How does the DERIVATIVE term slow down the signal?

Question

I look at PID controler and do not understand derivatve term. I know it's used in order to not let the signal to overshoot the set value. BUT when the sigal is arising, its derivative is positive. So when we add the positive derivative term, the signal gets even higher, so its derivative becomes even higher! I think that if the signal is rising too fast it should be slowed down by derivative term, which means that derivative term should produce NEGATIVE value to DECREASE the signal that controlls the object. How does it work in reality? Is there any good simulation online of how PID works? I've seen some simulations, but they weren't very informative.

One more question: I want to check if I understand the integral term properly. Let's assume our PV is arising and getting closer and closer to the SP. During that time integral part of the PID output is GROWING all the time, until the PV reaches SP, than the integral part has already grown so much that the PV overshoots the SP. So PV rises above SP and the integral part becomes smaller and smaller (but still bigger than zero), and finally it becomes so small, that it gets negative smaller than zero). That casues the PV to start decreasing. It gets smaller and smaller and again, integral part is so big (on the minus part this time) that the PV goes down too much and gets smaller than the SP. And that's why integral part causes oscillations. Was my example right? Is that how it really works?

Answer 1

$$ e = r - y$$

• $e \rightarrow$ error
• $r \rightarrow$ reference input
• $y \rightarrow$ actual output of the plant

If $e$ is increasing ($\left[\frac{d e}{d t}\right]>0$) this means that $y$ is becoming smaller and smaller compared to $r$. So the input needs to be increased to increase $y$.

I think your analysis of the integral term is right.