PWM above 16v for speed control? Will it work?

Question

I have recently ordered 1 x VNH3SP30 Motor Driver Carrier MD01B

I was just reading the charts and found below saying about the pwm,

VNH3SP30 and VNH2SP30 Comparison:   VNH3SP30        VNH2SP30
Operating supply voltage (Vcc)      5.5 - 36 V*     5.5 - 16 V
Maximum current rating              30 A            30 A
MOSFET on-resistance (per leg)      34 mΩ           19 mΩ
Maximum PWM frequency               10 kHz          20 kHz
Current sense                       none            approximately 0.13 V/A
Over-voltage shutoff                36 V*           16 V minimum (19 V typical)
Time to overheat at 20 A**          8 seconds       35 seconds
Time to overheat at 15 A**          30 seconds      150 seconds
Current for infinite run time**     9 A             14 A

* Manufacturer specification. In our experience, shoot-through currents make PWM operation impractical above 16 V.

** Typical results using Pololu motor driver carrier with 100% duty cycle at room temperature.

Does this mean if I want to PWM (vary the speed) of 24v actuators under load, will this work correctly with this VNH3SP30 board? or will it 'jump' voltage and not work How/What would you recommend please?

Thanks in advance, Galen

Answer 1

Shoot-through current refers to the condition where both switches/MOSFETs on one side of the H-Bridge are on simultaneously. Under normal conditions, the H-bridge is under one of the following conditions:

If, however, both switches on one side are on simultaneously, a huge current can flow (Only 34mΩ per leg, remember?), which is usually destructive. I'm not sure why both switches would be on - Possibly an issue with gate capacitance and switching time?

The datasheet says that the PWM pin is:

Voltage controlled input pin with hysteresis, CMOS compatible. Gates of low side FETs are modulated by the PWM signal during their ON phase allowing speed control of the motor.

You might get better luck with higher voltages if you modulated the PWM such that it was centered on the ON-pulses:

Motor on:     ----    ----     --      --
S-Left-Top__/------\________/------\________  
S-Rght-Bot___/----\___________/--\__________
S-Rght-Top__________/------\________/------\
S-Left-Bot___________/----\___________/--\__  

But you'll definitely want to look very carefully at Figures 4, 5, and 6 of that datasheet.

Answer 2

Is it possible that the author of this data sheet meant to say "flyback current" rather than "shootthrough current"?

During the "on" time of the PWM, two transistors are on and getting warm: current flows from the battery through one high-side transistor through the motor through one low-side transistor to ground.

During the dead band, the other two transistors have current flowing through them and getting warm, even though all four transistors are turned off. The parasitic inductance of the motor causes current to flow from ground through the internal protection diode of one low-side transistor through the motor through the internal protection diode of one high-side transistor to the battery.

So that parasitic inductance indirectly causes the voltage across each of the original two transistors (that were recently turned off) to be a significantly higher voltage than the battery voltage.

The voltage blocking rating of a transistor applies to the actual voltage across the pins of the transistor (which during flyback is significantly more than the battery voltage). So to keep the transistor within its rating -- i.e., in order to keep the transistor from self-destructing -- the battery voltage much be significantly less than the transistor rating.

Alas, practically all motor controller driver manufacturers simply copy the voltage blocking rating of a transistor directly to their advertisement. It's used as a marketing number because it sounds bigger and more impressive than the actual battery voltage you can actually use with it.

Answer 3

If you have total control of the four transistors comprising the H-bridge, you can mitigate against shoot-through by leaving a dead-band in the control signals. What this means is, for both sides of the bridge, before you turn a high-side driver ON, you turn the low-side driver OFF and allow a small amount of time to elapse. Same goes for turning off the high side and turning on the low side.

The reason you would do this is that turn-on and turn-off times are asymmetric. Usually it takes a device a little longer to turn off than to turn on. If you try to simultaneously change a side of the bridge from high drive to low drive (or the other way), there will be a brief interval of time during which both devices will be partially conducting, and this allows large currents to flow directly between the supply rails, heating the transistors in the process. For efficiency and longer component lifetimes, you want to keep only one transistor on a side fully conducting, the other completely not.

The effect is exacerbated by higher voltages by ohm's law; more voltage means more current, and the heating in the transistors is consequently worse. The effect is also made worse by higher switching frequencies. Every time the polarity is reversed, a little burst of shoot-through current occurs. At a few hundred Hz and under 12V it's not worth worrying about too much. An industrial controller working at 300+ bus volts and 10+ kHz pwm definitely has a concern. Fwiw, dead band times of 1 to 2 microseconds suffice to ensure that shoot through doesn't occur.