EV Digest 6108
Topics covered in this issue include:
1) Re: Pusher Trailer
by "Michael Perry" <[EMAIL PROTECTED]>
2) Re: EV pusher Trailer
by "Michael Perry" <[EMAIL PROTECTED]>
3) Re: EV pusher Trailer
by "Michael Perry" <[EMAIL PROTECTED]>
4) Re: Power Trailer, was EV pusher Trailer
by Danny Miller <[EMAIL PROTECTED]>
5) Re: Heaters
by mike golub <[EMAIL PROTECTED]>
6) Re: Ultra Capacitors, Aluminum Batteries
by Danny Miller <[EMAIL PROTECTED]>
7) Re: speedometer replacement
by "David" <[EMAIL PROTECTED]>
8) Re: EV controllers? the 4th option...
by Jack Murray <[EMAIL PROTECTED]>
9) Re: EV pusher Trailer
by "Peter VanDerWal" <[EMAIL PROTECTED]>
10) Re: EV controllers? the 4th option...
by "Arthur W. Matteson" <[EMAIL PROTECTED]>
11) Re: Solar EV power
by "steve clunn" <[EMAIL PROTECTED]>
12) Re: speedometer replacement
by "Stefan T. Peters" <[EMAIL PROTECTED]>
13) Re: Ultra Capacitors, Aluminum Batteries
by "Stefan T. Peters" <[EMAIL PROTECTED]>
14) Dual outlet opportunity charging
by mike golub <[EMAIL PROTECTED]>
15) Re: EV controllers? the 4th option...
by Danny Miller <[EMAIL PROTECTED]>
16) Re: EV pusher Trailer
by "Death to All Spammers" <[EMAIL PROTECTED]>
17) Re: EV controllers? the 4th option...
by "Arthur W. Matteson" <[EMAIL PROTECTED]>
--- Begin Message ---
Again, I think this "with or without motive power" could be an old tie-back.
The founder of Parker-Hannifin was towing his farm trailer (selling
cylinders town to town) when he started down a steep grade. His rig was
definitely not under power... the trailer pushed him down the mountain...
ending in disaster. I think that may well be what this refers to... even if
the truck "dies" this is still a trailer???
BTW, this idiot stunt one day because the world's largest supplier of
motion/control units.
In these states, it appears they are still considering a motor vehicle as
being the towing force. If a person has enough money to fight it, it may be
proved that the trailer itself could be the motive force... though it may
well be best to claim you were towing the trailer when things went wrong.
<g> (In this case, you'd be covered by your car's insurance. Play dumb and
collect.)
----- Original Message -----
From: "Mike Chancey" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Monday, November 06, 2006 3:17 AM
Subject: Re: Pusher Trailer
>
> It is however, legal in Kansas where the statute reads:
>
> 8-1479. "Trailer" defined. "Trailer" means every vehicle
> with or without motive power, other than a pole trailer, designed for
> carrying persons or property and for being drawn by a motor vehicle,
> and so constructed that no part of its weight rests upon the towing
vehicle.
--- End Message ---
--- Begin Message ---
These were different that those, John. We naturally had the "safety chains."
These are designed so the trailer comes loose from the hitch and your towing
vehicle now tows a trailer, often with its tongue digging into the pavement,
and whipping side to side (depending on the chain length) but is still
hooked to your vehicle. For these, there's no length specified, nor a weight
requirement. Many users choose a very light chain... so the towed vehicle
comes loose (breaks a link) at low force... rather than dragging the towing
vehicle to its death. Still attached to the vehicle, this can very quickly
become an out of control condition, even on very short chains. The only
recovery of such a vehicle I've heard of was at fairly low speed.
These were in addition that that requirement. (I still think the failure was
that the hitch was not ratched down completely, or at all.) But, these
chains kept the hitch near or on the ball. These were about 6" in length and
bolted to the hitch itself. So, if the hitch broke or wasn't tied down, all
the damage would be contained... you wouldn't have the trailer jumping into
the tailgate of the truck, or dropping below the bumper. This was additional
insurance.
Of course, the owner also required that everyone "lock" their trailer in
place, despite the chains. He didn't have any more failures, once the
couplers were replaced and chains installed... no more tests to see if the
chains were successful. As said, I think he simply forgot to lock the system
down... but w/ the requirement of a "bolt" being added to the hitch, the
problem never reoccurred. A bolt could not be put through a latch, unless it
latched *above* the ball... and that's fairly obvious. (I personally had
times when I couldn't get the latch to latch enough to get the bolt into
place... so had to "jerk" the town vehicle to get it to latch.)
----- Original Message -----
From: "John G. Lussmyer" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Monday, November 06, 2006 2:55 PM
Subject: Re: EV pusher Trailer
> At 12:41 PM 11/6/2006, Michael wrote:
> >For one of our utility trailers, we also used a couple short chains.
Thus,
> >even if the tongue unlatched itself, the chains would keep the hitch from
> >lifting completely off the ball. (My boss started doing that when a new
> >hitch tongue broke and a trailer/bulldozer tried to drive over the top of
> >his PU. <g>)
>
> Last I knew, the chains were required by law. Not using them is both
> illegal and stupid.
>
> --
> John G. Lussmyer mailto:[EMAIL PROTECTED]
> Dragons soar and Tigers prowl while I dream.... http://www.CasaDelGato.com
>
--- End Message ---
--- Begin Message ---
I am trying to figure your figures. Please excuse me if this doesn't burst
my bubble.
Your range is true. It takes 36.6 kWh to be the equivalent of one gallon of
gas. This "dooms" all EVs to never exist. Gas is just too cheap to compete
with an EV. (Sidebar: EVs at slow speeds, such as in town, get 3 times the
range per equivalent power.) Basically, it takes a lot more than 600 lbs of
LA batts to be the equivalent of a gallon of gas.
I do buy the fact of pulling an EV. What if you aren't pulling it? There's
no reason to pull the trailer when you don't need it. You drop the trailer,
and go on with your normal 32MPG rating. Having pulled a trailer of 1/2 ton,
I don't think the mileage drops as much as you assume, but granted, if you
pull this dead weight, you'll be seeing a drop. But why tow it, if it isn't
the primary force of your trip. (That's as dumb as pushing an EV with an ICE
and seeing similar drop in MPG.)
EV pushers should not be all that worse than converting a car to EV power.
Granted, you still have to hook up a vacuum pump to operate the brakes and a
DC converter to operate the lights... but you aren't pushing along all that
much greater load than trashing the car for full-EV power... and all the
weight would be in the trailer. In my thoughts, that could enable the
modification of more modern cars (rather than those ready for the scrap bin)
to EV power. As for motive power, you will be pushing along a few hundred
extra pounds. Considering the total load, not all that much extra weight,
especially if the pusher is designed for in town use, not for freeway. Drop
the pusher at home, if you want to take the high speed route. Take the slow
speed route for EV... at maybe an extra 5 minutes.
As for buying a used Hybrid... that's another matter. I don't know where you
live or your experience in buying them, but locally the ones with less than
60K are more expensive (per expected mile) than they are new.
...just my thoughts... especially as any pusher is currently (apparently)
illegal to operate...
----- Original Message -----
From: "Peter VanDerWal" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Monday, November 06, 2006 6:19 PM
Subject: Re: EV pusher Trailer
> Are you all STILL talking about electric pushers? Sorry to burst your
> bubble, but this is just a plain bad idea.
>
> It takes about 600lbs of lead-acid batteries to get the same range as one
> gallon of gasoline. Because EVs only carry the equivelent of one to
> three gallons of gasoline, they HAVE to be efficient. An electric pusher
> simply isn't efficient.
> Let's assume we try a small electric pusher with only 600lbs of batteries,
> 100+ lbs of motor, 75-100lbs of tranmission, 25-50 lbs of controller,
> circuit breakers, cables, etc. plus the weight of the trailer and we are
> talking about 1,2000 lbs, maybe more.
> Now let's assume we have a car with average fuel economy, approx 27 mpg.
> What does it's fuel economy drop to when pulling a 1,200 lb trailer?
> 19-20mpg, maybe worse?
> So you spend thousands to get a vehicle that can only get you 10 miles
> before you have to return, or you put up with poor fuel economy when you
> go further. Plus you have to put up with hualing a trailer around town.
> You can make the trailer heavier, but you are looking at diminishing
> returns, then next 600 lbs gets you maybe 15 miles and so forth. Plus
> it's too heavy for a small car to hual (most passenger vehicles only have
> a 1,000 towing capacity or less.)
>
> Spend the money on a used hybrid, you're far better off. Buy an old Prius
> and you can convert it to a plug in hybrid and get the same electric
> \range as above and WAY BETTER fuel economy when you run out of juice.
>
> Or get a SECOND car and convert it to electric. You'll have lower costs
> and a lot less work than making a pusher trailer and much better range and
> efficiency.
>
> ICE pushers make a certain sense for some situations. ELectric pushers
> don't really make any sense.
--- End Message ---
--- Begin Message ---
There is a point to the design if you want to try a radically different
power source like a Capstone Microturbine. They are supposed to have
good emissions and are fairly efficient at producing electricity with an
integrated DC generator head. However, the calculations I recall doing
some time ago suggested similar mpg to the original engine of an
"average" vehicle. Perhaps a bit better, on the order of maybe 20%, but
it was hard to say.
Ordinary generators, I have stated the same case as David on multiple
occasions. It will end up at best a bit poorer mpg (depends on what
vehicle you compare it to), and terrible emissions. It matters not
whether you want to use the gen to charge the batteries slowly while
parked or a big one to provide the motive power as you drive. At best,
if you make the case that using it 10% of the time makes the EV
practical and otherwise you wouldn't own an EV, there's sort of a value
in terms of fossil fuel consumption. In terms of pollution it's way
worse even if only a very minor portion of your driving is done off of
generator power.
Another note was that generators are inherently VERY loud. Adding a car
muffler to the exhaust has significant value, but exhaust noise is only
a portion of the problem.
Danny
David Roden wrote:
On 6 Nov 2006 at 20:31, Roland Wiench wrote:
There is another way to use a EV Trailer, is to have a engine generator with
enough power to drive the EV motor by itself ... It did 22.5 mpg ...
This scheme and its variants has been discussed quite frequently on this
list. It's very similar to what many EV neophytes hope to do with an APU.
But while such a machine may indeed be suited to driving an EV cross
country, it has some downsides for routine use.
--- End Message ---
--- Begin Message ---
Can the cheapo elements be cut into to make more air
flow, and then change the wiring so you get the right
amount of resistance?
--- John <[EMAIL PROTECTED]> wrote:
>
> On Sunday, November 5, 2006, at 06:40 PM, John
> Wayland wrote:
>
> > Hello to All,
> > <snip>
>
> > When ceramic elements started to show up in the
> first 'cube type'
> > desktop heaters and when EV suppliers started to
> offer better quality
> > ceramic cores with less restrictive heat fin grids
> so the car's blower
> > could push more air through the core, I switched
> to this type of
> > heater and have never looked back.
> >
> > See Ya......John Wayland
>
>
> Is the heater element that EV Parts sells a ceramic
> element?
>
> Assuming yes, what about it is ceramic?
>
> John O'C
>
>
____________________________________________________________________________________
Sponsored Link
Get an Online or Campus degree
Associate's, Bachelor's, or Master's - in less than one year.
http://www.findtherightschool.com
--- End Message ---
--- Begin Message ---
"supercapactior" and "ultracapacitor" have no official meaning, and are
essentially meaningless. They do not specify any particular technology
or size.
There are a number of researchers working on the idea now. Unlike
batteries, there do not seem to be hard limits as to what is possible in
terms of capacity/power to size/weight ratios. I mean, there are no
credible claims that a person can make a li-ion with 10x the capacity
per unit weight, or even 5x. But with capacitors, the technical
limitations are unclear. Somebody claiming a capacitor 50x larger than
anything seen before, I'd be skeptical but wouldn't call it impossible
and bullsh*t right off the bat.
The claim of being able to recharge in minutes is problematic. The part
which is unclear is even if the cap could take astronomical charging
currents, where would you get that much current to do it that fast?
Even hooking directly up to your house's 220V mains is nowhere near
enough current to charge that fast before you'd pop the breakers.
Danny
Roland Wiench wrote:
George F. Hamstra is correct, they are Super capacitors, not ultra
capacitors. They are a different chemistry.
I brought the subject of super capacitors up about four years ago, and most
everyone said that was impossible to drive a EV with a capacitor, because it
charge would only last seconds.
People back in the 50's said it was impossible for electric cars to go over
50 miles or to charge a battery in one hour.
If you want to order super capacitors design for EV's (not the type for
Hybreds) here is the address: [EMAIL PROTECTED]
You tell them what voltage and ampere you need plus the weight of EV, speed,
acceleration and etc.
For my EV, they said 15 modules of 17 volts each
would be equal to about 50ah battery at a cost of $40,000.00 which would
give me a range of 15 miles.
They normally used these for electric buses that have a drive loop of 10
miles and then they can quick charge them at the end of each loop.
No, I will not used these, I will go back to a battery that is design for
EV's that I had before. Not these 6 or 12 volts batteries. Something like
the Exide Tuder EV cells I had before.
Roland
--- End Message ---
--- Begin Message ---
SA was basically hacked in the late 90s using DGPS (differential GPS). SA
affected all receivers the same way, so 1 GPS unit was placed in a known
location. By using the indicated position and the known position the error
from SA was known and could be used by any other GPS in a fairly large area
to correct its reading. Also, SA error averaged to 0 over time, so just
record readings and average for a few hours for very accurate location.
----- Original Message -----
From: "Danny Miller" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Monday, November 06, 2006 8:06 PM
Subject: Re: speedometer replacement
The feature which reduced the accuracy of the GPS signal when GPS first
came out was called Selective Availability. It was disabled back in 2000
and there are no plans to return to it again. Aircraft now often depend
on fully accuracy GPS for navigation.
It made no sense to have it on during peacetime anyways. The intent was
to deny the enemy accurate GPS information during wartime. The bizzare
part is that during Gulf War I the troops did not have enough military
units which could remove the error signal and troops began using normal
GPS, and thus the military actually disabled SA for the area so any GPS
unit in the area had full accuracy.
SA was not cracked in its lifetime. Nowadays, I'd give it like 6 months
before hackers figured out the algorithm and figured out how to get the
keys to remove the SA error.
Danny
Lee Hart wrote:
Lawrence Rhodes wrote:
Not this one. It seems as useful as my speedometer. It's a great way
to find out the error of your speedometer and it's an amazing trip
odometer.
Isn't this because the Defense Dept. has enabled high accuracy due to the
war in Iraq? They need to use civilian GPS units because there aren't
enough of the military ones? This accuracy is likely to go away when they
get enough equipment, as they've found that the enemy is also using the
civilian GPS units.
--- End Message ---
--- Begin Message ---
With these figures it looks like you need to keep the module cooler
than 121F that should be possible.
I would also think 118amps could be OK as continuous draw, accelerating
you would use a lot more, but that is temporary.
Clearly for an IGBT module, the higher the voltage the better it is to
offset the voltage drop. A Mosfet is probably best for a DC controller
at 160v, a low-side doesn't need a voltage boost for the gate drive so
that makes it simpler, the 15v drive for IGBT is easy to produce.
These modules seem to be built for better temperature dissipation, can
you do better using your own heatsinking from a chip or set of chips?
The mosfet modules seem to be unobtainium,
such as http://rocky.digikey.com/WebLib/IXYS/Web%20Data/VMO580-02F.pdf
which looks good technically, at $261 x 2 its getting expensive.
Jack
>>>Lee Hart wrote:
This is a Powerex CM600HA-24H IGBT. It has a 2.5-3.4v on-state voltage,
and a junction-to-heatsink resistance of about 0.07 deg.C/watt. Assume
your heatsink is at 50 deg.C (121 deg.F), and has a thermal resistance
of 0.1 deg.C/watt (which is very good), your total thermal resistance is
0.17 deg.C/watt.
You don't want the IGBT junction to be over 100 deg.C for reasonable
life. So, the most you can have is 100-50= 50 deg.C from junction to
ambient. How many watts can you dissipate?
Power = 50 deg.C / (0.17 deg.C/w) = 294 watts
How much current is this with the nominal 2.5v drop?
Current = 294 watts / 2.5v = 118 amps
So, the most this part can dissipate on a continuous duty basis is only
118 amps. It would be even less when you add switching losses, or you
you were unlucky and got a part that had the maximum 3.4v on-state drop.
See what I mean about the huge gap between advertised marketing numbers
and real-world performance?
Eric Poulsen wrote:
Heck the power part is what I was concerned about.
Then, you just need some controllers to look at. I have several, but no
digital camera or way to post pictures. Can anyone post photos of some
typical controllers, so folks can see how they are built?
Here's how Curtis does it. View with a fixed-width font. All parts mount
to an aluminum extrusion shaped like this:
_ _
| | | |<--extrusion
| | | |
____| |____| |____
|__________________|
The bottom has an insulating sheet of plastic over it, and slides into
the box-shaped outside case. Six or eight screws with plastic insulating
washers screw the extrusion and case together.
____________________
| _ _ |
| | | | | |<--outside case
| | | | | |
| ____| |____| |____ |
||__________________||
| ================== | <--insulating sheet
--------------------
_| |_ _| |_ <--plastic washers
| | <--screws
= =
The transistors, diodes, and capacitors are mounted to a large PC board
that sits on top of the extrusion. The capacitors are all between the
two vertical parts of the extrusion. The transistors and diodes are
screwed to the left and right sides of the vertical parts of the extrusion.
================== <--PC board
| _|_ |
| | _ | | _ | |
=- | || || || || | -= <--screws
|_|| ||___|| ||_|
____| |_____| |____
|___________________|
^ ^ ^
| | |
diodes capacitors MOSFETs
The exposed surface of the MOSFET transistors is the drain. The exposed
surface of the diodes is the anode. These two are connected directly
together, so they are screwed directly to the extrusion, with no
insulation needed. The diodes and MOSFETs alternate lengthwise; not all
on one side. A piece of buss bar is screwed to the extrusion and comes
out the end as the M- terminal.
The B+ and B- leads are pieces of buss bars screwed to the top of the PC
board, and coming out the ends of the case. The A2 terminal has just two
diodes between it and the extrusion.
___ ___
B- busbar--> |___| |___| <--B+ busbar
=================== <--PC board
_ _
M- --> ||| | | |_|| <--A2 busbar, with diodes
busbar ||| | | |_|| between it and extrusion
____| |_____| |____
|___________________|
Finally, the logic board stands vertically along the left side. It
connects to the power board with a couple of pin headers or connectors.
This setup is simple, but has problems.
- The capacitor-diode-MOSFET loop is rather long.
- There aren't enough diodes to handle max motor current.
- The A2 diodes (used only for plug braking) are especially weak.
- The long busbars, connected at only one end, means that the
closer diodes, capacitors, and resistors hog more of the current.
- They depend on foil on the PC board to carry high current.
- Heatsinking is poor. Heat has to flow edgewise down the vertical
part of the extrusion, then through the plastic sheet, to get to
the case. Then it has to flow edgewise around the sides of the
case to get to the (negligible) fins, or through yet another
interface to get to the actual heatsink bolted to the bottom.
I was just going to use a PWM uController for the control section,
do all the tuning and limits in software.
Did you notice the thread on problems getting the Zilla's micro set up
and figuring out its error codes? This is the sort of thing people run
into with micros. What seems obvious and easy to the programmer can be
obscure and frustrating to a user.
Is a LEM sensor good enough for battery, motor (freewheel) current
sensing? It has the advantage of being isolated, but I'm concerened
about it's response.
I'd say no. A LEM sensor is expensive, and doesn't do anything we need
here. The current sensor has to respond *instantly*, which means it
shuts off the transistors *right now*. You can't wait for a micro to
respond.
Otmar says mosfets don't have to be matched, but diodes do. Is diode
matching simply a matter of making sure their forward drop is within
a certain range of each other?
MOSFETs behave like resistors. If they aren't matched, the current
divides between them according to their on-resistances. A 10% difference
means the current will split 45-55 instead of 50-50. The one that gets
10% more current will run hotter. But a MOSFET's resistance rises as it
gets hotter; this tends to narrow the imbalance to maybe 5%. Thus, you
can get away without matching if you size all the parts big enough.
Diodes and IGBTs are quite different. Their on-state voltage drop is
relatively constant with current, but varies between devices. The one
that happens to have the lowest voltage drop hogs most of the current; a
10% voltage imbalance can cause a 50% current imbalance. Worse, the
voltage drop goes down as the part gets hotter; this makes the imbalance
even worse. So, you have to carefully match diodes or IGBTs in parallel
to get anywhere near their rated current limits.
But in any case, you still have to worry about balancing issues. If one
part has a worse solder conneciton, or a poorer connection to the
heatsink, it will still die long before the others. With many devices in
parallel, this can easily start a chain reaction that causes them all to
fail.
Other than water sensitivity, what causes the Curtis failure modes?
Poor cooling. Too high a voltage. Trying to "lug" the motor at low speed
and high current. Too big a motor for the controller size. No precharge
resistor. Trying to use its plug braking feature with a heavy EV.
An advantage of this design is that all of the parts are simple,
inexpensive, and generic.
Yeah, and a PWM PIC is about $6. Considering the cost of the rest of
it ... not a big deal.
The important point is that the micro is not an essential part; it is an
"extra". You don't need a $6 PIC to get PWM. A $1 PWM chip can do it;
and it won't need programming, or regulated power supplies, or crash
from noise, or respond too slowly to save expensive transistors.
I'm keeping my eye on the 1200V, 600A IGBT modules on Ebay -- $85.
Trying to find a reason not to use it!
How about... You can't get more than about 300 amps out of an advertised
"600 amp" part. It will have a voltage drop of 2-3 volts, which would
produce twice as much heat as a MOSFET based controller. You aren't
going to be using a pack voltage anywhere near high enough to need a
1200v part. Most IGBTs switch so slowly that you won't be able to run
them at inaudible frequencies.
device I'm looking at (http://tinyurl.com/u7v93)
This is a Powerex CM600HA-24H IGBT. It has a 2.5-3.4v on-state voltage,
and a junction-to-heatsink resistance of about 0.07 deg.C/watt. Assume
your heatsink is at 50 deg.C (121 deg.F), and has a thermal resistance
of 0.1 deg.C/watt (which is very good), your total thermal resistance is
0.17 deg.C/watt.
You don't want the IGBT junction to be over 100 deg.C for reasonable
life. So, the most you can have is 100-50= 50 deg.C from junction to
ambient. How many watts can you dissipate?
Power = 50 deg.C / (0.17 deg.C/w) = 294 watts
How much current is this with the nominal 2.5v drop?
Current = 294 watts / 2.5v = 118 amps
So, the most this part can dissipate on a continuous duty basis is only
118 amps. It would be even less when you add switching losses, or you
you were unlucky and got a part that had the maximum 3.4v on-state drop.
See what I mean about the huge gap between advertised marketing numbers
and real-world performance?
Mr. Borges schematic is (literally) unreadable, scanned at too low
a resolution.
you can also get the original book.
Any idea what book it is? I just ordered "Motor Control Electronics
Handbook" by Valentine, at Otmar's suggestion
That's the one!
Problem is, most people *can't* contribute.
If they truly can't contribute anything, neither skill nor time nor
money, then they are not likely to get anything in return.
--
Ring the bells that still can ring
Forget the perfect offering
There is a crack in everything
That's how the light gets in -- Leonard Cohen
--
Lee A. Hart, 814 8th Ave N, Sartell MN 56377, leeahart_at_earthlink.net
--- End Message ---
--- Begin Message ---
> I am trying to figure your figures. Please excuse me if this doesn't
> burst my bubble.
>
> Your range is true. It takes 36.6 kWh to be the equivalent of one gallon
> of gas. This "dooms" all EVs to never exist. Gas is just too cheap to
> compete with an EV. (Sidebar: EVs at slow speeds, such as in town, get 3
> times the range per equivalent power.) Basically, it takes a lot more
> than 600 lbs of LA batts to be the equivalent of a gallon of gas.
>
I'm not sure what you are calculating. The 600lbs = 1 gal is a
generalized rule of thumb developed from peoples experience driving EVs,
predominately around town.
> I do buy the fact of pulling an EV. What if you aren't pulling it?
> There's no reason to pull the trailer when you don't need it.
> You drop the trailer,
> and go on with your normal 32MPG rating.
I guess I'm not getting your point. Are you saying that you'd drive half
way to work (or where ever) and then leave the trailer on the side of the
road for the rest of the trip?
Or not use the trailer any time you are going more than 8-9 miles away?
If you trip is only 10 miles or less each way, can I recomend a bicycle?
It's what I use for short range commutes.
> EV pushers should not be all that worse than converting a car to EV
> power. Granted, you still have to hook up a vacuum pump to operate the
> brakes and a DC converter to operate the lights... but you aren't
> pushing along all that much greater load than trashing the car for
full-> EV power... and all the weight would be in the trailer.
Do you happen to know of a trailer that already has a drive axle and
transmission installed? Most cars come with these.
So you have to build or modify a trailer to mount the transmission,
differential, drive axle, etc.
Stripping the ICE out of a car is a lot easier than building a trailer
from scratch (or massively modifying one)
After that all of the work is the same for a pusher trailer or a full up EV.
> In my thoughts, that could enable the
> modification of more modern cars (rather than those ready for the scrap
> bin) to EV power.
People modify modern cars. Granted the very recent ones are getting
complicated.
> As for motive power, you will be pushing along a few hundred
> extra pounds.
Plus two more wheels worth of rolling resistance, plus more aerodynamic drag.
A few hundred extra pounds makes a large difference when you are going up
hill. You might now notice it driving an ICE, but you really notice it in
an EV, especially if you have a watt meter.
The same goes for accelerating from a stop. City driving tends to have
LOTS of stops and goes
--
If you send email to me, or the EVDL, that has > 4 lines of legalistic
junk at the end; then you are specifically authorizing me to do whatever I
wish with the message. By posting the message you agree that your long
legalistic signature is void.
--- End Message ---
--- Begin Message ---
Like electrical resistances, thermal resistances are always *between*
two things. So if you already know the heatsink temperature [voltage],
you can figure out the ambient temperature [voltage] based on the power
dissipated [current] and the resistance.
But this resistance shouldn't be included here, because we aren't saying
the ambient temperature is 50*C. 118A would be right for an ambient
temperature of 50*C. Recalculating for a known 50*C heatsink:
Power = 50 deg.C / (0.07 deg.C/w) = 714 watts
How much current is this with the nominal 2.5v drop?
Current = 714 watts / 2.5v = 286 amps
That feels a lot better! But it's still wrong! Why you ask? The 2.5V
drop is for 600A, not for 286A. It's also for 25*C junction, which is
obviously not true.
Let's look at 350A. The graphs show less than 2V of drop for this
current at 100*C junction:
Current = 714 watts / 2.0v = 357 amps
So more than three times 118A would be okay for a typical module.
Sometimes you have to look a little more deeply into the datasheets!
- Arthur
> >>>Lee Hart wrote:
> This is a Powerex CM600HA-24H IGBT. It has a 2.5-3.4v on-state voltage,
> and a junction-to-heatsink resistance of about 0.07 deg.C/watt. Assume
> your heatsink is at 50 deg.C (121 deg.F), and has a thermal resistance
> of 0.1 deg.C/watt (which is very good), your total thermal resistance is
> 0.17 deg.C/watt.
>
> You don't want the IGBT junction to be over 100 deg.C for reasonable
> life. So, the most you can have is 100-50= 50 deg.C from junction to
> ambient. How many watts can you dissipate?
>
> Power = 50 deg.C / (0.17 deg.C/w) = 294 watts
>
> How much current is this with the nominal 2.5v drop?
>
> Current = 294 watts / 2.5v = 118 amps
>
> So, the most this part can dissipate on a continuous duty basis is only
> 118 amps. It would be even less when you add switching losses, or you
> you were unlucky and got a part that had the maximum 3.4v on-state drop.
>
> See what I mean about the huge gap between advertised marketing numbers
> and real-world performance?
--- End Message ---
--- Begin Message ---
There are a couple of things to keep in mind , you can't hook your house
battery to any of the traction pack batteries and still use it as the "
house" battery as this would in effect connect some part of the traction
pack to the frame of the car . You could have 2 house batteries and
completely disconnect one ( both + and -) then charge your low traction pack
from that while the other house battery runs your lights ect. . The other
thing I found , when you can charge one battery in your traction pack buy
its self , either from a solar panel or dc to dc , you may find your self
throwing your pack more out of balance buy using it only some of the time or
if you forget and leave it on , well , that's not good either.
Steve Clunn .
----- Original Message -----
From: "Storm Connors" <[EMAIL PROTECTED]>
To: "List EV" <[email protected]>
Sent: Monday, November 06, 2006 8:55 PM
Subject: Solar EV power
I have a 190 watt solar panel I'm thinking of mounting on my car. It could
easily charge my house battery but would produce more power than needed. I
was thinking of using an inverter and dumping the excess in the traction
pack. I realise it wouldn't be a real significant contribution, but at
least it wouldn't be wasted.
But from this list I learned about Lee Hart's battery balancer that
charges each battery individually, and Steve Clunn who charges his
lawnmowers from his truck pack. Would it be reasonable to have a large
house battery and use it to charge up each pack battery in turn? Charge up
house, when full charge up first traction battery. Charge up house and use
it to charge next traction battery. etc.
It should be impossible to overcharge the traction batteries this way. I
assume the transfer of charge would be rapid. Eventually the pack would be
balanced. I suppose that if the traction batteries were discharged it
would require a pretty healthy connector.
Is there anything here worth pursuing?
storm
--- End Message ---
--- Begin Message ---
Lock Hughes wrote:
Can't get changes in elevation on the speedo either
:)
tks
Lock
Toronto
Human/Electric Hybrid Pedestrian
--- Lawrence Rhodes <[EMAIL PROTECTED]> wrote:
Not this one. It seems as useful as my speedometer. It's a great
way to
find out the error of your speedometer and it's an amazing trip
odometer.
Lawrence Rhodes.
----- Original Message -----
From: "John G. Lussmyer" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Sunday, November 05, 2006 10:54 AM
Subject: Re: speedometer replacement
What about tunnels?
--- End Message ---
--- Begin Message ---
Roland Wiench wrote:
For my EV, they said 15 modules of 17 volts each
would be equal to about 50ah battery at a cost of $40,000.00 which would
give me a range of 15 miles.
What are the joule rating of these caps? When I looked it up, 3600
kilojoules is equal to 1KWh. Isn't a 240V, 50Ah pack around 12KWh? Did I
get something wrong here, or do you need about 43 megajoules to equal
such a pack?
--- End Message ---
--- Begin Message ---
First, all the outlets for block heaters here in
Fairbanks, AK are on 20 amp breakers.
Sometimes I'll park somewhere, and I can have access
to two 20 amp outlets.
I was wondering if I could connect each outlet to a
separate diode bridge and then parallel the output for
my 120v dc battery pack. (Take the +168v from each
bridge and each -168v from each bridge, and applying
them to my batteries).
Can that work?
thanks,
Michael
__________________________________________________________________________________________
Sponsored Link
Talk more and pay less. Vonage can save you up to $300 a year on your phone
bill.
Sign up now. http://www.vonage.com/startsavingnow/
--- End Message ---
--- Begin Message ---
You are not accurately describing the heatsink. Lee is correct there.
A heatsink is not a fixed temp. Well, they do have a temperature, but
it's not fixed. The effective characterization is an ambient temp and
the heatsink's thermal resistance.
0.1 deg C/W is realistic for a good heatsink, actually that's probably
the best. If the heatsink's fan is taking in 35C air in a hot car, the
package will see a heatsink temp of 50C at a power level of 150W. At
714W that heatsink's temp alone will be 106.4C, exceeding the max
junction temp at the heatsink face alone.
There are no great workarounds to the heatsink problem. 0.1C/W is
already for a very good heatsink, we're talking lots of fins and the
most airflow that is practical. It may be difficult to attain anything
like this by "working with what you have". Water-cooled can be better,
but they're still not zero. There's also the matter of the thermal
connection (grease, silicone thermal pad, etc) also has its own nonzero
thermal resistance. Here's a description:
http://www.electronics-cooling.com/Resources/EC_Articles/JUN95/jun95_01.htm
So take a look. The thermal resistance of the silicone insulator alone
(if you need insulation) is 0.78C/W if it's a full sq in. The best
listed in 0.07C/W for a full sq in of contact area.
I just did a brief websearch and those super-effective water-cooled CPU
coolers are talking about 0.4C/W to 0.22C/W, these are not as good as
the original 0.1C/W estimate. And the 0.22C/W was listed as the spec
for the waterblock, I think that's from water-to-mounting surface
resistance, but the water will be warmer than the ambient under load
because the radiator has its own thermal impedance.
Danny
Arthur W. Matteson wrote:
Like electrical resistances, thermal resistances are always *between*
two things. So if you already know the heatsink temperature [voltage],
you can figure out the ambient temperature [voltage] based on the power
dissipated [current] and the resistance.
But this resistance shouldn't be included here, because we aren't saying
the ambient temperature is 50*C. 118A would be right for an ambient
temperature of 50*C. Recalculating for a known 50*C heatsink:
Power = 50 deg.C / (0.07 deg.C/w) = 714 watts
How much current is this with the nominal 2.5v drop?
Current = 714 watts / 2.5v = 286 amps
That feels a lot better! But it's still wrong! Why you ask? The 2.5V
drop is for 600A, not for 286A. It's also for 25*C junction, which is
obviously not true.
Let's look at 350A. The graphs show less than 2V of drop for this
current at 100*C junction:
Current = 714 watts / 2.0v = 357 amps
So more than three times 118A would be okay for a typical module.
Sometimes you have to look a little more deeply into the datasheets!
- Arthur
>>>Lee Hart wrote:
This is a Powerex CM600HA-24H IGBT. It has a 2.5-3.4v on-state voltage,
and a junction-to-heatsink resistance of about 0.07 deg.C/watt. Assume
your heatsink is at 50 deg.C (121 deg.F), and has a thermal resistance
of 0.1 deg.C/watt (which is very good), your total thermal resistance is
0.17 deg.C/watt.
You don't want the IGBT junction to be over 100 deg.C for reasonable
life. So, the most you can have is 100-50= 50 deg.C from junction to
ambient. How many watts can you dissipate?
Power = 50 deg.C / (0.17 deg.C/w) = 294 watts
How much current is this with the nominal 2.5v drop?
Current = 294 watts / 2.5v = 118 amps
So, the most this part can dissipate on a continuous duty basis is only
118 amps. It would be even less when you add switching losses, or you
you were unlucky and got a part that had the maximum 3.4v on-state drop.
See what I mean about the huge gap between advertised marketing numbers
and real-world performance?
--- End Message ---
--- Begin Message ---
> > I am trying to figure your figures. Please excuse me if this doesn't
> > burst my bubble.
> >
> > Your range is true. It takes 36.6 kWh to be the equivalent of one
gallon
> > of gas. This "dooms" all EVs to never exist. Gas is just too cheap to
> > compete with an EV. (Sidebar: EVs at slow speeds, such as in town,
get 3
> > times the range per equivalent power.) Basically, it takes a lot more
> > than 600 lbs of LA batts to be the equivalent of a gallon of gas.
> >
>
> I'm not sure what you are calculating. The 600lbs = 1 gal is a
> generalized rule of thumb developed from peoples experience driving EVs,
> predominately around town.
>
Yeah, the 600 lbs is the same as the energy *output* from an ICE
burning a gallon of gas, so takes into account typical inefficiency.
--- End Message ---
--- Begin Message ---
You seem to have little idea of what's involved here.
Lee said the heatsink temperature was 50*C. Given that parameter, I
calculated the current correctly. Even if he had said that the ambient
temperature were 50*C, his current calculation is still too low because
it doesn't account for the lower voltage drop due to the lower-than-600A
current and the higher-than-25*C junction temperature.
Clearly, a brief websearch is not going to teach you much (at least if
you interpret the results incorrectly). The heatsink in question is
obviously larger than that of a CPU.
These are the heatsinks we use at work:
http://www.d6industries.com/HeatSinks.htm
Even the worst on that page is 0.15*C/W, and it can fit in a 5" cube!
This would barely hold the module anyway. The water-cooled heatsink I'm
using for my project is around 0.006*C/W, and could fit about four
modules. This would still be one-quarter of the thermal resistance Lee
quoted (but the value doesn't matter anyway, as I said above).
- Arthur
On Tue, 2006-11-07 at 01:59 -0600, Danny Miller wrote:
> You are not accurately describing the heatsink. Lee is correct there.
> A heatsink is not a fixed temp. Well, they do have a temperature, but
> it's not fixed. The effective characterization is an ambient temp and
> the heatsink's thermal resistance.
>
> 0.1 deg C/W is realistic for a good heatsink, actually that's probably
> the best. If the heatsink's fan is taking in 35C air in a hot car, the
> package will see a heatsink temp of 50C at a power level of 150W. At
> 714W that heatsink's temp alone will be 106.4C, exceeding the max
> junction temp at the heatsink face alone.
>
> There are no great workarounds to the heatsink problem. 0.1C/W is
> already for a very good heatsink, we're talking lots of fins and the
> most airflow that is practical. It may be difficult to attain anything
> like this by "working with what you have". Water-cooled can be better,
> but they're still not zero. There's also the matter of the thermal
> connection (grease, silicone thermal pad, etc) also has its own nonzero
> thermal resistance. Here's a description:
>
> http://www.electronics-cooling.com/Resources/EC_Articles/JUN95/jun95_01.htm
>
> So take a look. The thermal resistance of the silicone insulator alone
> (if you need insulation) is 0.78C/W if it's a full sq in. The best
> listed in 0.07C/W for a full sq in of contact area.
>
> I just did a brief websearch and those super-effective water-cooled CPU
> coolers are talking about 0.4C/W to 0.22C/W, these are not as good as
> the original 0.1C/W estimate. And the 0.22C/W was listed as the spec
> for the waterblock, I think that's from water-to-mounting surface
> resistance, but the water will be warmer than the ambient under load
> because the radiator has its own thermal impedance.
>
> Danny
>
> Arthur W. Matteson wrote:
>
> >Like electrical resistances, thermal resistances are always *between*
> >two things. So if you already know the heatsink temperature [voltage],
> >you can figure out the ambient temperature [voltage] based on the power
> >dissipated [current] and the resistance.
> >
> >But this resistance shouldn't be included here, because we aren't saying
> >the ambient temperature is 50*C. 118A would be right for an ambient
> >temperature of 50*C. Recalculating for a known 50*C heatsink:
> >
> > Power = 50 deg.C / (0.07 deg.C/w) = 714 watts
> >
> >How much current is this with the nominal 2.5v drop?
> >
> > Current = 714 watts / 2.5v = 286 amps
> >
> >That feels a lot better! But it's still wrong! Why you ask? The 2.5V
> >drop is for 600A, not for 286A. It's also for 25*C junction, which is
> >obviously not true.
> >
> >Let's look at 350A. The graphs show less than 2V of drop for this
> >current at 100*C junction:
> >
> > Current = 714 watts / 2.0v = 357 amps
> >
> >So more than three times 118A would be okay for a typical module.
> >Sometimes you have to look a little more deeply into the datasheets!
> >
> >- Arthur
> >
> >
> >
> >
> >> >>>Lee Hart wrote:
> >>This is a Powerex CM600HA-24H IGBT. It has a 2.5-3.4v on-state voltage,
> >>and a junction-to-heatsink resistance of about 0.07 deg.C/watt. Assume
> >>your heatsink is at 50 deg.C (121 deg.F), and has a thermal resistance
> >>of 0.1 deg.C/watt (which is very good), your total thermal resistance is
> >>0.17 deg.C/watt.
> >>
> >>You don't want the IGBT junction to be over 100 deg.C for reasonable
> >>life. So, the most you can have is 100-50= 50 deg.C from junction to
> >>ambient. How many watts can you dissipate?
> >>
> >> Power = 50 deg.C / (0.17 deg.C/w) = 294 watts
> >>
> >>How much current is this with the nominal 2.5v drop?
> >>
> >> Current = 294 watts / 2.5v = 118 amps
> >>
> >>So, the most this part can dissipate on a continuous duty basis is only
> >>118 amps. It would be even less when you add switching losses, or you
> >>you were unlucky and got a part that had the maximum 3.4v on-state drop.
> >>
> >>See what I mean about the huge gap between advertised marketing numbers
> >>and real-world performance?
--- End Message ---
