TVS diode across motor leads failing short. How to protect MOSFET from motor back EMF?

Question

I am having an issue with a TVS diode failing short on 3 out of 7 units. The intended purpose of putting this diode across the motor leads was to protect the MOSFET and relay from the motor's back EMF.

The MOSFET is intended to turn off and on the motor as a low side switch. The MOSFET gate is receiving a PWM signal that is ramped to a 100% duty cycle for soft starting the motor.

The relay is intended to select the direction the motor is turning.

I believe the issue is that the TVS diode that was selected was not specced high enough for max. power dissipation. The PWM is creating thousands of instances where the TVS is conducting and eventually it becomes so hot it fails.

I was looking at back-to-back Zener diodes to use instead of the TVS, I don't know how I should spec those, as the power dissipation for most is not significantly higher than the current TVS diode's power dissipation rating.

Could I add a resistor in series with the TVS/back-to-back Zener diode to limit the power dissipation? Or is there a better way to protect the MOSFET and other components from the motor's back EMF?

Or I could select a different MOSFET which can withstand the back EMF seen on the drain of the MOSFET. The measured (with oscilloscope) back EMF of the motor seen on the MOSFET (drain to source) is 69 V without the TVS in the circuit. This perplexingly does not kill the MOSFET and it is still operating okay.

TVS currently used: P6KE33CA

MOSFET currently used: NTBS2D7N06M7

Picture of circuit:

Thanks for any insights.

Part 2:

After some testing and with some consideration and help from a few replies I have a much better looking waveform:

Changed the circuit for a Schottky diode to absorb the back EMF of the motor since I can now add the diode on the non motor side of the direction selecting relay. And experimented with values for an RC snubber across the motor leads and found that a .15uF cap with a 2 Ohm Resistor worked well.

Circuit now:

Oscilloscope pictures: Ch1 - Yellow - using differential probe across motor terminals with 1/10 attenuation (note might have polarity of probe reversed) Ch2 - Purple - Mosfet Drain to Source Voltage

PWM of motor without Schottky or snubber: Zoomed in:

PWM of motor with Schottky, without snubber:

PWM of motor without Schottky, with snubber 0.15uF 2.0ohm:

PWM of motor with Schottky and with 0.15uF 2.0ohm snubber:

Summary: Having the Schottky diode on the opposite side of the direction selecting relay than the motor ensures the back emf of the motor is always in the forward direction of the Schottky. Spec'ing the Schottky for full motor amps and reverse standoff voltage of at least twice system voltage of 24V. Schottky protects the Mosfet as intended clamping the VDS of mosfet within safe operating conditions.

Having the Snubber helps smooth the ringing when the mosfet is turned off and on.

Thanks for all help and insight provided.

Answer 1

The TVS is will not work. The motor current needs a path to follow. With the FET on, the path is through the FET. When the FET turns off, there is no path until the TVS breaks over, which it will on every pwm cycle. This is periodic not transient.

The TVS average power dissipation is: $$P_{ave}=V_{brkdn}I_{motor}\frac{t_{off}}{t_{period}}$$

The peak drain voltage will be 24V +Vbreakdown.

To allow a path for the current, use a Shottky diode that will handle the maximum motor current. This will maintain the torque and limit the drain-source voltage to Vcc+a diode drop.

Position the diode to allow the motor current a direct path from the drain side to the 24V side.

Incidentally, the relay as drawn will not provide direction control.

Update:

To provide bi-directional pwm, both the supply and the pwm signal must be switched as shown below. Then a fly-back diode can be used on the pole side of the SPDT switch as shown.

The TVS is now used to protect against transients caused by changing direction while the motor still has current. The fly-back diode prevents any inductive kick-back from occurring by routing the current back to the motor, thus the TVS will not be engaged under normal operation.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

You are correct in that it means your TVS is undersized. Not surprising if your PWMing things since the TVS basically needs to be on par with the motor in a sense, and motors tend to be larger than most TVS diodes.

You can add an RC snubber in parallel with the TVS to help take up some of the energy. You could even replace the TVS with an RC snubber. RC snubbers can be made much hardier and also help with EMI whereas a TVS does nothing for EMI.

However, an RC snubber is incapable of clamping the flyback voltage to a fixed level. It will only slow the dv/dt (this is the part that helps with EMI that a TVS won't do) and in doing so decreases the flyback voltage peak. This is different than clamping the flyback voltage to a fixed level.

From there, the energy might still be too much for TVS because when it conducts it shunts energy away above the clamping voltage away from the the RC snubber. That's when you add a series resistor to the TVS so the TVS can't clamp so sharply which lets the RC snubber do more for longer.

Varistors will do it too mainly because you can buy them in massive sizes. But they are similar to RC snubbers in that they do not clamping. They also don't really decrease the dV/dt in my understanding (but they do more about it than a TVS diode will just by loading down the coil).

Answer 3

Imagine that the PWM is at a level where 4 A were commonly flowing while "on."

• During this time, when the FET opens, that energy (V*I) is still inside the coils of the motor.
• An "unsupported" electro-magnetic field almost instantly collapses, causing the voltage to invert and increase to whatever limits it. Since the power remains the same, the current must decrease proportionally.
• The TVS begins limiting the increase of this voltage when it reaches 32 Volts or so, some nanoseconds later.
• If there is enough energy available, it can push this all the way to 50V or so. Lets say 48 V for easy math.
• Now 48 V is twice 24 V, so the available current will be half.
• 48 V * 2 A = 96 Watts of energy in the TVS.
• If the duty cycle is only 10%, then the poor device still has to dissipate peak 9.6 Watts of power, of which most are just too small to dissipate this much heat.

One solution to this thermal issue is to use several TVS in series. Instead of one 32 V TVS, use three 10 V TVS. With a third the voltage dropped across each TVS, each dissipation will be a third (net dissipation will be the same however.)

This is still wasting quite a bit of power though. Why not just use a typical power diode as a flyback arrestor?

This might have to be added at the power input to the switch, to be valid for a reversible configuration. Note, this means that the motor cannot be energized and spinning when the relay is actuated - a voltage spike will be produced because there's a (relatively large) amount of time where the relay contacts are in-between positions.

Then, when the FET turns off, the voltage will only be able to invert and go to a volt or so. This will have the effect of "keeping more of the current in the motor windings" between pulses.

Since only 1 V * 4 A is 4 W, and 10% of this is 400mW, this wastes significantly less power and can be handled by a 1W diode (just get a 5W one; extra margin is good.)

Still another option is to use another FET in place of the diode altogether, and actively "short" the motor during pulses. This sounds counter-intuitive, but has the effect of having less voltage drop than a diode, thus wasting even less energy and keeping more of it in the motor. Of course that requires some rather precise timing to prevent voltage spikes and "shoot-through" when two FETs are on at the same time.

Of course this also depends on which type of motor is being used; this may work well for a universal or brushed DC motor, but might not be ideal for a brushless type.

Answer 4

Your circuit designer misunderstood what a Transient Voltage Suppressor (TVS) is for. The TVS are quite reasonably failing.

They should be removed from the circuit and replaced with another alternative, based on your characteristics like the motor specs, typical load, relay switching time vs motor powered time etc. For back emf clamping, this is a clamp diode across the motor, anode to -ve and cathode to +ve.

A TVS is for dissipating (a) high voltage and (b) very infrequent energy from ESD events, such as from human body static discharge reaching a circuit through a connector/cable.

They are not intended for dissipating power produced by a circuit in normal operation. They can dissipate high peak energy at a very low repetition rate which causes very low average power dissipation. Your part is spec'd for 1/1000 duty cycle and you're using it far above that.

They sometimes get seen (and used) as general-purpose bi-directional Zeners, which they are not. They have an imprecise conduction threshold voltage.

From the Littelfuse PK6E series webpage:

The P6KE Series is designed specifically to protect sensitive electronic equipment from voltage transients induced by lightning and other transient voltage events.

600W peak pulse capability at 10—1000µs waveform, repetition rate (duty cycles): 0.01%

As used, the TVS diodes are failing because they cannot conduct the heat through their package. They can do for the very brief and infrequent pulses they are designed to cope with.