What's internal resistor value(R1 & R2 & R3) of the AD8479?
We want to know the internal resistor value of the AD8479.
Because need to calculate the Vx and Vy.
This design is single supply so need your support.
Your Vx and Vy should be 1V and 14V instead of 1V to 9V in computing the maximum Vcm since, its value can swing to within 1V of your supply(15-1 and 0+1). Therefore your Vcm+ = (60x14)-(59x2.5) and Vcm- = (60x1)-(59x2.5) is +692.5V to -87.5V respectively.
If you want your Vcm range to be +400 to -400 you need to adjust both your Vs and Vref, since your Vs is only at 15V, and setting the Vref at midsupply will only give you +397.5 to -382.5. For your application use Vs = 16V, Vref = 8V. Your Vcm will be +428V to -412V.Regards,Goz
Actually you do not need the resistor values on the diff amp to know the values of Vx and Vy. The equation for the common-mode voltage is already given in the datasheet, which also you have attached on your question. Vx and Vy can be any value 1V of the rail.For example in the datasheet your supply is 10V and Vref is 5V. Vx and Vy can be 1V to 9V. Substituting their values on the equation Vcm+ = (60x9)-(59x5) and Vcm- = (60x1)-(59x5) you will get a maximum allowable common-mode voltage rang of +245V to -235V respectivelyMay I know .what your set-up conditions are so we can confirm your Vcm?Regards,Goz
Thanks for your reply.
Our set-up conditions are Vs: 15V and Vref: 2.5V. That we want to do common-mode voltage rang of +400V to -400V
So our supply is 15V and Vref is 2.5V. Vx and Vy can be 1V to 9V. Substituting their values on the equation Vcm+ = (60x9)-(59x2.55) and Vcm- = (60x1)-(59x2.5) we will get a maximum allowable common-mode voltage rang of +392V to -87.5V respectively. Right? So we need increase voltage of the reference?
Retrieving data ...