0
\$\begingroup\$

Below shows a piezoelectric sensor connected to a charge amplifier:

enter image description here

The text claims that the output voltage Uout is to a good approximation only charge q which is going to be measured divided by the feedback capacitor value Cf.

The text then says the amplification is independent of the capacitance of the sensor and the cable without explaining why.

I dont understand why Cg and Cc have no effect on the output. (?)

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

When an op-amp is operating closed-loop with negative feedback within its linear region of operation (all input/output specifications met). The two input terminals will have an equal voltage due the feedback action of amplifier A. This action creates a virtual ground at the opposing input terminal of the amplifier.

With no potential being developed across the cable's terminals, no current flows through \$C_g\$ or \$C_c\$. All the signal current flows through the feedback network of amplifier A.

\$\endgroup\$
7
  • \$\begingroup\$ That’s with infinite amplifier gain. With finite gain, input capacitance will affect your frequency response and stability. \$\endgroup\$
    – DavidG25
    Commented Oct 17, 2018 at 18:29
  • \$\begingroup\$ But in the last page of this document endevco.com/news/newsletters/2015_07/tp320.pdf it says about PE type in limitations section “Capacitive loading from long cable run” Im confused I thought PE has no issue with the capacitance. What do you guys think? \$\endgroup\$
    – cm64
    Commented Oct 17, 2018 at 18:33
  • \$\begingroup\$ On the other hand for one of the advantages of IEPE types it says “Drives long cables” :( \$\endgroup\$
    – cm64
    Commented Oct 17, 2018 at 18:36
  • \$\begingroup\$ @cm64 You need to compare actual specifications of charge amplifier versus IEPE \$\endgroup\$
    – sstobbe
    Commented Oct 17, 2018 at 18:46
  • \$\begingroup\$ Do they mean long cable is problematic for PE because it picks up more interference due to being high impedance? \$\endgroup\$
    – cm64
    Commented Oct 17, 2018 at 18:47
0
\$\begingroup\$

Because 1) the voltage across a capacitor is equal to the charge divided by the capacitance, and 2) in the configuration shown, unless the amplifier is saturated the difference between the two inputs will be very small, since the gain is large, and 3) the +input (the lower input) is tied to ground.

So the - input (the upper) will be at approximately zero volts and the output voltage will simply be the capacitor voltage, which follows the relationship shown.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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