changes to be made in the PWM circuit

Discussion in 'General Electronics Chat' started by itsmyturn, Mar 24, 2005.

  1. itsmyturn

    Thread Starter Member

    Feb 15, 2005
    16
    0
    I am using the PWM circuit of the below link to control the speed of a line-following robot. Here i have replaced the 100K pot (VR1) with an LDR(light dep. resistor) of the robot. But it is not giving any speed variation. Instead , it is giving some unusual results like: when the supply is given, the drop across the resistor R8 is zero but the motor is running. But i have one confidence that the resistance of LDR is not being compatible with the circuit. Can anybody help me out how to DEBUG this circuit.......

    PWM circuit
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The circuit you have selected should work fine for controlling the speed of the motor.

    The device being used to drive your motor in this circuit is a FET or field-effect transistor. Unlike a BJT (bipolar transistor), this device is controlled by the voltage applied to the gate. The input impedance into the gate of an FET is very high and so there is no voltage drop across the series resistor since the current flowing through R8 is extremely small. The measurement that would be more useful would be the voltage between the gate and the source terminal or ground in this case. The FET will turn off with voltages less than a few volts. The FET will be on when the gate is +4 or +5 volts.

    You say you replaced the 100K potentiometer with your LDR (light-dependent resistor). Since the pot is a three terminal devices that leaves me wondering if the LDR is between the input of the opamp and ground or between the input of the opamp and the positive supply. Can you clarify this point?

    To help determine if you have an LDR value related problem, can you also provide any specifications for the LDR or the part number perhaps?
     
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