#### athman abed

Joined Nov 11, 2018
4
I want 0-16 kN to equal 0-5V (1024 on the analog input).The loadcell gives a value of 2 mV/V (on the full range), so with 5 V and 2000 kg (19620 kN), the output will be 10 mV.Which means that the input voltage to the instrumentation amplifier will be 1/1962 mV/N.Maximum input voltage will be 1/1962*16000(My desired max load) = 8.15 mV.8.15 mv should be amplified to 5 V, so the optimal gain is 5000/8.15 = 613.5.The Rg resistor on the AD623 determines the gain with the formula Rg = (100000/G-1) where G = desired gain.The optimal resistor is therefore 100000/613.5-1 = 163.2 Ohm

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#### ericgibbs

Joined Jan 29, 2010
12,573
hi athman,