MOSFET "Backwards" Current / Motor anti-generate

It seems that whenever you have voltage at the supply, the FWD of the = PMOS=20 will conduct and the H-bridge will be powered. At least as you show it = with=20 the drain to the supply and the source to the control resistors. But = when=20 you turn it ON, the gate will be maybe 5 volts more negative than the=20 source, so it will just possibly allow a little higher voltage to the=20 motors.

What you probably meant to do was to have the source at the supply and = bring=20 the control resistors to the supply side rather than the load as shown. = This=20 will turn the supply on and off as you desire, but it will not prevent=20 backflow from the motor as generator from driving the supply. Actually = this=20 is usually a good idea, and you could add a capacitor from the source to =

ground, and a diode to isolate the supply from the charging current.

What will work is a diode in series from the drain to the load, to block =

reverse current flow. Or you might be able to use a PNP transistor, but = I=20 think you will still need the diode to keep from forward-biasing the C-B =

junction.

I'd try an LTSpice simulation and see what happens. I don't feel like = doing=20 it, or thinking about it in more detail. Good luck!=20

No, I didn't mean to have the source at the supply -- that's what I have now, and it's failing at Job 1, which is to keep from backfeeding the supply.

Having the thing fail to block the supply when the motor isn't generating is just fine -- the H bridge is off until I turn it on, and I intend to turn the PMOS on hard before the H bridge goes on. It's preventing the motor from ramming power backwards through my circuit that I'm trying to avoid.

If I put a diode in there it'll be the biggest power consumer on the whole board, by a mile. So that's right out.

So you've got a half or H-bridge, 100% conduction angle (whether the current is forward or backward at any given time), and a filter inductance (including the motor itself)?

Then presumably, operation is such that:

1. Under normal, powered operation, the motor's [average] terminal voltage is less than the supply voltage. Peak operation might reach 100% duty cycle, but presumably you aren't running there for long. Since after all, if you're maxing it out on a continuous basis, you must've chosen an insufficiently sized power supply or motor for the application.

1. As a result, it stands to reason that, when generating, the motor's terminal voltage will not exceed the forward voltage. Thus, since you've presumably dimensioned the supply sufficiently in excess of forward voltage, the motor won't "bang into" the power rails when generating, unless your driver specifically makes it do so.

2. Of course, as long as the driver keeps switching, the motor inductance also serves as a boost inductance, so you can draw power from the motor, which is producing a lower voltage than your supply.

If this is the case, then source to supply is correct for both PMOS (S = +V) and NMOS (S = -V). It's just a large CMOS inverter.

If this isn't the case, and you do need to allow for conditions where the motor is not delivering current, yet its terminal voltage must be allowed to rise arbitrarily high (above the supply rails), then you've made yourself a problem. Might use back-to-back IGBTs or MOSFETs to turn off the motor when it's desirable to ensure its "out-of-circuit-ness". Note that these MOSFETs must stand off generating voltage less the total supply (measured from motor common to +/-V for half bridge, or +V to -V for full bridge), or: Vds >= Vgen - Vsupply. They also have to be rated for enough current to minimize losses; on the upside, they don't need to switch quickly.

Tim

-- Deep Friar: a very philosophical monk. Website:

How high is the BEMF when the motor is turned off? Is there a lot of = energy=20 stored in the motor due to an inertial load, or is it possible that the=20 shaft could be driven at a higher speed to produce a higher generated=20 voltage? There will normally be some generated spikes during the = dead-time=20 of the PWM switching of the H-bridge, which is usually dealt with by = using=20 snubbers, or even by adding a capacitor across the motor.

If you want to eliminate any possibility of reverse power flow, you can = turn=20 on both bottom (or both top) MOSFETs in the bridge, for braking mode, = which=20 will quickly stop the motor by dissipating the energy. As long as there = is=20 not a large inertial load or a chance of generation due to external = drive of=20 the motor shaft (as in an EV rolling downhill), the power should be of=20 reasonable level.

Another method is to use an N-channel MOSFET and a high side gate = driver.=20 But that would not be an easy PCB patch.

Sometimes ya just gotta punt!

Paul=20

There is a possibility that Bubba might grab the output shaft when the thing is supposed to be off and give it a spin.

Or that Bubba might turn the output shaft faster enough that the motor will generate more than the supply.

I need a method to turn the _supply_ to the H-bridge OFF at _any time_.

That won't work when the rest of the circuit is powered down.

That won't work for the same reason that a P-channel won't -- because power MOSFETs have intrinsic diodes that let the current run "backwards" when the FET is off.

To reiterate:

My question is, can I run a MOSFET "backwards", reliably, all day.

Right. That's my situation. And I'm asking about the FET in the back to back pair that has the current running "backwards" (because the top FETs in my half-bridges are the 'forward' FETs.

How kosher is it to do this? What unexpected problem am I going to have?

I say yes, I've used back to back fets for 230V AC, fast enough to turn off if shorted

here's another example (fig4)

-Lasse

Tim Wescott a écrit :

So you're after a "no drop" diode and don't want to switch the PSU rail?

In that case it's perfectly OK. Your mosfet can be driven because the source voltage is defined by the mosfet internal diode.

Obviously, don't forget to switch the MOS "off" when the motor is in source mode...

d,

=A0|

=A0 =A0| =A0H-bridge

o---.

=A0 =A0|

=A0 =A0o----.

=A0 =A0| =A0 =A0O

=A0 =A0| =A0 / \

=A0 =A0 =A0| =A0| =A0 | motor

=A0 =A0 =A0| =A0 \ /

=A0| =A0 =A0O

=A0 =A0 =A0o----'

=A0 =A0 =A0|

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=A0 =A0|

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=A0=3D=3D=3D

=A0GND

Yes, you can use the FET backwards like that--it's fine.

Your particular circuit won't work--whenever power's applied, the motor load pulls the FET gate low whether you want it low or not.

You need to pull the gate up to the supply voltage rather than the load voltage.

-- Cheers, James Arthur

Okay, I jumped the gun--I thought you were switching power to the H- bridge.

Power is applied to the H-bridge via the intrinsic diode. Works if the H-bridge transistors are guaranteed 'off.'

ing a load.

Fine.

Can that produce open circuit voltages >Vgs(br)? If yes, you need a Vgs clamp.

See above.

=A0 =A0 =A0| =A0H-bridge

o---.

=A0 =A0|

=A0 =A0o----.

=A0 =A0| =A0 =A0O

=A0 =A0| =A0 / \

=A0 =A0 =A0| =A0| =A0 | motor

=A0 =A0 =A0| =A0 \ /

=A0| =A0 =A0O

=A0 =A0 =A0o----'

=A0 =A0 =A0|

-o--'

=A0 =A0|

=A0 =A0|

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=A0GND

-- Cheers, James Arthur

Simply use 2 PMOS mosfets in series drain-to-drain. Just don't tie the gates together; they'll need seperate levels. Also make sure not to exceed Vgs limits.

The back-to-back MOSFETs form a solid state switch that is equally good = on=20 AC or DC, in both directions, so it won't prevent the motor from = generating=20 current into the source if its voltage is higher. But you can add a = current=20 sensor to turn off the switch if the current changes polarity and tries = to=20 charge the source. I made a simple LTSpice circuit. But the trick is how = to=20 drive the gates of the two MOSFETs without using an isolated supply. It=20 might be possible with an NMOS and a PMOS, with the PMOS on the positive =

side and the NMOS on the negative side. Yes, it's possible, and here it = is:

I used a +/- 15V square wave for the motor as generator, and I varied = the=20 supply voltage to show when the generator signal starts driving through = the=20 switch when it's on. Some of it also comes through when the source = voltage=20 is less than 3.5V, but that can probably be fixed. There's probably a = point=20 at which the generated current doesn't matter.

The NMOS back-to-back circuit does not have any problem:

Paul=20

OK. That's what I figured, but I wanted to check.

Yup. And switch the thing on when the motor is being driven (unless I like toasty transistors).

Sure. That's the usual supply polarity reversal protection circuit, when you can't stand a diode drop: PFET with gate grounded, source to + supply, drain to the rest of the circuit.

Cheers

Phil Hobbs

Thanks Phil. I figured it was perfectly OK.

Absolutely fine! I've never seen a MOSFET whose body diode wasn't rated for the same current as the channel. As long as the gate is "on", the channel will conduct equally well in both directions.

If the channel's resistance is sufficiently low so as to keep voltage drop under Vf, the diode will never even see injected charge -- which isn't a concern for "on/off" duty, but will keep "recovery time" negligible, if for some reason you needed it to reverse quickly.

The only time it's not acceptable is when Rds(on) is insufficient to keep the diode off, and the charge injection through the body diode disrupts other junctions -- this can only occur in monolithic devices, where the injected charge turns on a parasitic VDD-to-VSS SCR structure. The gain of this SCR is intentionally very low, so that it can't be triggered at normal levels (supposedly up to 100mA peak, on input pins, for 74HC logic), which isn't to say it's impossible.

Tim

If you don't want regen power at all, then put a series diode after the PMOS switch. For that matter, you don't even need the PMOS switch if you don't ever need regen energy.

One could even put a clamping R circuit (DBr) around that diode as part of a power supply over voltage and switch in the DBr when needed to drain it off.

have a good day.

Jamie

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=20

IIRC back to back power fets is one way of making DC SSRs.

?-)

I use N-channels, "bass-ackwards", to prevent the body diode from conducting when doing charge-pump gimmicks to generate a negative supply.

(Integrated MOSFET's and most discrete's are symmetric in their behavior except for the body diode.) ...Jim Thompson