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For AFM tapping mode, what is the physical meaning of RMS amplitude? Everytime after tuning, RMS amplitude value is about 1.7V. However, when tip is engaging towards to the surface, this value decreases step by step and finally rest on about 1.2V.
According to my knowledge, tuning is to find the proper drive frequency of the tip. How would tuning affect the RMS amplitude? Sometimes if I do not do tuning, RMS amplitude is away from 1.7V. When scanning, I just get fake image.
Thanks so much!
Xiaoqian
Xiaoqian,
The 1.7V RMS is when your tip is oscillating in free space. As you approach your surface the oscillations are reduced resulting in a smaller amplitude. The tip should approach the surface until you reach the 'Amplitude Setpoint', which in your case appears to be 1.2V.
You are correct, tuning is to find the proper drive frequency. Changing the drive frequency away from resonance will reduce the magnitude of the tip oscillation.
Rob
Hi Xiaoqian,
Following up on Rob's answer, the Cantilever Tune will sweep through a range of frequencies to find the cantilevers resonance frequency. If using AutoTune, it will then setup the oscilation amplitude at the resonance frequency, which will produce the resulting RMS amplitude read by the detector. So, Cantilever Tune will set both the "Drive Frequency" and "Drive Amplitude" parameters, which are both signals applied to the piezo in the tip holder driving the cantilever. If you have set the Target Amplitude of 1.7 volts (or 2 volts with an Peak Offset of ~15%), then the live RMS amplitude signal on the microscope will show a RMS of 1.7 volts after AutoTune. As Rob said, this is your free air amplitude. Once the probe engages, the oscillation amplitude is reduced by the presence of the sample (and other factors such as air damping, various forces, ..), and the result in a reduced amplitude, which is 1.2volts in this case. At that point, your Amplitude Setpoint will be set to 1.2 volts and you control the RMS amplitude of the probe by adjusting this parameter.
I hope this helps.John
Rob,
Thanks so much for your answer.
Hi John,
This helps a lot. I am more clear about this question. Thanks.