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'[EE]: AUVic needs help from EE with Motor Control '
2002\09\08@002601 by Donovan Parks

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face
Hello,

I am contacting this list on behave of the a University of Victoria student club known as the AUVic.  We are attempting to build an autonomous underwater vehical and as such we are designing a medium current (24V @ 30A) motor controller.

We have just finished the design of the motor controller and would like to get in contact with a knowlegable individual who has some experience with motor control circuits so we can clarify some design issues (or anyone who thinks they can help).  Namely, we are uncertain about the following to design choices:

   1) placing a zener diode between the gate and source of our MOSFETs to ensure Vgs stay within safe level even in the face of transient voltage spikes caused by lead inductance and the gate capacitance.  We do have a gate resistor so perhaps the zener is unnecessary, but perhaps not.  The zener is attached directly to the source but on the drive side of the gate resistor as   we've encounter some documentation that suggests a zener diode may lead to increased ringing if attached directly across the gate and source.  Is this zener needed? Are we using it properly?

   2) placing a schottky diode across the gate capacitor so that is is forward biased when the MOSFET is turning off.  The rational for this diode is to reduce the RC time constant caused by the gate capacitance and gate resistor so the MOSFET can be turned off faster (a desirable feature given the motor controller IC we are using).  Our analysis of the circuit indicates this diode should be fine (that is, we have not created a low impedance path between our parallel MOSFETs), but we would like a second opinion.

   3) placing bi-directional TVS between the drive and source of the MOSFETs.  Are they needed or will the intrisic body diodes be enough (the real problem here is we don't know how to calcualte how long it will take the body diode or TVS to turn on - the datasheet of both just indicate it will be dominated by other factors, but we don't know what the other factors are)?

I realize there is not enough information here to answer these question.  I'm hoping to find an individual or two who are willing to help so I can send these people our schematic and design document.

Regards,
Donovan Parks
AUVic Team Captain

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2002\09\08@003458 by Spehro Pefhany

picon face
At 09:25 PM 9/7/02 -0700, you wrote:
>Hello,

Donavan: Send your schematic (PDF or GIF preferred, zipped) and I'll have a
look at it as
time permits.

Please indicate operating frequency, if not shown. Inductance in the source
lead can cause
problems, as you indicate. If you have a gif or JPG photo of the layout,
pls. send that too.
Turn on of the body diode is not a problem, trr can be.

Best regards,



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2002\09\08@102842 by Peter nicol

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Are you using a "H Bridge" driver?

Some very useful application notes for the HIP4080/81/82 devices at

http://www.intersil.com/design/Parametric/productinfo.asp?pn=HIP4081AIP

Recommends 10 Ohm inline resistors (driver to MOSFET gate),  Schottkys for
faster discharge of the MOSFET Gates, but nothing about Zeners or additional
Flywheel Diodes when using the 4081.

Word of caution, watch the Back EMF, it a killer...


{Original Message removed}

2002\09\08@112058 by Doug Butler

picon face
My experience is sonar transmitters, FETs driving transformers at 10 -
500kHz.  I have seen people put all sorts of crap between the driver chips
and the FET gates, but I have found almost none of it necessary.  The guys
who design the driver chips know what they are doing, just follow the data
sheet!  Sometimes I add a large series cap and pull down resistor (.1uF +
100k) in case the processor locks up, to prevent the PCB from catching fire
if one FET is left on.  I have seen this happen!

Doug Butler
Sherpa Engineering


> {Original Message removed}

2002\09\08@183321 by Donovan Parks

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face
Hello,

We are indeed using an h-bridge drive.  We are using the LT1162 from Linear
Technologies (we looked at the HIP4080/81/82, but it isn't avaliable from
digi-key so went with the LT1162).

Peter: thanks for the link.

In what manner is the back EMF a killer?

Regards,
Donovan Parks
AUV2003 Team Captain


> Are you using a "H Bridge" driver?
>
> Some very useful application notes for the HIP4080/81/82 devices at
>
> http://www.intersil.com/design/Parametric/productinfo.asp?pn=HIP4081AIP
>
> Recommends 10 Ohm inline resistors (driver to MOSFET gate),  Schottkys for
> faster discharge of the MOSFET Gates, but nothing about Zeners or
additional
> Flywheel Diodes when using the 4081.
>
> Word of caution, watch the Back EMF, it a killer...
>
>
> {Original Message removed}

2002\09\09@022722 by Peter nicol

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face
When using PWM you will find a lot of "Back EMF" this is because you turning
the inductive load on and off every pulse, if you turn off all MOSFET gates
then your "H Bridge" becomes a bridge rectifier.

Rule of thumb is your Back EMF may be 7 times your line voltage!

Regards

Peter Nicol.

{Original Message removed}

2002\09\09@032250 by Donovan Parks

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face
Hello,

Yes, I understand.  We have designed the motor controller to never turn off
all the MOSFETs.  On the "off" part of the PWM signal we turn on both
low-side MOSFETs so there is a path for current to flow.  This should reduce
the Back EMF generated (or more accurately the voltage caused by changing
the current through the inductive load) as the current is not rapidly
brought to zero. Thoughts?

Regards,
Donovan


> When using PWM you will find a lot of "Back EMF" this is because you
turning
> the inductive load on and off every pulse, if you turn off all MOSFET
gates
> then your "H Bridge" becomes a bridge rectifier.
>
> Rule of thumb is your Back EMF may be 7 times your line voltage!
>
> Regards
>
> Peter Nicol.
>
> {Original Message removed}

2002\09\09@081944 by Olin Lathrop

face picon face
> Rule of thumb is your Back EMF may be 7 times your line voltage!

Oh no, here comes the voodoo physics again.

Unless the motor is being mechanically driven by the load, its back EMF
can't be higher than the supply voltage.  I'll assume that when you said
line voltage you really meant motor power supply voltage.

The back EMF comes from the motor acting like a generator.  The back EMF is
proportional to the speed.  This is what limits the speed of unloaded
motors.  Even with a perfect motor with a fixed voltage applied to it, the
speed will increase until the back EMF matches the applied voltage.  At this
point the motor speed can't increase any further because the voltage to
drive the motor is the applied voltage minus the back EMF.

It is therefore impossible with a perfect unloaded motor to get it to run at
at speed at which the back EMF exceeds the applied voltage.  Losses and
loads on real motors only mean that the back EMF can't even reach the
applied voltage.


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2002\09\09@082809 by Olin Lathrop

face picon face
> On the "off" part of the PWM signal we turn on both
> low-side MOSFETs so there is a path for current to flow.

Maybe I misunderstand what you are doing, but this sounds like a bad idea.
You need to let the motor "free wheel" between PWM pulses.  If not, you will
be drawing current from its back EMF which will appear like viscous
friction.  In other words, you will be shorting it while it's acting like a
generator.

In fact, you have to make sure that whatever current path you allow for the
flyback pulses doesn't also short the back EMF.

Believe it or not, I actually ended up picking up a project where someone
had stupidly done that.  Their complaint was that they couldn't get the
motor to run fast enough.  That's because the coils being driven at any one
time were fighting against the other coils that were shorted generators.


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2002\09\09@123859 by Spehro Pefhany

picon face
At 08:27 AM 9/9/02 -0400, you wrote:


>Maybe I misunderstand what you are doing, but this sounds like a bad idea.
>You need to let the motor "free wheel" between PWM pulses.  If not, you will
>be drawing current from its back EMF which will appear like viscous
>friction.  In other words, you will be shorting it while it's acting like a
>generator.

It's okay. His PWM frequency is high enough that the current is
essentially constant, assuming a conventional motor.

You do have to worry about this if the frequency is too low in comparison
with the motor inductance.

Best regards,

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2002\09\10@010610 by Donovan Parks

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Hello,

The h-bridge is being driven at 20kHz and we are using a PM DC motor.  I am
indeed "shorting" the motor during the off time of the PWM signal.  The
rational for this is to reduce the power dissipated in the MOSFETs.  If the
MOSFET is not turned on then the power dissipated is greater due to the
relatively large voltage drop across the body diode.  As Spehro mentioned,
the inductive nature of the motor allows this to work as long as the
frequency is high enough (and we have seen motor controllers that do this
for the reason I mentioned).

Thanks for the thoughts though.  Keep them coming.

Regards,
Donovan Parks
AUVic 2003 Team Captain

> >Maybe I misunderstand what you are doing, but this sounds like a bad
idea.
> >You need to let the motor "free wheel" between PWM pulses.  If not, you
will
{Quote hidden}

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2002\09\10@061106 by John

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Hello Olin, Peter, Donovan & PIC.ers,

Aren't you guys talking past each other?

   {
   ............
   I know that you believe you understand what
   you think I said, but I am sure you realise
   that what you heard is not what I meant.

           Richard Nixon
   .........
   }

The e.m.f. back-generation as described by Olin...  ok....

I think Peter was actually referring to inductive kick, experienced
when attempting to force high dI/dt upon an unsuspecting
motor, coil, etc.
Different animal, same day.


{Quote hidden}

<el snippo>
...........

>> When using PWM you will find a lot of "Back EMF" this is because you
>>turning the inductive load on and off every pulse, if you turn off all
>>MOSFET gates then your "H Bridge" becomes a bridge rectifier.
>>
>> Rule of thumb is your Back EMF may be 7 times your line voltage!
>>

           best regards,   John


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Cellphone no:   082 741 6275
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2002\09\10@073112 by Olin Lathrop

face picon face
> The h-bridge is being driven at 20kHz and we are using a PM DC motor.  I
am
> indeed "shorting" the motor during the off time of the PWM signal.  The
> rational for this is to reduce the power dissipated in the MOSFETs.  If
the
> MOSFET is not turned on then the power dissipated is greater due to the
> relatively large voltage drop across the body diode.  As Spehro mentioned,
> the inductive nature of the motor allows this to work as long as the
> frequency is high enough (and we have seen motor controllers that do this
> for the reason I mentioned).

That's fine if your pulse frequency is indeed high enough so that you are
always seeing flyback current between pulses.  However, I thought this was a
brushless DC motor, so each winding will be off during part of the cycle.
That should be much longer than it takes the flyback current to die down.
What are you doing with the winding during that time?


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2002\09\10@102246 by Madhu Annapragada

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I figured I might put down what I understand about switching behavior of
FETs used as bridges. This might get you started in the right direction.
Here are some calculations I ran from the data on the datasheet for the
IRF1404.
First the Data in the data sheet that you will need:
(1) RDS_On (Rd)= 4mOhms
(2) Total Gate Charge(Qg) = 200nC
(3) Turn_on_Delay_Time(Tdon) = 17ns
(4) Rise_Time (Tr)= 140ns
(5) Turn_Off_Delay_Time(Tdoff)=72ns
(6) Fall_Time(Tf) = 26ns
(7) Reverse_Recovery_charge (Qrr) = 270nC (for the internal diode)
(8) Reverse_Recovery_Time (Trr) = 110ns (for the internal diode)

I am going to assume the following:
(1) Your Bus voltage (Vb) = 24V (change this if you need to )
(2) Your Load Current (Iload) = 100A
(3) Switching Frequency (f)= 20KHz.
(4) The amount of current your FET driver can source(Ig) is 1.5A. (this is
critical. If you are using a FET driver to drive your bridge make sure that
the driver can supply enough current to turn the FET on in the time you
want it to turn on ; current_needed = gate_charge/rise_time.

Now the simplification (assume no stray inductances in the bridge and a
perfect board layout)
From the above values I get the following:
(1) RDS_On at a Junction Temp (RDS_on_Tj) of say 100degC = RDS_on * 1.3 =
5.2mOhms (from Figure 4 on the data sheet)
(2) Actual Rise time (Tar) = Qg/Ig = 133ns. Since the rise time of the FET
is greater than this the Actual rise time will be equal to Tr.
(3) Fall time is the same as Tar (unless your FET driver can source a
higher current for turn off which is sometimes the case)
(4) Di/Dt for the rise time = Load current (Iload) / Tar = 0.714 A/ns.
(5) Dv/Dt for the fall time = Bus voltage / Fall time = 0.18V/ns (this is
much less than the device dv/dt so you are okay here)
(6) Reverse recovery current (Irrm) = 2 * Qrr/Trr = 4.9A.
(7) Turn on losses = [0.55 * Vb * (Iload + Irrm)^2 * f * 2E-9] / Di/Dt =
8.475 Watts
(8) Turn off losses = Vb^2 * ILoad E-9 * f / (dv/dt) = 6.667 Watts
(9) Conduction Losses = RdsonTj * ILoad^2 = 64 Watts.
(10) Total Losses are approximately 80 Watts per cycle of your PWM
waveform.

Now for designing your gate drive and your heatsink.
Use the total losses and a Tj of 100 degC to compute the heat sink needed.
This is the easy part. Wakefield has a number of app notes on selecting the
right heat sink given the amount of power you want to dissipate. In order
to figure out if the FET, and your FET driver will do the job look at a
single switching sequence.
(1) Let us look at the point when the top FET in one leg and the bottom FET
in the other leg have been conducting.
(2) Your input waveform now transitions from say high to low for the top
leg. The input waveform for the bottom leg will still be low if you have a
deadtime built into the system to prevent shoot through. The drain current
in the TOP FET of one leg now starts to fall with a rate of Di/Dt. This
induces a voltage of magnitude Ldi/dt where L is the inductance of your
motor winding at the center of the leg. As soon as the voltage at the
center of the leg reaches one diode drop below the body diode of the bottom
FET of the same leg, the load current will be supported by the body diode
of the bottom FET.
(3) After the deadtime is up, the Top FET of the other leg is turned on the
bottom FET of the first leg is also turned on. This is where the reverse
recovery time of the body diode comes into play. The body diode of the
bottom FET has to go into reverse bias (i.e. the minority carriers in the
body-drain region have to be swept away and the depletion region
established) before the top FET can block voltage. IF this process takes
too long, then you will have shoot through currents. To keep things simple
one would not want the Trr of the body diode to be any longer than the dead
time (preferably at least 50% less than the dead time). I would recommend
the app note from IOR AN947 that explains this behavior very well. If you
find that the reverse recovery time is too long, then you will have to get
an external diode that has a faster reverse recovery time or a smaller
reverse recovery charge.

The thing to keep in mind is that you want to minimize your switching
losses. The obvious way to do this is to increase the amount of current
into the gate of your FET. However, this would mean that when you switch ,
you will have to sweep away a lager amount of minority carriers in the
body -drain region to re-establish the depletion region. So your reverse
recovery time will be greater and your motor performance will suffer. The
other consequence of switching too fast is the dv/dt of one FET will induce
a spike in the gate of the other FET due to the parasitic Gate-Source
capacitance. If your dv/dt is too large, then the induced spike might have
enough amplitude to turn on the FET when it should be off. I have found
that experimenting with various values of the gate drive resistance  (use a
parallel schottky to make sure you are not increasing turn off ) is the
final answer to this multidimensional optimization.
In some of my designs I have used a HIP4086 (3ph driver) and an
APT10M19BVFRR to get about 25A of current (RMS) at 48VDC to run a 3ph bldc
motor. I ended up using a dead time of 500ns and a gate drive resistance of
20ohms (with a schottky in parallel).
Good luck
Madhu
{Original Message removed}

2002\09\10@134721 by Mike Singer

picon face
I'd consider slightly different approach, I mentioned few weeks
ago. As far as I understand  Donovan has power supply with
some fixed voltage and DC motor, which is to rotate at variable
speed.
You know these DC/DC converter controllers for about
$1. They have a potential divider in the loopback. Why not
replace bottom resistor of the loopback by digital potentiometer
or by just a set of resistors with open collectors, controlled by
PIC. Let DC/DC converter do all this dirty job to handle FET and
inductance at its 1Mhz with 85-95% efficiency. Variable voltage
to feed DC motor could be achieved by varying bottom resistor
value of the loopback.
I didn't try this due to the lack of time. May be somebody could
try this approach and tell the List about the results.
Three such channels could rather cheaply produce 3-phase AC
to feed brushless EC motors, by the way.

Thank you.
Mike.


Olin Lathrop wrote:
>> The h-bridge is being driven at 20kHz and we are using a PM DC motor.
I am
>> indeed "shorting" the motor during the off time of the PWM signal.
The
>> rational for this is to reduce the power dissipated in the MOSFETs.
If the
>> MOSFET is not turned on then the power dissipated is greater due to
the
>> relatively large voltage drop across the body diode.  As Spehro
mentioned,
>> the inductive nature of the motor allows this to work as long as the
>> frequency is high enough (and we have seen motor controllers that do
this
>> for the reason I mentioned).
>
>That's fine if your pulse frequency is indeed high enough so that you
are
>always seeing flyback current between pulses.  However, I thought this
was a
>brushless DC motor, so each winding will be off during part of the
cycle.
>That should be much longer than it takes the flyback current to die
down.
>What are you doing with the winding during that time?

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2002\09\10@172923 by Peter L. Peres

picon face
On Mon, 9 Sep 2002, Olin Lathrop wrote:

>> Rule of thumb is your Back EMF may be 7 times your line voltage!
>
>Oh no, here comes the voodoo physics again.

A loaded circuit Q of 7 brings luck. Avoid loaded Q's of 13.

>Unless the motor is being mechanically driven by the load, its back EMF
>can't be higher than the supply voltage.  I'll assume that when you said
>line voltage you really meant motor power supply voltage.

The back EMF of the motor is not higher than the supply, the kickback from
switching the phases brutally off is.

>The back EMF comes from the motor acting like a generator.  The back EMF is
>proportional to the speed.  This is what limits the speed of unloaded
>motors.  Even with a perfect motor with a fixed voltage applied to it, the
>speed will increase until the back EMF matches the applied voltage.  At this
>point the motor speed can't increase any further because the voltage to
>drive the motor is the applied voltage minus the back EMF.

That's correct but the switching spikes are much higher as you know. Also
as you know the motor has a Ri and each coil in it is fed curent through
Ri when on. This drops some vlotage. When the phase is turned off the coil
will try to continue to drive the same current but nothing will be
opposing it so the voltage drop on Ri is low. Thus the EMF due to coil
reaction is always higher than the drive voltage even without kickback
considered.

The same problem exists with generators which tend to output incredible
voltages when unloaded abruptly (see load dump).

The op's did not mention whether their supply is from a source that can
accept reverse charging. If it does, then the 'bridge' formed by the
turned off FETs will actually charge the battery with a current pulse.
This is convenient in most cases. The efficiency of this can be increased
by turning the appropriate FETs on briefly.

Peter

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2002\09\11@005827 by Donovan Parks

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Hello Olin and company,

Hmmm... first, we are planning to use a PM DC motor.  Is this the same as a
brushless DC motor?

Given that it is, then we have not considered doing anything with the
windings during the off time.  What should we be doing?

From what we have read it appears to be a poor idea to let the motor "free
wheel" during the off time of the PWM pulse as then there is no path for the
inductive current to follow which results in a large voltage being produced.
Is this correct?  The way we understand it, is that if you have a current
going through an inductive load and you quickly turn off all the MOSFETs
then the voltage produced will be extremely large due to the rapid change in
current (v = L * di/dt).  However, if instead you turn on both the upper or
lower MOSFETs during the off time of the PWM signal then the current has a
path to follow and the di/dt term will be much lower (i.e. the current is
still being reduced as you are no longer supplying a voltage, but the
current cycles around for a finite amount of time).  That doesn't really
sound right now that I've written it out though.

So much to learn!

Regards,
Donovan Parks


> > The h-bridge is being driven at 20kHz and we are using a PM DC motor.  I
> am
> > indeed "shorting" the motor during the off time of the PWM signal.  The
> > rational for this is to reduce the power dissipated in the MOSFETs.  If
> the
> > MOSFET is not turned on then the power dissipated is greater due to the
> > relatively large voltage drop across the body diode.  As Spehro
mentioned,
> > the inductive nature of the motor allows this to work as long as the
> > frequency is high enough (and we have seen motor controllers that do
this
> > for the reason I mentioned).
>
> That's fine if your pulse frequency is indeed high enough so that you are
> always seeing flyback current between pulses.  However, I thought this was
a
> brushless DC motor, so each winding will be off during part of the cycle.
> That should be much longer than it takes the flyback current to die down.
> What are you doing with the winding during that time?

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2002\09\11@005907 by Donovan Parks

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face
Hello,

We are planning to drive the motors at 12 or 24V from some NiCd batteries.
The maximum continuous current draw is around 12A for the motors we are
using (peak current around 100A).  Three questions:

   1) Can NiCd accept "reverse charging" (they are obviously rechargable,
but I'm guessing this isn't exactly the same thing)?
   2) Can NiCd accept this sort of heavy current draw?
   3) How can I find the answers to this questions for myself? (I've read
several websites on batteries, but they did not address these issues)

Regards,
Donovan Parks
AUV 2003 Team Captain

{Quote hidden}

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2002\09\11@030457 by Donovan Parks

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face
Hello Madhu,

Thanks for the detailed calculations.  It will take me a few days to digest
this, but I'm sure I'll have a few questions at that time.  Can you
recommend any books and/or links that may help me understand some of this
information?  Thanks.

Regards,
Donovan Parks
AUVic Team Leader


{Quote hidden}

that
{Quote hidden}

the
> right heat sink given the amount of power you want to dissipate. In order
> to figure out if the FET, and your FET driver will do the job look at a
> single switching sequence.
> (1) Let us look at the point when the top FET in one leg and the bottom
FET
> in the other leg have been conducting.
> (2) Your input waveform now transitions from say high to low for the top
> leg. The input waveform for the bottom leg will still be low if you have a
> deadtime built into the system to prevent shoot through. The drain current
> in the TOP FET of one leg now starts to fall with a rate of Di/Dt. This
> induces a voltage of magnitude Ldi/dt where L is the inductance of your
> motor winding at the center of the leg. As soon as the voltage at the
> center of the leg reaches one diode drop below the body diode of the
bottom
> FET of the same leg, the load current will be supported by the body diode
> of the bottom FET.
> (3) After the deadtime is up, the Top FET of the other leg is turned on
the
> bottom FET of the first leg is also turned on. This is where the reverse
> recovery time of the body diode comes into play. The body diode of the
> bottom FET has to go into reverse bias (i.e. the minority carriers in the
> body-drain region have to be swept away and the depletion region
> established) before the top FET can block voltage. IF this process takes
> too long, then you will have shoot through currents. To keep things simple
> one would not want the Trr of the body diode to be any longer than the
dead
{Quote hidden}

induce
> a spike in the gate of the other FET due to the parasitic Gate-Source
> capacitance. If your dv/dt is too large, then the induced spike might have
> enough amplitude to turn on the FET when it should be off. I have found
> that experimenting with various values of the gate drive resistance  (use
a
> parallel schottky to make sure you are not increasing turn off ) is the
> final answer to this multidimensional optimization.
> In some of my designs I have used a HIP4086 (3ph driver) and an
> APT10M19BVFRR to get about 25A of current (RMS) at 48VDC to run a 3ph bldc
> motor. I ended up using a dead time of 500ns and a gate drive resistance
of
> 20ohms (with a schottky in parallel).
> Good luck
> Madhu
> {Original Message removed}

2002\09\11@053458 by Roman Black

flavicon
face
Donovan Parks wrote:
>
> Hello,
>
> We are planning to drive the motors at 12 or 24V from some NiCd batteries.
> The maximum continuous current draw is around 12A for the motors we are
> using (peak current around 100A).  Three questions:
>
>     1) Can NiCd accept "reverse charging" (they are obviously rechargable,
> but I'm guessing this isn't exactly the same thing)?
>     2) Can NiCd accept this sort of heavy current draw?
>     3) How can I find the answers to this questions for myself? (I've read
> several websites on batteries, but they did not address these issues)


My suggestion is to NOT use the NiCds as part of the
recirculation current path. They are a chemical device
and do wear out, besides having lousy charging efficiency.

You should decouple the bridge and recirculating current
with LARGE filter caps. The pulses of current are supported
by the caps, NOT the batteries. These caps should be close
to the bridge which in turn should be close to the motor.
You need low ESR electros, one cheap way to do this is to
parallel a lot of smaller electros.

The batteries can be separated from the caps by a low-ohms
resistor which spares the batteries from much of the
pulse current and also doubles as a current measurement
resistor, ie will show average current from/to the batteries.

You ratio of peak motor current to average curent is 100:12
which seems high and somewhat unnecessary? If these motors
are for submarine propulsion can't you soft-ramp them
under PIC control? This will keep your h-bridge cost
lower and reliability higher, it also reduces drain from
the batteries upon starting and current forced back into
the batteries upon deceleration.

There are two main methods of handling recirculating
current, one involves just switching the FETs off and
letting the body diodes pass the current back to the
filter caps. This is better when you need high rates of
change of the current, like fast accelerations and
decelerations. In your case you are probably better off
with the other method (which you suggested) of turning
on the bottom pair (or top pair) of FETs and keeping
the current as slow-decay (ie in the motor).

Have you tried a google search for "high power h-bridge
design" ?? :o)
-Roman

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2002\09\11@074137 by Olin Lathrop

face picon face
> Hmmm... first, we are planning to use a PM DC motor.  Is this the same as
a
> brushless DC motor?

That depends on what "PM" stands for.

> Given that it is, then we have not considered doing anything with the
> windings during the off time.  What should we be doing?
>
> From what we have read it appears to be a poor idea to let the motor "free
> wheel" during the off time of the PWM pulse as then there is no path for
the
> inductive current to follow which results in a large voltage being
produced.
> Is this correct?

There seems to be some confusion because there are two different kinds of
voltages produced by a motor coil when it is shut off.

The first is solely because the coil is an inductor.  You get kickback
voltage like you would from any inductor that you try to shut off current
to.  This is a relatively short lived pulse that needs to be dealt with,
else it will destroy the driving electronics.  Since you want the coil to
see the average current produced by the PWM, it is best to let the kickback
current continue thru the coil with as little external voltage drop as
possible.

The second is the back EMF of the motor acting like a generator.  This is
mostly an issue during the off phase of the coil as the motor turns, not the
off phase between PWM pulses.  The back EMF must not be shorted, else it
will create apparant viscous drag on the motor rotation.  In other words,
your circuit for one of the coils is dumping energy into the motor to turn
it, but another coil is draining this energy right back out again.

The ballance between these two can be tricky.  The maximum back EMF is
proportional to rotation speed, and can never exceed the driving voltage.
The kickback pulse voltage can get very high if you let it.  That's because
the coil is really kicking back a finite amount of energy, and the voltage
versus time is function of the external circuit.

One possible solution is to clip the kickback voltage at a level just a
little greater than the largest possible back EMF voltage.  This sounds
simple, but there is often no easy place to dump this voltage efficiently.
The power in all the kickback pulses can be significant, so you don't want
to waste it.  I've seen a circuit that had the kickback voltage create a
higher supply voltage (just like a boost converter, which it is if you think
about it), then drop that supply voltage back onto the main power supply
with a buck regulator.

Some other solutions envolve knowing when the off phase is and changing the
output drive circuit accordingly.

Another solution uses more switching elements, but walks the energy from one
coil to the next as the motor phase advances.  This drives each coil full on
during its on phase, without PWM.  To get variable speed, you vary the DC
supply driving the whole circuit, typically with a buck regulator.

All in all, this is not a "light" subject, and you need to spend some time
reading some references.  I'm sure there have been volumes written about
motor control somewhere.


*****************************************************************
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(978) 742-9014, http://www.embedinc.com

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2002\09\11@074558 by Olin Lathrop

face picon face
>     1) Can NiCd accept "reverse charging"

They hate that.

>     2) Can NiCd accept this sort of heavy current draw?

NiCD can do pretty decent sustained currents.  The battery spec sheet should
answer this question.

>     3) How can I find the answers to this questions for myself?

The spec sheets for the specific batteries you are using?


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2002\09\11@161759 by Madhu Annapragada

picon face
Unfortunately, I do not know of any single book I can recommend for this
stuff. Most of what I know about BLDC and BDC design comes from data
sheets, Horowitz&Hill and a lot of bench work. I can tell you though that
International Rectifier, Motorola, National Semiconductor, OnSemi, Seimens
amoung others have excellent application notes on Mosfets and Motor
control. Takes time to understand the physics but once you have it the
stuff is very logical and easy to design with.
Madhu
{Original Message removed}

2002\09\11@171804 by Peter L. Peres

picon face
The simple explanation for phase switching:

When you turn on a phase you supply current to it until it reaches the
desired magnetic field strength. Then any current that still goes through
it becomes heat. When you wish to turn a phase off you have an interest to
remove the field asap (or not). The simple way is to turn the driver off.
The field will collapse and produce a terrific inductive kickback,
returning most of the energy put into it (Wl = L * I^2 / 2). A H bridge
driving a phase needs protection diodes to route this kickback current
safely (and to prevent it from becoming a destructive voltage).  They are
already built in and they cause the kickback to be rectified and to try to
charge the battery through the supply rails. This is good (it is also used
as energy recovery method in electric vehicles and some electric rail
engines, during regenerative braking). You can help the efficiency of the
process by driving the H bridge in the opposite direction, from when you
stop driving the phase, until the current through the coil goes to zero.
This will effectively help the rectifer bridge formed by the MOSFET bulk
diodes to be more efficient and to heat up less. There are more elaborate
schemes like this. The reverse forcing (one of the names by which this
scheme is known) of phases also speeds up the removal of the field and
helps achieve higher speeds.

Peter

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2002\09\11@171808 by Peter L. Peres

picon face
On Tue, 10 Sep 2002, Donovan Parks wrote:

>Hello,
>
>We are planning to drive the motors at 12 or 24V from some NiCd batteries.
>The maximum continuous current draw is around 12A for the motors we are
>using (peak current around 100A).  Three questions:
>
>    1) Can NiCd accept "reverse charging" (they are obviously rechargable,
>but I'm guessing this isn't exactly the same thing)?

Yes it is, just make sure that the batteries are charged at start, else
the high peaks could reverse empty cells and destroy the battery.

>    2) Can NiCd accept this sort of heavy current draw?

Yes, but check cell ratings from manufacturers. I have used 'D' size NiCd
cells to draw 25A for a few msec so it's no big magic. The batteries used
by the RC electro flight crowd are stressed to >50A for 3-10 minutes and
last tens of cycles like this, with quick charge afterwards often at 4C
and above. 12A is certainly no problem. Batteries WILL get very hot when
used like this, and you should prevent this (forced cooling, oil bath,
etc).

>    3) How can I find the answers to this questions for myself? (I've read
>several websites on batteries, but they did not address these issues)

Look at peak current and at peak charge current. The latter can be deduced
from some chargers data sheets rather than from the manufacturers,
although the manufacturer might cooperate with data if you ask their reps
(tell them what you need it for).

Usually you will have LC filters on the power lines to the motors, which
will equalize the peak pulses to values bearable by batteries. Be aware
that measuring motor current in such a configuration is 'interesting'.

Peter

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