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BME 552 HW2 2009 SOLUTION

# BME 552 HW2 2009 SOLUTION - BME 552 Homework Set 1 Assigned...

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Unformatted text preview: BME 552 Homework Set 1 Assigned Feb 2, 2009 Due Feb 9, 2009 1. Voltage response for current and voltage pulse. Plot the voltage response for a current pulse 1 ms, 100 uA. Assume Rp=1 MOhm. For now, okay to use monophasic pulse, positive current. a. Electrode radii of 50, 250, 500 microns b. Tissue resistivity of 50, 150, 300 Ohm‐cm (nine total plots for (a +b) combinations) c. It was shown in class how pseudo capacity effectively increases the Pt capacitance by 10X, show this effect. (pick 2 cases above to plot) d. Now use a voltage pulse of 3 V for 1 ms. Plot the current through the tissue versus time, repeating a,b,c variations. (11 total plots) e. Comment on i. Voltage response to this current pulse applied to the smallest electrode diameter ii. Current produced by constant voltage stimuli iii. Under which conditions above is a 1V water window limit maintained? Plots are below. Part d, equation is I(t)= V/(Rs+Rp(1‐e(‐t/RC)) Part e i. This results in a very high voltage, due to the small capacitance of this electrode, except when pseudocapacity is included, which decreases voltage significantly ii. There is in general a high peak (due to decreased impedance caused by short cirtcuit capacitor) followed by steady current, equal to V/(Rs+Rp). The graphs do not show the actual current peak, which will be V/Rs, due to my spreadsheet start V=3 at t=0.01 ms, but the general shape is shown. Pseudocapacity increase the capacitance and therefore more current must flow to charge the capacitor. iii. Only when the pseudo capacity is include, do any of my plots show adherence to the water window limit (see part c plots) 2. Read the Rose 1988 paper on iridium oxide (not platinum) to answer this question. Using the CV (from the paper) below, explain why an anodic bias allows the most charge injection. In terms of power usage, why is a bias inefficient? Biasing the electrode in the anodic region (positive voltage) allows the cathodic current pulse to use more of the charge storage capacity before violating the water window. Referring to the CV, if a 0.6V bias is used, then more of the hatched area (charge storage capacity) can be used when a cathodic pulse is applied. Anodic bias is inefficient because net current will always flow at the interface to try and return the electrode to its equilibrium potential, but this current happens to be low with iridium oxide. Plot for a, b ‐50 um, 50 Ohm‐cm Plot for a, b 50 um, 150 ohm‐cm Plot for a,b 50 um, 300 ohm‐cm Plot for a,b 250 um, 50 ohm‐cm Plot for a,b 250 um, 150 ohm‐cm Plot for a,b 250 um, 300 ohm‐cm Plot for a,b 500 um, 50 ohm‐cm Plot for a,b 500 um, 150 ohm‐cm Plot for a,b 500 um, 300 Ohm‐cm Part 1 C I picked 250 um, 300 Ohm‐cm (left below) and 50 um, 300 Ohm‐cm (right below) Part 1D I(t)=V/(Rs+Rp(1‐e(‐t/RpC))), plot this equation 50 um, 50 Ohm‐cm 50 um, 150 Ohm‐cm 50 um, 300 Ohm‐cm 250 um, 50 Ohm‐cm 250 um, 150 Ohm‐cm 250 um, 300 Ohm‐cm 500 um, 50 Ohm‐cm 500 um, 150 Ohm‐cm 500 um, 300 Ohm‐cm With pseudocapacity, I picked 250 um, 300 Ohm‐cm (left below) and 50 um, 300 Ohm‐cm (right below) ...
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