(Image 1 is for the general idea, image2 is for questions 1 and 2, image 3 is for the 3rd question)

In the setup shown below, the tuning fork is driven by an AC voltage source being connected to one prong and its current output is measured by a current amplifier connected to the other prong.

I looked up the specifications of a current amplifier such as this. It says "input impedance (DC, Z // 15 pF) 50 Ω - 70 kΩ."

So my question is the following,

  1. If the input impedance of the current amplifier is low enough, can we basically treat it like GND and just not connect the ground side of the AC voltage? This would allow me to connect one prong of the fork to the AC voltage source via one cable

  2. I am not sure how to read the spec of the input impedance. Is it saying 50 Ω for DC and 70kΩ under different conditions?

  3. This diagram shows the tuning fork electrically excited with each of the two wires connected to either of the prongs. (one wire, one prong) Then, there are cables from preamplifier to the tuning fork (again one wire, one prong). I am not sure how this configuration would work. Wouldn't current be split and hence the output of the current amplifier will not give you the accurate reading?

Image1 Image2 Image3

  • \$\begingroup\$ Do you just want to understand why/how it works? or what? ..... 50 Ohm driver and 75k||15pF preamp. Is this for Non-contact-atomic force microscopy (ncAFM) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 10 '18 at 21:23
  • \$\begingroup\$ @TonyEErocketscientist I apologize for laying out my questions in a confusing way. Question 1. In the second image, they are basically treating V- of the preamp as GND, which allows them to apply AC voltage across the tuning fork by connecting one prong to the voltage source using one cable. How valid is this? (I understand V+ and V- are basically at the same voltage but V+ is grounded whereas V- is not) Is this acceptable? 2. when you are given something such as "input impedance (DC, Z // 15 pF) : 50 Ω - 70 kΩ," how would you make sense of 50 Ω - 70 kΩ? \$\endgroup\$ – Blackwidow Aug 10 '18 at 21:31
  • \$\begingroup\$ 50 out 70k in. .... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 10 '18 at 21:37
  • \$\begingroup\$ @TonyEErocketscientist Is it 50Ω for output impedance because of impedance matching? \$\endgroup\$ – Blackwidow Aug 10 '18 at 21:43
  • \$\begingroup\$ Yes, so changing ground paths could alter that \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 10 '18 at 21:47

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