Charging System Capacity -- Power for Accessories

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lyd
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Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 1:52 pm

Quigg and I just mostly completed an uncharacteristically on-topic discussion over on the OT in exile forum, about the amount of surplus electric power available on the VOL. Since that is exactly the place where it is likely to do the most people the least good, I figured I'd try to move the gist of it over here.

Hope it helps.

lyd
Last edited by lyd on March 22nd, 2008, 2:08 pm, edited 1 time in total.
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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 1:55 pm

I finally tracked down the output specs in the service manual. They were way in the back in an index. Someone just sent me a new copy, I don't think I have this page in the one I had been using.

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The same page has wattage for the lights.

Image

Max cooling fan load is 5A @ 12V per the service manual.

Not sure what the ignition coils draw, or what the normal maintenance load of the battery is. I need some help with this, quigg, 'cause I am not understanding something correctly. I will come back to this.

So, okay. Let's make apples and apples, out of what we have so far.

Headlight 60W/12V = 5A
Frnt. running lights 2W/12V = .2A
Tail light 5W/12V = .4A

Total minimum running load (less the stuff I don't know) = 5.6A

Front turn signal 21W/12V = 1.8A
Rear turn signal 21W/12V = 1.8A
Brake light 21W/12V = 1.8A
Fan on max ........ = 5A

Total maximum running load (less the stuff I don't know) = 15.4

And what's available at the unlikely RPM of 5K...

375W/15.5V = 24.2A
375W/14V = 26.8A
375W/12V = 31.3A

Okay, quigg, now help out here, 'cause this is where I get stuck. First of all, do I want to assume I have 31 amps to work with, because all I need is 12V and dropping to that is no problem, or do I want to assume that I have closer to 25, because the system really wants to be making between 14 and 15?

Second is all this business about ignition and battery charging. Take a look at this.

Image

I'm getting really lost here. Peak primary coil voltage of 200V at, split the difference and say 4 ohms, is 50 amps! Can that be right? And there's two coils, so you have to double that? 100A for ignition? Doesn't seem possible if the unloaded max output of the generator is 70V. From the schematic, it looks like the ignitor is fed straight from the generator -- that's the pickup coil? -- and from there to the ignition coils -- so is that circuit 3-phase AC? Is that where I am going wrong here? Argh.

If the pickup coil is 1.5V @ 160 Ohms, it can supply 0.009375A... is that my ignition load? Argh again.

I've thought myself into a corner and gotten all confused. Help get me back on track, man.

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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 1:57 pm

Okay, wait, the pickup coil is just for timing information... I knew that much already. Told you I had completely confused myself.

I still can't figure out what I should be looking at to determine normal operating current draw into the coils. Argh a third time. How does 200V get to the primary?

I know that gets stepped up to something like 20KV in the secondary... is the primary only getting fed 12v and stepping it up to the 200? If that's the case, then the load into the coil is around 6 amps? Times two for 12 amps?

The circuit is fused at 15 amps, so this is sounding at least more likely...

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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 1:58 pm

Quigg wrote:

Okay. The ignition first; You cannot determine current flow (or power consumption) of a coil from its pure ohmic resistance. A coil pretty much is a short ckt to DC, just a long wire, it resists a “change” in current, it is a reactive component. This resistance to a change in current plus the resistance of the conductor is the impedance and the impedance can be very large compared to the resistance of the wire since the signal to it is a pulse at a frequency relative to engine rpm. Those numbers for primary and secondary coil resistance of only of value when you trouble shooting. They tell you if the coil winding is burned or shorted internally.

Next; Lighting;
Turn signals bulbs are rated by their draw full on but turn signals are on/off. An incandescent lamp is very low resistance when it is cold and very high after it heats up. You can look at it as a temp. sensitive variable resistor. When you first energize an incandescent lamp the current is very high. You can consider this as equivalent to inrush current in a motor. By turning the lamp on and off the average current is higher than the wattage rating of the lamp would lead you to believe and you have two of them working at one time. Look at the way turn signals are fused as compared to the headlamps.

Maintenance current. In order for the battery to maintain its charge it must maintain 12.6V. If system voltage falls below that the battery will begin to help the system by discharging. If the counter emf of the system is greater than fully charged battery voltage the system charges the battery. One reason you want to keep the system voltage over 14V is economy. Back to the lighting group A lamp at 12 volts draws more current than one at 14V and useful life is extended also. It’s the same with motors (fan for example). The igniter develops the primary voltage. The pick up coil is generates a timing signal for the igniter and can be ignored as a load. What we need is a number for igniter power consumption.

The igniter is fused at 15A. this would lead you to believe 5A to 10 A is a reasonable guess.

The only place you have three-phase is in the alt./rectifier ckt. Everything else is 14V DC.



Okay. Whew. Thank you, I was flailing.

Ignition: If that's the case, how the hell do I determine the load for spark? (Theoretically, I mean. I'll have fun testing all of this when my thingy arrives and it warms up a bit outside.) EDIT: I see you answered this one above -- we need ignitor current. Hmm.

Lighting: Um. Yeah, makes sense. I neglected to consider the flashy nature of the signals. I just looked up inrush on miniature lamps, and the suggested calculation was 10*rated current. 18 amps per signal! 36 amps when turning, 72 amps with the 4-ways on! That seems extreme. Now, I also see the duration of this draw is only about 30ms - 40ms. I don't know what the break time on the fuses are, but that circuit is only fused at 15 amps, so either it is greater than 40ms or these numbers are wrong. Thoughts? EDIT 2: Is it possible that the filaments are hot enough after the first blink that this big inrush only happens once on each turn? Also, the fronts are always getting .2A in their running-light mode, so they should be hotted up to a materially higher resistance already, no? Even though the turn filament is not on, there is heat in the bulb...

Battery Maintenance Current: I understand this from the voltage angle, but I am unclear on the current angle. CEMF in the battery drops to the point where it begins accepting voltage from the system, ok, but what determines the current flow? If I am using a smart charger, the charger is managing the current that the battery is allowed to pull, but that doesn't happen here as far as I can tell.

lyd
Last edited by lyd on March 22nd, 2008, 2:02 pm, edited 1 time in total.
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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 2:00 pm

Quigg wrote:

Well the simple-minded answer to your first question is to forget the spark ckt and look at the igniter/ ignition as a black box. You don’t care where the current goes after the igniter gets it. The hard way is very involved. First, because engine rpm is a variable the calculation for coil impedance gets hairy. The basic formula is, I = 2 pi FL, I think, where I = impedance and L is the inductance (or the reactive component) and F = is the frequency (rpm). You also need the parameters for the input signal; peak voltage, duration( also varies with rpm) and such other things. Once you have this info you can determine the average primary current at some rpm. A pain in the ass and not very useful.

One thought before we get to lamps. The fuse rating has nothing at all to do with the load. It is the wiring that serves the load that the fuse is rated for. Yet the engineers don’t waste wire so the ckt rating (fuse) can be an indicator for normal draw. Try looking at turn signals as never really getting hot, even though they do, and think of them as having an average temp which will always be below the rating temp and so the current is higher on average. (It's the area under the curve kind of thing.) The numbers you have for inrush are really scary!

I forgot the battery thing. System voltage minus battery voltage or say, 14V –12.5V = 1.5 Volts driving force across the internal resistance of the battery = charging current. Internal resistance gets into a whole nuther thing and mostly beyond my ken. Those battery tenders are voltage controlled, current limited, usually pulsed chargers, designed to prevent over charging. You bike charging system wants to quickly bring the battery up from a discharged state and maintain it there. It will do whatever it can to maintain system voltage without regard for the health of your battery. I never expose a battery in a very low state of charge to the bike charging system for that reason, unless there is no other choice.


Hey, I really appreciate you helping me work through this, Quigg. You're not the king for nothing. ;-) Let me know if you lose patience with this whole discussion, I don't want to impose. But in the meantime...

...let's back up a minute and revisit those first calculations based on all this new information.

Headlight 60W/12V = 5A
Frnt. running lights 2W/12V = .2A
Tail light 5W/12V = .4A
Ignition ..... = 5A

Total minimum running load (less the stuff I don't know) = 10.6A

Headlight 60W/12V = 5A
Front turn signal 21W/12V = 1.8A
Rear turn signal 21W/12V = 1.8A
Brake light 21W/12V = 1.8A
Fan on max ........ = 5A
Ignition ..... = 10A
Battery charging = 2A

Total maximum running load (less the stuff I don't know) = 27.4

Plus some indicator LED's, LCD display, spedo backlight, etc. call it 1.5A.

Let's forget the inrush issue for the moment, just for the sake of discussion, and say 28 Amps of load with everything going.

And what's available at the unlikely RPM of 5K...

375W/15.5V = 24.2A
375W/14V = 26.8A
375W/12V = 31.3A

Okay, we just barely have our head above water here, at 5K RPM/12V!. Let's turn off the signals and get off the brake, since those are intermittent loads and could reasonably put us into draining the battery by design.

22.6 Amps with the battery drawing 2A to charge.

Get the battery topped off, and we have 20.6.

How much current do you think we have at a more reasonable cruising RPM, say around 3500? The output curve is not going to be linear, so without a published spec this is probably going to have to wait for measurement.

Still, it is starting to look like we have something well under 10 amps left to work with, and that is drawing maximum output at all times, which afaik is not a good idea as I mentioned earlier. Even calling that 10A necessary buffer, untouchable, we are hitting the system at about 75% of maximum 5K RPM/14V capacity, without adding a singe, even factory, accessory! Drop down to street speeds, stop lights, etc, and it seems really high odds that we discharging the battery most of the time. (I'm going to stop holding the brake at lights!)

Add the factory light bar, two 25W lights so say about another 4A, and I don't see how it can possibly keep the battery charged under normal conditions.

I am freaking out. Is this thing just under-engineered? This can't be right...

To make sure apples are still apples, I'll re-figure everything for 14v across the board. Ooops! Those running lights are a second filament, and still on when you are signaling... more current. Ditto for the tail. New table in post to follow.

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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 2:06 pm

Quigg wrote:

Dammit Dave you’re starting to scare me!

I have to think about this some more. I have to be missing something.

So, during normal riding conditions on the stock bike we can assume we will be somewhere between your calculated min. and max load. At times at max and at time at min. At times of max load the battery helps and at time of less than max it will recover. That seems to say the system is designed to operate as stock with a little left over for maybe one accessory. Well engineered for stock, under-engineered as a cruiser.

Lyd, we either have to soup up the generator or put a windmill on the sissy bar. Your numbers are right. Your logic is flawless. On my Gold Wing, Honda chose to go with an external automotive type alternator. I’m going to check on its specs to get an idea of their idea of the power needs of a full tourer and get back to you.

[...]

A little perspective.

The Gold Wing's alternator is rated 1KW @ 2400 RPM! ( About 75A peak) This may represent overkill for a cruiser but the Wing really doesn't require that much more minimum power than a cruiser and I would think all that excess capacity is what Honda thinks is necessary for touring accessories. The wing has two headlamps and a few more running lights, a stock radio, and more dash lights but that may represent only 10 to 15 Amps more than the Volusia in stock dress.


Damn, quigg. So how do we hack an external generator onto the flywheel? (Gotta be producing power when we're standing still, right? Otherwise we could just spin something off the tire. ;-)

Here's what I wound up with after cleaning up my numbers.

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This is a little bit more encouraging, until you stop to consider the fact that we haven't even begun to figure electrical inefficiency into this -- how many of our precious amps are being turned into heat along the way? -- not to mention your inrush voltages. And of course this whole thing is still assuming 5,000 RPM.

This is just ugly.
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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 2:07 pm

Quigg wrote:

My 01 was stock but for a suzuki lightbar and a cigarette lighter. It usually started with only a couple cranks so I had no idea how badly I underestimated the lack of reserve. I understand now why folks were having battery problems.

I think this experience tells us to consider these numbers when buying a bike.

I wish I had an answer. I doubt there is an easy one or even a hard one.

You may be left with the two battery system and charging at resturants at lunch time.


Honestly, if there is nothing wrong with these numbers (I'll still test the reality when I get the chance, that'll be fun and interesting to do anyway) then I am going to abandon all my plans to geek out this bike with a lot of gadgets and high-power lights. It's just the wrong platform for it.

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Re: Charging System Capacity -- Power for Accessories

Postby lyd » March 22nd, 2008, 2:59 pm

Okay, back in real time now, and hopefully quigg will follow me over here to continue this...

I realized that table I made is kind of misleading, because it doesn't really take into account the way the current flows when the battery drops below nominal SoC. It really looks more like this.

Image

Below 13 volts or so, the battery is going to suck up all available current (per quigg's statement earlier) trying to return to SoC, 14.5v or so. So while there is more amperage on tap when everything is running at 12v, there is none at all available. It is a death spiral, where the lower the battery gets the more current the regulator allows the generator to produce, until you are running at max capacity. If that state of affairs continues for any length of time the battery is going to be hosed, first because you are slamming way too much charging current into and then because it will be run flat.

Based on this, it seems we need to allow a couple of amps for battery charging at all times, given the alternative, and our numbers (which are still overly optimistic for the reasons discussed above) for surplus power drop even further.

Think this is a close to accurate picture yet?
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Re: Charging System Capacity -- Power for Accessories

Postby annieh » March 22nd, 2008, 5:57 pm

All that just to figger out how to keep your GeePeeESs working? :roll:
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Re: Charging System Capacity -- Power for Accessories

Postby Quigg » March 22nd, 2008, 5:59 pm

Couple things; Think of the gen/regulator first as a 14 V battery with infinite capacity connected in parallel with your battery. What that means is that it can deliver enough current to the system to develop 14 V no matter the load. (Max load is a dead short and that can happen.) You have a constant voltage system.

The battery (generator) sees the bike battery as a load. It has resistance. The electrolyte resistance, the terminal resistance, and the battery terminal voltage appear to the charging system as a simple resistor. The battery voltage subtracts from the applied voltage and the charging current is determined by that difference voltage divided by the total internal resistance of the battery. Example: 14V applied and say 12.5 V battery voltage = 1.5V across , say, 1.5 ohms internal resistance results in 1A charging current. As the battery comes up in charge that difference voltage is reduced and so the charging current is reduced.

Now think of the gen/regulator as a normal 14 V battery with a limit on the amount of current it can supply. When total current demand exceeds that limit the system voltage will drop. If applied voltage drops to 13V, the difference voltage is now .5V and the resulting charging current is about 300 ma. So in effect the battery doesn’t hurt the system that much when the system voltage falls to 13V. If system voltage falls below battery voltage the battery will help the system by discharging. You can determine the amount of discharge current the same way you did the charging current. Round numbers. Available power = 25A @ 14V. = Max load of 2.8 ohms. Call that load 2.0 ohms because we are in trouble, overloaded. Just for shits and giggles I’ll say that drops the gen output voltage to 12 V. The difference is .5V across 2 ohms or 250ma discharge current.

Again think of the gen/regulator as a battery. If that battery has an internal resistance of, say 1 ohm and a terminal voltage of 14 volts open ckt, a current of 10 amps will develop a cemf in the battery of 10 Volts!!! This results in the available system driving voltage dropping to only 4V. I’m pulling these numbers out of my ass but you get the idea. Is a poor analogy, generators don’t work exactly like that but it’s close enough.


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