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)}

$k_{N}=\frac{1}{4}Yw\left( \frac{t}{L}\right) ^{3}=\frac{1}{4}\left(160000\right) 28\left( \frac{3}{225}\right) ^{3}=2.\,654\,8$

$f[Hz]=\frac{t\beta_{i}^{2}}{4\pi L^{2}}\sqrt{\frac{Y}{3\rho}} =\frac{\left( 3\ast10^{-6}\right) \left( 1.875\right) ^{2}}{4\pi\left( 225\ast10^{-6}\right) ^{2}}\sqrt{\frac{\left( 160\ast10^{9}\right) } {3\ast2330}}=79318.Hz$

$2\left( \frac{wh}{tL}\right) ^{2}=2\left( \frac{28\ast18.5}{3\ast 225}\right) ^{2}=1.\,177\,8$ so $k_{lat}~k_{tor}.$

$k_{tor}=k_{N}\frac{1}{2}\left( \frac{L}{h}\right) ^{2}=2.65\frac{1} {2}\left( \frac{225}{18.5}\right) ^{2}=195.\,99$

$k_{lat}=\frac{1}{4}Y\frac{tw^{3}}{L^{3}}=\frac{1}{4}\left( 160000\right) 3\left( \frac{28}{225}\right) ^{3}=231.\,26$

For a MPP311

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

$k_{N}=\frac{1}{4}Yw\left( \frac{t}{L}\right) ^{3}=\frac{1}{4}\left( 160000\right) 30\left( \frac{4}{440}\right) ^{3}=0.901\,58$

f $[Hz]=\frac{t\beta_{i}^{2}}{4\pi L^{2}}\sqrt{\frac{Y}{3\rho}} =\frac{\left( 4\ast10^{-6}\right) \left( 1.875\right) ^{2}}{4\pi\left( 440\ast10^{-6}\right) ^{2}}\sqrt{\frac{\left( 160\ast10^{9}\right) } {3\ast2330}}=27655.Hz$

$2\left( \frac{wh}{tL}\right) ^{2}=2\left( \frac{30\ast19.5}{4\ast 440}\right) ^{2}=0.220\,96$ so k_lat < k_tor.

$k_{tor}=k_{N}\frac{1}{2}\left( \frac{L}{h}\right) ^{2}=0.9\frac{1} {2}\left( \frac{440}{19.5}\right) ^{2}=229.\,11$

$k_{lat}=\frac{1}{4}Y\frac{tw^{3}}{L^{3}}=\frac{1}{4}\left( 160000\right) 4\left( \frac{30}{440}\right) ^{3}=50.\,714$

For a MPP211

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

$k_{N}=\frac{1}{4}Yw\left( \frac{t}{L}\right) ^{3}=\frac{1}{4}\left( 160000\right) 30\left( \frac{4}{215}\right) ^{3}=7.\, 727\,6$

f $[Hz]=\frac{t\beta_{i}^{2}}{4\pi L^{2}}\sqrt{\frac{Y}{3\rho}} =\frac{\left( 4\ast10^{-6}\right) \left( 1.875\right) ^{2}}{4\pi\left( 215\ast10^{-6}\right) ^{2}}\sqrt{\frac{\left( 160\ast10^{9}\right) } {3\ast2330}}=1.\,158\,2\times10^{5}Hz$

$2\left( \frac{wh}{tL}\right) ^{2}=2\left( \frac{30\ast19.5}{4\ast 215}\right) ^{2}=0.925\,43$

so $k_{lat}~k_{tor}.$

$k_{tor}=k_{N}\frac{1}{2}\left( \frac{L}{h}\right) ^{2}=7.7\frac{1} {2}\left( \frac{215}{19.5}\right) ^{2}=468.\,02$

$k_{lat}=\frac{1}{4}Y\frac{tw^{3}}{L^{3}}=\frac{1}{4}\left( 160000\right) 4\left( \frac{30}{215}\right) ^{3}=434.\,68$

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 $[Hz]=\frac{t\beta_{i}^{2}}{4\pi L^{2}}\sqrt{\frac{Y}{3\rho}} =\frac{\left( 4\ast10^{-6}\right) \left( 1.875\right) ^{2}}{4\pi\left( 115\ast10^{-6}\right) ^{2}}\sqrt{\frac{\left( 160\ast10^{9}\right) } {3\ast2330}}=4.\,048\,4\times10^{5}Hz$

$2\left( \frac{wh}{tL}\right) ^{2}=2\left( \frac{30\ast19.5}{4\ast 115}\right) ^{2}=3.\,234\,6$ so k_lat > k_tor.

$k_{tor}=k_{N}\frac{1}{2}\left( \frac{L}{h}\right) ^{2}=50.4\frac{1} {2}\left( \frac{115}{19.5}\right) ^{2}=876.\,45$

$k_{lat}=\frac{1}{4}Y\frac{tw^{3}}{L^{3}}=\frac{1}{4}\left( 160000\right) 4\left( \frac{30}{115}\right) ^{3}=2840.\,5$