Ttelmah Guest
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Posted: Wed Aug 01, 2007 9:56 am |
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Just divide it by at least two. How best to divide it, will depend on the source impedance. If the impedance is low, a simple resistive potential divider. If the impedance is higher, then an op-amp divider will be needed. Generally, assuming this is something reasonably 'high accuracy', you will presumably use an external reference voltage, and depending on the real output range (some '0-10v' sensors, will output a higher voltage than 10v, as a fault/error signal), you should consider this, when calculating the division to be used.
Best Wishes |
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