2.2 Robot drives

Figure 2.1: motors driver board



2.2.1 Introduction

To control the direction of D.C motor we need the H-bridge driver circuit using relays or transistors act like switches or integrated circuits like L298 or L6203.

In our project we use two D.C motors to provide the vehicle with four movement directions: forward and backward, rotate left and rotate right. The speed of the motors controlled by pulse width modulation (PWM). Also, One D.C motor used to move the main arm backward and forward.

For the stepper motor used to rotate the camera, we can use software from PC or microcontroller to produce the required sequence and an integrated circuit like UCN5804.

2.2.2 SPEED CONTROL

2.2.2.1 PWM

Pulse width modulation is a technique for reducing the amount of power delivered to a DC motor. This is typically used in mechanical systems that will not need to be operated at full power all of the time.

Instead of reducing the voltage operating the motor (which would reduce its power), the motor's power supply is rapidly switched on and off. The percentage of time that the power is on determines the percentage of full operating power that is accomplished and so reducing the speed.

Figure below illustrates this concept, showing pulse width modulation signals to operate a motor at 75%, 50%, and 25% of the full power potential.

Figure 2.2 Pulse width Modulation (PWM)

We produced the PWM signal in our project from the PIC microcontroller to control the speed of the vehicle.

2.2.2.2 Pulse Rate of stepper

The speed of the stepper motor is controlled by the width of the pulses that supplied to the integrated circuit (UCN5804B).

In our project we produced these pulses by (555) integrated circuit like shown in figure 2.2 below:

Figure 2.3 (555) pulser

Where:

R₁ = R₂ = 2 KΩ , C = 10 μF

 Substituting these values to find period, duty cycle, t_on, and t_off we obtain:

Period  = C [( R₁ + 2 R₂) ln(2)]

            = 10 [(2 + ( 2*2)) ln(2)] =  0.04159 sec

duty cycle =  ton / period = (R₁ + R₂ ) / ( R₁+ 2R₂)

                = (2 + 2) / (2 + 4) = 0.6667 sec

 ton = C ( R1 + R2) ln(2)

       = 10 (4) * ln(2) = 0.02773 sec

t_off  = C R2 ln(2)

      = 10 * 2 * ln(2) = 0.01386 sec

Obviously, the smaller the step angle is, the more accurate the motor. But the number of pulses stepper motors can accept per second has an upper limit. Heavy duty steppers usually have a maximum pulse rate of 200 or 300 steps per second. Some smaller steppers can accept a thousand or more pulses per second, but they don't usually provide very much torque.