Chapter4 ConnectingSignals
©NationalInstruments Corporation 4-13 NI6013/6014 User Manual
Youmust reference the source to AIGND. The easiest way is to connect
the positiveside of the signal to the positive input of the PGIA and connect
the negativeside of the signal to AIGND as well as to the negative input
oft he PGIA, without any resistors. This connection works well for
DC-coupled sourceswith low source impedance (less than 100 Ω).
However,for larger source impedances, this connection leaves the
differentialsignal path significantly off balance. Noise that couples
electrostatically onto the positiveline does not couple onto the negative
line,because itis connected to ground. Hence, this noise appears as a
differential-modesignal instead of a commo n-mode signal, and the PGIA
does not reject it.In this case, instead of directly connecting the negative
line to AIGND, connect it to AIGND through a resistor that is about
100 times the equivalentsource impedance. The resistor puts the signal
pathnearly in balance, so that about the same amount of noise couples onto
both connections, yielding better rejection of electrostatically-coupled
noise. Also, this configurationdoes not lo ad downthe source (other than
the veryhigh inp ut impedance of the PGIA).
Youcan fully balance the signal path by connecting another resistor of the
samevalue between the positive input and AIGND, as shown in Figure 4-5.
This fully balanced configurationoffers slightly better noise rejection
buthas the disadvantage of loading the source down with the series
combination (sum) of the two resistors. If, for example, the source
impedance is 2 kΩand each of the two resistors is 100 kΩ,the resistors
loaddown the source with 200 kΩand produce a –1% gain error.
Bothinputs of the PGIA require a DC path to ground in o rder for the PGIA
towork. If the source is AC coupled (capacitively coupled),t he PGIA needs
aresistor between the positiveinput and AIGND. If the source has low
impedance, choose a resistor that is large enough not to significantly load
thesource but small enough no tto producesignificant input offset voltage
asa result of input bias current (typically 100 kΩto1MΩ).In thiscase,
you can tiet he negativeinput directly to AIGND. If the source has high
output impedance, you should balance the signal path as previously
described using the same valueresistor on both the positive and negative
inputs. Youshould be aware that there is some gain error from loading
downthe source.