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I'm using a DAQ from National Instruments. Testing my LabView software I noticed that when I plug a 1.5V DC battery in the analog input channels if I connect an RC filter (R=56 kΩ, C=10 uF) to the batter then my DAQ instead of measuring the expected voltage it measures a lower value, 1.22V, as if there was a load of about 300 kΩ. The declared imput impedance is instead 144 kΩ and this happens only if I measure the voltage in differential mode, if I measure it in RSE (Reference Single Ended) that doesn't happen (i.e. I read 1.5V). Can you explain why?

Furthermore: in the lower schematic, what is the purpose of the +2.5 Vref and of the resistors?

usb-6009

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  • \$\begingroup\$ Have you measured the input impedance of your DAQ? Perhaps it's 300 kΩ. \$\endgroup\$ – Harry Svensson Nov 17 '17 at 12:08
  • \$\begingroup\$ The declared imput impedance is 144 kΩ, but if I look to the analog input circuitry (the pic above) I see something different. Furthermore why should this phenomenon be present only in differential mode? \$\endgroup\$ – Alessandro Zunino Nov 17 '17 at 12:10
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The input impedance is indeed 144 kΩ:

\begin{equation} R_{in} = 127 k\Omega + (30.9 k\Omega || 39.2 k\Omega) = 127k\Omega + \frac{(30.9)(39.2)}{30.9 + 39.2} k\Omega \approx 144.28k\Omega \end{equation}

Hence, in steady-state (capacitor is an open circuit), your RC filter creates a voltage divider with the input impedance of the analog front end. If you are measuring in differential mode, the input impedance is doubled, at approximately 289 kΩ, which explains your reading of 1.22V (battery voltage was about 1.46V).

The 30.9kΩ and 39.2kΩ create a voltage divider with the 2.5V reference, effectively biasing the measurement to 1.398V (for use with a single supply instead of traditional bipolar supplies).

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