5 – 5
Transpector SPS Operating Manual
The following equations summarize the potentials applied to the rods:
[1]
[2]
V
= RF amplitude
f
= RF frequency
t
= time
U
= DC potential
PZ
= pole zero.
The RF component removes the low-mass ions from the beam. Ions of sufficiently
low mass have their motions remain in phase with the applied RF. These ions will
gain energy from the field and oscillate with increasingly large amplitudes.
Eventually, as they travel along the length of the rods, they will strike one of the
rods and be neutralized. High mass ions are focused by the RF component to an
area close to the quadrupole’s long axis, the Z axis.
The DC component is superimposed on the RF to remove high mass ions from the
beam. The DC field deflects the high mass ions toward the negative poles,
opposing the focusing effects of the RF field. Eventually, these high mass ions
strike the negative rods and are neutralized. By a suitable choice of DC-to-RF ratio,
the mass filter can be made to discriminate against both high and low mass ions.
The ion energy directed along the Z axis of the mass filter is dependent on the
difference between the potential at which the ions were formed (the anode voltage)
and the pole zero. The ion energy is usually only slightly modified by the electric
field (the
fringing
field) between the source exit aperture and the quadrupole.
Imbalances in the amplitude of the two phases of RF applied to the rod pairs, and
DC voltages are also applied, resulting in a further modification of the ion energy.
The mass of the ions passed by the filter is determined by the RF amplitude, the
RF frequency, and the quadrupole radius. (See
[3]
V
= peak-to-peak RF amplitude in Volts
M
= mass of the ion in atomic mass units (AMU) per electron charge
f
= RF frequency in megahertz
r
0
= quadrupole radius in centimeters.
For example, a 200 AMU singly-charged ion would pass through a quadrupole with
nominal 6.35 mm (1/4 in.) diameter rods (an
r
0
of 0.277 cm), operating at
1.78 MHz, at a peak-to-peak RF amplitude of approximately 700 Volts.
X
V
2
ft
cos
U PZ
+
+
=
Y
V
2
ft
+
cos
U
–
PZ
+
=
V
14.438Mf
2
r0
2
=
