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Hi sorry for the poor schematic. I want to use a MOSFET H-Bridge to drive a linear actuator based on outputs from a microcontroller. The actuator I'm using has inbuilt limit switches so when it reaches its maximum point the switch will be toggled and there will be no current across the motor. For the program I'm writing to the microcontroller I need to know the full range of motion of the actuator, and so I was thinking of having the gate of an NPN BJT connected to the drain of M1 with its emitter going back to the microcontroller so that as long as there is current across the motor a signal will be sent back to the microcontroller, and when the limit switch is toggled the signal being sent back will stop and I will know the range of motion from that. Is this a viable option or does M1 need to be grounded for the circuit to operate as intended. If it is a viable option, how should I reduce the voltage and current so that the BJT I use can handle it?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ you may disappointed without linear feedback servo. consider any suitable encoder. Then you can optimized a,v,X profiles and minimized error against desired control. Torque , load may be not consistent and will be temperature dependant \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 12 '16 at 5:09
  • \$\begingroup\$ But as far as control is concerned you want to shunt coil on end, not open current, in order to stop immediately. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 12 '16 at 5:15
  • \$\begingroup\$ Even optical feedback would prevent and end-stop. Maybe link the LM specs and explain purpose will get you better advice \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 12 '16 at 14:29
  • \$\begingroup\$ Thanks for responding I've switched to using an IC that incorporates current feedback which I should've just done in the first place :p \$\endgroup\$ – Lokso Sep 13 '16 at 6:15
  • \$\begingroup\$ how do you retract motor if limit switch is open? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 13 '16 at 6:23
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Just to not leave this thread in the lurch, unanswered, I will try here: Most microcontrollers have an analog-to-digital converter, many also have a built-in comparator. In either case, a low-value sense resistor in the mosfet source provides a signal that is proportional to the current through the transistor and therefore the transistor load. The concept of a current-sense resistor is illustrated here as the "low-side option". A typical value of this current-sense resistor is .1 or .01 ohms. The signal output may need to be amplified a bit to properly feed the analog-to-digital converter or comparator.

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