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Commit 13375a02 authored by Pascal Engeler's avatar Pascal Engeler
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Fixed parameter calculation, added explanation

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%% Cell type:code id: tags:
``` python
import numpy as np
import matplotlib.pyplot as plt
```
%% Cell type:markdown id: tags:
# Units
The units that suit the situation are
Mass: ug
Voltage: V
Voltage: uV
Current: nA
Capacitance: mF
Charge: C
Charge: nC
Time: ms
Frequency: kHz
Length: um
%% Cell type:code id: tags:
``` python
#Drum parameters:
drum_diameter = 1000. #um
drum_thickness = 0.3 #um
electrode_thickness = 0.05 #um
electrode_area = 200.*200. #um^2
drum_gap = 1. #um
drum_resonance = 10. #kHz
drum_Q = 10000. #no units
```
%% Cell type:code id: tags:
``` python
#Physical constants
rho_si3n4 = 3.44e-6 #ug/um^3
rho_au = 1.93e-5 #ug/um^3
coulomb_k = 9.0e12 #um/mF
permittivity = 8.85e-15 #mF/um
permittivity = 8.85e-15 #mF/um or nN/(mV)^2
```
%% Cell type:code id: tags:
``` python
#Drum mass
m_si3n4 = (0.5*drum_diameter)**2*3.141592653589793*drum_thickness*rho_si3n4
m_au = 4.*electrode_area*electrode_thickness*rho_au
m = m_si3n4 + m_au
print(f'Si3N4 Mass: {m_si3n4}')
print(f'Au Mass: {m_au}')
print(f'Drum Mass: {m}')
```
%% Output
Si3N4 Mass: 0.8105309046261666
Au Mass: 0.1544
Drum Mass: 0.9649309046261666
%% Cell type:code id: tags:
``` python
#Drum capacitance
capacitance = permittivity*electrode_area/drum_gap #nC/uV
print(f'Capacitance: {capacitance} nC/uV')
```
%% Output
Capacitance: 3.54e-10 nC/uV
%% Cell type:code id: tags:
``` python
#Electrode charge
def charge(volt):
return capacitance * volt*1e6
```
%% Cell type:code id: tags:
``` python
print(f'Charge at 5V: {charge(5.)} pC')
```
%% Output
Charge at 5V: 0.00177 pC
%% Cell type:code id: tags:
``` python
def force(volts):
nanocouls = charge(volts)
return coulomb_k * nanocouls*(-nanocouls) / drum_gap**2
```
%% Cell type:code id: tags:
``` python
print(f'Force at {drum_gap} um and 5 V: {force(5.)} nN = {force(5.)*1e-6} mN')
```
%% Output
Force at 1.0 um and 5 V: -28196100.0 nN = -28.196099999999998 mN
%% Cell type:code id: tags:
``` python
#Damping coefficient
c = 2.*3.141592653589793*drum_resonance / drum_Q
```
%% Cell type:code id: tags:
``` python
#Print relevant parameters
print(f"Gap: {drum_gap} um")
print(f"1/Gap^2: {1./drum_gap**2} um^2")
print(f"2/Gap^3: {2./drum_gap**3} um^3")
print(f"Capacitance: {capacitance:.8E} nC/uV")
print(f"Coulomb_k * Capacitance^2 / m: {coulomb_k*capacitance**2/m:.8E} um/mF")
print(f"Prefactor k_pf: {coulomb_k * permittivity**2 * electrode_area**2 / (4 * m * drum_gap**4)}")
print(f"Drum Mass: {m:.8E} ug")
print(f"[F] = nN")
print(f"Damping c: {c:.8E} kHz")
print(f"w^2: {(2.*3.141592653589793*drum_resonance)**2/m:.8E} kHz^2")
```
%% Output
Gap: 1.0 um
1/Gap^2: 1.0 um^2
2/Gap^3: 2.0 um^3
Capacitance: 3.54000000E-10 nC/uV
Coulomb_k * Capacitance^2 / m: 1.16883395E-06 um/mF
Prefactor k_pf: 2.922084873105368e-07
Drum Mass: 9.64930905E-01 ug
[F] = nN
Damping c: 6.28318531E-03 kHz
w^2: 4.09132068E+03 kHz^2
%% Cell type:markdown id: tags:
# Force per unit mass
The force per unit mass is now
$\text{F} / \text{m} = \text{Hooke} + \text{Damping} + \text{Drive} + \text{Coupling}$,
with
$\text{Hooke} = - \omega^2 \, x,$
$\text{Damping} = - c \, \dot{x},$
$\text{Drive} = k_{\text{pf}} \, V^2 \, (1. + \frac{x}{\text{gap}}),$
$\text{Coupling} = \sum_i k_{\text{pf}} \, t_i^2 \, (1. + \frac{x + x_i}{\text{gap}}),$
where
$k_\text{pf} = \frac{k_c\,\varepsilon^2\,A^2}{4\,m\,d^4}$
and V, t_i are the driving voltage and the coupling electrode i voltage.
The derivation of the coulomb term goes via assuming that our electrodes form parallel capacitors, which is valid when
$1\,\text{um} \approx \text{gap} << \text{linear dim} \approx 200\, \text{um}$.
Then the capacity is given by $C = \frac{\varepsilon\,A}{d^2}$, which depends on the gap between the two electrodes, $d = \text{gap}$.
%% Cell type:code id: tags:
``` python
```
......
%% Cell type:code id: tags:
``` python
import numpy as np
import matplotlib.pyplot as plt
```
%% Cell type:markdown id: tags:
# Units
The units that suit the situation are
Mass: ug
Voltage: uV
Current: nA
Capacitance: mF
Charge: nC
Time: ms
Frequency: kHz
Length: um
%% Cell type:code id: tags:
``` python
#Drum parameters:
drum_diameter = 1000. #um
drum_thickness = 0.3 #um
electrode_thickness = 0.05 #um
electrode_area = 200.*200. #um^2
drum_gap = 1. #um
drum_resonance = 10. #kHz
drum_Q = 10000. #no units
```
%% Cell type:code id: tags:
``` python
#Physical constants
rho_si3n4 = 3.44e-6 #ug/um^3
rho_au = 1.93e-5 #ug/um^3
coulomb_k = 9.0e12 #um/mF
permittivity = 8.85e-15 #mF/um or nN/(mV)^2
```
%% Cell type:code id: tags:
``` python
#Drum mass
m_si3n4 = (0.5*drum_diameter)**2*3.141592653589793*drum_thickness*rho_si3n4
m_au = 4.*electrode_area*electrode_thickness*rho_au
m = m_si3n4 + m_au
print(f'Si3N4 Mass: {m_si3n4}')
print(f'Au Mass: {m_au}')
print(f'Drum Mass: {m}')
```
%% Output
Si3N4 Mass: 0.8105309046261666
Au Mass: 0.1544
Drum Mass: 0.9649309046261666
%% Cell type:code id: tags:
``` python
#Drum capacitance
capacitance = permittivity*electrode_area/drum_gap #nC/uV
print(f'Capacitance: {capacitance} nC/uV')
```
%% Output
Capacitance: 3.54e-10 nC/uV
%% Cell type:code id: tags:
``` python
#Electrode charge
def charge(volt):
return capacitance * volt*1e6
```
%% Cell type:code id: tags:
``` python
print(f'Charge at 5V: {charge(5.)} pC')
```
%% Output
Charge at 5V: 0.00177 pC
%% Cell type:code id: tags:
``` python
def force(volts):
nanocouls = charge(volts)
return coulomb_k * nanocouls*(-nanocouls) / drum_gap**2
```
%% Cell type:code id: tags:
``` python
print(f'Force at {drum_gap} um and 5 V: {force(5.)} nN = {force(5.)*1e-6} mN')
```
%% Output
Force at 1.0 um and 5 V: -28196100.0 nN = -28.196099999999998 mN
%% Cell type:code id: tags:
``` python
#Damping coefficient
c = 2.*3.141592653589793*drum_resonance / drum_Q
```
%% Cell type:code id: tags:
``` python
#Print relevant parameters
print(f"Gap: {drum_gap} um")
print(f"1/Gap^2: {1./drum_gap**2} um^2")
print(f"2/Gap^3: {2./drum_gap**3} um^3")
print(f"Capacitance: {capacitance:.8E} nC/uV")
print(f"Coulomb_k * Capacitance^2 / m: {coulomb_k*capacitance**2/m:.8E} um/mF")
print(f"Prefactor k_pf: {permittivity * electrode_area / (4. * m * drum_gap**2):.8E} mFum/ug")
print(f"Drum Mass: {m:.8E} ug")
print(f"[F] = nN")
print(f"Damping c: {c:.8E} kHz")
print(f"w^2: {(2.*3.141592653589793*drum_resonance)**2/m:.8E} kHz^2")
```
%% Output
Gap: 1.0 um
1/Gap^2: 1.0 um^2
2/Gap^3: 2.0 um^3
Capacitance: 3.54000000E-10 nC/uV
Coulomb_k * Capacitance^2 / m: 1.16883395E-06 um/mF
Prefactor k_pf: 9.17164116E-11 mFum/ug
Drum Mass: 9.64930905E-01 ug
[F] = nN
Damping c: 6.28318531E-03 kHz
w^2: 4.09132068E+03 kHz^2
%% Cell type:markdown id: tags:
# Force per unit mass
The force per unit mass is now
$\text{F} / \text{m} = \text{Hooke} + \text{Damping} + \text{Drive} + \text{Coupling}$,
with
$\text{Hooke} = - \omega^2 \, x,$
$\text{Damping} = - c \, \dot{x},$
$\text{Drive} = k_{\text{pf}} \, V^2 \, (1. + \frac{x}{\text{gap}}),$
$\text{Coupling} = \sum_i k_{\text{pf}} \, t_i^2 \, (1. + \frac{x + x_i}{\text{gap}}),$
where
$k_\text{pf} = \frac{\varepsilon_0\,A}{4\,m\,\text{gap}^2}$
and V, t_i are the driving voltage and the coupling electrode i voltage.
Note that we choose coordinates where drums from both, sublattice A and B, are displaced in positive direction when they move closer to their respective driving electrode.
%% Cell type:code id: tags:
``` python
```
......
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