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Explanation of the Analog Input Circuitry on the 6008/6009



Hardware: Multifunction DAQ (MIO)>>Portable>>USB-6008, Multifunction DAQ (MIO)>>Portable>>USB-6009

Problem:
In the USB-6008/6009 user manual, I noticed a figure that describes the analog input circuitry. I am confused by the picture. I see a voltage of approximately 1.4 V on my test panels if I have no signal connected to the input, but the voltage reading is correct when I connect a constant voltage source. How does this circuit work?

Solution:


Figure 1: USB-6008/USB-6009 Input Block Diagram

The resistor network scales the +/-10 V signal to +/- 1.2 V and level shifts it by 1.23 V. The purpose of this network is to fit the input signal into the range of the analog to digital converter (ADC), which is 0 to 2.5 V. The circuit is included in the manual to make the input design clear since it is very different from the input circuitry of a typical DAQ board. If no signal is connected, the voltage divider across the resistors outputs a value of approximately 1.4 V, which we can see on a MAX test panel.

Related Links:
Product Manuals: USB-6008/6009 User Guide and Specifications
KnowledgeBase 3RH97BRM: USB-6008/6009 Analog Input Reads Approximately -1.4V Offset with RSE Connections

Attachments:


6009_input.GIF


Report Date: 10/27/2005
Last Updated: 02/26/2010
Document ID: 3QQBGL0Q

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