I chose to use the L6202 H-bridge IC because it has built in dead time as well a thermal shutdown protection. However I've already broken multiple of these from what appears to be current spikes when switching motor directions.

During operation the DC motor will be powered with 48V, and draws around 200-400 mA when moving. Right now we are only giving it full speed either direction, no PWM.

I've set up the circuit as basic as I can following the datasheet reccomendations.

  • Vs: connected to 48V
  • SENSE: connected to ground
  • Bootstrapping capacitors: 47nF/50V
  • VREF: 10 nF capacitor to ground
  • IN1/IN2: 3.3V high / 0V low
  • EN: connected direct to 3.3V

I powered the motors with a power supply (24V, 1A current limit) for testing and it worked fine. The chip has a built in 40ns dead time which I thought should be more than enough considering the switching is happening on the scale of nanoseconds. However, I started adding a delay because I saw current spikes on the power supply exceeding 1A (power supply was entering constant current mode). That made me realize the 40ns dead time was probably not enough. I added a 50ms delay between one direction and the other. When we tried running the motors in operation with 48V power supply (unsure of the current limit), a 500ms delay between switching directions worked, but with a 250 ms delay, the chip was destroyed.

I also thought maybe we needed to add freewheeling diodes to help current flow when we make sudden changes to the motor speed, but it seems to me that this is already built into the chip as part of the MOSFET structure.

When the chip is destroyed it appears to short the two outputs, Vs and gronud all together which is very strange considering that the chip has built in thermal protection that is supposed to shut it down at 150C junction temperature. When the chip is cooled down again Vs and ground are still shorted together along with the outputs and the chip is no longer usable.

Am I doing something wrong and/or is there a way to mitigate this problem and continue using this chip? 250 ms seems like a ridiculously long time to wait for a motor to switch direction.


1 Answer 1


During operation the DC motor will be powered with 48V

The data sheet first line says: -


Additionally, figure 47 (Bidirectional DC Motor Control) says Vs = 36 volts max.

The absolute maximum rating is 52 volts and it is likely that back emfs are pushing up the 48 volts supply beyond the absolute maximum limit. You are running too close to the absolute limit this device is intended form. Choose something that is 60 volt+ rated or implement a better supply that has energy dissipation facilities when the motor is being reversed.

  • \$\begingroup\$ It should be noted that there are other solutions besides a power supply which can sink current. a) use a slow decay PWM mode instead of fast decay when decelerating the motor. This dissipates energy in the windings instead of regenerative braking. b) add a braking resistor that is switched on when the voltage rises above a given threshold, keeping the voltage within limits. \$\endgroup\$
    – jms
    Dec 12, 2019 at 9:36
  • \$\begingroup\$ Good point - the overvoltage could explain why the chip is getting destroyed. I thought the purpose of the freewheeling diodes was to dissipate this extra energy and clamp the voltage? What are some other ways in hardware that I can reduce the back emfs ? \$\endgroup\$
    – VanGo
    Dec 12, 2019 at 16:10
  • \$\begingroup\$ You have to rely on the MOSFET flyback diodes dumping energy into the supply lines and if the supply has more bulk capacitance then it won't get lifted so much. Maybe also, when you implement a reverse you temporarily shunt the supply with a moderate value of resistance so that it dissipates the reversing energy. \$\endgroup\$
    – Andy aka
    Dec 12, 2019 at 16:14
  • 1
    \$\begingroup\$ Bulk capacitance is fundamental in stopping the bus voltage rising. Not having it is asking for trouble. Look at how the flyback diodes in your bridge have no option other than to dump excess energy to the supply. The resistor value is dependent on the energy dump when reversing and cannot be determined on paper without a lot more detail. \$\endgroup\$
    – Andy aka
    Dec 12, 2019 at 17:23
  • 1
    \$\begingroup\$ They may be charged to (say) 48 volts but they are also a dumping ground for energy and although they must let the voltage rise a little bit when "taking" the flyback energy, the very bulk size of these usually means that the voltage rise is significantly less than that voltage rise should there be very poor bulk capacitance. Most supplies are good at pushing current into a load when the voltage is being dragged down by the load but, because series control elements are used they are remarkably poor at dealing with things when an over voltage occurs. \$\endgroup\$
    – Andy aka
    Dec 13, 2019 at 8:07

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