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170 5 Creation of the Signal
The capacitances, calculated by Eq. (5.33), are therefore
C
11
= C
22
= 0 C
12
= C
21
=
2
πε
0
l
ln(b/a)
. (5.39)
For the chosen parameters we find C
12
= 20 pF. In the final step we impress the
induced currents as ideal current sources on the equivalent circuit of the tube as
shown in Fig. 5.10b. The wire is read out by an amplifier with input resistance
R
in
, which we choose to be 50
Ω
. In the equivalent circuit, the wire and the tube
have become pointlike nodes. This is only admissible if the tube is ‘electrically
short’, i.e., if the shortest wavelength
λ
= c/ f
0
processed by the readout electronics
with bandwidth limit f
0
is large compared to the tube length l. For the assumed
parameters this translates into f
0
300MHz. We use a value of 1.6 MHz in the
following. In a case where l is not much smaller than c/ f
0
we would have to consider
the tube to be a transmission line. In practical applications it is not convenient to set
the tube to negative high voltage −U, as shown in Fig. 5.6, but rather to ground the
tube and set the wire to positive voltage U. The signal −I
ind
1
induced on the tube can
be neglected because the current is ‘leaving’ to the ground node and is not entering
the amplifier.
Choice of the circuit parameters. The capacitor C decoupling the amplifier
from the potential U must be large compared to C
12
so that the current signal goes
into the amplifier instead of flowing into C
12
. But it must not be so large that the
energy E = CU
2
/2 it stores would destroy the amplifier in the case of a spark. Let
us take the typical value of 1 nF for C.
The resistor R
L
connects the wire to the voltage supply. It must be large in com-
parison to the impedance of C and R
in
in series so that only a small fraction of
the signal is lost into the power supply. If we want 99% of the bandwidth to find
this impedance sufficiently small, we may evaluate it at 1% of the bandwidth limit
leaving a value of
R
L
>
1
2
π
f
0
×0.01
+ R
in
≈ 10
4
Ω
(5.40)
in our example. In addition, this resistor is a source of thermal current noise which
is proportional to 1/
√
R
L
, so in this respect as well one chooses the resistor as large
as possible. The upper limit for R
L
is defined by the maximum voltage drop that
one allows for a given chamber current. With a primary ionization of, e.g., 125 elec-
trons and a gas gain of 5 ×10
5
at a rate of 100 kHz, we find a chamber current
of 1
μ
A, which leads to a tolerable voltage drop of 1 V for an R
L
value of 1M
Ω
.
Values for R
L
of 100 k
Ω
to several M
Ω
are typical; let us make R
L
= 1M
Ω
for
definiteness.
The relation between the induced current I
ind
1
and amplifier input current I
A
can
be determined by evaluating the equivalent circuit diagram in Fig. 5.10b. Using the
methods of Chap. 6 we change to the Laplace domain and find
I
A
(s)=
sR
L
C
1+ s(R
L
C + R
in
C + R
L
C
12
)+s
2
R
in
C
12
R
L
C
I
ind
1
(s). (5.41)