Physics and Engineering of Radiation Detection

8.1. Preamplification 467

fact most of the radiation detectors can be modeled as current sources. So how
do we relate this current to the detector’s voltage that we amplify by a voltage
sensitive preamplifier? This can be done if we take into account the fact that this
instantaneous current actually gets integrated on detector’s capacitance. If at any
instant in timeta currentisflows through the detector, then the total chargeQ
accumulated on the detector’s capacitanceCdfromt=0tot=t 0 will be given by

Q=

∫t 0

0

is(t)dt. (8.1.5)

Of courseCdhere is thecombineddetector and stray capacitance. Hence the pream-
plifier’s output voltage can be expressed in terms of detector’s instantaneous current
as

Vout=

A

Cd

∫t 0

0

is(t)dt. (8.1.6)

This expression implies that the output voltage has a direct dependence on ca-
pacitive load of the amplifier. This capacitive loadCdtogether with the amplifier’s
input impedanceRadetermines how fast the capacitor discharges through the time
constant of the circuitRaCd. The integrated current and the time constant deter-
mine the shape of the voltage pulse.
Since we are integrating the current to convert it into voltage,Cdshould discharge
slower than the charge collection timetd. The criterion

tdRaCd, (8.1.7)

is therefore sometimes used to decide if a voltage sensitive preamplifier is suitable
for a detector.

8.1.B CurrentSensitivePreamplifier

In certain applications it is desirable to measure the instantaneous current flow-
ing through the detector. This can be done through a current sensitive preampli-
fier, which converts the detector’s instantaneous current into a measurable voltage.
Therefore, this device can also be called a current-to-voltage converter. A current
sensitive amplifier can be constructed in the same way as a voltage sensitive ampli-
fier with the exception that the input impedance in this case must be kept at the
minimum to allow the current to flow through the amplifier (see Fig.8.1.2(a)).
The currentiaflowing into the preamplifier of Fig.8.1.2(a) is related to the source
currentisby

ia=

Rs

Rs+Ra

is. (8.1.8)

This equation implies that the requirement, that the current flowing into the
preamplifier is approximately equal to the source current, can be fulfilled by making
preamplifier’s input impedance very small as compared to source impedance, i.e.,

RaRs =⇒ia≈is
Since a linear current sensitive preamplifier is simply a current-to-voltage con-
verter, therefore its output voltage is proportional to the source current.

Vout∝is (8.1.9)