Basic Principles of RF Superconductivity and Superconducting Cavities

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Basic Principles of RF Superconductivity and Superconducting
Cavities
Peter Schm¨user
September 12,2003
1
2
Pill Box CavitySimplest practical model of accelerating cavity:hollow cylinder
Neglect beam pipes,then field pattern inside resonator and all cavity parameters can be calculated
analyticallyField pattern:for particle acceleration we need longitudinal electric field on the axis
⇒ choose TM (transverse magnetic) eigenmode of cavityElectric and magnetic field in a pillbox cavity for the accelerating mode TM
010
Use cylindrical coordinates (r,θ,z),search for eigenmode with cylindrical symmetry (independence of θ)
and with longitudinal electric and azimuthal magnetic field
Wave equation for electric field

2
E
z∂r
2
+
1r
∂E
z∂r
=
1c
2

2
E
z∂t
2
(1)3
For harmonic time dependence E
z
(r) cos(ωt)

2
E
z∂u
2
+
1u
∂E
z∂u
+E
z
(u) = 0 with u =
ωrc
(2)
Bessel equation of zero order,solution J
0
(u)
E
z
(r) = E
0
J
0
￿
ωr c
￿
(3)
For perfectly conducting cylinder of radius R
c
:E
z
(R
c
) = 0 ⇒ J
0
(ωR
c
/c) = 0
First zero of J
0
(u) is at u = 2.405.This defines the frequency of the lowest eigenmodefundamental modeoraccelerating modef
0
=
2.405c2πR
c

0
=
2.405cR
c
(4)In cylindrical cavity:frequency independent of length L
c4
Magnetic field is computed from
∂E
z∂r
= µ
0
∂H
θ∂t
Fields in the fundamental TM
010
mode
E
z
(r,t) = E
0
J
0
(
ω
0
rc
) cos(ω
0
t),
H
θ
(r,t) = −
E
0 µ
0
c
J
1
(
ω
0
rc
) sin(ω
0
t).(5)Electric and magnetic fields are 90

out of phase.Magnetic field vanishes on the axis,maximum value
close to cavity wall5
red curve: E
z

blue curve:
B


Stored energy
Integrate energy density (ε
0
/2)E
2
(at time t = 0) over volume of cavity
U =
ε
02
2πL
c
E
2
0
￿
R
c
0
J
2
0
(
ω
0
rc
)rdr
=
ε
0 2
2πL
c
E
2
0
￿

0
￿
2
￿
a
0
J
2
0
(u)udu(6)where a = 2.405 is the first zero of J
0
.Using
￿
a
0
J
2
0
(u)udu = 0.5(aJ
1
(a))
2
U =
ε
02
E
2
0
(J
1
(2.405))
2
πR
2
c
L
c
(7)
Power dissipation
Consider first copper cavity:•rf electric field vanishes at cavity wall,hence no losses•magnetic field penetrates into wall with exponential attenuation,induces currents within skin depth
δ =
￿2µ
0
ωσ
(8)6
Here σ is the conductivity of the metal
Copper at room temperature and 1 GHz δ = 2µm
The current density in the skin depth is
j =
H
θδDissipated power per unit area
dP
dissdA
=
12σδ
H
2
θ
=
12
R
surf
H
2
θ
(9)7
Important parameter of rf cavities:surface resistanceR
surf
=
1σδ
(10)
The power per unit area has to be integrated over the inner surface of cavity
Integration is straightforward for the cylindrical mantle where H
θ
= const.To compute power dissipation in the circular
end plates evaluate
￿
a
0
(J
1
(u))
2
udu = a
2
(J
1
(a))
2
/2 with a = 2.405
Total power dissipation in cavity walls
P
diss
= R
surf

E
2
0 2 µ
2
0
c
2
(J
1
(2.405))
2
2πR
c
L
c
(1 +R
c
/L
c
) (11)
Quality Factor Q
0
Very important parameter of a resonating cavity
The quality factor is defined as the number of cycles needed to dissipate the stored energy (except for
factor of 2π)
alternative definition:resonance frequency f
0
divided by half width Δf of resonance curve
Q
0
= 2π ∙
U f
0P
diss
=
f
0Δf
(12)8
Using (7) and (11) we get:
Q
0
=
GR
surf
with G =
2.405 µ
0
c2(1 +R
c
/L
c
)
(13)geometry constant Gdepends only on cavity shape but not on the material,typical value is G = 300 Ω.
Accelerating field,peak electric and magnetic fields
A relativistic particle needs a time c/L
c
to travel through the cavity.During this time the longitudinal
electric field changes
The accelerating field E
accis the average field seen by particle
E
acc
=
1L
c
￿
L
c
/2
−L
c
/2
E
0
cos(ω
0
z/c)dz,V
acc
= E
acc
L
c
.(14)
Choosing a cell length of one half the rf wavelength,L
c
= c/(2f
0
),we get E
acc
= 0.64 E
0
for a pill
box cavity Peak electric field E
peakat the cavity end plate,here E
peak
= E
0Peak magnetic field B
peakis near cylindrical wall
For a pillbox cavity
E
peak
/E
acc
= 1.57,B
peak
/E
acc
= 2.7 mT/(MV/m).(15)
Pillbox cavity with beam pipes:peak fields increase by 20 −30%9
Shunt Impedance
Represent the cavity by a parallel LCR circuit,parallel Ohmic resistor is calledshunt impedanceRelation between peak voltage in equivalent circuit and accelerating field in cavity:
V
0
= V
acc
= E
acc
L
c
Dissipated power in LCR circuit
P
diss
=
V
2
02R
shunt
Identify this with the dissipated power in the cavity,eq.(11)
Then the shunt impedance of the pillbox cavity is 1R
shunt
=
2L
2
c
µ
2
0
c

3
(J
1
(2.405))
2
R
c
(R
c
+L
c
)

1R
surf
The ratio of shunt impedance to quality factor is an important cavity parameter
(R/Q) ≡
R
shunt Q
0
=
4L
c
µ
0

3
(J
1
(2.405))
2
2.405R
c
(16)
(R/Q) is independent of the material,depends only on the shape
Typical value for 1-cell cavity (R/Q) = 100 Ω 1
R
shunt
is often defined by
P
diss
=
V
2
0R
shunt
then (R/Q) is a factor of 2 larger 10
Superconductivity BasicsShort introduction into:•Type I and type II superconductors•Hard and soft superconductors•Superconductors in microwave fields11
12
13
14
15
Superconductors in microwave fieldsCopper cavity:surface resistance is given by
R
surf
=
1δσ
(17)
In case of superconductor:skin depth δ must be replaced withLondon penetration depthλ
L
≈ 50 nm ￿ δ ≈ 2000 nm
Big question:what is the conductivity σ?If we take σ → ∞we get R
surf
= 0.That would be
nice but is wrong!
According to the Bardeen-Cooper-Schrieffer (BCS) theory of superconductivity the supercurrent is carried
by Cooper pairsResponse of a superconductor to an ac field described byTwo-Fluid Model:•Cooper pairs are superfluid•Unpaired electrons are normal fluid,yield conductivity σ
n
Study response of two fluids to a periodic electric field
normal current obeys Ohm’s law (and dissipates power)
J
n
= σ
n
E
0
exp(−iωt) (18)16
Cooper pairs are accelerated m
c
˙v
c
= −2e E
0
exp(−iωt)
Supercurrent density
J
s
= i
n
c
2 e
2m
e
ω
E
0
exp(−iωt) (19)
Supercurrent is 90

out of phase with electric field,hence no power dissipation
Write for total current density J = J
n
+J
s
= σE
0
exp(−iωt)
with a complex conductivity:
σ = σ
n
+iσ
s
with σ
s
=
2 n
c
e
2m
e
ω
=

0
λ
2
L
ω
(20)
Surface resistance:real part of the complex surface impedance
R
surf
= Re
￿
1 λ
L

n
+iσ
s
)
￿
=

L

σ

2
n

2
s
(21)
Important observation:σ
2
n
￿σ
2
s
at microwave frequencies hence disregard σ
2
n
in the denominator
⇒R
surf
= σ
n
/(λ
L
σ
2
s
)
Surprising result:The microwave surface resistanceis proportional to the normal-stateconductivityConductivity of normal metal given by classic Drude expression
σ
n
= n
n
e
2
￿/(m
e
v
F•n
n
density of single (unpaired) electrons17
•￿ mean free path of single electrons•v
F
the Fermi velocity
Unpaired electrons are created by thermal breakup of Cooper pairs
Energy gap E
g
= 2Δ between the superconducting (BCS) ground state and the free electron states
By analogy with the conductivity of an intrinsic (undoped) semiconductor we get
n
n
∝ exp(−E
g
/(2k
B
T))
and hence
σ
n
∝ ￿ exp(−Δ/(k
B
T)).(22)
Using 1/σ
s
= µ
0
λ
2
L
ω and Δ = 1.76k
B
T
c
we finally obtain for the BCS surface resistanceR
BCS
∝ λ
3
L
ω
2
￿ exp(−1.76 T
c
/T)(23)
This formula displays two important aspects of microwave superconductivity
- the surface resistance depends exponentially on temperature
- it is proportional to the square of the rf frequency
For niobium:small correction needed,replace λ
L
by Λ = λ
L
￿ 1 +ξ/￿18
2 3 4 5 6
7
R
S
[nΩ
]
T
c
/T
R
res
= 3 nΩ
1000
100
10
1
R
BCS
Surface resistance of a 9-cell TESLA cavity plotted as a function of T
c
/T.The residual resistance of
3 nΩ corresponds to a quality factor Q
0
= 10
11
Residual resistance
Residual resistance caused by impurities,frozen-in magnetic flux or lattice distortions
R
surf
= R
BCS
+R
res
(24)19
R
res
is temperature independent,amounts to a few nΩ for a clean niobium surface
Heat conduction in niobium and heat transfer to liquid helium
Heat produced at inner cavity surface must be guided through cavity wall to liquid helium bath
Thermal conductivity of Nb drops strongly at T →0
Very pure Nb with largeresidual resistivity ratio RRRneeded
RRR = R(300K)/R(10K)Measured heat conductivity in niobium with RRR = 270 resp.500 as a function of temperature20
110
1
10
100
1000
8
6
4
2
2
0
RRR=500
RRR=270
λ
[W/mK]
T[K]
Low frequency cavities (350-500 MHz):small BCS surface resistance at 4.2 K,effective cooling by normal
liquid helium
Due to f
2
dependence of BCS resistance:at higher frequency,cooling with superfluid helium at 1.8 - 2
K is better
Note:Kapitza resistanceat superfluid helium-niobium interface leads to temperature jump
Maximum Field in SC Cavities
Magnetic field of microwave must stay below the critical magnetic field of superconductor
Situation is clear for a type I superconductor such as lead:
at T = 2 K one has B
c
= 80 mT →E
acc
≤ 20 MV/m
For type II superconductors the situation is not that clear.Magnetic flux moving in and out of the sc
produces heat.Flux pinning is undesirable since the magnetic hysteresis again leads to heat generation in
a microwave field.
Consequence:a hard superconductor like NbTi or Nb
3
Sn is not well suited for rf cavities,at the large B
c2
of 10 - 20 Tesla the heat generation would be untolerable.
What about niobium?This superconductor is of type II,but close to the boundary of type I.The
critical fields at 2 Kelvin are approximately•B
c1
≈ 160 mT•B
therm
c
≈ 200 mT•B
c2
≈ 350 mT21
Very safe limit:B < B
c1
⇒E
acc
≤ 40 MV/m:no magnetic flux enters the sc
Experimental observation E
acc
> 40 MV/m has been achieved repeatedly,the best value was 45 MV/m
Question:how far above B
c1
can one go?
Hint
The following pages contain further interesting material on sc cavities which,however,is beyond the scope
of a 1-hour lecture at CAS22