LM3S5D56 Datasheet, PDF (1027/1146 Pages) Texas Instruments – Stellaris® LM3S5D56 Microcontroller

Stellaris^® LM3S5D56 Microcontroller
Figure 20-2 on page 1027 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 20-2. 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 QEI Timer
Load (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 required 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 clear and
4 for CAPMODE set)
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 is clear) 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 because 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.
Important: Reducing constant factors at compile time is the best way to control the intermediate
values of this equation and reduce the processing requirement of computing this
equation.
January 23, 2012
Texas Instruments-Production Data
1027