Here it comes.... a crash course in Analogue to Digital Convertors...
The signal needs to be digitised into 0 and 1's for the computer.
The device that does this is an A/D (Analogue to Digital) Convertor
These are expensive to make and they have a range of numbers over which they can work.
This range is expressed in bits (6009 = 14 bits)
In our case 14 bits = 16384 combinations or increments
The range 16384 has to be divided into the maximum voltage input of the A/D
To get the resolution divide the maximum input voltage range by the number of bits
The above means that if you were to have a fixed input voltage range then the step size would be to big for some measurements or the input range too small for others. What you do is to scale the input capacity by ranges..
In the case of the 6009 there are 8 input ranges available from ±20.0 to ±1.0
These ranges allow the best resolution over the users input voltage range.
Following on a bit if you select the 1 v input range and input 1.1v to the A/D, the counts on the A/D have reached the maximum number and won't go further.
So in summary, depending on how you configure the 6009 you can have an input greater than the scaled range and thus 'saturate the input', you need to make sure that the input range is suitable for your intended configuration.
As far as the impedances are concerned it should be fine and with the ranges available on the 6009 you should not need scaling resistors.
P.S for an Engineering student, if you understand differential mode your doing well!
Message Edité par Conseils le 05-20-2006 10:08 AM