I would like to build the potentiostatic circuit in Application Note AN-1798 (Figure 2):
However, I cannot determine what parts to use for Q1 and R6.
Electronic – Potentiostat circuit components
operational-amplifier
Related Solutions
You have a serious input and output voltage range problem : the input signal must stay within input voltage range (V- to V+ -2V) and the expected output voltage must stay within output voltage range (V- to V+ - 3.5V). Here is what I suggest :
simulate this circuit – Schematic created using CircuitLab
This way :
- The input voltage swings around ground, and stays far from the power rails.
- The output can swing from -9V to about +5V
As Kaz noted, the output drive of LM358 is probably not strong enough for headphones ; you may want to add an output booster stage between the LM358 out and the pot.
If you want to stay with single battery, you will need to create a ground at an intermediate value (I'd suggest around 3V since that op-amp output swing is limited from V- to about 3.5V below V+). For that you can replace the two batteries with a single and create the ground from a resistive divider and an op-amp (the second half of your lm358) as buffer :
It's not in either of those modes, but it's being used as a differential amplifier:
It doesn't amplify a single signal, either inverting or non-inverting - it amplifies the difference between two signals - in this case the difference in voltage between the two sides of R8.
Best Answer
R6: value is not very influential, as the signal from the sensor WE terminal is expected to produce a particular current, with the op amp circuit acting to translate that current to a voltage. The op amp will adjust its output to a voltage that causes the same current to flow in R4, with the feedback action setting the op amp neg input to 0V. So output voltage will be WE current x R4 value (10k). So actually no R6 is needed, but it's a good idea to have some resistance there which would limit current if the sensor wire for some reason contacted an untoward voltage. 1k or 10k should be fine.
The P-channel JFET is there only to create a short circuit across sensor RE and WE terminals when the circuit is powered off, as explained in the application note. When powered on, the +V applied to the JFET input should switch it off, making it effectively disappear from the circuit. So we want a JFET which switches on adequately with VGS = 0, and switches off adequately with VGS=12.
The JFET off characteristics are the primary concern, as leakage between D and S, or G and S would add to or subtract from the sensor current. The degree of concern depends on the current expected from the sensor, which presumably varies by application. The article discusses a 500 nA signal, in which case leakages of 5 nA constitute 1%, and 50 nA is 10%.
FWIW, the P-channel JFET, being normally on, is not nearly as popular as N-channel, so there is some lack of variety readily available. Looking at Jameco, the 2N5460 looks like it might be good enough for < $1. However, given the concern in the article for low-leakage ("bias current") op amps, the same concern may well apply to the JFET, so you might look for a lower-leakage JFET -- with the 2N5115 popping up as one that has significantly lower leakages (VGS-OFF 3-6V at 1nA IDS, IGSS < 500pA. No doubt there are others.
Hope that helps.