Designing a current limiting circuit for my project

Question

I've got a small DC brushless motor for which I would like to set a current limiting circuit as a protection against over-current.

I thought I could pull it off with a simple fuse (or those current limiting fuse-like chips) but I can't. The fuse would only work to protect against short circuits because I want to set my limit just 20% (max!) above the nominal current draw of the motor.

I didn't design the motor electronics, and my competence in electronics is not great but I was basically wondering if I could use a circuit similar to those:

_{(source: wikimedia.org)}

and if so how I would select the correct components (like transistors and resistors)?

The nominal current draw for the motor is 1.5A. I'd like the circuit to cut the power off when the current is 1.8A or more.

The reason I need something like this is because the load on the motor can vary greatly and would sometimes bring the current draw to 1.8A and more (which I consider unsafe - it's very subjective but that's what I'd like to do...)

Any idea?

edit: clarification: I want to switch power off when the current exceeds a certain value, which has to be about 20% max higher than the nominal power draw of my motor.

edit: how about that second schematic?

Answer 1

Your first circuit appears to be a current limiter. If the resistor values are properly chosen, the circuit will clamp the collector current (the current through the transistor and thus your load) at some value. It will not cut power.

The second circuit image you posted is not a complete circuit. The box labelled "Drive Circuit" represents some logic block that you would have to design. It takes feedback and directly turns power on and off to the load. That could be done any number of ways.

If you want something that acts more like a fuse that can be easily reset, something like a latching relay in series with the load could work. I'd probably use a comparator to trigger the relay when the voltage across a shunt resistor went above a threshold.

^{simulate this circuit – Schematic created using CircuitLab}

The relay is a latching type. Rshunt is an appropriately chosen shunt resistor. R1 and R2 are chosen such that they divide to the voltage that triggers the comparator when the current through the load is 1.5A or higher: Vtrigger = 1.5A * Rshunt = Vcc * R2/(R1+R2)

Here are some considerations:
1. Rshunt should be small so it does not steal too much power from the drive circuit.
2. This circuit assumes Vcc is a stable voltage source. If it's weak and changes voltage under load, the comparator circuit will not trigger correctly.
3. If you're driving a motor, there might be a temporary current surge on start up that exceeds 1.5A for a short period of time. If that's allowable, you may need to build a low-pass filter into the non-inverting input of the comparator.
4. If you choose a double-coil latching relay (not shown), you can rig a button up to the second coil to reset the relay after it triggers.
5. The coil of the relay may not be compatible with Vcc.