Maker Pro
Maker Pro

Sunday Musings: regenerative MOSFET driver

T

Tim Williams

Jan 1, 1970
0
It's always so annoying to invert yourself a new power supply rail to
supply those high-side MOSFETs. And how much does it need? Average?
Peak? Microamperes or amperes?

Well the gate is a capacitor, why not recover some energy from it? Hmm,
could run a boost converter, but running one for sub-microsecond bursts
would be weird... How about just tacking an inductor on it, eh? Well,
then it'll resonate from its initial position to something swinging up and
down around a power supply rail. Let's clamp that flyback with a diode,
dumping the inductor's energy into the supply rail. With some keen drive,
I can see it running pretty low average current. Instead of a lossy
resistor, you have an inductor controlling rise/fall time. With a CMOS
sort of driver, I bet you could get consumption under a miliamp, using a
weedy flying capacitor supply for it.

Tim
 
W

whit3rd

Jan 1, 1970
0
It's always so annoying to invert yourself a new power supply rail to
supply those high-side MOSFETs. And how much does it need? Average?
Peak? Microamperes or amperes?
There's a trick that would work if the switching is frequent, that
uses a small transformer in the MOSFET source. When the switching
occurs, it puts a pulse of current onto the secondary (which drives
the
gate). Using a very small transformer, the source current
saturates the core and the gate drive doesn't run away.

So, if there's ANY switched MOSFET in the system, you can parasite
a small transformer onto it and use as many secondaries as required
for the high-side switches' little boost supplies.
 
G

gearhead

Jan 1, 1970
0
It's always so annoying to invert yourself a new power supply rail to
supply those high-side MOSFETs. And how much does it need? Average?
Peak? Microamperes or amperes?

Well the gate is a capacitor, why not recover some energy from it? Hmm,
could run a boost converter, but running one for sub-microsecond bursts
would be weird... How about just tacking an inductor on it, eh? Well,
then it'll resonate from its initial position to something swinging up and
down around a power supply rail. Let's clamp that flyback with a diode,
dumping the inductor's energy into the supply rail. With some keen drive,
I can see it running pretty low average current. Instead of a lossy
resistor, you have an inductor controlling rise/fall time. With a CMOS
sort of driver, I bet you could get consumption under a miliamp, using a
weedy flying capacitor supply for it.

Tim
I guess flying capacitor is just another way of saying charge pump.
And the idea is you want to get away with using a charge pump that can
only supply maybe a few dozen milliamps, when you really need more
than that for the drive.
Then you have to dump the spike from the inductor onto the charge
pump's output cap, not back into the supply rail. But you knew that.
 
J

James Arthur

Jan 1, 1970
0
It's always so annoying to invert yourself a new power supply rail to
supply those high-side MOSFETs. And how much does it need? Average?
Peak? Microamperes or amperes?

Well the gate is a capacitor, why not recover some energy from it? Hmm,
could run a boost converter, but running one for sub-microsecond bursts
would be weird... How about just tacking an inductor on it, eh? Well,
then it'll resonate from its initial position to something swinging up and
down around a power supply rail. Let's clamp that flyback with a diode,
dumping the inductor's energy into the supply rail. With some keen drive,
I can see it running pretty low average current. Instead of a lossy
resistor, you have an inductor controlling rise/fall time. With a CMOS
sort of driver, I bet you could get consumption under a miliamp, using a
weedy flying capacitor supply for it.

Tim

There ain't much c * Vgs^2 energy to recover. OTOH you can make a
snubber circuit that diverts energy into a capacitor, then use a
simple buck arrangement to return that energy to the supply rail.
That's been done, and improves efficiency a bit.

While you're at it, the snubber cap provides an above-the-rail voltage
that could be used as a high-side driver's supply. (Not previously
done, AFAIK) The converter would have to run inefficiently for a few
cycles to build up the bootstrap voltage, but you'd usually be soft-
starting anyway, then you'd be ready to party.

Cheers,
James Arthur
 
J

James Arthur

Jan 1, 1970
0
There ain't much c * Vgs^2 energy to recover. OTOH you can make a
snubber circuit that diverts energy into a capacitor, then use a
simple buck arrangement to return that energy to the supply rail.
That's been done, and improves efficiency a bit.

While you're at it, the snubber cap provides an above-the-rail voltage
that could be used as a high-side driver's supply. (Not previously
done, AFAIK)

Ah, done, patented (1988):

http://www.freepatentsonline.com/4760324.html

"A high efficiency switching power supply includes a non-dissipative
snubber circuit which provides dual polarity power utilized by a
driver of the switching transistor."

Cheers,
James Arthur
 
J

John Larkin

Jan 1, 1970
0
It's always so annoying to invert yourself a new power supply rail to
supply those high-side MOSFETs. And how much does it need? Average?
Peak? Microamperes or amperes?

Well the gate is a capacitor, why not recover some energy from it? Hmm,
could run a boost converter, but running one for sub-microsecond bursts
would be weird... How about just tacking an inductor on it, eh? Well,
then it'll resonate from its initial position to something swinging up and
down around a power supply rail. Let's clamp that flyback with a diode,
dumping the inductor's energy into the supply rail. With some keen drive,
I can see it running pretty low average current. Instead of a lossy
resistor, you have an inductor controlling rise/fall time. With a CMOS
sort of driver, I bet you could get consumption under a miliamp, using a
weedy flying capacitor supply for it.

Tim

One regenerative gate-drive configuration is the blocking oscillator.
All sorts of fun things can be done with this.

John
 
T

Tim Williams

Jan 1, 1970
0
James Arthur said:
There ain't much c * Vgs^2 energy to recover.

Idunno. Maybe not energy, but in terms of power, it's huge. I could pump
30 VA's into a whack of huge-channel FETs like the IRFZ46 -- 55V and about
as many amps, that's a wide channel and a wide gate on top of it. Toss
about ten in parallel for, say, a car starter switch and you've got a whole
mess of gate charge, and even more to switch it at a usual rate (like
100ns, not that you would want to switch 500A in that short time). And
it's all wasted power, that gate switching, if you use gate resistors.
This is what I was musing about.

Tim
 
T

Tim Williams

Jan 1, 1970
0
John Larkin said:
One regenerative gate-drive configuration is the blocking oscillator.
All sorts of fun things can be done with this.

Ah, I can see that being interesting.

I think.

How about...resonant class E?

Tim
 
J

James Arthur

Jan 1, 1970
0
Idunno. Maybe not energy, but in terms of power, it's huge. I could pump
30 VA's into a whack of huge-channel FETs like the IRFZ46 -- 55V and about
as many amps, that's a wide channel and a wide gate on top of it. Toss
about ten in parallel for, say, a car starter switch and you've got a whole
mess of gate charge, and even more to switch it at a usual rate (like
100ns, not that you would want to switch 500A in that short time). And
it's all wasted power, that gate switching, if you use gate resistors.
This is what I was musing about.

Tim

But modern FETs are so beautiful that that IRFZ46 you cite has a max.
Qg of just 72nC. Ten parts in parallel switching at 100kHz would only
use 10 * 72nC * 100kHz * 10V = 0.72 watts of gate drive, in exchange
for the ability to switch as much as about 7kW.

Let's double-check that. If we switch 14A @ 50V per FET, we get
rds(on) of 0.0165 ohms * (14 amps)^2 * duty cycle, or not more than
3.2 watts of conduction losses. If we can drive ~1A peak (per FET) we
can move the gate through the switching region in ~250nS, making for
switching losses of roughly 14A * 50V * 250nS * 100kHz = 17.5 watts
per FET.

Total loss per FET is then about 20.7 watts, giving a die temperature
rise of 20.7 watts x ~2 deg C/W = 42 deg. C. Applying the
manufacturer's derating factor of 0.72 W/C, each part would still be
rated to dissipate 107W - (0.72 W/C * 42 C) = 77 watts at 25 deg. C.
That's well under our 20.7 watt actual dissipation, leaving us room to
operate with a max. heat sink temperature of up to 103 deg. C without
exceeding the die limits.

20.7W of losses out of 7kW gives an efficiency of 97.0%.

Knock that 14A i(ds) down to 10A, drive the FETs harder, keep it cool,
and the thing'll run forever.

Not bad at all for 3/4 watts of drive.

Cheers,
James Arthur
 
Top