Capacitance calculation of comb-drive actuator

I just want to make sure that I am performing this calculation correctly...

From this COMSOL study, I gathered the following info...

$$C_t = N\frac{\varepsilon_o\varepsilon_rl_ot}{d}$$

$$C_f = \varepsilon_o\frac{WL}{G}\left(1+\frac{G}{\pi W}\left(1+\ln\left(\frac{2\pi W}{G}\right)\right)\right)\left(1 + \frac{G}{\pi L}\left(1+\ln\left(\frac{2\pi L}{W}\right)\right)\right)$$

$$C = C_t + C_f$$

& performed the calculation for $$\N=1\$$...

import math

l_o = 90E-6 #Finger overlap
t = 5E-6 #Finger thickness
N = 1 #Number of finger overlaps
L = 90E-6 #Finger overlap
W = 5E-6 #Finger width
G = 5E-6 #Finger gap
e_o = 8.854E-12 #Vacuum permittivity
e_r = 11.7 #Relative permittivity
g = G/math.pi

C_t = N*e_o*e_r*l_o*t/G
C_f = (e_o*W*L/G)*(1 + g/W + (g/W)*math.log(2*math.pi*W/G))*(1 + g/L + (g/L)*math.log(2*math.pi*L/W))
C = C_t + C_f
print(C_t)
print(C_f)
print(C)


& got...

9.323262e-15
1.6703194931600559e-15
1.0993581493160055e-14

But from Table III of referenced paper, the values should be...

2.8819e-14
0.1403e-14
3.022e-14

I don't think I'm calculating fringe field capacitance correctly because my answer is independent of # fingers. What else am I doing wrong?

• It's really unclear to me what values you have calculated (as listed in your question). – Andy aka Dec 10 '19 at 8:41
• @Andyaka Read the 3 print statements. – Landon Dec 10 '19 at 9:13
• For calculating Ct you divide by G but the formula says d. Are these the same? – Swedgin Dec 10 '19 at 9:23
• @Swedgin Yes, that much I'm confident about. – Landon Dec 10 '19 at 9:31