Questions on Electric Vehicle high power electronics

[Byron Jeff]

On Tue, May 27, 2008 at 12:56:27AM -0400, Mike Reid wrote:
>
>
> In the April 28 issue of Design News, the cover article is on battery
> technology.
>
> They show that gasoline packs 80 times more energy than lithium-ion
> batteries and 250 more times than lead-acid batteries.

That's well known. One number that I saw in an article (can't quote it off
the top of my head) pointed to a gallon of gas having the equivalent of 15
kWh worth of usable energy at its typical 25% efficiency. That's the same
capacity of a lead-acid pack weighing over 1500 lbs.

>
> http://www.designnews.com/article/CA6551948.html
>
> Above is a link to a number of articles on their website about designing
> electric cars.
>
> The April 28 article is titled, "It's Not a Slam Dunk."
>
> They asked five experts to estimate the specific energy and cost of EV
> batteries as they stand today.  Here are the approximate averages of their
> responses:
>
> Battery                 Specific Energy (W-hr/kg)               cost
> ($/kW-hr)
>
> Lithium-ion                             140
> $770
>
> Nickel-metal hydride            110
> $850
>
> Lead-acid                               50
> $100
>
> $= US dollars (check the web hourly for how much it's falling:)
>
>
> I didn't search for the above mentioned article but it's worth finding and
> reading.

I believe that I had seen it. I liken driving an EV to a vehicle that has a
1 gallon gas tank and can only be refilled at home.

It has its challenges. However, my daily commute takes less than a gallon
of gas to pull off. So there is hope.

BAJ
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[Byron Jeff]

Peter,

I think I accidentally deleted your post about the AC inverter paper.

However in my searching I found a really useful motor control handbook by
Richard Valentine:

http://tinyurl.com/6yujn9

And the whole book is previewable under Google Books.

So I have some decent reading material now.

Thanks for the suggestion,

BAJ
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[Cristóvão Dalla Costa]

On Sun, May 25, 2008 at 1:04 PM, Byron Jeff <byronjeff@...> wrote:

> > This provides both drive and regenerative braking
> > in a brutal way, and is the 'old' (and tried) way of doing it.
>
> Series wound motors. Not going to do regen braking. That energy is just
> going to be lost.
>


Then why bother at all? A significant part of the energy used to move cars
and trucks ends up as waste heat in the brakes. IIRC something in the order
of 60% -- specially with a heavy vehicle in city driving where speeds < 80
km/h and aerodynamic drag is not the dominant factor.

Now factor the increased weight of the lead batteries you want to carry over
the empty truck/car and you just waste that much more energy. Consider how
heavy the batteries will be and the waste accelerating and stopping that
mass that will just end up as heat on the brakes.

I'm sorry, but if you're going to all that effort to make an economical
vehicle, either go all the way and do the regenerative braking, or don't. A
half-assed solution is probably going to be a headache in the long run.

I have a pdf of a paper of the Catholic University of Chile where they
converted a truck to electric power using batteries and a ultracapacitor
bank for regenerative braking. I don't know if I can post it, but just
google "regenerative braking university chile" and a downloadable pdf will
be among the first links.

HTH
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[mikecreid]

In the April 28 issue of Design News, the cover article is on battery
technology.

They show that gasoline packs 80 times more energy than lithium-ion
batteries and 250 more times than lead-acid batteries.

http://www.designnews.com/article/CA6551948.html

Above is a link to a number of articles on their website about designing
electric cars.

The April 28 article is titled, "It's Not a Slam Dunk."

They asked five experts to estimate the specific energy and cost of EV
batteries as they stand today.  Here are the approximate averages of their
responses:

Battery Specific Energy (W-hr/kg) cost
($/kW-hr)

Lithium-ion 140
$770

Nickel-metal hydride 110
$850

Lead-acid 50
$100

$= US dollars (check the web hourly for how much it's falling:)


I didn't search for the above mentioned article but it's worth finding and
reading.



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[plpeter]

Peter <plpeter2006 <at> yahoo.com> writes:
> Jeff, please take a look at this document. It contains several interesting
> items, including a drive scheme and DOT road testing schedules:
>
>
http://scholar.lib.vt.edu/theses/available/etd-102497-12366/unrestricted/final.pdf

You can find older posts in archives, like here:

http://article.gmane.org/gmane.comp.hardware.microcontrollers.pic/142456

Peter


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[Byron Jeff]

On Tue, May 27, 2008 at 06:07:54AM -0400, Michael Rigby-Jones wrote:

>
>
> > -----Original Message-----
> > From: piclist-bounces@... [mailto:piclist-bounces@...] On
> Behalf
> > Of Byron Jeff
> > Sent: 25 May 2008 11:33
> > To: piclist@...
> > Subject: [EE] Questions on Electric Vehicle high power electronics
> >
> >
> > 1) Still looking for a donor truck. Probably will end up with on off
> > craigslist. With a blown engine, it won't be more than $500.
>
> Starting with a very heavy American truck and then making it much
> heavier with lead acid batteries doesn't strike me as a sensible
> starting point for an electric vehicle.  Why not start with something
> lightweight to start with and then you don't need as many batteries or
> such a powerful motor?  The electronics then become cheaper/simpler.

A couple of points.

First is that I'm only looking at compact trucks. Vehicles in the range of
the Chevy S10, Ford Ranger, Mazda B2200, Nissan Frontier, and the like.
They have curb weights in the mid to high 2000 lbs, which isn't too
terribly bad for a vehicle. You then get to strip out all the associated
ICE components, which will give you about 400 lbs back, putting you in the
low 2000s. The curb weight of the converted truck should still come in
under 4000 lbs.

The tradeoff with very light vehicles and lead acid is both maximum weight
and drivability. Trucks are actually designed to carry heavy loads, whereas
lightwight cars are not.

The second problem is that every less battery you carry has a double effect
on range. The first is that you lose the energy capacity of the battery.
The second is that the remaining batteries have to work harder to power the
car, which ups their amp draw. Due to an effect called the Peukert effect,
the faster you draw power from a battery, the less total energy you can
extract. So point in fact you really want to up the voltage.

Now you can combat that by using 12V batteries instead of 6V golf cart
batteries. But when you double the voltage, you half the energy capacity,
presuming that the 12V and 6V batteries are about the same size and weight,
which they often are. Reducing your energy capacity by half cuts your range
by half.

The only positive side is what you outlined. Carrying less weight will get
you further. However, the gains in weight you gain are heavily offset by the
losses in energy capacity and efficiency you lose. In short a lightweight
truck carrying a heavier battery pack will get you further than a
lightweight car carrying a lighter battery pack.
 
As for simplicity, trucks are much easier to convert than cars. They have
both a bed, and space under the bed, to carry batteries. You don't have to
worry about having batteries in the passenger compartment. And as outlined
before they are designed to carry the weight. Stock compact trucks have a
rated payload in the ballpark of 1500 lbs, and that's before any suspension
upgrades.

LionEV, which specializes in Lithium battery conversions, shows an example
of converting a Ford Ranger here:

http://www.lionev.com/DIY_Ranger.html

BAJ
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[Byron Jeff]

On Tue, May 27, 2008 at 07:01:29AM -0400, Peter wrote:
>
> 1979: http://www.econogics.com/ev/lep.htm
>
> Full schematics and driving instructions. Wetware controller.

The controller is too simple. In today's age of cheap power electronics,
there's no reason to go with a two stage contactor driven motor.

So I'd like to get back to original point. Valentine's book, of which I
need to order a copy, gave an excellent set of rules of thumb:

1. Due to inductance effects, mount the power electronics on the motor.

2. A freewheeling diode is necessary to combat kickback and back EMF.

3. Use opto drivers between the control electronics and power circuitry.

4. You have to find balance between audio noise, power losses due to
switching, and switching rise/fall times.

All excellent advice.

I guess I'll wait for the book to be delivered, then start back with some
more pointed questions.

BAJ
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[Michael Rigby-Jones]

> -----Original Message-----
> From: piclist-bounces@... [mailto:piclist-bounces@...] On
Behalf
> Of Byron Jeff
> Sent: 25 May 2008 11:33
> To: piclist@...
> Subject: [EE] Questions on Electric Vehicle high power electronics
>
>
> 1) Still looking for a donor truck. Probably will end up with on off
> craigslist. With a blown engine, it won't be more than $500.

Starting with a very heavy American truck and then making it much
heavier with lead acid batteries doesn't strike me as a sensible
starting point for an electric vehicle.  Why not start with something
lightweight to start with and then you don't need as many batteries or
such a powerful motor?  The electronics then become cheaper/simpler.

Mike

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[plpeter]

1979: http://www.econogics.com/ev/lep.htm

Full schematics and driving instructions. Wetware controller.

Peter


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[Byron Jeff]

On Tue, May 27, 2008 at 09:39:11AM -0400, Michael Rigby-Jones wrote:

>
>
> > -----Original Message-----
> > From: piclist-bounces@... [mailto:piclist-bounces@...] On
> Behalf
> > Of Byron Jeff
> > Sent: 27 May 2008 09:26
> > To: Microcontroller discussion list - Public.
> > Subject: Re: [EE] Questions on Electric Vehicle high power electronics
> >
> >
> > The second problem is that every less battery you carry has a double
> > effect
> > on range. The first is that you lose the energy capacity of the
> battery.
> > The second is that the remaining batteries have to work harder to
> power
> > the
> > car, which ups their amp draw
>
> Surely with a lighter car they wouldn't have to work so hard?

It's a multivariable knapsack problem. It's not easy to solve at all.

BAJ
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[Byron Jeff]

On Tue, May 27, 2008 at 09:18:18AM -0400, Sean Breheny wrote:
> Hi Byron,
>
> While I still think you are underestimating the motor drive
> electronics, it is clear that you have done some good homework.

Thanks.

> However, I think you should try to restate what you say below
> mathematically. Then I think you will find that the equations for
> range vs. battery mass and performance vs. battery mass will both have
> optimal points. In other words, I don't think that adding more and
> more batteries will continue to get you much more range. I think it
> will at least level off if not begin to drop again at some point.

True. But that marginal point is well above the 144-156V bank that I'm
planning on implementing.

My emprical example is the "Red Beastie" found here:

http://www.evalbum.com/037

It's a compact truck that's carrying 40 lead acid batteries in a 120V by
450Ah pack. The 16 extra batteries it carried represented a 1000 lbs
additional load. But the expected range is close to double of what I expect
to get.

> Also, you say that going to a higher voltage reduces your total
> energy. That's not true. If you have a 6V 100AH battery, it is about
> the same size as a 12V 50AH battery.

That's correct.

> If you design your system
> properly for each power source, you should be drawing half the current
> from the 12V setup compared to the 6V setup and get comparable usage
> time (i.e. range).

That presumes that you're going to double the voltage of the 12V pack over
the 6V one. But because of the motor and the control electronics, it's
generally no possible to do either. So you end up with packs of
approximately the same voltage. That means you have 1/2 the number of
batteries in the 12V pack as the 6V one. So the total energy is reduced.

The bottom line is that you'll exceed the gross vehicle weight rating and the
voltage rating of the motor well before you reach the marginal dropoff of
carrying more lead. And you can carry more lead without exceeding the
voltage rating of the motor by carrying 6V batteries.

BAJ
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[Sean Breheny]

Hi Byron,

While I still think you are underestimating the motor drive
electronics, it is clear that you have done some good homework.
However, I think you should try to restate what you say below
mathematically. Then I think you will find that the equations for
range vs. battery mass and performance vs. battery mass will both have
optimal points. In other words, I don't think that adding more and
more batteries will continue to get you much more range. I think it
will at least level off if not begin to drop again at some point.

Also, you say that going to a higher voltage reduces your total
energy. That's not true. If you have a 6V 100AH battery, it is about
the same size as a 12V 50AH battery. If you design your system
properly for each power source, you should be drawing half the current
from the 12V setup compared to the 6V setup and get comparable usage
time (i.e. range).

Sean


On Tue, May 27, 2008 at 4:25 AM, Byron Jeff <byronjeff@...> wrote:

> A couple of points.
>
> First is that I'm only looking at compact trucks. Vehicles in the range of
> the Chevy S10, Ford Ranger, Mazda B2200, Nissan Frontier, and the like.
> They have curb weights in the mid to high 2000 lbs, which isn't too
> terribly bad for a vehicle. You then get to strip out all the associated
> ICE components, which will give you about 400 lbs back, putting you in the
> low 2000s. The curb weight of the converted truck should still come in
> under 4000 lbs.
>
> The tradeoff with very light vehicles and lead acid is both maximum weight
> and drivability. Trucks are actually designed to carry heavy loads, whereas
> lightwight cars are not.
>
> The second problem is that every less battery you carry has a double effect
> on range. The first is that you lose the energy capacity of the battery.
> The second is that the remaining batteries have to work harder to power the
> car, which ups their amp draw. Due to an effect called the Peukert effect,
> the faster you draw power from a battery, the less total energy you can
> extract. So point in fact you really want to up the voltage.
>
> Now you can combat that by using 12V batteries instead of 6V golf cart
> batteries. But when you double the voltage, you half the energy capacity,
> presuming that the 12V and 6V batteries are about the same size and weight,
> which they often are. Reducing your energy capacity by half cuts your range
> by half.
>
> The only positive side is what you outlined. Carrying less weight will get
> you further. However, the gains in weight you gain are heavily offset by the
> losses in energy capacity and efficiency you lose. In short a lightweight
> truck carrying a heavier battery pack will get you further than a
> lightweight car carrying a lighter battery pack.
>
> As for simplicity, trucks are much easier to convert than cars. They have
> both a bed, and space under the bed, to carry batteries. You don't have to
> worry about having batteries in the passenger compartment. And as outlined
> before they are designed to carry the weight. Stock compact trucks have a
> rated payload in the ballpark of 1500 lbs, and that's before any suspension
> upgrades.
>
> LionEV, which specializes in Lithium battery conversions, shows an example
> of converting a Ford Ranger here:
>
> http://www.lionev.com/DIY_Ranger.html
>
> BAJ
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[Michael Rigby-Jones]

> -----Original Message-----
> From: piclist-bounces@... [mailto:piclist-bounces@...] On
Behalf
> Of Byron Jeff
> Sent: 27 May 2008 09:26
> To: Microcontroller discussion list - Public.
> Subject: Re: [EE] Questions on Electric Vehicle high power electronics
>
>
> The second problem is that every less battery you carry has a double
> effect
> on range. The first is that you lose the energy capacity of the
battery.
> The second is that the remaining batteries have to work harder to
power
> the
> car, which ups their amp draw

Surely with a lighter car they wouldn't have to work so hard?

Mike

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[Alan B. Pearce]

>Starting with a very heavy American truck and then making it much
>heavier with lead acid batteries doesn't strike me as a sensible
>starting point for an electric vehicle.  Why not start with something
>lightweight to start with and then you don't need as many batteries or
>such a powerful motor?  The electronics then become cheaper/simpler.

Yeah, I was thinking along the lines of a Ford Ka with 4 wheel electric
motors for a commute vehicle.

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[plpeter]

Peter <plpeter2006 <at> yahoo.com> writes:
1979 'brute force' Renault 12 conversion: http://www.econogics.com/ev/lep.htm

I am posting this again as the bit dog ate my posting.

Peter


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[Byron Jeff]

On Tue, May 27, 2008 at 04:58:48PM -0400, Gerhard Fiedler wrote:

> Byron Jeff wrote:
>
> >> If you design your system properly for each power source, you should be
> >> drawing half the current from the 12V setup compared to the 6V setup
> >> and get comparable usage time (i.e. range).
>
> > That presumes that you're going to double the voltage of the 12V pack over
> > the 6V one.
>
> Why? You can parallel them also.

That requires a committment to parallel strings from the beginning and you
have to add batteries in pairs.

So then exactly what does it buy you? By the time you've added enough
batteries to match your 6V, you have the same number of batteries anyway.

So it ends up being a wash. Except for the fact that 6V batteries have
thicker plates and so they are more robust.

>
> > But because of the motor and the control electronics, it's
> > generally no possible to do either. So you end up with packs of
> > approximately the same voltage. That means you have 1/2 the number of
> > batteries in the 12V pack as the 6V one. So the total energy is reduced.
>
> Or instead of 24 6V batteries in series, you use 2 x 12 12V batteries (12
> in series, 2 parallel strings, or 12 sets of 2 parallel batteries in
> series).
> Should get the same energy density, or a higher one. A 12V battery
> is nothing more than 2 6V batteries in one case. With less casing, the
> overall energy density of the pack should be (a little) higher with 12V
> batteries.

See above.

The only reason to go with 12V batteries is if you plan to lessen the
number of batteries you plan to carry. If you're not, then 6V are the
superior choice.

How did we get here? I was hoping to focus this discussion on design issues
on high power motor electronics, not vehicle choice (it's going to be a
pickup truck), battery type (it's going to be lead acid), battery voltage
(it's going to be 6V batteries), regenerative braking (the motor is going
to be a DC series wound motor, so no regen), or costs (it's a hobby build
so I'm not interested in why there's a nearly 10x markup of a commercial
product over the cost of components).

There are seriously thought out reasons for each of these choices. I'll
keep discussing them, but the original intent of the thread is virtually
dead. Again it seems to me a an integrated EV controller solution that
incorporates charging, PWM motor control, monitoring, and user interface
would have some utility due to system integration. Motor control and user
interface both need to know the current, charging and user interface both
need to know the voltage. A lot of EVs actually have two or three shunts in
the power loop to give info to the controller, user display, and charger
for example.

Let's start with a simple question. The PowerEx IGBT module 1200V @ 600A is
commonly available. According to the data sheet:

http://theelectrostore.com/datasheets/cm600ha24h.pdf

It integrates a reverse connected freewheeling fast recovery diode. My
question is that a motor controller must integrate such a diode to keep
spikes down when the IGBT turns off. Controllers such as the ones outlined
in Valentine's book show that diode across the motor. Is it sufficient to
have that diode across the IBGT? The presumes that the IGBT is going to be
mounted on the motor to keep the leads as short as possible.

BAJ
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[Gerhard Fiedler]

Byron Jeff wrote:

>> If you design your system properly for each power source, you should be
>> drawing half the current from the 12V setup compared to the 6V setup
>> and get comparable usage time (i.e. range).

> That presumes that you're going to double the voltage of the 12V pack over
> the 6V one.

Why? You can parallel them also.

> But because of the motor and the control electronics, it's
> generally no possible to do either. So you end up with packs of
> approximately the same voltage. That means you have 1/2 the number of
> batteries in the 12V pack as the 6V one. So the total energy is reduced.

Or instead of 24 6V batteries in series, you use 2 x 12 12V batteries (12
in series, 2 parallel strings, or 12 sets of 2 parallel batteries in
series). Should get the same energy density, or a higher one. A 12V battery
is nothing more than 2 6V batteries in one case. With less casing, the
overall energy density of the pack should be (a little) higher with 12V
batteries.

Gerhard

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[Sean Breheny]

Hi Byron,

Can you point me to the actual diagram showing the diode directly
across the motor? The main point of that diode is to protect the IGBT
(and possibly also to preserve the brushes in the motor and keep
efficiency up). It is best, usually, to put it closest to the IGBT as
possible. This minimizes the inductance which is unclamped when the
IGBT turns off (the current will continue to flow through the motor
and its supply wires).

I suspect that the reason why the book shows the diode across the
motor is that the diode integrated into the IGBT is in the wrong
direction if you are only using a single IGBT (instead of an H
bridge). In this case, it probably assumes that there is also a diode
inside the IGBT module.

There are a number of ways to do PWM. For example, you can have an H
bridge or just a single IGBT (depending on whether you need to reverse
direction). You can rely on the freewheeling diodes to conduct, or you
can actively short across them on the legs of the H bridge which
conduct during that part of the cycle. You can turn one pair of IGBTs
on during the "on" part of the cycle, and then turn the other pair on
for the "off time" (this way, 50% PWM=0 torque). You can also short
across the motor during the "off" part of the cycle. Each of these has
their advantages and disadvantages in each application.

I like to view a typical motor drive as a type of switching power
supply (buck converter). The PWM and the motor inductance transform a
high voltage/ low current (at the battery) into a low voltage/ high
current (in the motor windings). Incidentally, in motors which DO
regen, this also works in reverse and automatically forms a boost
converter to charge the batteries.

You should consider having a capacitor bank to lessen the ripple
current seen by the batteries. This reduces the (negative) effects of
the battery resistance, wire resistance, and wire inductance (wire
here being the leads from the batteries to your controller).

To a certain extent motor inductance is your friend as it allows you
to reduce the PWM frequency and still maintain a low current ripple in
the motor (reduces I2R losses). Lower PWM freq allows lower IGBT or
FET switching losses (since the switching element usually has a
constant transition time and this gets repeated more times per second
at a higher PWM freq). On the other hand, lower PWM frequencies may be
annoyingly audible and high motor inductance can in some cases
actually reduce motor performance (because the winding inductance acts
to limit current and cause phase shifts which make the pre-defined
commutation points less optimal).

Beware of the specs of modules such as this. They are usually very
unrealistic. For example, most of the specs are given for a Tj=25C,
which could only possibly happen (during continuous operation) if you
had a LN2 cooled heatsink! I would bet that if you work out the
details for this module, with a realistic heatsink and ambient
temperature, it can only do something like 400A continuous.

Beware also that an IGBT is essentially a FET driving a BJT.
Therefore, they share some of the pathologies of each. They are
sensitive to voltage spikes on the gate (which can punch through it).
They also have a (fairly constant) collector emitter saturation
voltage when fully on instead of an on resistance like a FET. They are
susceptible to second breakdown (unequal current sharing in different
parts of the BJT junctions) which limits the safe operating area to
something smaller than a rectangle. They are also slower than FETs.

The main reason for choosing an IGBT over a FET is the higher Vce max
voltage for IGBTs. If you can get away with using FETs, I'd recommend
those instead.

Bear in mind, for example, that at 400A continuous, this module is
going to be dissipating more than 400W. This is not including
switching losses.

Sean


On Tue, May 27, 2008 at 2:37 PM, Byron Jeff <byronjeff@...> wrote:

> Let's start with a simple question. The PowerEx IGBT module 1200V @ 600A is
> commonly available. According to the data sheet:
>
> http://theelectrostore.com/datasheets/cm600ha24h.pdf
>
> It integrates a reverse connected freewheeling fast recovery diode. My
> question is that a motor controller must integrate such a diode to keep
> spikes down when the IGBT turns off. Controllers such as the ones outlined
> in Valentine's book show that diode across the motor. Is it sufficient to
> have that diode across the IBGT? The presumes that the IGBT is going to be
> mounted on the motor to keep the leads as short as possible.
>
> BAJ
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[Byron Jeff]

On Wed, May 28, 2008 at 12:46:57AM -0400, Sean Breheny wrote:
> Hi Byron,
>
> Can you point me to the actual diagram showing the diode directly
> across the motor?

There are samples in the images on this page:

http://www.simreal.com/content/CustomMotorDriver

> The main point of that diode is to protect the IGBT
> (and possibly also to preserve the brushes in the motor and keep
> efficiency up). It is best, usually, to put it closest to the IGBT as
> possible. This minimizes the inductance which is unclamped when the
> IGBT turns off (the current will continue to flow through the motor
> and its supply wires).
>
> I suspect that the reason why the book shows the diode across the
> motor is that the diode integrated into the IGBT is in the wrong
> direction if you are only using a single IGBT (instead of an H
> bridge). In this case, it probably assumes that there is also a diode
> inside the IGBT module.

I thought the diode needed to be reverse connected. The PowerEx module
calls it a reverse connected ultra fast recovery diode. It's connected with
the cathode facing the collector, just like the diode in the image on the
web page.

> There are a number of ways to do PWM. For example, you can have an H
> bridge or just a single IGBT (depending on whether you need to reverse
> direction).

Single direction. I'm planning on keeping the transmission and the clutch
so the transmission can reverse the car.

> You can rely on the freewheeling diodes to conduct, or you
> can actively short across them on the legs of the H bridge which
> conduct during that part of the cycle.

> You can turn one pair of IGBTs
> on during the "on" part of the cycle, and then turn the other pair on
> for the "off time" (this way, 50% PWM=0 torque).

No H-bridge. It's a single IGBT module.

> You can also short
> across the motor during the "off" part of the cycle. Each of these has
> their advantages and disadvantages in each application.
>
> I like to view a typical motor drive as a type of switching power
> supply (buck converter). The PWM and the motor inductance transform a
> high voltage/ low current (at the battery) into a low voltage/ high
> current (in the motor windings). Incidentally, in motors which DO
> regen, this also works in reverse and automatically forms a boost
> converter to charge the batteries.

Interesting.

> You should consider having a capacitor bank to lessen the ripple
> current seen by the batteries. This reduces the (negative) effects of
> the battery resistance, wire resistance, and wire inductance (wire
> here being the leads from the batteries to your controller).

The suggestions that I've seen is using motor run capacitors such as the
ones that you often see in air conditioning units.

> To a certain extent motor inductance is your friend as it allows you
> to reduce the PWM frequency and still maintain a low current ripple in
> the motor (reduces I2R losses). Lower PWM freq allows lower IGBT or
> FET switching losses (since the switching element usually has a
> constant transition time and this gets repeated more times per second
> at a higher PWM freq). On the other hand, lower PWM frequencies may be
> annoyingly audible and high motor inductance can in some cases
> actually reduce motor performance (because the winding inductance acts
> to limit current and cause phase shifts which make the pre-defined
> commutation points less optimal).

I got that from what I've read in Valentine's book preview. I need to
locate a copy of the book locally if possible.

> Beware of the specs of modules such as this. They are usually very
> unrealistic. For example, most of the specs are given for a Tj=25C,
> which could only possibly happen (during continuous operation) if you
> had a LN2 cooled heatsink! I would bet that if you work out the
> details for this module, with a realistic heatsink and ambient
> temperature, it can only do something like 400A continuous.

That's fine because there's no way I'm going to be driving the motor with
400A continuous. If the continuous draw is more than 100A, then I'll be out
of juice so fast, that I'll be sitting on the side of the road.

It's overengineered for the application both the peak voltage and peak
current.

> Beware also that an IGBT is essentially a FET driving a BJT.
> Therefore, they share some of the pathologies of each. They are
> sensitive to voltage spikes on the gate (which can punch through it).
> They also have a (fairly constant) collector emitter saturation
> voltage when fully on instead of an on resistance like a FET. They are
> susceptible to second breakdown (unequal current sharing in different
> parts of the BJT junctions) which limits the safe operating area to
> something smaller than a rectangle. They are also slower than FETs.
>
> The main reason for choosing an IGBT over a FET is the higher Vce max
> voltage for IGBTs. If you can get away with using FETs, I'd recommend
> those instead.

Thanks for the design info. The primary reasons for looking at IGBTs are
cost, availability, packaging, and the fact the similar controller
solutions use them. Plus they are overengineered for the application so
it's less likely I'll actually smoke them.

Every application I've seen that have used FETs ended up paralleling a set
of them to get the requisite current capacity.

I'm really looking for simple, cheap, reliable, and available. I know I can
pick up 3 or 4 of the IGBTs off Ebay for around $100 and be reasonably
assured that they will work in the application.

If anyone can suggest a FET module that has similar characteristics, then
I'd be happy to take a look at them.

> Bear in mind, for example, that at 400A continuous, this module is
> going to be dissipating more than 400W. This is not including
> switching losses.

Again the continuous amps are going to be less than a quarter of that. Heat
sinking is going to be a design issue I still have to deal with.

Thanks for the overview. It really helps.

BAJ
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[Byron Jeff]

On Wed, May 28, 2008 at 10:22:35AM -0400, Apptech wrote:
> > No H-bridge. It's a single IGBT module.
>
> Byron,
>
> Do you have the IGBT module already?

Nope.

> And what is the part number.

PowerEx CM600HA-24H

> If you haven't got it yet I may be able to supply something.
> What specs (current and voltage etc) or part number?

You can find a sample here:

http://tinyurl.com/648p67

These are 600A 1200V modules. They are way overengineered for the
application which is going to be 144V with a max amperage of 350-400A in
short pulses.

Ebay folks seem to have a ton of them available for a reasonable price.
Enough to purchase a couple of hot spares if necessary.

BAJ
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