Other Parts Discussed in Thread: AFE4403
Hi. I am designing boards which contain MSP, AFE4403 and so on.
Anyway, I want to make an "Active Optodes" which has the similar meaning with active electrode in bio-potential field.
To operate properly with the AFE4403 board, the output signal of the board should be current.
And I found the post
I have several questions in this post.
1. I think it will act like a differentiator. Because of the serial capacitor, only the deviation of the phototransistor output voltage will make current through AFE4403 inputs. Isn't it? I=Cdv/dt.
2. How strong is the input (INN, INP) biasing current? My "active optodes" output signal will be transimpedance (amplifier + resistor feedback) voltage output. To change the voltage to current, I added a resistor between the "active optodes" output and AFE4403 input (*instead of capacitor which can be seen in the above-mentioned post). In the datasheet, the input bias will be set to 0.9V. But when the active optodes transimpedance voltage output increases, input bias does not stay at 0.9V. I think the input impedance of the AFE4403 is not that low. And it seems that the input bias current increases when the gain of the AFE4403 TIA decreases. In case of 500kohm, the 200nA current can move the input bias point.
3. Any other methods to make an "active optodes"? As you see, the signal flow is like this.
TIA in the active optodes (Current -> Voltage)
Resistor (Voltage -> Current)
TIA in AFE4403 (Current -> Voltage)
Is there other option? For example, current-current amplifier.
ps. I'm measuring the near-infrared spectroscopy signal on the scalp which has input current 1nA ~ 20nA.
