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TLV1805-Q1: Drive an electromagnetic break

Part Number: TLV1805-Q1
Other Parts Discussed in Thread: TLV1805, TIDA-020023, TPS1H200A-Q1, ISO1212, CSD87502Q2

Hi, I have to drive an electromagnetic break of a brushless dc motor, it has a rated voltage of 24Vdc and a switch to sense if it's manually activated on not (as it has to be active in order to drive the motor).

I was thinking of the break as a solenoid, and I found TIDA-020023 reference design of TLV1805 in order to drive the break and sense the feedback current, this last parameter would help me to sense the manually activation of the break, because if the break is released, no current would be sensed. 
My question is, am I right in selecting this design as a start? Also, I don't need the proportional control, but I suppose that would mean only, that I don't have to change the duty cycle based in the feedback information.

I would appreciate any help or suggestions. Thanks!

  • Hello Diana,

    Is the brake a physical solenoid engaging a braking surface? Or another winding that you "short" to stall the rotor (like a shorted generator winding).

    Can you give an example? Or a datasheet for the motor?

  • Hi, thanks for your answer, I think the brake is a solenoid that engage a braking surface, when it's connected to power supply it disengages. The brake is like the one in this video and it's from a wheelchair BLDC motor. Like the one in the video, the brake has a micro switch to sense if the brake is activated before driving the motor. I hope my explanation is ok. Thank you very much.

  • Hello Diana,

    Okay...I did a quick search on these brakes.

    The brake has a microswitch in series with the brake coil, and the microswitch opens when the manual release lever is applied. You need to detect when the coil is "open".

    Do you have access to the "ground" side of the coil? The simplest would be to use a "low-side" current detect with a shunt resistor. When the coil is energized, there will be a voltage (<100mV)  generated across the shunt. Since you do not need "proportional" measurement, then the amplifier is not really needed and it can be directly measured with a comparator.

    If you do not have access to the ground side of the coil, then it will have to be measured on the "high" side (battery side, as shown in your appnote). This is a little more complicated.

    Is your supply voltage also 24V? What is the coil current (coil resistance - most seemed to be around 46 ohms, or ~ 500mA). How low is the battery voltage allowed to drop (for design margin).

    We also have some high-side drivers, such as the TPS1H200A-Q1, that have these functions built-in, including the MOSFET and open-load fault detect (which is what you are looking for)..

  • Hello Paul, thank you for your answer, I don't have access to the ground side of the coil, but I've got the two wires to apply the 24V, coil resistance is about 52 ohms - 450mA aprox. I applied 24V to the wires and the brake disengage successfully. Someone gave me a solution using an optocoupler in reverse mode, with the coil and a resistor connected to the high side, so when I apply 24V to the coil and the switch is closed, current will flow through the low side of the optocoupler and with a pull up resistor the MCU would sense a low transition. 

    Could you give me your opinion about this solution?

  • Hello Diana,

    I was assuming you did not have access to the switch terminals – all you had was the 24V input and ground to the brake assembly.

    If you have access to the “positive” side of the coil, before the switch, then the simplest thing is to look at the coil voltage directly. This can be done by dividing down the 24V to whatever the GPIO pin of the processor can accept, or, drive an LED in a optocoupler if isolation is needed. This sounds like what you are planning to do – put the LED side of the optocoupler across the coil (with a 3.3k series resistor), and the photosensor side between GND and a pull-up resistor to the processor voltage. Though Iam not sure what you mean by optocoupler in “reverse”.

    Are there safety requirements? This method will tell you if there is voltage applied the coil, but cannot detect an open coil.

  • Hello Paul, thanks a lot for your help. 

    You are right about the term in "reverse" I was mistaken about that. The idea is to use the optocoupler  the way you described. 

    What do you mean with an open coil? Do you mean broken? If there is voltage applied through an open coil I would still detect it at the photosensor side of the opto?

    The break is attached to a wheel's motor, what I need to detect is if the switch is closed or not in order to drive the wheel only if the switch is closed. Actually, I have two wheels, each one with it's own break.


  • Hi Diana,

    If the coil was "open", as in broken with no continuity, such as a burnt winding or broken connector, then the LED would still light since the LED and coil are in parallel. The only way to make sure the coil was conducting is to run a current through it to make sure the path is not open.

    How are you switching the coil voltage? A relay? Switch? MOSFET?

    One thing you could do is put the optocoupler LED between the brake supply lead and +24V. That would put the LED in series with the coil when the power switch is open (off). If the coil is present and the microswitch is closed, the LED should light. The few mA that the LED needs would not be enough to energize the coil, but you could tell it is there. But then you would not be able to tell if the coil was actually actuated when "on" as the power switch would short out the LED.

  • Hi Paul, thank you again for your reply.

    I was thinking in switch the coil voltage with a MOSFET.

    It's a good idea what you suggest, that way I would know if the switch is close before actuating the coil, then can I use the MOSFET to actuate it?

    I've been searching components and I found CSD87502Q2 (2 MOSFETs) and ISO1212(Optocoupler), would those be appropriate?

    Thanks and regards.

  • Hello Diana,

    Even though it is our device, the optocoupler is a bit overkill for this application. I was thinking more of the simpler, more common 2N11 type LED + phototransistor optocoupler.

    The MOSFET should be a higher voltage P-Channel MOSFET. The Vds should be at least twice the supply voltage for safety (48V, so ~50 or 60V) at a minimum of 5A.

    The flyback diode is critical - it clamps the high-voltage inductive kick that occurs when the power is turned off. That diode shoudl be placed as close to the coil as possible to keep the currents off the circuit nodes.

    Here is the TINA simulation. Since it is a simple DC circuit, just run the "DC Operating Point", and it will show the votlages on the nodes.


    This is outside the scope of this forum, so I will have to back-out. The above should be enough to get you started in your investigations for the final design!

  • Thank you Paul, it's been a long discussion and it has been really helpful for me.

    Thank you very much for your support.