Anti reflecting coating, absorption of photons, shunt resistance, series resistance, ohmic contact between electrodes and the n-type, p-type layers connection..

1. Angle at which it is inclined

2. Depending on the hemisphere we stay, the direction that the panel has to be faced.

3. The distance of the panel from the ground level, so that it avoids hot spots, which is a major factor that affects the efficiency.

4. Free from shadows by trees etc.

5. Solar radiation at that position.

6. Proper battery system, and electical components etc...

There are numerous design variables for a PV system. One of the most important (which is largely responsible for the difference of efficiency between a single cell and a module or array) is current matching between all cells. Mismatched current usually results for non-uniform illumination, which usually occurs due to shading and soiling for a non-concentrating PV system. In more drastic cases, hotspots may form. A single hotspot could bring down the entire system's efficiency significantly. That's why bypass diodes are usually employed as a precaution.

Other important factors influencing system-level efficiency include frontal contacts geometry and spacing, balance-of-plant components efficiency and durability (e.g., batteries and inverter), MPP tracking, AR coatings, and system orientation.

High Dinesh,

If you want to know the efficiency with which solar radiation is converted into electric energy, you divide the electric energy produced over an hour expressed in kWh by incoming global radiation expressed in W/m². You have to know the surface area of your PV panels in m² as well. When your PV panels produce 10kWh electric energy in an hour with a surface of 1m² at a global radiation level of 1000W/m² and as AC at 220 Volt, then your efficiency is:

(10kWh/(1hx1m²))/(1000W/m²) or 10%.

That's not so bad for a PV panel. It is a typical value for a grid connected PV panel system.

What you do here is divide useful electric power per m² (AC at 220 V) by radiation power per m² as measured py a good pyranometer.

Cheers,

Frank

Dear Frank,

Your comments are really helpful and informative to understand not only D bu also AC supply efficiency of solar electrical system.

Thanks and Regards

Dinesh

There is a little mistake in the calculations of Mr. Frank Veroustraete.

The hourly efficiency of the solar cells used in a photovoltaic panel with 20% efficiency (a good efficiecy) and an area of 1 m² will produce 200 W under standard test conditions (STC). STC specifies a temperature of 25 °C and an irradiance of 1000 W/m2 with an air mass 1.5.

Best regards

Felix Sanz-Adan

The question is a bit unspecific, which is reflected in the answers. Do you want have information on how to optimally design the components, or how to optimally design a system using components currently on the market? Are we talking about a grid-connected or an autonomous system?

here are several ways to increase the efficiency using :

1- increasing PV array efficiency.

2- maximum power point tracking controllers that increases the conversion efficiency.

3- increasing the incident sun radiance using sun tracking controllers.

4- efficient  DC-AC inverter control.

5. Controlling the impedance Load

6. Cooling down the modules temperature

Dear Mahmoud Dhimish,

Thanks a lot for your answer. Your comments are really helpful and informative to understand the same.

regards,