Internal circuitry voltage of AD8220

Question

How to calculate V- and V+ voltage (see figure below) for Instrumentation amplifier AD8220? 
 
 I try to calculate V- and V+ voltage with the following equation (V -NI =0.5V, V +NI =1V, R G =49.4k): 
 G=1+(2*R F )/R G =2 
 V CM =(V +IN +V -IN )/2=0.75V 
 V DIFF =V +IN -V -IN =0.5V 
 V + =V CM +1/2*G*V DIFF =1.25V 
 V - =V CM -1/2*G*V DIFF =0.25V 
 Simulation (with perfect OPAMPS) gives the same results 
 
 But the diamond plot tool ( LINK ) gives the following result 
 
 Whit equation for V- and V+ voltage it would be much easier to calculate resistors for input voltage divider as shown in figure below.

GLADION · Accepted Answer

Hi Silverbeard , 
 
 Good day. I agree with your calculations and derivations of the internal nodes. However, the diamond plot tool considers the internal circuitry of AD8220. 
 
 Considering the schematic diagram, both the inputs of the op-amp A1 and A2 are tied to a voltage Vb proportional to the -Vs. The value of this voltage depends on the design of the part itself and differs per generic. For AD8220, that is 1.8 V. This voltage shifts your internal node to (0.25+1.8 = 2.05 V and 1.25+1.8 = 3.05 V). You may use the diamond plot tool to know this value, you have to set the Vcm and Vdiff to 0 V to determine this value and consider it in your calculation. Hope this helps. 
 Regards, Gilbeys)

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Internal circuitry voltage of AD8220

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