Controlling Infineon Profit BTS441TG

Hi,
I'm developing a project using Microcontroller, BLDC motor driver and BLDC motor. This setup supplied by 24VDC Battery. And battery always mounted to the system. BLDC motor driver supplied by 24VDC Battery directly and I don't want to power on BLDC motor driver always on. So I want to switch on/off Motor driver with BTS441TG N-Channel MosFET and control it from Arduino with digital output for switch on/off. I read the datasheet and connect BTS441TG to the circuit as attachment. In the datasheet BTS441TG ground resistor optional and I do not connect 150R resistor to the BTS441TG's ground pin.

Schematic

So When the system is power on and mcu digital output set to high for BTS441TG enable pin, BLDC motor driver swithc on and can be controlled by mcu. Nominal current for BLDC motor 8.8A and everything is ok but low to enable pin BTS441TG burning :frowning:
Do you have any experience with BTS441TG or any other profet MosFET. Can you give me any reference design about it?

Did you switch off the BTS441TG when the motor was running?

Klaus_K:
Did you switch off the BTS441TG when the motor was running?

Yes, I'm. BLDC driver on and motor running. And I switched off BTS441TG

According the BTS441TG datasheet there is a maximum load inductance. When you switch off the motor the energy stored in the magnetic field will create a back current. There is a limit to how much the device can handle.

You could test your circuit with a non-inductive load to confirm whether this is the true issue.

No need to confirm that, you must use a free-wheel diode everytime, so do it. The slow switching speed of this device may be helping somewhat, but the correct approach is free-wheel diode (always when switching inductance, whether MOSFET, BJT, relay or switch)
.

This means a suitable rated diode (ie 10A pulse rating) backwards across the load to tame the back EMF.
There is no back-current, the current continues, its the EMF that becomes backward. Without the diode the inductive energy tries to destroy the switching device, with the diode it safely and slowly dissipates in the diode.

The device needs some moderate heatsinking at 9A, as its a pretty high 20 milliohm on-resistance, more suited
to 3 to 5A loads than 9A. The 21A rating in the datasheet is theoretical, as you'd have to water-cool it to handle that.

Note this is an intelligent switch, meaning its not designed for PWM at all, but here that's not an issue.

Be sure to have a fuse protecting the wiring to your circuit. If the circuit fails short-circuit you don't want the wiring bursting into flames.

Thank you for all responses. I connect 150R resistor to GND pin and ground it is working without burning :slight_smile: And after that I read the datasheet and I think BTS441TG already has a freewheel diode. Do you think it is still required a freewheel diode? If your answer yes, Diode that in this link does suitable? It is 45V 20A and fast

Yes, you need an additional diode to protect the BTS441TG. As the datasheet states there is a limit of inductive load supported by the device itself. Additionally, your experiment with the resistor showed that your circuit works for non-inductive loads.

I am not sure about the characteristic. I guess it needs to be as fast or faster than the BTS441TG.

A zener clamp doesn't prevent inductive spikes, it just caps the maximum voltage. Inductive spikes generate tons of EMI and are best eliminated fully.

Klaus_K:
I am not sure about the characteristic. I guess it needs to be as fast or faster than the BTS441TG.

Any diode has AFAIK a very fast 'on' time. It's the 'off' time that differs between types of diodes.
And only 'on' time is relevant for protection, 'off' time could only be relevant with PWM.
Leo..