using LDR with a DC motor

Discussion in 'The Projects Forum' started by miketest, Nov 24, 2012.

  1. miketest

    Thread Starter New Member

    Nov 9, 2012
    12
    0
    Hey guys, im kinda new with electronics and i just wanna ask if there is a way i can simply use an ldr(light detecting resistor) to turn a dc motor on or off.

    My main project is gonna be an elevator where i can use the ldr as a sensor to stop the elevator's motor when it reaches the top floor.

    I've been using this guide (http://www.buildcircuit.com/darklight-sensor-using-transistor/ ) and the schematic diagrams from it for the led lights that I installed along with this project. It worked and I tried using that with a motor. Sadly, its not doing anything for me. The motor can run but doesnt turn off at all when i try to block the light. Btw im using a 1.5v reversible dc motor.

    Will it even work or do i need to look at another way of doing this? If so,, what's the simplest thing i can do to do it?(I dont think i can use any ic's or relay for this class 'cause its not an electronics class but just a math class that im doing it for)
     
  2. miketest

    Thread Starter New Member

    Nov 9, 2012
    12
    0
    shameless bump..... doing some more researchr i realize that i will probably need a relay for this... if so,, how can i use that in a circuit
     
  3. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Yes you need a relay unless the motor draws very little current.
    Check out the link below:
    You would need a small relay (Radio Shack has these) which has a 5VDC coil. The relay goes in place of the LED and 330 ohm resistor.

    A 1n4001 or 1n4002 diode (Shack also) should be connected in parallel with the relay coil with the cathode (band) connected to V+ to protect the transistor.

    If your motor is 1.5vdc then you need some way to drop the voltage.
    A resistor would work, but needs to be calculated from motor current
    If you know that post it, or Google OHMS LAW.

    You can also use more of the 1n4001 diodes in series with the motor.
    each one drops approximately 0.7volts. Try it.
     
  4. miketest

    Thread Starter New Member

    Nov 9, 2012
    12
    0
    Thanks for the reply... this is the circuit diagram that i have come up with http://tinypic.com/r/16h99p4/6 will this work??


    other questions.. would it still be the same if i used a 9v dc motor and a 9v battery?? Would having an inductor connected with the motor allow it to have a slow start?(im trying to make it so that when i turn the motor off,, the elevator wouldn't drop instantly).
     
  5. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Very important - the 1N4007 diode is backwards, the band or cathode must be connected to V+. As it is now it will conduct through the transistor with no current limiting - bad news for the transistor. The diode only needs to be 1n4001 or 2 with the low voltages.

    Connect the wiper of pot to one end of pot to make it variable. Al larger value pot 1k-10k would be better.
    Then the circuit should work for On/Off.

    This adds a wrinkle. Various ways, but relay not the best and not an inductor.

    Please post current draw specs for new motor and more detail. Do you need slow start and stop etc.
     
    Last edited: Nov 26, 2012
  6. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Got to thinking - you probably could use a braking resistor which would be switched in across the motor when it was off and the elevator was dropping. You would need a SPDT relay which has a NC and NO contact to
    accomplish that. . Your drawing shows a SPST NO relay.

    Very wild guess based on a small 9v motor 220 - 470 ohms at 2 watt resistor.
    I do math only when necessary, have no formula for the above.;)
     
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