightning at night. someone, in his answer, also noticed that photoelectric cells are highly angle dependent.

Hence my first thought: why the photoelectric cells are not designed to be hemispheric?

EDIT1: people misunderstood, so I'm correcting: I mean the panel would be flat, parallel to the ground for example, but each photoelectric cell would be hemispheric. Thus getting optimum direct radiance over time from the sun as also irradiance from the whole sky. As side-effect, light from every point and also because cells are highly reflective, including sunlight would also bounce multiple times from one cell to the others around before dissipating instead of being reflected toward the sky. So does it worth it?

share|cite|improve this question
1  
Mostly for cost reason. You can make them in any shape you like, but sunlight is basically parallel, so there is no point in making the panels curved. If you always want them to point at the light source, then you have to change their angle, which, in some applications, is also done, but it doesn't really do much for efficiency (because of the cosine dependence of angle) and it's expensive. – CuriousOne Jan 4 at 7:22
1  
    
@curiousone, that's the point: with hemispheric cells, you just put them on an horizontal panel and don't waste energy to move it. Also, you get radiance from the whole sky, and other side effect is that you also get the light from multiple bounces, including bounces from parallel sunlight. So would it somewhat compensates? – Eddy Khemiri Jan 4 at 8:09
1  
If you make the panels round then you are wasting energy all the time because the incidence angle is only correct on one location on the panel and wrong in all others. You aren't getting much power from the entire sky, that's just a tiny bit of scattered light from the atmosphere, but nothing but direct sunlight will generate a useful amount of energy. If you can, get yourself a small solar cell and a multimeter and try for yourself. – CuriousOne Jan 4 at 8:20
    
@curiousone, you misunderstood I think: the panel is flat but would contain thousand of small hemispheric cells. – Eddy Khemiri Jan 4 at 8:24

The reason is simple. If you make it hemispherical, and shine sunlight vertically on it, less light will reach the photoelectric cells ( due to the formation of shadows )! So, if we put it horizontally, there will be no shadows and most of the light will be used for the desired purpose and in the other case, much of the light is wasted!

share|cite|improve this answer
2  
This is incorrect. If you make it hemispherical, you will be no worse off than if you left it flat, as the cross-section area (the disk at the "top" of the hemispheric hole) is the same as for a flat cell. Yes, you'll get low-angle sunlight, but it may in fact bounce around the hole, increasing the chance of photon absorption (think black-body cavity). If you were thinking of a hemispherical "bump," I believe the cross-section remains the same, but you will have crappy incidence angles as you suggested. – Carl Witthoft Jan 4 at 14:13
    
I think the thought process behind flat panels is that it's more efficient to use shaped mirrors instead of shaping light panels. – user1306322 Jan 4 at 15:44

The photocell itself is far easier to produce when flat, although bendable sheets (allegedly for clothing!) are in development.

However, as the other answers point out, curving the photocell doesn't help. The most common way of maintaining the cell's perpendicularity to the sun is with external servos to rotate the array. There is an alternative, used in some digital cameras: mount a small hemispherical lenslet over each element (solar cell or camera pixel) to partially focus light onto the sensitive area. For cameras, this has the added advantage that photons reflected off the pixel are likely to suffer total internal reflection at the lenslet's surface and then get a "second chance" to be absorbed at the pixel.

For solar cells, I suspect the difficulty in maintainence of such lenslets overwhelms and possible collection advantage.

share|cite|improve this answer
    
A lenslet array would dramatically reduce the overall efficiency. The purpose of this arrangement in a sensor is to provide a direction agnostic measurement of intensity. It isn't at all concerned with absolute transmission intensity - it merely normalizes (to some degree) the transmission intensity as a function of AoI. The majority of light hitting a lenslet array will be at non-normal incidence and the area-average reflection coefficient will naturally be higher. – J... Jan 4 at 16:21
    
@J... Not necessarily true. In a reasonably fast lens system, there is a converging cone of light for each pixel. The total-internal-reflection thing does work: I've used (and designed) such systems back in my Adaptive Optics days. – Carl Witthoft Jan 4 at 16:43
    
Total internal reflection doesn't help for the light that never entered in the first place. Add a lenslet array and you will lose a lot of light at first surface reflection. – J... Jan 4 at 17:55

I've been studying solar panels so I hope I can clarify your doubts. The amount of solar irradiance which actually interest your surface (the panel) is dependent on the angle between your surface and the sun beams. This means that it is both affected by the titl angle of the panel and by the sun daily and seasonal position. Solar panels angles are then usually optimized (in terms of their tilt angle) in order to absorb the maximum amount of solar irradiance over the year. As CuriousOne tried to explain, if you have emispheric cells it means that you don't have a titl angle for the panel, but you rather have infinite tilt angles for any infinitesimal point of the emispheric surface! Only one point of the surface will get the maximum irradiance, while others will be penalised, even if neglecting shadowing effects and other stuff. I've understood that you are talking about a FLAT panel made of many emispheric cells, but you would have the problem I've explained for any single emispherical cell on the panel.

share|cite|improve this answer
    
Of course, it would be also economically much more costly, and impossible to realize with monocristalline sylicon, which currently covers 95% of the market. It would be possible only with thin films technologies or other new ones. – Franz Eskö Jan 4 at 11:02
    
I do understand, but what about all the reflected lights? Can we neglected them? It's a key point: cells are highly reflective, so even with parallel beams, I think that half of them would be reflected toward nearby cells, no? – Eddy Khemiri Jan 4 at 11:28
    
Why do you say "cells are highly reflective"? Actually the glass cover reflects part of the incident solar radiation, but the reflection coefficient vary depending again on the angle, and is not so high except for very tilted angles. So, yes, you'll have a portion of reflected rays but totally neglectable as compared to the energy you would lose changing from a flat panel to this design. Reflection by means of hemispheric design is in fact used only with mirror tecnologies with high reflection coefficient, mostly solar thermal technolgies. Look for Solar Dish for example – Franz Eskö Jan 4 at 14:19
    
Think about it: if cells were "higly reflective", rays would be reflected a first time, and then reflected again on the next cell, and so on possibly escaping out of the "hemisphere" in the end. Think of a flat panel, now: if it really was "highly reflective" it would not capture any solar irradiance at all – Franz Eskö Jan 4 at 14:22

Easier to manufacturer. To add a curve to panel to capture the sun's photons better would very fractional and not be cost efficient to build vs added energy.

share|cite|improve this answer

protected by Qmechanic Jan 4 at 19:29

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.