Physics and Engineering of Radiation Detection

5.1. Semiconductor Detectors 287

this value must be multiplied by the permittivity of free space 0 =8. 854 × 10 −^12
C^2 N−^1 m−^2.
Moving back to our derivation, for the sake of simplicity, let us write the equation
5.1.54 in one dimension as
d^2 Φ
dx^2

=−

ρ(x)



, (5.1.56)

Although the charge density has a continuous profile inside the region, but to sim-
plify the calculations we can approximate this with a step profile given by (see also
Fig.5.1.25)

ρ(x)=

eND :0≤x<xn n-side

−eNA : −xp<x≤0p-side.

(5.1.57)

HereNDandNAare the donor and acceptor impurity concentrations respectively,
eis the usual unit electrical charge, andxpandxnrespectively are the depths of
junction on p and n sides. It should be noted that this charge density profile is not
always a good approximation, specially in the so calledfully depleteddetectors or
when the applied bias is very small; the two extremes corresponding to very large
and very small depletion regions respectively. On the p-side (−xp<x≤0), it
becomes
d^2 Φ
dx^2

=

eNA



. (5.1.58)

eND

xn

−xp x

Charge Density

−eNA

Figure 5.1.25: Realistic (dotted line) and ide-

alized (solid line) charge density distributions

in a pn junction.xpandxnare the depths of

depletion regions on p and n sides respectively.

In the majority of semiconductor detectors only

one side is heavily doped making the depletion

region very large on the opposite side.

Integrating this once gives us the electric field profile on the p-side

E(x)=−

dΦ

dx

=−

∫

d^2 Φ

dx^2

dx

= −

eNA



∫

dx

= −

eNA



x+A. (5.1.59)

To determine the integration constantAwe note that the electric fieldEmust vanish
at the edge of the depletion region, i.e.,E(−xp) = 0. This gives

E(x)=−

dΦ

dx

=−

eNA



(x+xp)for−xp<x≤ 0 (5.1.60)