Nanotechnology/AFM/Overview of properties of various cantilevers

Examples of AFM cantilevers

For a BS-75kHz

L=225 w=28 t=3 h=17+3/2 (is actually trapezoidal)}

$k_N=\frac{1}{4}Yw{\left(\frac{t}{L}\right)}^3=\frac{1}{4}\left(160000\right)28{\left(\frac{3}{225}\right)}^3=2.6548$

$f[Hz]=\frac{t{\beta }_i^2}{4\pi L^2}\sqrt{\frac{Y}{3\rho }}=\frac{(3\ast 10^{-6}){(1.875)}^2}{4\pi {(225\ast 10^{-6})}^2}\sqrt{\frac{(160\ast 10^9)}{3\ast 2330}}=79318.Hz$

$2{\left(\frac{wh}{tL}\right)}^2=2{\left(\frac{28\ast 18.5}{3\ast 225}\right)}^2=1.1778$ so $k_{lat}{k}_{tor}.$

$k_{tor}=k_N\frac{1}{2}{\left(\frac{L}{h}\right)}^2=2.65\frac{1}{2}{\left(\frac{225}{18.5}\right)}^2=195.99$

$k_{lat}=\frac{1}{4}Y\frac{t{w}^3}{L^3}=\frac{1}{4}\left(160000\right)3{\left(\frac{28}{225}\right)}^3=231.26$

For a MPP311

Specs: 13 kHz, 0.45 N/m : L=440 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)

$k_N=\frac{1}{4}Yw{\left(\frac{t}{L}\right)}^3=\frac{1}{4}\left(160000\right)30{\left(\frac{4}{440}\right)}^3=0.90158$

$f$ $[Hz]=\frac{t{\beta }_i^2}{4\pi L^2}\sqrt{\frac{Y}{3\rho }}=\frac{(4\ast 10^{-6}){(1.875)}^2}{4\pi {(440\ast 10^{-6})}^2}\sqrt{\frac{(160\ast 10^9)}{3\ast 2330}}=27655.Hz$

$2{\left(\frac{wh}{tL}\right)}^2=2{\left(\frac{30\ast 19.5}{4\ast 440}\right)}^2=0.22096$ so $k_{lat}<k_{tor}.$

$k_{tor}=k_N\frac{1}{2}{\left(\frac{L}{h}\right)}^2=0.9\frac{1}{2}{\left(\frac{440}{19.5}\right)}^2=229.11$

$k_{lat}=\frac{1}{4}Y\frac{t{w}^3}{L^3}=\frac{1}{4}\left(160000\right)4{\left(\frac{30}{440}\right)}^3=50.714$

For a MPP211

Specs: 50 kHz, 1.5 N/m : L=215 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)

$k_N=\frac{1}{4}Yw{\left(\frac{t}{L}\right)}^3=\frac{1}{4}\left(160000\right)30{\left(\frac{4}{215}\right)}^3=7.7276$

$f$ $[Hz]=\frac{t{\beta }_i^2}{4\pi L^2}\sqrt{\frac{Y}{3\rho }}=\frac{(4\ast 10^{-6}){(1.875)}^2}{4\pi {(215\ast 10^{-6})}^2}\sqrt{\frac{(160\ast 10^9)}{3\ast 2330}}=1.1582\times 10^{5}Hz$

$2{\left(\frac{wh}{tL}\right)}^2=2{\left(\frac{30\ast 19.5}{4\ast 215}\right)}^2=0.92543$

so $k_{lat}{k}_{tor}.$

$k_{tor}=k_N\frac{1}{2}{\left(\frac{L}{h}\right)}^2=7.7\frac{1}{2}{\left(\frac{215}{19.5}\right)}^2=468.02$

$k_{lat}=\frac{1}{4}Y\frac{t{w}^3}{L^3}=\frac{1}{4}\left(160000\right)4{\left(\frac{30}{215}\right)}^3=434.68$

For a MPP111

Specs: 200 kHz, 20 N/m : L=115 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)

$k_N=\frac{1}{4}Yw{\left(\frac{t}{L}\right)}^3=\frac{1}{4}\left(160000\right)30{\left(\frac{4}{115}\right)}^3=50.497$

$f$ $[Hz]=\frac{t{\beta }_i^2}{4\pi L^2}\sqrt{\frac{Y}{3\rho }}=\frac{(4\ast 10^{-6}){(1.875)}^2}{4\pi {(115\ast 10^{-6})}^2}\sqrt{\frac{(160\ast 10^9)}{3\ast 2330}}=4.0484\times 10^{5}Hz$

$2{\left(\frac{wh}{tL}\right)}^2=2{\left(\frac{30\ast 19.5}{4\ast 115}\right)}^2=3.2346$ so $k_{lat}>k_{tor}.$

$k_{tor}=k_N\frac{1}{2}{\left(\frac{L}{h}\right)}^2=50.4\frac{1}{2}{\left(\frac{115}{19.5}\right)}^2=876.45$

$k_{lat}=\frac{1}{4}Y\frac{t{w}^3}{L^3}=\frac{1}{4}\left(160000\right)4{\left(\frac{30}{115}\right)}^3=2840.5$