Asha-dee Celestine November 3rd 2009

Asha-Dee Celestine November 3rd 2009

AE 523: Nanoscale Contact Mechanics Lab #1: AFM 1 Instrument set-up: 1. What were the sum and difference signal values of the photodetector when it was aligned? What do these values represent? The sum signal value was 4 V while the difference signal value was 0.2 V. The sum signal value represents the sum of the signals obtained from both photodiodes while the difference signal value represents the difference or the error signal. 2. What were the cantilever drive frequency and resonance frequency? The cantilever drive frequency was 180.96 kHz while the resonance frequency was 181.0 kHz. 3. What was Ao /Aresonance? A0= 0.95 V Aresonance =1 V.

Therefore Ao /Aresonance = 0.95

4. What was the true Asp upon engaging the tip on the specimen surface? The true Asp upon engaging the tip on the specimen surface was 1.401 V. This value was lowered to 1.35 V, however, to obtain better tracking of the surface.

Sample: 1. What are the dimensions of the features of the calibration grating (lateral and vertical)? The trench measured 5x5 µm while the height of each step was 200 nm.

Asha-Dee Celestine November 3rd 2009

Imaging: 1. Collect a 40x40 µm image at 0.5 Hz. Collect the height and the phase images. Set the z-scale to 500 nm.

Fig. 1: Trace and Retrace Plot for 40x40 µ image at 0.5 Hz

Fig. 2: Height and Phase images for 40x40 µm image at 0.5 Hz

Asha-Dee Celestine November 3rd 2009

2. Obtain an image of the calibration grating cross-section at a clean specimen area.

Fig. 3: Cross-section of calibration grating at 0.5 Hz

3. How well are the steps being imaged? What are possible reasons for any discrepancies? The steps are imaged fairly well at this frequency with slight differences (a few µm) in heights and widths due to creep in the PZT as it moves across the surface. The trace and retrace plots also show some discrepancies when scanning the middle portion of the grating. This discrepancy may be caused by hysteresis in the displacement-voltage loop. 4. Collect a 40x40 µm image at 2.0 Hz. Collect a height and a phase image. Set the z-scale to 500 nm. The image collected here is not as sharp as that collected at 0.5 Hz due to the faster scan rate which reduces the amount of time available for the cantilever to achieve steady state at any particular point on the surface.

Asha-Dee Celestine November 3rd 2009

Fig. 4: Height and Phase images for 40x40 µm image at 2.0 Hz

5. Select a 12x12 µm area that includes at least one whole trench and a step and save images at 0.5 Hz and 2.0 Hz.

Fig. 5: Height and Phase images of 12x12 µm image at 0.5 Hz

Asha-Dee Celestine November 3rd 2009

Fig. 6: Height and Phase images of 12x12 µm image at 2.0 Hz

6. Obtain cross-sections of the 12x12 µm image at both scan rates. Save screen shots. (a) (b)

Fig. 7: Cross-sections of 12x12 µm image at (a) 0.5 Hz and (b) 2.0 Hz

1. Zoom to the left of a vertical step and take a 1x1 µm image at 0.5 Hz. Use this image to determine the tip cone angle and its radius.

Asha-Dee Celestine November 3rd 2009

Fig. 8: Height and Phase images of 1x1 µm image at 0.5 Hz

Effect of integral gain on step imaging: 1. Obtain a single image of an 8 x 8 µm area at 0.5 Hz with the following integral gains: a) 0.3 b) 0.05 c) 0.1 d) 0.3 e) 0.5 f) 2.0

Fig. 9: Height and Phase images of 8x8 µm image at 0.5 Hz with varying integral gains

Asha-Dee Celestine November 3rd 2009 2. Obtain cross sections for each value of the integral gain. (a) Integral gain = 0.3 0.05

(b) Integral gain =

(c) Integral gain = 0.1

(d) Integral gain = 0.3

(e) Integral gain = 0.5

(f) Integral gain = 2.0

Fig. 10: Cross-sections of 8x8 µm image at 0.5 Hz for varying integral gains

Asha-Dee Celestine November 3rd 2009

Analysis 1. For a conical tip, provide an estimate for the cone angle and the tip radius.

Fig. 11: Section analysis screen shot from Nanoscope to determine cone angle

Asha-Dee Celestine November 3rd 2009

Fig. 12: Section analysis screen shot from Nanoscope to determine tip radius

Half cone angle of Tip Width Radius tip (φ)

Fig. 13: Measurements needed to determine cone angle and tip radius

From the analysis done in Nanoscope: i) 1-φ = 41.995 o => Half cone angle of tip, φ = 48.0 o ii) Horizontal distance between markers ≈ Tip radius = 15.625 nm

Asha-Dee Celestine November 3rd 2009 1. Compare the cross-sections obtained for the 40x40 µm images at 2.0 Hz and 0.5 Hz. What are the step heights and widths at different (vertical) locations of the step? Are they different and why? What is the effect of scan rate on the ability of the scanner to obtain accurate images? Explain in terms of the operation of non-contact AFM.

Fig. 14: Cross-section analysis of 40x40 µm image at 0.5 Hz

Asha-Dee Celestine November 3rd 2009

Fig. 15: Cross-section analysis of 40x40 µm image at 2.0 Hz

The width measured at the 0 nm mark for the images taken at 0.5 Hz and 2.0 Hz are the same at 5.156 µm. As the tip moves deeper into the trench, however, we see a substantial difference between the width obtained at 0.5 Hz, and that obtained at 2.0 Hz. The value obtained at 0.5 Hz (4.688 µm) is much closer to the actual width of 5 µm than that obtained at 2.0 Hz (3.594 µm). The higher scan rate prevents the tip from achieving steady state frequency as it images the trench and so before it has time to settle in and measure the full trench, it is being lifted out, thus registering a lower value for the width. 2. Discuss any imaging artifacts seen in the 40x40 µm images. In the images obtained at 2.0 Hz one can clearly observe artifacts from the conical tip in the distorted angle of the sidewalls of the vertical steps. This is not as pronounced at the lower sampling rate (0.5 Hz) because the PZT scanner has more time to achieve steady state. 3. How does the integral gain affect the quality and accuracy of imaging? Increasing the integral gain results in sharper, higher quality images of the surface since the rise time and steady state errors are reduced in the control system. 4. What is the effect of low integral gain? What is the effect of very high integral gain?

Asha-Dee Celestine November 3rd 2009 As can be seen from Figs. 9 & 10, when the integral is low, the image is not very sharp and the distortion is significant. When the gain is increased to a high value, however, the image is sharp but susceptible to noise (unstable). Also with high gains the overshoot value increases. 5. Use the 40x40 µm and 12x12 µm images taken at 0.5 Hz to calculate the relative error in the instrument calibration in the x and z directions. Which gain did you choose as your optimum gain? Why?

Fig. 16: Cross-section analysis of 40x40 µm image at 0.5 Hz

Asha-Dee Celestine November 3rd 2009

Fig. 17: Cross-section analysis of 12x12 µm image at 0.5 Hz

Relative error in x direction=|width of 12x12-width of 40x40|width of 12x12=4.734-4.6884.734 =0.97% Relative error in z direction= |height of 12x12-height of 40x40|height of 12x12 =|197.63-205.83|197.63 =4.15% The gain chosen is 0.5. This value gives a fairly sharp image which is not affected by noise. 6. Describe in detail (using schematics) the operation of NC-AFM. In this form of AFM, the cantilever is modulated very near the sample surface at its resonance frequency, ωo. As the tip approaches the sample surface the gradient of the force interactions between tip and surface changes and this results in a change in the resonance frequency of the cantilever. For non-contact AFM, this force gradient usually increases thus lowering the frequency of the cantilever. The change in resonance frequency is monitored by the controller which seeks to restore the original frequency (or amplitude) by adjusting the vertical position of the PZT scanner. As a result, the surface topography is imaged.

Laser diode

Photo-detector

Asha-Dee Celestine November 3rd 2009

PZT scanner AFM Cantilever Sample tip

Fig. 18: Schematic of Non Contact AFM