I'm working on a ph probe, following this gentleman's guide.
I'm having trouble understanding the op amp voltage shift and hope someone can help.
The first part (top circuit) has a gain of 5.6 - I think I get this so far.
The second part should shift the output by +2.5v - because the ph probe will deliver a positive or negative charge depending on the ph. I'm having trouble understanding this part.
It seems to me that if the probe was reading -0.4v, the first part would provide -0.4 * 5.6 = -2.24v at aout. The voltage divider gives +1.25v on pin 5. The gain on the second part is 2.
So wouldn't we end up with 2 * (1.25 - 2.24) = 6.98v at pin 7?
First stage gain 5.7 = (4.7 + 1.0) / 1.0
The gain of the second stage is -1, not 2. You'll get -0.4V in going to -2.28V,
and the second stage flips that about the 1.2V reference to 4.68V
Aahh that's interesting MartT. But first thing is I don't understand why the gain of the second stage is -1. Isn't the gain G2=1+(R4/R3)=2?
Second thing is the objective of the circuit is to convert -0.4v to +0.4v from the probe to 0v to 5v for consumption by the arduino adc. So 4.68v from -0.4v wouldn't be a good result. (should have explained this in the first post sorry).
The gain of the second circuit is -1 because the signal is going into the -ve input and so that first formular does not apply. It is an inverting amplifier and the gain is simply the ratio of the two resistors.
There is no need to have two circuits for your application you can do it with just the one circuit if you like. What is the exact circuit responce you require?
This circuit, with a TL072, needs at least a +7volt -5volt supply.
Two 9volt batteries will last a long time with this chip....
The first opamp is easy to understand. Just a non-inverting 5.7x amp.
The diagram shows a polarised cap. Can't do that since reversed polarity can be expected here.
I would use a 1uF MKT cap.
The second opamp has a bias voltage on the +input.
The negative input wants that too.
Since R3/Aout is normally ground level (Ph7), the only way to get that voltage is from R4.
The output will be twice the bias voltage with a Ph of 7 (ratio of R3 and R4).
Same story for C2. 1uF MKT.
R5 just protects the Arduino pin.
Leo..
Thanks very much for the replies. This is such a great forum and I really appreciate it.
What I was expecting was to see the ph probe output of -0.4v to +0.4v scaled and offset to 0v to 5v for consumption by the arduino.
I still don't really get the values that the author has used, but what through me was that the output is inverted. I was looking at the result for -0.4v and could make sense of it.
The voltage divider should produce 1.25volt (R6:R7 = 3:1) for 2.5volt center voltage (sensor = ovolt).
With the original values of 24k/76k it's 1.2volt.
Maybe the author didn't want Ph7 in the center.
The gain (R2) is indeed a bit too high for +0.4/-0.4 input voltage.
Leo..
