LM3S1968 Datasheet, PDF (605/709 Pages) List of Unclassifed Manufacturers – Microcontroller

Stellaris^® LM3S1968 Microcontroller
counting commences on a new time period. The number of edges counted in a given time period
is directly proportional to the velocity of the encoder.
Figure 17-2 on page 605 shows how the Stellaris quadrature encoder converts the phase input signals
into clock pulses, the direction signal, and how the velocity predivider operates (in Divide by 4 mode).
Figure 17-2. Quadrature Encoder and Velocity Predivider Operation
PhA
PhB
clk
clkdiv
dir
pos -1 -1 -1 -1 -1 -1 -1 -1 -1
rel +1
+1
+1
+1 +1 +1 +1 +1 +1 +1 +1
+1
+1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
+1
+1
+1
The period of the timer is configurable by specifying the load value for the timer in the QEILOAD
register. When the timer reaches zero, an interrupt can be triggered, and the hardware reloads the
timer with the QEILOAD value and continues to count down. At lower encoder speeds, a longer
timer period is needed to be able to capture enough edges to have a meaningful result. At higher
encoder speeds, both a shorter timer period and/or the velocity predivider can be used.
The following equation converts the velocity counter value into an rpm value:
rpm = (clock * (2 ^ VelDiv) * Speed * 60) ÷ (Load * ppr * edges)
where:
clock is the controller clock rate
ppr is the number of pulses per revolution of the physical encoder
edges is 2 or 4, based on the capture mode set in the QEICTL register (2 for CapMode set to 0 and
4 for CapMode set to 1)
For example, consider a motor running at 600 rpm. A 2048 pulse per revolution quadrature encoder
is attached to the motor, producing 8192 phase edges per revolution. With a velocity predivider of
÷1 (VelDiv set to 0) and clocking on both PhA and PhB edges, this results in 81,920 pulses per
second (the motor turns 10 times per second). If the timer were clocked at 10,000 Hz, and the load
value was 2,500 (¼ of a second), it would count 20,480 pulses per update. Using the above equation:
rpm = (10000 * 1 * 20480 * 60) ÷ (2500 * 2048 * 4) = 600 rpm
Now, consider that the motor is sped up to 3000 rpm. This results in 409,600 pulses per second,
or 102,400 every ¼ of a second. Again, the above equation gives:
rpm = (10000 * 1 * 102400 * 60) ÷ (2500 * 2048 * 4) = 3000 rpm
Care must be taken when evaluating this equation since intermediate values may exceed the capacity
of a 32-bit integer. In the above examples, the clock is 10,000 and the divider is 2,500; both could
be predivided by 100 (at compile time if they are constants) and therefore be 100 and 25. In fact, if
they were compile-time constants, they could also be reduced to a simple multiply by 4, cancelled
by the ÷4 for the edge-count factor.
July 15, 2014
605
Texas Instruments-Production Data