Several motors on one belt: master-slave and droop

Long or heavily loaded conveyors are driven by more than one motor: one at the head and one at the tail, or several spread along the run. Those motors are mechanically coupled through the belt itself, and that coupling is the whole problem. This lesson is course sections 5.2.1 and 5.2.2, with Tutorial 5.1(b, c) and Lab V.

Why coupled drives fight

Suppose each drive runs its own speed loop to the same 2.5 m/s. Their speed measurements are never exactly equal: a slightly worn pulley, a small gear-ratio difference or a different slip is enough. The drive that believes the belt is too slow pushes harder; the other, which believes it is fast enough, backs off. Their integrators push in opposite directions until one sits at its current limit and the other takes almost nothing, or is even dragged into generating, with power circulating through the belt and overloading the drives.

The control problem is to make the motors share the load in a fixed proportion while holding one line speed.

Two strategies

Both are standard functions in networked conveyor VFDs exchanging references over a fieldbus.

Master-slave (speed-torque). One drive, the master, runs in speed control and sets the line speed. The others, the slaves, run in torque control and follow the torque reference the master broadcasts. They take an assigned share of the load without contesting the speed.

Droop. Every drive runs in speed control, but its speed reference is lowered slightly in proportion to its own load:

$$\omega^*_i = \omega^*_0\left(1 - k_d\,\frac{T_i}{T_\text{rated}}\right) \tag{5.4}$$

A drive that takes more than its share slows down a little and so sheds load to the others until they balance, exactly as parallel generators share load through frequency droop.

Try it

The widget is Lab V: the Tutorial 5.1 conveyor on two 15 kW drives, with drive 2's speed feedback reading 0.5 % low and the load rising by 30 % at 10 s. Start with no scheme, then try master-slave and droop.

No schemeMaster-slaveDroop

TwoThree

Drive 2 speed-feedback error

0.50 %

A worn pulley or a small gear-ratio difference: drive 2 believes the belt is slower than it is.

Load surge +30 % at 10 s

• Drive 1
• Drive 2

A drive sits at its current limit while the other takes almost nothing: the drives are fighting.

Split before the surge

35.5 / 64.5 %

Split at 40 s

33.9 / 66.1 %

Belt speed at 40 s

2.500 m/s

Speed error

0.00 %

ω*_i = ω*_0 (1 − k_d T_i / T_rated) ; F_rated = P η / v = 5.4 kN

What the widget shows, with the default 0.5 % error:

Scheme	Split before the surge	Split at 40 s	Belt speed at 40 s
No scheme	35 / 65 %	34 / 66 %, drive 2 at its 150 % limit	2.500 m/s
Master-slave	50 / 50 %	50 / 50 %	2.500 m/s
Droop, $k_d$ = 3 %	46 / 54 %	46 / 54 %	2.421 m/s (−3.2 %)
Droop, $k_d$ = 10 %	49 / 51 %	49 / 51 %	2.222 m/s (−11 %)

With three drives and no scheme, the third drive is dragged to about −20 % of its rating: it generates while the others motor.

Predict first

Master-slave, and the fieldbus link is lost at 5 s with the slave freezing its last reference. The load surges at 10 s. Who takes the extra load?

Both drives equallyThe master aloneThe slave alone

Choosing

The choice is a general control trade-off, summarised in the course's Table 5.1:

	Master-slave	Droop
Drive modes	one in speed, the others in torque	all in speed
Sharing accuracy	high (stiff)	moderate, load-dependent
Line-speed error	none	rises with load (typically 2 to 5 %)
Communication	required and time-critical	not required
Loss of one drive	fails if the master is lost	degrades gracefully
Commissioning	a master must be designated	only $k_d$ to set
Best suited to	precise, tightly coupled belts	long or distributed drives

FoundationStart here if this is new to you

Two people carry a long plank. If each tries to walk at their own idea of the right pace, one ends up dragging the other. Either one leads and the other simply pushes as hard as the leader does (master-slave), or each agrees to slow down a little whenever they feel they are carrying more than their share (droop). The first needs them to talk; the second works in silence, but the pair walks a little slower when the plank is heavy.

ExplorerGo deeper: derivations and open questions

The droop equilibrium. With integral speed loops, the steady state of each drive satisfies $v(1 + e_i) = v_0 (1 - k_d F_i / F_\text{rated})$, where $e_i$ is its speed-feedback error, and the forces add up to the resistance. Solve for $v$, $F_1$ and $F_2$ and check the table above. (This is what the Lab V checker uses.)

Three drives. Switch the widget to three drives. Does master-slave still give an even split? With droop, which drive carries most, and why?