Homework Chapter 06

EEE 165

CSUS

Instructor: Russ Tatro

Chapter 6 Photovoltaic Devices S.O. Kasap, Optoelectronics and Photonics, Principles and Practices , 2001 Solutions to Chapter 6 Homework.

Problems 3 part a) only, 4 3 a.

4

Remember to scale incident light power for the IV graph. I’ = 22.5 mA V’ = 0.45 V Pdelivered = 10.1 mW Input sun-light power Pin = 0.4 W η = 2.5% Isc = 25 mA For n = 1 Voc = 0.532V For n = 2 Voc = 0.514 V

6.3

Solar cell driving a load 2

a A Si solar cell of area 4 cm  is connected to drive a load R as in Figure 6.8 (a). It -2 has the I-V  characteristics in Figure 6.8 (b) under an illumination of 600 W m . Suppose -2 that the load is 20 Ω and it is used under a light intensity of 1 kW m . What are the current and voltage in the circuit? What is the power delivered to the load? What is the efficiency of the solar cell in this circuit? Solution -2

The solar cell is used under an illumination of 1 kW m . The short circuit current a has to be scaled up by 1000/600 = 1.67. Figure 6Q3-1 shows the solar cell characteristics scaled by a factor 1.67 along the current axis. The load line for R = 20 Ω and its   characteristics at P  which is the operating point P . intersection with the solar cell I −V Thus,  I ′  ≈ 22.5 mA and V ′  ≈ 0.45 V The power delivered to the load is  P out  = Ι′ V′   = (22.5×10 )(0.45V) = 0.0101W, or 10.1 mW. -3

This is not the maximum power available from the solar ce ll. The input sun-light  power is  P in = (Light Intensity)(Surface Area) -2

2

= (1000 W m )(4 cm The efficiency is  P  η  = 100 out  P in which is poor.

= 100

0.010 0.4

= 2.5 0 0

× 10-4 m2/cm2) = 0.4 W

 I (mA) V ′ 

0 0.2

0.4

V  0.6

 –10  I-V for a solar cell under an illumination of 1000 Wm-2.

 –20

 I ′ 

 P   M 

 –30

Figure 6Q3-1

The load line for R  = 20 ž ( I-V  for the load)

-2

6.4 Open circuit voltage A solar cell under an illumination of 100 W m  has a short circuit current I sc of 50 mA and an open circuit output voltage V oc, of 0.55V. What are the short circuit current and open circuit voltages when the light intensity is halved? Solution

The short circuit current is the photocurrent so that at  I  sc2

I   = I  sc1  2  = (50  I 1 

 50 W m−2   = 25 mA mA)  100 W m −2 

Assuming n = 1, the new open circuit voltage is V oc2

= V oc1 +

nk  BT  e

ln 

I 2 

 = 0.55 + 1( 0.0259)ln(0.5) = 0.532 V  I 1 

Assuming n = 2, the new open circuit voltage is V oc2

= V oc1 +

nk  BT  e

ln 

I 2 

 = 0.55 + 2(0.0259)ln(0.5)= 0.514 V  I 1 