neophyte question: How to measure "open circuit" voltage

[jamin100]

Hi,

I'm trying to measure the voltage induced by a peizoelectric cable (applying pressure to cable creates voltage/current on cable).  But when I measure the voltage  accross the cable I get a "background" signal that looks like an exponentially decaying voltage, like a large capicitance discharging. I can measure the real voltage across the cable, it appears on top of the decaying voltage, but soon the voltage reaches what I assume is the max low value for my board (BNC 2120, floating point measurement), -10 volts. I can get the voltage to go back to zero by touching the leads but this just sets the process in motion again.  This decaying voltage characteristic is also present when nothing is connected to the leads, which makes sense, hence my question about measuring open circuits. Needless to say I would like the voltage of an open circuit, or accross my cable to be 0. How can I make this happen?

Thanks in Advance.

[WM_John_Weeks]

I haven't heard of a piezzoelectric cable, but I used to use piezzoelectric transducers. On the chance that a piezzoelectric cable works like a piezzoelectric transducers, I offer this:


Since piezzoelectric transducers are insulating ceramics, they provide no path for the bias currents flowing into the DAQ device inputs. You will need to use a bias resistor to ground if you are using differential input mode.

Further, the signal that a piezzoelectric transducer generates is really an amount of charge, the voltage is dependent on the capacitance they work into, and the leakage resistance of the entire system.

Really you should be using a charge amplifier and feeding the output of the amplifier into your DAQ device.

Here are a couple of links:

http://digital.ni.com/public.nsf/allkb/ADACA8045C6ACD0A86256E1B0072C9B2 "Why Am I Reading a Floating Voltage for My Piezoelectric?"

http://zone.ni.com/devzone/conceptd.nsf/webmain/01f147e156a1be15862568650057df15?OpenDocument#3 "Field Wiring and Noise Considerations for Analog Signals"


Hope this helps. Hope I'm not off-base on my interpretation of your problem.