Question

The crystal puller shown in figure1.3 has the following characteristics: (1) the motor speed varies as 10 rpm/volt, (2) the reducer/puller causes a rise of the vertical shaft of 0. mm/s for each 10 rpm of the motor shaft, and (3) the tach outputs 0.4 volts/rpm. a. For a steady pull rate of 1.0 cm/min, find the required motor voltage. V 1.7 and the required reference voltage, v b. Find the required ramp voltage, v,(1), if the pull up should ramp from 0.0 cm/min to 2.0 cm/min in 2.5 minutes

Answer 1

a.

Given pull rate is 0.1mm/s for each 10 rpm of motor shaft

So, for 0.1 mm/s -- 10 rpm

therefore for 1 cm/min -- $\small \frac{1000}{60}$ rpm

and it is given that for 1 volt motor speed is 10 rpm

therefore for   $\small \frac{1000}{60}$ rpm --    $\small \frac{5}{3}$ volt

or V_m  = 1.67 volt (equivalent)

It is given that, tach output is 0.4 volts for 1 rpm

therefore for $\small \frac{1000}{60}$ rpm ---- output is $\small \frac{400}{60}$ volt or 6.67 volt(equivalent)

Error voltage V_e = V_r - V_s

Therefore V_r = V_e + V_s

The Value of V_e will be decided by design of compensator if it is required to be zero so

V_r = V_s = 6.67 volt

b.

From previous solution, required motor speed for 1 cm/min is $\small \frac{1000}{60}$ rpm

therefore for 2 cm/min, speed would be $\small \frac{2000}{60}$ rpm

so Tach output would be as previously, 6.67 X 2 = 13.34 volt

So for  V_e = 0

The Ramp voltage V_r(t), for change from 0.0 cm/min to 2.0 cm/min in 2.5 mins would be

$\small \left ( \frac{13.34-0}{2.5} \right )t$

or 5.34t (where t is in minutes)