CSD19538Q2: part burned out

Hello BROWN_P,

Thanks for your interest in TI FETs. In your circuit diagram, the gate of the top freewheeling FET is left floating. This is never a good idea as the gate can float up and turn on the FET. If you're going to use the FET in this manner, I would recommend adding a gate-to-source resistor (10k - 100k) to make sure the top FET is actually off. The waveforms show that there is still current flowing thru the FET when it is OFF. Is that correct? The waveforms show ~50V across the FET and ~170mA of current.

From your description, the FET operates OK at lower duty cycles but fails at higher duty cycles. It may be possible that the FET is going into thermal runaway and failing at higher duty cycle. Have you measured the temperature rise of the FET?

All of TI's 80V and 100V FETs require minimum VGS >=6V. If the gate drive voltage is 5V, then the FET is operating in very steep portion of the Rds(on) vs. VGS curve as shown on page 1 of the datasheet. Keep in mind, this curve is for a typical device and a slight increase in VTH can cause a large increase in Rds(on). We do not have any 80V or 100V FETs that can be driven from 5V.

My understanding is the freewheeling diode is required in this type of motor drive application.

Best Regards,

John Wallace

Thank you so much for the quick reply.
I saw it as an answer and started testing right away. Surprisingly, the gate was floating at 10V. Why is that so? This happened when I connected the brake, that is, when the inductor was connected.

As you said, I connected a 100k resistor to gate and ground, but both q1 and q2 fets burned out. In my opinion, there shouldn't be 10v in the gate in the first place, but when the brake is connected, 10v appears. Is the input voltage leaking toward the gate through the brake coil?

What's even more strange is that I slowly increased the voltage applied to the brake from 30v, and it was fine up to 39.9v, but from 40v, the gate voltage increased linearly from 0v, then rose to 10v, and burned out within a few seconds.

Hi BROWN_P,

It is not surprising when the gate is left open that it floats up to 10V. Are you using the same FET on the top as the bottom? A 100kΩ resistor to GND should be enough to hold the gate low. You might try to reduce the resistor to 10kΩ and connect it between gate and source of the top FET. I still do not know why the FETs are failing in your application. The most common causes of FET failures are over voltage, over current or thermal runaway due to excess power dissipation. Have you contacted the brake manufacturer to get their recommendation for driving it?

Thanks,

John

Thank you for your quick reply
Why is 10v displayed even though the gate voltage is not applied to the top fet?
(At this time, 10k and 100k resistors were installed between the top FET and GND.
The top and bottom use the same FET. Break recommendation is 200mA.
As you said, if you connect the gate and source terminals of the top FET with a resistor, since the brake is connected to the source terminal of the top FET, that voltage seems to be applied to the GATE. is not it? In this case, the gate voltage of the upper FET exceeds 20V.

Hi BROWN_P,

Can you tell me how you are driving the main (bottom) FET? Is the gate of the top FET being driven at all? If not, why use a FET and not a diode? There must be a path for leakage current to flow and charge the gate of the top FET. The abs max VGS = +/-20V for the CSD19538Q2. Adding a gate-source resistor is not going to cause cause VGS to exceed abs max. We want VGS = 0V to make sure the FET is off. Have you measured VGS instead of gate to GND?

Thanks,

John