(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known da

Show that for a PV cell finding a maximum of power leads to the following equation:

##(1 +\frac{qV_{max}}{kT})\exp\left(\frac{qV_{max}}{kT}\right) = 1 + \frac{I_{sc}}{I_{rs}}##

##I_{sc}## is short circuit current and ##I_{rs}## is reverse saturation current.

2. Relevant equations

3. The attempt at a solution

I get pretty close, but I've missed something out. The current is given by:

##I = I_{rs}(\exp(V/V_t)-1)-I_L = I_{rs}(e^{qV/kT}-1)-I_L##

For maximum power ##P_{max} = I_{max}V_{max}## and ##I_{max}## is the short circuit current, which occurs at ##V=0##. Subbing ##V=0## into the current equation gives ##I_{sc} = -I_L##, so

##I = I_{rs}(e^{qV/kT}-1)+I_{sc}##

Max power is at ##\frac{dP}{dV}=0## so given ##P=IV##:

##\frac{dP}{dV} = I_{rs}(e^{\frac{qV_m}{kT}}-1)+\frac{qI_{rs}V_m}{kT}(e^{\frac{qV_m}{kT}})+I_{sc}##

##\left(1+\frac{qV_m}{kT}\right)e^{\frac{qV_m}{kT}} = -\frac{I_{sc}}{I_{rs}}##

That's very close to what I'm looking for, I'm missing a ##1## on the RHS and the sign of the fraction is wrong, have I gone wrong somewhere? I've looked and really can't spot it, thanks for any help!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Max power of a photovoltaic cell -- where did I go wrong?

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**