Condensers

Lenses have been at focus in previous subsections due to the fact that they can be used to mathematically treat a large portion of ion optics used in the off-line set-up. How-ever, there remains one more category of ion optical elements that have an important role in the system and remain undiscussed. These are condensers. In its most simple form a condenser consists of two conducting parallel plates.

y

x x_d

L

θ E

Figure 8:Parallel plate condenser as a beam deflector

If a voltage is applied between the plates of a condenser and a beam of particles tra-verses through the condenser as in figure 8, the beam is deflected by an angle θ. This angle can be solved starting with forces acting on the particles. The particles experience a force due to the electric field,

qE=my,¨ (2.35)

where y = y(t). This differential equation can be solved fory, ˙yand ¨y. Let us impose a set of boundary conditions that simplify the problem. Let us require y(0) = 0 and

˙

y(0) =0. This leads to

¨ y = ^qE

m (2.36)

˙ y = ^qE

m t (2.37)

y= ^qE

2mt². (2.38)

Let V₀ be the voltage at which the incoming particles have been initially accelerated.

Thus their velocity is v0 = ^p2qV0/m. Expressing time as t = x/v0 equations (2.37) and (2.38) can be written atx =Las

y˙L = ^EL

2V0v0 (2.39)

y_L = ^EL

2

4V0

. (2.40)

To an outside observer the beam coming out of the condenser would seem like coming from a single point on the x axis. This point is called the virtual deflection center [1].

The position of the deflection center can be solved by inspecting the slope formed by v0 and ˙y, i.e. the angle at which the beam seems to have intersected the optical axis.

This slopey⁰can be written using equation (2.39) as y⁰ = ^y˙L

v0

= ^EL

2V0. (2.41)

This result allows writing the position of the virtual deflection centerxcenter as xcenter = ^yL

y⁰ = ^L

2. (2.42)

It is worth noting that the use of slope y⁰ in solving for the position xcenter is justi-fied since we are dealing with a virtual deflection center rather than a real one. It is tempting to think that using a slope solved using a derivative of position to determine another position is incorrect. However, this procedure is justified and it is what makes the deflection center virtual. The angleθbetween the exiting beam and the optical axis can be expressed as

tanθ = ^yL

L/2. (2.43)

A useful approximation is to assume thatθ is small enough so that tanθ= ^sinθ

cosθ ≈ ^θ

1 = ^yL

L/2. (2.44)

Using this relation along with equation (2.40) leads to an expression for the angleθ, θ = ^EL

2V0

. (2.45)

Parallel plate condensers are used in the off-line set-up as beam deflectors and also as a way of moving the beam parallel to the optical axis. This requires two condensers to be used in series. The first condenser deflects the beam by an angleφ. After this the beam is allowed to drift for a distance and then finally there is another condenser which deflects the beam back to its initial angle compared to the optical axis. In addition to adjusting the beam position parallel to the optical axis this system can, naturally, be used to give an additional net deflection. This system has come to be known as double xy-plates at the IGISOL facility.

There is one more type of condenser that is necessary in the off-line set-up. This is a cylindrical condenser. In general these can be used as energy spectrometers and thick lenses, but in the case of this work, one is used to bend a beam of particles from a vertical beam line to a horizontal beam line. Here only a part of the properties of cylindrical condenser shall be discussed with a primary focus in determining necessary voltages for bending particles of given energy.

Let us start with determining voltages to be applied to the condenser plates in order to bend a particle that arrives at the middle of the entrance gap parallel to the vertical axis.

This kind of situation is presented in figure 9. In such a case the force exerted on the particles by the electric field must be equal to the centrifugal force. This requirement can be written as

Here the last equality is merely the definition of electric field. PotentialV correspond-ing to this field can solved by integratcorrespond-ing equation (2.47). The potential becomes

V(r) = −2V0ln r

rc, (2.48)

where V0 = _Ekin/q, i.e. the acceleration voltage, andrc is the radius of the center line of the condenser. Expanding in a Taylor series the potential becomes

V(_r) =−_2V0ln

Figure 9:Schematic of a cylindrical condenser

If the cubic term and the ones after it are now cut off and a new variable,ρ = _r^r

Now all that remains before voltages for the condenser plates can be solved is to ex-press the position of the electrodes using the new variable ρ. Let us denote the radial distance between electrodesd. This leads to an expression

r± =rc±^d

Plugging this into equation (2.52) results in potentials



and finally in the voltage between the electrodes V+−V− =2dV0

rc . (2.56)

There is one cylindrical condenser in the off-line set-up. It is 400 mm in radius and has electrode separation of 19 mm. The system is designed to operate using an acceleration voltage of 30 kV. Using these values equation (2.56) gives a voltage of 2850 V to be applied between the electrodes. A more thorough discussion on condensers along with the derivation presented above can be found in [1].