MOSFET protection diode question.

[qoole]

Hi all,

I am starting out in the world of speed controllers.
I have currently set up a circuit as described in the attached image.

The PWM signal comes from an Arduino running at 62khz. I chose that frequency because it was the only one that didn't turn the motor into a voice-coil that was driving all the neighbourhood cats and dogs (and me) mad!

As you can see, I'm using an N-Channel MOSFET in a bootstrapped high-side configuration.

This circuit works perfectly when the motor is unloaded. As soon as I put load on the motor (making the current increase) I release the magic blue smoke from the protection diode (D2 in that picture).
My question is this.... Why!?
Is it because the reverse recovery of the 1N4004 is too slow and the reverse current generated by the inductor that is the motor is surging through it and it's staying open long enough for current to shoot through it in reverse and thus cook itself?
Or is it just the amount of current coming back off the motor that is doing it?
Could I solve the issue by replacing it with a Schottky diode (if I had any)?


Question for future reference:
One thing I notice about Schottky diodes is that they have a relatively low reverse bias voltage. I've seen them up to about 70V. If, for argument sake, my motor controller ends up driving a 60V motor (or over-driving a 48V motor at 60V). Is that 10V overhead enough? Would that be safe to use?

Thanks,

Alex

[richard.cs]

I suspect the motor current is too high, remember that at the moment of switch-off all the current passes through that diode. Details of the motor please.

Unrelated: that source-follower means there will always be a few volts wasted across the mosfet. Unless you have a good reason for it put the mosfet in the negative side.

[DanielS]

I second what Richard said.

You are smoking your diodes because they are under-rated (all the kinetic energy in the motor will be dissipated in the diode, windings and mechanical losses, which can be quite a bit of energy) and using an open-drain N-MOS is the preferred way of building switched circuits since common-source makes the gate much easier to drive: in many cases, you can drive it directly with 2.5-3.3V logic.

The situation when you turn off power to the motor is actually worse than what Richard said: when you apply power to the motor, your current is: (source voltage - back-EMF - brush losses) / winding resistance but when you cut power, your current becomes (back-EMF - brush loss - diode) / winding resistance... if your back-EMF is 8V, your brush loss is 2V, diode is 1V and your winding is 0.2 ohms, your diode ends up seeing 25A during electro-braking instead of the 5A you might be driving your motor with at 12V.

[mzzj]

I second what Richard said.

You are smoking your diodes because they are under-rated (all the kinetic energy in the motor will be dissipated in the diode, windings and mechanical losses, which can be quite a bit of energy) and using an open-drain N-MOS is the preferred way of building switched circuits since common-source makes the gate much easier to drive: in many cases, you can drive it directly with 2.5-3.3V logic.

The situation when you turn off power to the motor is actually worse than what Richard said: when you apply power to the motor, your current is: (source voltage - back-EMF - brush losses) / winding resistance but when you cut power, your current becomes (back-EMF - brush loss - diode) / winding resistance... if your back-EMF is 8V, your brush loss is 2V, diode is 1V and your winding is 0.2 ohms, your diode ends up seeing 25A during electro-braking instead of the 5A you might be driving your motor with at 12V.

Think again. When you cut the power to the motor the polarity of the mechanical back-emf is same as during running and the diode is not conductin in that direction.

To orginal poster: yes, the 4004 is too slow for 64 khz and too wimpy if you motor current is more than 1 amp.

70v scotky diode sounds pretty marginal but for example Onsemi has schottky diodes up to 200 volts. Plenty of other possible diodes like bog-standard MUR420-460

[DanielS]

Think again. When you cut the power to the motor the polarity of the mechanical back-emf is same as during running and the diode is not conductin in that direction.

The overall back-EMF has the same polarity but the junk during brush switch-overs when the source is off (assuming OP uses a DC motor) might not.

A better option might be an RCD/LCD snubber network to eat that junk.

[qoole]

Hi all,

Thanks for the replies!

I suspect the motor current is too high, remember that at the moment of switch-off all the current passes through that diode.

Ah, I didn't realise this was quite how things worked, of course it seems obvious now. If the motor is drawing 3A under load when the shaft rotates and the polarity reverses all that current has to rush out of the motor (via the diode in this case). Makes so much sense. The back rush is proportional to the amount of current the motor is drawing.
I rather naively assumed the only things that needed to be uprated for a high-power motor were going to be the MOSFETs.
The motor I'm using is a very high-power wheelchair motor (which is going to be the the end use for this project, to control an electric wheelchair).

Unrelated: that source-follower means there will always be a few volts wasted across the mosfet. Unless you have a good reason for it put the mosfet in the negative side.

Why would there be a few volts wasted? I'm driving the gate hard open. With ((2xGateVoltage)-D1 loss). The reason for running it that way round is when I promote this up to a full H-Bridge all four quadrants will be N-Channel MOSFETs! P-Channel MOSFETs have worse ratings and higher RDSon than equivalent N-Channel ones.

Thanks,

Alex

[richard.cs]

Why would there be a few volts wasted? I'm driving the gate hard open. With ((2xGateVoltage)-D1 loss). The reason for running it that way round is when I promote this up to a full H-Bridge all four quadrants will be N-Channel MOSFETs! P-Channel MOSFETs have worse ratings and higher RDSon than equivalent N-Channel ones.

When Q1 is off the gate will (eventually, your frequency is a bit high to be using resisive get pullups) rise to about 11.3V through D1 and R2, if the gate threshold voltage was 4V say then the source will never rise past 11.3-4=7.3 Volts. It looks like you may have intended something else with D1 and C1, there are configurations that allow the output PWM to pump a capacitor up higher than the supply but this is not one of those.

If you added a pair of series diodes from ground to the top of C1 (keeping D1 or it'll never start) and connected a capacitor between their midpoint and the source of Q2 then if would behave like your current circuit for the first few cycles but C1 would evenutally charge up to around 20V giving you plenty of gate drive.

[qoole]

It looks like you may have intended something else with D1 and C1,

Ah, there is a mistake in my doodle. The bottom leg of C1 should be connected to the SOURCE pin of the MOSFET not GND.
Therefore when the mosfet starts conducting and the source pin heads towards VCC the bottom leg of C1 goes up with it pushing the voltage at the other pin up relative to ground. D1 is now reverse biased giving the charge nowhere to go other than towards the gate pushing the gate pin higher than VCC. When the PWM signal rises the transistor conducts pulling the gate pin low allowing C1 to recharge back up to (VCC-D1 drop) ish. Yes, depending on duty cycle C1 will have less time to recharge but so long as you never hit 100% for too long it will always retain enough charge.

Thanks,

Alex

[wraper]

Why would there be a few volts wasted? I'm driving the gate hard open. With ((2xGateVoltage)-D1 loss). The reason for running it that way round is when I promote this up to a full H-Bridge all four quadrants will be N-Channel MOSFETs! P-Channel MOSFETs have worse ratings and higher RDSon than equivalent N-Channel ones.

Thanks,

Alex

Because it is not fully open, there bi several volts drop. MOSFET needs at least couple of volts between source and gate to to so. Therefore it cannot fully open if you connect gate to source voltage. To do so, you need separate voltage source which voltage that least 5-10V higher than on source. That Protection diode is not necessarily needed, it produces huge waste of power trying to stop the motor. You can just return reverse voltage to power supply through higher voltage MOSFET internal protection diode if it can handle the current + varistor/RC snubber across motor leads. 1k gate resistor also is not good for 62kHz PWM as gate won't charge instantly. If you take, for example, MOSFET with 10nF gate capacitance, there will be 16 kHz low pass filter on it's gate. So this scematic basically becomes a sawtooth generator, not normal PWM. 62 kHz likely is to high for a motor, there should be significant torque loss. You need to charge and discharge gate fast, not only discharge. So better use such schematics. That charge pump you mentioned in the post above will fix fully opening (half assed way due to back EMF), but not other problems.

[richard.cs]

Ah, there is a mistake in my doodle. The bottom leg of C1 should be connected to the SOURCE pin of the MOSFET not GND.

This scheme will work but as you point out there will be a duty cycle limit. As wraper notes you still have a problem with the switching frequency and the 1k pullup.