thats helpful but I'm not an opamp wizkid so lets see if I am getting this right.
Your suggesting i use a summing amplifier and make setpoint and feedback voltage "directions" opposite so that the system naturally tends to nullify itself and so stop the motor where it is needed. so far so good, what exactly is the capacitor in the feedback loop for?, my understanding of integrators is that they "add over time" the input voltage, or will this do the opposite ? subtract from the top down so that the drive reduces to stop overstoot ?
I will have only a fixed 12V supply so a +/- supply will just mean adding extra circuitry, would i be safe with a virtual ground and +/-6V ?
maybe time to get the breadboard out
The feedback cap _is_ the integrator. You know from basic circuit theory, that if you drive a current into a cap, the voltage of the cap rises. That is the definition of capacitance: C = As/V of ampere-seconds per volt. So a whopping 1 farad cap would show a 1V increase in voltage for every second you drive a 1 amp current into it.
Google for inverting amplifier and you will get explanation for the way this works. But the idea is simple: assuming a symmetrical bipolar case where the non-inverting input is grounded , an inverting amp will create a virtual ground on the inverting input. The output will act so as to drive the inverting input to match the non-inverting one. Think of it this way: the inverting input will stay at the same voltage as the non-inverting one. If you present a voltage via a resistor to the inverting input, a current will flow since the other end of the resistor is at "ground" potential (the virtual ground). That current has to go somewhere and it isn't going to go into the amp input, as these are real high impedance. But there is the feedback path from virtual ground to amp output. The current can go there. The amplifier drives the output to sink or source this current as needed, and the resulting output voltage is then subject to the impedance of the feedback path. A simple example: +/- supply, amp positive input is grounded. input resistor is 1 kohm and feedback resistor is 10 kohm. You feed in + 1V via the input resistor. Now a 1 mA current starts flowing into the virtual ground and the amp inputs are no longer in balance. This imbalance drives the output to the negative direction (your positive signal went into the negative input). Now the input current continues from virtual ground to amp output through the 10kohm feedback resistor. To achieve balance again, and recalling Ohm's law U=RI, the output must swing all the way to -1mA*10kohm = -10 volts.
If you replace the resistor with a cap, the difference is that instead of U=RI, you get U=It/C. Now you have the quantity of time as part of the output voltage equation and that is a sign of either integrating or differentiating action. Using the same numbers as before, a 1mA current into the virtual ground, and a 100uF cap from virtual ground to output, the amp output would slew 10 V/s. This is exactly what an integrator does.
I can draw you a simple schematic in a minute, to illustrate better.
Edit: schematic attached.
This is the general idea. Can be implemented by any reasonable quad op amp, such as the ubiquitous LM324 - dirt cheap. No component values indicated, that will be your homework. If you do this, then make sure e.g. trimpot R6 is good quality cermet or similar, that _won't_ break its slide connection after a few years. Or else.
As this is NOT a PID system, with human input that's not being tracked real time.
It is not clear which of the 2 systems you are talking about. The flap control can be just a simple positioning loop with target window control. The valve controller on the other hand is (a PID loop), although there is hardly need for the D term. The fact that there is human input has nothing to do with the nature of the controller, that is entirely a different choice and issue.
Just a "P".
Again, sorry no. With P term only, you always get a static error. Whether that matters or not, is another thing but a valve, especially a sticky one, presents a relatively big static friction load rendering pure P law control inefficient. If you tune the P term high enough to break the friction in a bad case, you will get an oscillator. Sometimes an integrator is a bit difficult to tune to avoid oscillation, and in that case a measured dose of setpoint feedforward usually does the trick.
But the point is that the process determines what control terms are desirable, not the nature of the entity that controls the setpoint.
As i said somewhere we have the motor drive with built in position pot, and we use water valves using a similar system. I just need to control it. I have no idea how many are required.
My reply was targeted to this alternative.