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What would I use for this function in a circuit?

Former Member

If I were to have a motor, and stopped it with my fingers, what component would shut the motor off if it experiences a certain amount of load? This is on a normal DC motor, but i could also use a stepper motor. Thanks! Feel free to ask for more information.

  • 0

    With a normal DC motor its current will increase as it is loaded and slows down.

    So you could use a current sensor to trigger at shutoff.

    An over current breaker or fuse or even a PTC fuse might work, but these may not be precise enough or fast enough for many applications.

    A stepper motor is somewhat similar if the steps are at high enough frequency, where the armature never stops rotating, (usually above 300 Hz) but the pulsed nature of the voltage and current make measurement a bit more difficult.

    One way to detect a stepper load is to determine if the motor "slips". It should move a precise angle for every step, if it starts to "miss" steps it is slipping. This could be monitored by rotary encoder pulses which get compared to drive signal pulses.

  • 0 in reply to dougw

    I'm relatively new to the electronics scene, so what I shold do is measure the current draw of the motor and determine what it rises to under the load I want it to stop at? After I accomplish this, how do I use the sensor to trigger the shutoff? Would I use some sort of microcontroller? Thank you and sorry for my illiteracy in electronics.

  • 0 in reply to Former Member

    A microcontroller is not needed, but detailed design depends on many factors such as how big is the motor, what voltage and current are involved, how sensitive does it need to be, etc..

    Here is an example using a couple of transistors and a relay: (the motor gets hooked up at the "LOAD" terminals)

    Low Battery Cut-off and Overload Protection Circuit. | Homemade Circuit Designs Just for You

    Note R1 must be able to handle the full motor current and a small enough value that at full current there is only .7 volts across it.

  • 0

    Hi Mathew,

     

    I see that Doug has covered things very well while I was drawing up this circuit. I will post anyway just in case it is of any help to you.

    Here is a little circuit that illustrates how this can be accomplished. The potentiometer R-1 determines the level of the current necessary to trip the relay. As long as the voltage at pin 4 of the comparator is higher than the voltage at pin three the comparator is low and the connection to the motor is made by the relay RL-1. As you increase the load to the motor the current in the resistor R-2 increases and this also increases the voltage at pin 3 of the comparator. When the voltage at pin 3 becomes higher than pin 4 of the comparator the comparator will turn on the MOSFET Q-1. When the MOSFET Q-1 turns on it will energize the relay which will pull in and turn off the motor. The second thing that RL-1 does when it energizes is to turn itself on. We say that the relay latches. The relay will remain energized and the motor will remain unconnected until the power is turned off or the reset switch SW-1 is pushed. I have not put in resistor or component values as they would depend on things I do not know about your motor or your needs. There may be simpler ways to do this but this is the first circuit that I came up with.

    John

     

    image

    If you click on the schematic it will get a little bit bigger.

    John