In 4-quadrant switching amplifiers, it's quite common to run current
through a MOSFET "backwards" when it is on -- i.e., to run an NMOS with
the current traveling from source to drain, instead of drain to source.
Yet all the data sheets only show 1st-quadrant conduction, with current
(for an NMOS) running from drain to source.
Which boils down to my real problem: I'm working on a motor control
board, and I done messed up. The circuit I have has a PMOS transistor
that is supposed to block current from the motor when it is in generator
mode. In a moment of supreme blondness (extra-supreme, if you figure
that I was, at the same time, counseling a customer not to forget that
power FETs have intrinsic diodes), I laid the thing onto the schematic
with the source toward the + supply and the drain toward the "load",
neglecting the intrinsic diode.
So, can I just turn it around, as I show here? There will be four modes
of operation: (1), the board will be on, the H-bridge will be unpowered,
and the PMOS will not be activated. (2), the board will be on, the PMOS
activated, and the motor will be driving a load. (3), the board will be
on, the motor will be driven by the external "load" and will be
generating, and the PMOS will be off, to prevent "back feeding" the power
supply. (4), the board (and PMOS) will be off but the motor will be
driven, acting as a generator.
So, in "normal" operation the current will always be going through the
PMOS "backward" -- the only time that the source of the PMOS will be
higher than the drain will (hopefully) be when the PMOS is turned off and
blocking current.
Is this gonna work? Is it even remotely normal?
pmos
supply o-----+^+--------o------------.
||| | |
=== | | H-bridge
| | .--o---.
| .-. | |
| | | | o----.
| | | | | O
| '-' | | / \
| | | | | | motor
| | | | \ /
'--------o | | O
| | o----'
| | |
.-. '---o--'
| | |
| | |
'-' |
| |
| ===
| GND
|/
on o-------------|
|>
|
|
|
|
===
GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
MOSFET "Backwards" Current / Motor anti-generate
Started by ●July 10, 2012
Reply by ●July 11, 20122012-07-11
"Tim Wescott" wrote in message=20 news:A7mdnSHRgrGffWHSnZ2dnUVZ_hudnZ2d@web-ster.com...> I laid the thing onto the schematic with the source toward > the + supply and the drain toward the "load", > neglecting the intrinsic diode.[snip]> So, in "normal" operation the current will always be > going through the PMOS "backward" -- the only time > that the source of the PMOS will be higher than the > drain will (hopefully) be when the PMOS is turned off > and blocking current.> Is this gonna work? Is it even remotely normal?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
Reply by ●July 11, 20122012-07-11
On Tue, 10 Jul 2012 23:02:36 -0400, P E Schoen wrote:> "Tim Wescott" wrote in message > news:A7mdnSHRgrGffWHSnZ2dnUVZ_hudnZ2d@web-ster.com... > >> I laid the thing onto the schematic with the source toward the + supply >> and the drain toward the "load", >> neglecting the intrinsic diode. > [snip] > >> So, in "normal" operation the current will always be going through the >> PMOS "backward" -- the only time that the source of the PMOS will be >> higher than the drain will (hopefully) be when the PMOS is turned off >> and blocking current. > >> Is this gonna work? Is it even remotely normal? > > It seems that whenever you have voltage at the supply, the FWD of the > PMOS will conduct and the H-bridge will be powered. At least as you show > it with the drain to the supply and the source to the control resistors. > But when you turn it ON, the gate will be maybe 5 volts more negative > than the source, so it will just possibly allow a little higher voltage > to the motors. > > What you probably meant to do was to have the source at the supply and > bring the control resistors to the supply side rather than the load as > shown. This will turn the supply on and off as you desire, but it will > not prevent backflow from the motor as generator from driving the > supply. Actually this 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 think you will still need the diode to keep from forward-biasing the > C-B junction.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. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by ●July 11, 20122012-07-11
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. 2. 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. 3. 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: http://webpages.charter.net/dawill/tmoranwms "Tim Wescott" <tim@seemywebsite.com> wrote in message news:A7mdnSHRgrGffWHSnZ2dnUVZ_hudnZ2d@web-ster.com...> In 4-quadrant switching amplifiers, it's quite common to run current > through a MOSFET "backwards" when it is on -- i.e., to run an NMOS with > the current traveling from source to drain, instead of drain to source. > > Yet all the data sheets only show 1st-quadrant conduction, with current > (for an NMOS) running from drain to source. > > Which boils down to my real problem: I'm working on a motor control > board, and I done messed up. The circuit I have has a PMOS transistor > that is supposed to block current from the motor when it is in generator > mode. In a moment of supreme blondness (extra-supreme, if you figure > that I was, at the same time, counseling a customer not to forget that > power FETs have intrinsic diodes), I laid the thing onto the schematic > with the source toward the + supply and the drain toward the "load", > neglecting the intrinsic diode. > > So, can I just turn it around, as I show here? There will be four modes > of operation: (1), the board will be on, the H-bridge will be unpowered, > and the PMOS will not be activated. (2), the board will be on, the PMOS > activated, and the motor will be driving a load. (3), the board will be > on, the motor will be driven by the external "load" and will be > generating, and the PMOS will be off, to prevent "back feeding" the power > supply. (4), the board (and PMOS) will be off but the motor will be > driven, acting as a generator. > > So, in "normal" operation the current will always be going through the > PMOS "backward" -- the only time that the source of the PMOS will be > higher than the drain will (hopefully) be when the PMOS is turned off and > blocking current. > > Is this gonna work? Is it even remotely normal? > > pmos > > supply o-----+^+--------o------------. > ||| | | > === | | H-bridge > | | .--o---. > | .-. | | > | | | | o----. > | | | | | O > | '-' | | / \ > | | | | | | motor > | | | | \ / > '--------o | | O > | | o----' > | | | > .-. '---o--' > | | | > | | | > '-' | > | | > | === > | GND > |/ > on o-------------| > |> > | > | > | > | > === > GND > (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) > > > -- > My liberal friends think I'm a conservative kook. > My conservative friends think I'm a liberal kook. > Why am I not happy that they have found common ground? > > Tim Wescott, Communications, Control, Circuits & Software > http://www.wescottdesign.com
Reply by ●July 11, 20122012-07-11
"Tim Wescott" wrote in message=20 news:-fadnZXyVsu9bmHSnZ2dnUVZ_hudnZ2d@web-ster.com...> 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.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
Reply by ●July 11, 20122012-07-11
On Wed, 11 Jul 2012 00:49:21 -0400, P E Schoen wrote:> "Tim Wescott" wrote in message > news:-fadnZXyVsu9bmHSnZ2dnUVZ_hudnZ2d@web-ster.com... > >> 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. > > How high is the BEMF when the motor is turned off? Is there a lot of > energy stored in the motor due to an inertial load, or is it possible > that the shaft could be driven at a higher speed to produce a higher > generated voltage? There will normally be some generated spikes during > the dead-time of the PWM switching of the H-bridge, which is usually > dealt with by using snubbers, or even by adding a capacitor across the > motor.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_.> If you want to eliminate any possibility of reverse power flow, you can > turn on both bottom (or both top) MOSFETs in the bridge, for braking > mode, which will quickly stop the motor by dissipating the energy. As > long as there is not a large inertial load or a chance of generation due > to external drive of the motor shaft (as in an EV rolling downhill), the > power should be of reasonable level.That won't work when the rest of the circuit is powered down.> Another method is to use an N-channel MOSFET and a high side gate > driver. But that would not be an easy PCB patch.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. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●July 11, 20122012-07-11
On Tue, 10 Jul 2012 22:51:34 -0500, Tim Williams wrote:> 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.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? -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●July 11, 20122012-07-11
On Jul 11, 7:55=A0am, Tim Wescott <t...@seemywebsite.com> wrote:> On Wed, 11 Jul 2012 00:49:21 -0400, P E Schoen wrote: > > "Tim Wescott" =A0wrote in message > >news:-fadnZXyVsu9bmHSnZ2dnUVZ_hudnZ2d@web-ster.com... > > >> 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. =A0It'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. =A0So that's right out. > > > How high is the BEMF when the motor is turned off? Is there a lot of > > energy stored in the motor due to an inertial load, or is it possible > > that the shaft could be driven at a higher speed to produce a higher > > generated voltage? There will normally be some generated spikes during > > the dead-time of the PWM switching of the H-bridge, which is usually > > dealt with by using snubbers, or even by adding a capacitor across the > > motor. > > 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_. > > > If you want to eliminate any possibility of reverse power flow, you can > > turn on both bottom (or both top) MOSFETs in the bridge, for braking > > mode, which will quickly stop the motor by dissipating the energy. As > > long as there is not a large inertial load or a chance of generation du=e> > to external drive of the motor shaft (as in an EV rolling downhill), th=e> > power should be of reasonable level. > > That won't work when the rest of the circuit is powered down. > > > Another method is to use an N-channel MOSFET and a high side gate > > driver. But that would not be an easy PCB patch. > > 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. >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) http://cds.linear.com/docs/Datasheet/4412fa.p= df and (fig33) http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LI= TERATURE/DATASHEET/CD00134336.pdf -Lasse
Reply by ●July 11, 20122012-07-11
Tim Wescott a écrit :> In 4-quadrant switching amplifiers, it's quite common to run current > through a MOSFET "backwards" when it is on -- i.e., to run an NMOS with > the current traveling from source to drain, instead of drain to source. > > Yet all the data sheets only show 1st-quadrant conduction, with current > (for an NMOS) running from drain to source. > > Which boils down to my real problem: I'm working on a motor control > board, and I done messed up. The circuit I have has a PMOS transistor > that is supposed to block current from the motor when it is in generator > mode. In a moment of supreme blondness (extra-supreme, if you figure > that I was, at the same time, counseling a customer not to forget that > power FETs have intrinsic diodes), I laid the thing onto the schematic > with the source toward the + supply and the drain toward the "load", > neglecting the intrinsic diode. > > So, can I just turn it around, as I show here? There will be four modes > of operation: (1), the board will be on, the H-bridge will be unpowered, > and the PMOS will not be activated. (2), the board will be on, the PMOS > activated, and the motor will be driving a load. (3), the board will be > on, the motor will be driven by the external "load" and will be > generating, and the PMOS will be off, to prevent "back feeding" the power > supply. (4), the board (and PMOS) will be off but the motor will be > driven, acting as a generator. > > So, in "normal" operation the current will always be going through the > PMOS "backward" -- the only time that the source of the PMOS will be > higher than the drain will (hopefully) be when the PMOS is turned off and > blocking current. > > Is this gonna work? Is it even remotely normal? > > pmos > > supply o-----+^+--------o------------. > ||| | | > === | | H-bridge > | | .--o---. > | .-. | | > | | | | o----. > | | | | | O > | '-' | | / \ > | | | | | | motor > | | | | \ / > '--------o | | O > | | o----' > | | | > .-. '---o--' > | | | > | | | > '-' | > | | > | === > | GND > |/ > on o-------------| > |> > | > | > | > | > === > GND > (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) > >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... -- Thanks, Fred.
Reply by ●July 11, 20122012-07-11
On Jul 10, 10:08=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:> In 4-quadrant switching amplifiers, it's quite common to run current > through a MOSFET "backwards" when it is on -- i.e., to run an NMOS with > the current traveling from source to drain, instead of drain to source. > > Yet all the data sheets only show 1st-quadrant conduction, with current > (for an NMOS) running from drain to source. > > Which boils down to my real problem: I'm working on a motor control > board, and I done messed up. =A0The circuit I have has a PMOS transistor > that is supposed to block current from the motor when it is in generator > mode. =A0In a moment of supreme blondness (extra-supreme, if you figure > that I was, at the same time, counseling a customer not to forget that > power FETs have intrinsic diodes), I laid the thing onto the schematic > with the source toward the + supply and the drain toward the "load", > neglecting the intrinsic diode. > > So, can I just turn it around, as I show here? =A0There will be four mode=s> of operation: =A0(1), the board will be on, the H-bridge will be unpowere=d,> and the PMOS will not be activated. =A0(2), the board will be on, the PMO=S> activated, and the motor will be driving a load. =A0(3), the board will b=e> on, the motor will be driven by the external "load" and will be > generating, and the PMOS will be off, to prevent "back feeding" the power > supply. =A0(4), the board (and PMOS) will be off but the motor will be > driven, acting as a generator. > > So, in "normal" operation the current will always be going through the > PMOS "backward" -- the only time that the source of the PMOS will be > higher than the drain will (hopefully) be when the PMOS is turned off and > blocking current. > > Is this gonna work? =A0Is it even remotely normal? > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 pmos > > =A0supply =A0o-----+^+--------o------------. > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0||| =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 ==A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D=3D=3D =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0==A0 =A0| =A0H-bridge> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 .--=o---.> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 .-. =A0 =A0 =A0 =A0| =A0==A0 =A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 | | =A0 =A0 =A0 =A0| =A0==A0 =A0o----.> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 | | =A0 =A0 =A0 =A0| =A0==A0 =A0| =A0 =A0O> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 '-' =A0 =A0 =A0 =A0| =A0==A0 =A0| =A0 / \> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 | ==A0 =A0 =A0| =A0| =A0 | motor> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 | ==A0 =A0 =A0| =A0 \ /> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0'--------o =A0 =A0 =A0 =A0 | =A0 =A0 ==A0| =A0 =A0O> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 =A0 =A0 | ==A0 =A0 =A0o----'> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 =A0 =A0 | ==A0 =A0 =A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0.-. =A0 =A0 =A0 =A0'--=-o--'> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| | =A0 =A0 =A0 =A0 ==A0 =A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| | =A0 =A0 =A0 =A0 ==A0 =A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0'-' =A0 =A0 =A0 =A0 ==A0 =A0|> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 =A0 =A0 =A0==A0 |> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 =A0 =A0 =A0==A0=3D=3D=3D> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 =A0 =A0 =A0==A0GND> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |/ > =A0 =A0 =A0 =A0on o-------------| > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D=3D=3D > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0GND > (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)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
