Here, a real-world example: Laser interferometer data taken at a tooling point. In this example, the stage stack includes three axes (X, Y, and θ) and the move is performed on the bottom-most axis. The vertical scale is 100 nm per division, while the horizontal scale is 50 msec per division; the payload mass is 4 kg.
Select figure to enlarge.

The magnitude of the problem can be readily calculated. To do so we simply divide the friction (in Newtons) by the stiffness, in Newtons per meter, to obtain the error in meters. This result can then be doubled if we want to consider bi-directional motion. Taking advantage of the stiffness formula provided above, and using the more familiar Hertz value for the servo bandwidth:

Error (in meters) = F/m·π2f02

where F = Friction, N-m

m = Mass, kg

f0 = Servo bandwidth, Hz

If friction is 2 N, moving mass 1 kg, and servo bandwidth 50 Hz, the friction-boundary value is a whopping 81 microns. For moves of this size or smaller, the proportional term might as well be turned off.

These values are typical of mechanical bearing stages; friction would be a bit greater for a recirculating bearing stage, and a bit smaller for a crossed roller stage — but not by much. For the micron and sub-micron sized moves required during alignment operations, our servo loop simply doesn't work. This is clearly not an acceptable situation.

In the above example, we assume a direct drive for our mechanical bearing stage. It's worth asking if the use of a leadscrew can improve the situation. In fact, there are numerous reasons to avoid leadscrews in high-precision mechanisms, but let us see how they address the issue of friction. Reformulating the above equation to reflect the case of a leadscrew-based system, the angular error, in meters, is:

Error = LT/2J · π3f02

Where L = Screw lead - — advancement per revolution, m

T = Torque, Nm

ℱ = Total rotary moment of inertia, kg-m2

Rotary inertia is dominated by the motor rotor, followed by the leadscrew, and in a distant third, reflected payload inertia. If we plug in typical values (a screw lead of 0.002 m, leadscrew torque of 0.05 N-m, total rotary inertia of 5 × 10-5 kg-m2, and a servo bandwidth of 50 Hz) the friction boundary is 13 microns.

Conclusion: The mechanical advantage of leadscrews helps a bit here, but the inability to make moves smaller than 13 microns is of little value.

Integrator term help?

In a typical servocontroller PID loop, however, there are two additional terms present. The D term supplies a force or torque that opposes motion, proportional to velocity. While this term provides damping for stability, it is of no use once motion has ceased at the friction boundary, and the force it produces is of the wrong sign, anyway. If conventional stages with friction are to close to final position at all, they must turn to the last of the three terms in the PID loop — the I or integrator term.

The good news is that the integrator term, unlike the proportional term, gets the picture, and slowly sums the errors of past samples to produce a growing output command that eventually get the system to zero steady-state position error.