Question
1. One customer used TS2620N21E11+AD2S1210, and the resolver's parameter is below. if the resolver's input signal is 7Vrms, then SIN and COS input ofAD2S1210 will be bigger than the limitations. He want to know how construct the resistor divider to attenuate the amplitude.2. When the input signal of the resolver is 7Vpp, and he found his AD2S1210's DOS often alarm, and he used AU6802N1 before and haven't found similarproblem. He think it may be result from the high input impedance of AD2S1210.
Answer
Standard differential resistor divider – 3 resistors are needed. One in series connected to Sine input other one in series connected to SinLOterminal and 3rd resistor connected between Sine and SinLO inputs. Same configuration for Cosine channel.Attached is resistor divider structure which could be used in front of the AD2S1210 device. Please be aware that input amplitude shouldn’t be greaterthan 4 Vpp therefore to get 3.33 V on the input you can use exact the same 3 resistors. You can modify values slightly to get other voltages on theinputs but 3.3 V is ok. Please explain to the customers that precision resistors should be used to not introduce amplitude mismatch between thechannels. At least 1 % resistors are recommended.
For example 5 kOhm resistors Rx, Rx and Rz.
