0
\$\begingroup\$

A capacitively based MEMS affinity glucose sensor.

image of the sensor

link to the IEEE paper, from where I've got this doubt.

The gold electrodes are separated with some distance and the permalloy is vibrating with the help of solenoid(it generates a time-varying magnetic field[solenoid is not shown in the image]). Glucose comes inside through the semipermeable membrane and makes the liquid in which the cantilever is there more viscous which reduces the vibration in the cantilever. We measure the capacitance created by the gold electrode.

Question: How does the capacitance changes when the vibration of the permalloy slows down?

\$\endgroup\$
  • \$\begingroup\$ @brhans I suspect the solenoid driving frequency is fixed, the most you can do is dampen the vibration, thus reduce the amplitude, not the frequency. \$\endgroup\$ – crasic Aug 20 '18 at 21:52
  • \$\begingroup\$ @crasic I think you're right about the fixed frequency of the cantilever as it is driven by a fixed frequency of the solenoid. \$\endgroup\$ – Arshdeep Singh Aug 20 '18 at 21:57
2
\$\begingroup\$

Increased viscostiy reduces the amplitude. Not the frequency. (Doesn't slow down.)

For vibration, the capacitance between electrodes changes with sinusoidal time-dependence. Probably the average capacitance doesn't vary. But the peaks in capacitance will be much smaller as viscosity increases.

\$\endgroup\$
  • \$\begingroup\$ I think it reduces the amplitude but also the frequency because it'll become difficult for the cantilever to vibrate and the rate of vibration(frequency) decreases! is it? and also according to the guy who commented on my question we measure frequency. \$\endgroup\$ – Arshdeep Singh Aug 20 '18 at 21:49
  • \$\begingroup\$ Without knowing the details of the solenoid, there is no reason to expect the frequency to change if the solenoid driving frequency is fixed. \$\endgroup\$ – crasic Aug 20 '18 at 21:52
  • \$\begingroup\$ @ArshdeepSingh no, it's pretty unlikely that the frequency would change, unless it was close to some resonant frequency of the system (even then, probably not) \$\endgroup\$ – BeB00 Aug 21 '18 at 0:49
-1
\$\begingroup\$

formula for capacitance

Maybe this formula helps. I dont know exactly if it does but Capaitance change by the factors on the right hand side..

\$\endgroup\$
  • \$\begingroup\$ The distance between the plates varies with the phase of the vibration so the capacitance signal is also varying, the amplitude of the distance determines the viscosity. Thus it is using the amplitude of the change in capacitance to determine viscosity. \$\endgroup\$ – crasic Aug 20 '18 at 21:49
  • \$\begingroup\$ Consider reading the OP's question entirely before coming up with an answer :) No reason to read the title like.... "How do we measure capacitance BOOM! \$ \displaystyle C=\frac{\epsilon_0 A}{d}\$.... I got this...." Also, when answering questions, try to be as confident as possible. If you know the answer, stick by it. Don't second guess yourself by saying, "maybe" and "I don't know". If you don't know the answer, try not to answer. Instead learn what other people have to say. \$\endgroup\$ – KingDuken Aug 20 '18 at 21:55
  • \$\begingroup\$ That being said, I believe you might know what the OP is asking. Distance is certainly important for capacitance, you're right about that. However, this isn't a steady state calculation. \$\endgroup\$ – KingDuken Aug 20 '18 at 22:01
  • \$\begingroup\$ It might not be a steady state calculation but this equation tells the spontaneous change in the capacitance. Watever the steady state calculation might be the capacitance will remain inversely propotional to the distance between the plates thats what I was actually sharing.I believe you people focus on down rating more than information sharing.... \$\endgroup\$ – Sohaib Imran Bhatti Sep 2 '18 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.