Question

Atomic Force Microscopes (AFMs) are instruments that measure the topography of samples with nanometer resolution. A tip mounted at the end of a cantilever scans over the sample: bumps and valleys are detected as changes in photocurrent in a balanced photodetector as a laser beam is reflected from the cantilever onto the photodetector itself. When the cantilever is deflected up due to a bump, more photocurrent is generated in the A-half photodetector, while when it is pulled down in the presence of a valley, more photocurrent is generated in the B-half photodetector. You have just found an old custom-built AFM in the lab you work in. This was rigged up by a previous technician. Your supervisor wants you to make it work again. The old technician left a calibration sample and a detailed sketch of the wiring she made for the electronic parts of the instrument (Figure 2b). In her notes, she left a comment that she did not have many resistors, so she had to mix and match what she had available, but she noted the name and value of each resistor). RA = 1 kΩ, R1A = 50Ω, R2A = 100Ω, RB = 2 kΩ, R18 = 1 kΩ, R2b = 3 kΩ, R4 = 1.2 kΩ, Rs = 200Ω, R6 = 600Ω, R = 400Ω.

a) Find the values of the currents I4 and I8. To do that, you can use the calibration sample. The calibration sample has a bump of unknown height, which you know produces 3 times as much current in the detector A than in B (i.e., I4 = 3I8) with a corresponding positive output signal Vo = 50 V.

Answer 1

To find the values of the currents I4 and I8, we can use the given information that I4 is three times greater than I8. Using Kirchhoff's junction rule, we set up an equation and solve for the values of I4 and I8. The calculated values are 1.66667 x 10^-4 A for I4 and 5.55555 x 10^-5 A for I8.

Step-by-step explanation:

To find the values of the currents I4 and I8, we can use the given information that I4 is three times greater than I8. Let's assume I8 = x, then I4 = 3x. According to Kirchhoff's junction rule, the current entering a junction is equal to the sum of the currents leaving the junction. In this case, the junction is between R18 and R4. So, I4 + I8 = I18 + I6.

Using the resistor values given, we can write the equation as 3x + x = V/R18 + V/R6. Simplifying further, we get 4x = V/R18 + V/R6. Rearranging the equation, we have 4x = V(1/R18 + 1/R6).

Substituting the given values for V and the resistors, we get 4x = (50 V)(1/1000 Ω + 1/600 Ω). Solving for x, we find that x = 5.55555 x 10^-5 A. Therefore, I4 = 3x = 1.66667 x 10^-4 A and I8 = x = 5.55555 x 10^-5 A.