b)

c)

e) 4 LA,,\\(.&.

f) N-= P..'T

:fr (x-= x ") =

'(..J t c1. I�)

I +juJ,, fc(

4.2) The electrical power dissipated in the external resistor is also the electrical power output of the solar cell:

DLKTqV

oDout VIeIIVRIP D 12

Recall that:

R

VI

IeII

D

LKTqV

oD

1

And since the external quantum efficiency is 100%, we must have:

inc

LP

qI

The above can be used to calculate LI given incP . We need to maximize the power output. The power output depends on the voltage DV which in turn depends on the external resistor R . There is an optimal value of the resistor that gives the maximum output power. So we set: 0dRdPout This gives,

Re

KT

qI

dR

dV

R

VV

dR

dV

dV

dI

dR

dVIV

dR

dV

dV

dI

dR

dP

KTqVo

DDD

D

D

DD

D

D

out

D 1

0

The optimal value of the resistor is related to the voltage via the above relation. In general the voltage DV is not related to the resistor R by the above relation – only the optimal value of the resistor is related to the corresponding voltage DV by the above relation. a) We use the relation:

DLKTqV

oDout VIeIIVRIP D 12

and plot it as a function of R . Note the above only explicitly depends on the voltage DV but not R . But we also know that the optimal value of R must be related to the corresponding voltage by the relation:

R

eKT

qI KTqVo D 1

So for every R we can calculate the voltage using the above relation and then calculate outP , and then

plot outP as a function of R and find the value of R that maximizes outP . For incP equal to 0.1 mW, I

get maximum outP for R equal to ~5.8 k and a maximum conversion efficiency incout PP of 8%. b) For incP equal to 1 mW, I get maximum outP for R equal to ~710 and a maximum conversion

efficiency incout PP of 10.3%. c) For incP equal to 10 mW, I get maximum outP for R equal to ~84.5 and a maximum conversion

efficiency incout PP of 12.6%.

d) The optimal power conversion efficiency is better for larger incP and, therefore, larger LI . The optimal power conversion efficiency stems from a competition between the power dissipation in the external resistor and the forward biasing of the diode from the potential drop across this resistor. The

power dissipation RI2 varies quadratically with the current whereas the potential drop IR across the

resistor varies linearly with the current. This suggests that increasing the current I , and therefore LI ,

ought to be a better strategy for increasing the power output. Of course, the problem is that the forward bias current goes exponentially with the potential drop so one does not win as much. In actual solar cell systems, solar concentrators are often used to increase incP per cell. e) The reasoning in part (d) suggests that minimizing the forward bias current is a good strategy for improving the power conversion efficiency. If the temperature is lowered, everything else being equal, the forward bias current will increase. Therefore, lowering the temperature is not a good idea for improving the power conversion efficiency in our simple model. f) The simplest thing to do would be to reduce the oI of the diode by design (while maintaining 100%

external quantum efficiency). Reducing oI would reduce the unwanted forward bias current for the cell.

ECE 3150 Homework 5 Solutions 5.2) a)

b)

c) V 83.0log2

i

adpnB

n

NN

q

KT

d) The potential drop in all regions must add up to B . Therefore:

Box

dpaox

ox

dnd

s

dnd xqNt

xqNxqN

22

22

Also note that the equality of the total positive and negative charge in the structure implies that:

dpdn

dpadnd

xx

xqNxqN

The latter follows from ad NN . So we get:

Boxox

dnd

s

dnad txqNxNNq

2

2

Solving gives: dnx 59.8 nm

e) If a voltage V is applied, then the potential drop in all regions must add up to VB . Therefore,

VtxqNxNNq

Boxox

dnd

s

dnad

2

2

The potential drop in every region will therefore increase or decrease with the applied voltage. If the potential drop on the N-side becomes equal to n2 then the surface of the N-side right next to the oxide will become inverted; i.e. the hole density there will become as large as the bulk electron density, which is approximately dN . To see this clearly note that on the N-side:

KT

d

iKT

KT

q

eN

nq

ebulkpp

xxq

expxp

bulksurfacebulksurface

surface

12

2

12

If n 2surfacebulk , then:

dd

d

d

KT

q

i

KT

d

i NN

N

N

nennq

eN

np

2

2

22

surface

Similarly, if the potential drop on the P-side becomes equal to p2 then the surface of the P-side right

next to the oxide will become inverted; i.e. the electron density there will become as large as the bulk hole density, which is approximately aN . Since the structure is symmetric, the hole inversion layer on the N-side and the electron inversion layer on the P-side will take place at the same applied voltage. The easiest way to solve this problem is to first find dnx as a function of the applied voltage V using:

VtxqNxNNq

Boxox

dnd

s

dnad

2

2

and then figuring out at what voltage value does the potential drop on the N-side, given by:

s

dnd xqN

2

2

equals V83.02 n . Below is a matlab plot of the potential drop on the N-side as a function of the

applied voltage V.

When V equals 1.31 Volts, the potential drop on the N-side becomes equal to 0.83 Volts.

0 0.5 10

0.2

0.4

0.6

0.8

1

Applied Voltage (V)

Po

ten

tial

Dro

p o

n t

he

N-s

ide