THE PID CONTROLLER — Part 1 •••••
For some types of control, this is acceptable; for
others, it is not. You wouldn’t want this type of control for
a servo motor — bad things would happen! Just imagine
— the motor would be full power in one direction and, the
next moment, full power in the other direction. You can
see where the term bang-bang comes from. That servo
won’t last long!
The PID controller takes control systems to the next
level. It can provide a controlled — almost intelligent —
drive for systems. We will now examine the individual
components of the PID system. This step is necessary to
understand the entire PID system. Please don’t skip this
section; you must know how the individual components
function to understand the whole system.
components of both
time and knowledge.
Obviously, we all
started as babies
with virtually no
time, we have integrated knowledge
into our brains.
In our PID con-
troller, we are integrating voltage as time progresses. A
schematic of an integrator circuit is shown in Figure 2.
The output voltage is described mathematically by the
Vout = -(1/RC) * (area under curve) + initial charge on
What Is Proportional?
This one is easy. The proportional component is
simply gain. We can use an inverting op-amp, as shown
in Figure 1. In this op-amp circuit, the gain is set by
the values of the resistors. We have the following mathematical relationship:
Vout = -Vin * Rf /Ri
What Is Integral?
Integral is shorthand for integration. You can think of
this as accumulation (adding) of a quantity over time. For
example, you are now integrating this information into
your store of knowledge. Your store of knowledge has
Area is a component of voltage and time. Let’s
examine the operation of an ideal integrator. We can
simplify the math by making the 1/RC term equal to 1
(i.e., let R=100 KΩ and C= 10 µF). Figure 3 illustrates the
input/output relationships of the integrator. From Time 0
to 2 seconds, have a 2 V square wave applied to the input
of the integrator. The output of the integrator at the
end of this time period is - 4 V (remember the circuit is
inverting). The integrator has accumulated a 2 V signal
for 2 seconds. The area is equal to 4. From T2 to T4,
there is no voltage applied to the integrator. The output is
unchanged. In the remainder of this diagram, you can see
that the integrator output changes polarity when the input
signal changes polarity.
The previous discussion assumed an ideal integrator.
Real capacitors will have some leakage and will tend to
discharge themselves. Also, real op-amps may charge
the capacitor with no input present. If the circuit is built
as drawn, it will likely saturate after a few minutes of operation. To prevent this saturation, add a resistor in parallel
to the capacitor. For our purposes, we are not concerned
about the saturation. We will be using the integrator with
other circuits to control the charge on the capacitor.
NUTS & VOLTS