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Hi everyone,
As you all know, Coffee (Dean) passed away a couple of years ago. I am Dean's ex-wife's husband and happen to have spent my career in tech. Over the years, I occasionally helped Dean with various tech issues.
When he passed, I worked with his kids to gather the necessary credentials to keep this site running. Since then (and for however long they worked with Coffee), Woodschick and Dirtdame have been maintaining the site and covering the costs. Without their hard work and financial support, CafeHusky would have been lost.
Over the past couple of weeks, I’ve been working to migrate the site to a free cloud compute instance so that Woodschick and Dirtdame no longer have to fund it. At the same time, I’ve updated the site to a current version of XenForo (the discussion software it runs on). The previous version was outdated and no longer supported.
Unfortunately, the new software version doesn’t support importing the old site’s styles, so for now, you’ll see the XenForo default style. This may change over time.
Coffee didn’t document the work he did on the site, so I’ve been digging through the old setup to understand how everything was running. There may still be things I’ve missed. One known issue is that email functionality is not yet working on the new site, but I hope to resolve this over time.
Thanks for your patience and support!
'09 610.. stranded with a stuffed battery.
- Thread starter BrandonR
- Start date
mxracernumber1
Husqvarna
A Class
Motosportz;111364 said:I believe the system in the snowmobiles is different. The husky will not start without good current to the fuel pump. I think
The Japanese MX bikes are EFI, kickstart, no battery.
BrandonR
Husqvarna
A Class
I finally got around to testing a capacitor on the 610, and it works. I got a $5.30 35V 4700uF capacitor from Radio Shack, wired it into the electrical system with the battery disconnected, jump started it and the bike runs perfectly smooth without a battery in the system. Disconnect the capacitor and the bike starts missing, the lights on the dash start flashing etc.. reconnect it and right back to smooth running. There are two wires to the negative post on the battery, the smaller blue wire has to stay connected to chassis ground at all times or the bike will shut off, the larger wire connects the battery to chassis ground. To simplify things I moved the small wire to the stud on the regulator/rectifier. This way there's only the two cables going to the battery now. In the event of a battery failure the procedure is to simply disconnect the negative cable and zip-ty it away from the battery terminal then jump start the bike. I doubt you could bump start it but I'll probably find a big hill the next chance I get and see what happens.
I can't figure out how to embed on this forum so here's a link..
http://www.youtube.com/watch?v=F1C14A2rVbc
I can't figure out how to embed on this forum so here's a link..
http://www.youtube.com/watch?v=F1C14A2rVbc
Brandon, unless you have a reasonable idea what you are doing, and preferably access to an oscilloscope too, you are asking for trouble with your experiments. At least that is my opinion.
That 4700 uF is an electrolytic and has a reasonably high esr at the higher frequencies (equivalent series resistance). At the very least put a 1uF ceramic disk in parallel with it to cut out some of the higher frequency noise. A 0.47 uF and 4.7uF in parallel with that electrolytic would help even more. Total cost should be under $1.
The electrolytic is not 'quick' enough to absorb the faster, higher frequency voltage spikes.
That 4700 uF is an electrolytic and has a reasonably high esr at the higher frequencies (equivalent series resistance). At the very least put a 1uF ceramic disk in parallel with it to cut out some of the higher frequency noise. A 0.47 uF and 4.7uF in parallel with that electrolytic would help even more. Total cost should be under $1.
The electrolytic is not 'quick' enough to absorb the faster, higher frequency voltage spikes.
BrandonR
Husqvarna
A Class
Let's model redline for a moment as the highest frequency output from the stator, I think the 610 stator has 12 poles, so the waveform cycles 6 times per revolution, at 8200 rpm that makes 820hz AC single-phase, assume full-wave rectification and we get 1640hz, or .00164Mhz. Find me a mhz+ waveform on the DC rail in the system and I'll get concerned.
If the core issue was voltage above battery voltage then the battery would be a poor choice to control that since it would absorb far less of the over-voltage, the line voltage would spike out of control because a lead acid battery is extremely slow to absorb power. That's what the voltage regulator is there for, what the regulator can't do though is fill in the dip in voltage as the waveform heads down to zero.
I am speculating that the core problem is the unfiltered rectified waveform dropping to 0 volts between cycles. The battery keeps that voltage up by supplying stored energy from it's reserve, because it has a large reserve that dip is almost entirely eliminated, a capacitor has a much smaller reserve so there will most likely be a larger dip between cycles but as long as it has enough capacity to keep the voltage above the minimum operating threshold of the electronics (most likely 5 volts or less) they will be perfectly happy. The computer is most likely running at 10+mhz so if it's using the 12v rail to supply a reference voltage anywhere it's quite likely that without any filtering that it is seeing low voltage for hundreds to thousands of CPU cycles every time the waveform heads back down to 0, somewhere it the ROM there's a countdown threshold for how many cycles to wait for voltage to get back to adequate, once that number is exceeded the computer shuts down the ignition. This line of thought is in line with the observed evidence (dash displays a low battery warning before shutting down)
If the core issue was voltage above battery voltage then the battery would be a poor choice to control that since it would absorb far less of the over-voltage, the line voltage would spike out of control because a lead acid battery is extremely slow to absorb power. That's what the voltage regulator is there for, what the regulator can't do though is fill in the dip in voltage as the waveform heads down to zero.
I am speculating that the core problem is the unfiltered rectified waveform dropping to 0 volts between cycles. The battery keeps that voltage up by supplying stored energy from it's reserve, because it has a large reserve that dip is almost entirely eliminated, a capacitor has a much smaller reserve so there will most likely be a larger dip between cycles but as long as it has enough capacity to keep the voltage above the minimum operating threshold of the electronics (most likely 5 volts or less) they will be perfectly happy. The computer is most likely running at 10+mhz so if it's using the 12v rail to supply a reference voltage anywhere it's quite likely that without any filtering that it is seeing low voltage for hundreds to thousands of CPU cycles every time the waveform heads back down to 0, somewhere it the ROM there's a countdown threshold for how many cycles to wait for voltage to get back to adequate, once that number is exceeded the computer shuts down the ignition. This line of thought is in line with the observed evidence (dash displays a low battery warning before shutting down)
BrandonR;131875 said:
A lifetime ago I designed automotive test equipment and add on products. I've *no* idea how the bikes compare and I've long since sold the oscilloscope and the rest of the test equipment.
In an automotive charging circuit without a battery to filter the output there are high frequency voltage spikes of hundreds of volts, those spikes are indeed in the MHz range if you do an fft on them.
Again, I've no idea how the bikes compare. If someone wanted to post an oscilloscope display of an unfiltered charging circuit in a bike that would be great. I'd imagine if anyone has a scope it would be seymore.
lankydoug
Husqvarna
Pro Class
Once a 12 v battery is fully charged it is equal to a 1/4 ohm resistor. However if you were to put a 1/4 ohm resistor or any other device in line instead of a battery and it should fail and become an open circuit it would most likely fry anything and everything that the charge system can find a ground through. My advise is buy a battery and ride.
BrandonR
Husqvarna
A Class
lankydoug;132334 said:
It's hard to buy a battery when you're stranded 50 miles out in the woods or desert. The 610 is e-start nobody is talking about running it on any regular basis without a battery we are talking about making it limp out without problems when the battery dies without warning.
lankydoug
Husqvarna
Pro Class
BrandonR;132413 said:
A high amp rated 1/4 ohm resistor would get you home but I would not suggest leaving it in there if it was to go open the voltage regulator would have no way to control the voltage which usually results in it going WFO until it fries everything on the bike
BrandonR
Husqvarna
A Class
Coffee;132429 said:
Now that sounds crazy.
The voltage regulator is set to about 14 volts, a 1/4 ohm resistor would mean shunting 50+ amps, 700+ watts (over double the max output of the stator) through a big resistor. And exactly how would a big resistor store energy to cover the waveform collapse to zero?
lankydoug
Husqvarna
Pro Class
A 1/4 ohm resistor simulates a charged 12 volt battery and keeps the charge circuit a closed loop so the regulator can work properly. It must be a high amp rated resistor or it will get hot and burn open and the regulator can't control the voltage. This was an old trick to verify if the charge system was working. If the charge system is working it will only charge what the components are drawing. If not the system will die due to no charge. A discharged or shorted battery has little to no resistance. The resistor goes in series between the pos cable and the pos terminal of the battery.
I should clarify that if the battery failed due to a massive internal short I'm not sure if anything will get you back besides another battery.
I should clarify that if the battery failed due to a massive internal short I'm not sure if anything will get you back besides another battery.
BrandonR
Husqvarna
A Class
lankydoug;132484 said:
A $5.28 capacitor will, that's the whole point of what Coffee and I have been discussing. The cap simulates a battery to the extent that it can. My contention is that the waveform collapse is the root cause of the bike being unable to run with a failed battery. Basic power supplies always have some sort of filtering capacitor in them. I am working on the hypothesis that the 12vDC side of the system can essentially be modeled as a simple AC-DC power supply.
Installing a big resistor as essentially a voltage divider will reduce the current flowing through a failed battery preventing the battery from cooking, also preventing the failed battery from drawing down line voltage but there's a lot of faith you have to put into what's left of an already failed battery, the 610 has a 300+ watt stator, if the battery continues to fail you will end up with that resistor becoming a 300 watt heater and the bike will shut off anyways so you've essentially gained nothing but you have spent a chunk of money on an expensive rare component.
From a simplistic view the battery can be compared to a 600k Farad capacitor. About half that energy is unusable due to the chemistry, and the rest is mainly there for running the starter. What the capacitor test shows is that the electronics in the bike actually don't need that much storage to operate. Coffee's legitimate concern is that the capacitor may not be clamping high frequencies if they exist but neither of us has the tools on hand to prove it one way or the other. Personally I'm happy I found a simple way to get out of a shitty situation that could strike at any time.
lankydoug;132484 said:
When you said 1/4 ohm I was assuming you were referring to the Thévenin equivalent circuit output impedance (Rth).
Assuming a perfectly working voltage source of 12 - 14 volts, yes shunting it to ground will get hot. P=E^2/R=12^2/0.25 = 576 watts.
Placing a 1/4 ohm resistor in place of the battery is not something I would recommend.
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Attachments
If you take Brandons idea for a capacitor and install a zener diode in line with it wouldn't that filter out any potential high voltage spikes? Since the voltage reg. is still in the circut I don't think you even need a heat sink for the diode.
rajobigguy;132731 said:
Typical zener diodes are too slow to respond to the high frequency spikes (if there are any), voltage regulators are also typically slow. If anything it would need to be about a 15V zener across (in parallel) the capacitor to absorb anything above 15V
Brandon is correct with what he is trying to do, I simply suggested a couple of additional ceramic capacitors (which respond quickly) in parallel with the electrolytic.
lankydoug
Husqvarna
Pro Class
Have you tested your capacitor idea? A 1/4 ohm resistor in series with a dead battery simulates a not dead or charged battery, that's it. that info may help you in your quest. It does not guard against a dead short to ground. Lets face it batteries in rough service are prone to failure at any time and so is every other component on the bike. If a guarantee of not being stranded is what you are after short of having a spare bike in your backpack I suggest not leaving the house. On a lighter note if you want a guarantee that your battery will never fail just start carrying one in your backpack... it works every time.