Comments (5)
Hi Il Taek,
I haven't changed the range because I haven't needed to yet. I added the function over winter break as an exercise for myself in bit masking. I currently leave it in Range 1 as that's the default and it has the lowest output impedance, but I don't know how much that changes things since we're using the AD8606 as a buffer.
With your suggestion to using 2Vpp or 0.2Vpp for the potentiometer, could you explain more? I'm not very familiar with digital potentiometers so I don't know the effect choosing a particular range for a certain vpp value will have.
Thanks.
from drive.
Thanks.
- The function was good for practice.
- Yes, the AD8606 OP-AMP circuit reduces the output impedance.
- Generally, digital potentiometer endures the 5mA output current as a maximum. (You have to look up the accurate value, and it depends on each chip.) Our first design uses 2Vpp pulse with 5.8mA current, which exceeds the maximum current. If we use 0.2Vpp pulse with 0.15mA current, we can use AD5933 with the potentiometer. (The amount of current might differ because the spec sheet based on when the supply voltage is 3.3V.)
- Besides, we might need to test how the result changes if we change the "range."
from drive.
Oh I get what you mean now, I'll test the range function when I get in tomorrow.
from drive.
Based on my recent test (with the circuit with MUX,) It works pretty well except more accumulating error. In other words, the error increases faster when the resistance (impedance) is farther from the calibration resistor.
Based on the calculation of AN-1252, the current 332 Ohm feedback resistor can accommodate the resistance from 378 Ohm. In other words, the minimum limit is 378 Ohm. I assume that the increased error stems from this fact.
((3.3/2-0.2)*378/(0.173+3.3/2-0.173))
from drive.
I see, when calibrating I'll use the maximum current with the lowest output impedance I can get away with. Thanks for the heads up!
from drive.
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from drive.