Honda Twins - Three Phase Alternator Project

Sonreir said:
Very respectable.

So for the lights we are only drawing 1.5 amps per hour. Shouldn't the stock ac generator be keeping up?


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If that were all the power that were being used, then absolutely.

Stock alternator is good for about 110W when new, maybe 90 or 100 now.
 
So on our bike we have one module of a Dynatek Dyna S for electronic ignition maybe 50 watts? The panel maybe 2 watts? The lights 18 watts. Anything else to consider?


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Tail light should be about it. Everything else would be momentary such as the horn and turn signals.

Might be worth running an Ammeter between the negative terminal of the battery and ground (inline, disconnect the battery and use the Ammeter to complete the circuit) and performing a measurement.

With everything turned on and the bike not running, you should see about 6 amps. Most multimeters will read up to 10A, so you might be able to get away with that.
 
Sonreir said:
Tail light should be about it. Everything else would be momentary such as the horn and turn signals.

Might be worth running an Ammeter between the negative terminal of the battery and ground (inline, disconnect the battery and use the Ammeter to complete the circuit) and performing a measurement.

With everything turned on and the bike not running, you should see about 6 amps. Most multimeters will read up to 10A, so you might be able to get away with that.
So what is the power factor of a typical motorcycle charging system to determine how many watts a system will generate?


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Manufacturers generally try to design/select an alternator the produces just enough to run the system. Anything over what's needed is just waste.

I recall reading that the Honda twins were 110W and started in 1974 they went up to 120W. Rotor changes, maybe? Not too sure on the specifics.

Generally, the alternator should have about 20W leftover for charging the battery.

On a CB360, headlight is 35W, coils are about 60W together, 7W for the tail light, and then another 9W for all the gauge bulbs. That's 111W already, but that number will drop when the bike is running because the coils pull less power when the engine is turning. I can't remember what the exact dwell is, but assuming it's 120° of crank rotation, we can drop the power requirements for the coils to about 50W when the riding. This brings us to 101W, with about 19W leftover for charging.
 
I am trying to get to an exact correlation based upon actual VAC generated. Here is the calculator and they have a power factor box for a range of 0-1 http://www.rapidtables.com/calc/electric/Volt_to_Watt_Calculator.htm
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Wattage = Volts x Amps.

Power factors don't usually apply to motorcycles because we're generally concerned about things on the DC side of the equation. There are some losses when rectifying from AC to DC, but they're usually less than 10% with modern electronics.
 
So the higher the VAC the more rapid the charge? So let's say that we have 35 vac times 6 amps = 210 watts and let's say we produce 70 vac times our 6 amp requirements and we will produce 420 watts therefore charging in 1/2 of the time?


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So a 100w system / 35 VAC=2.87 amps to sustain a current charged battery


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Not that simple, unfortunately.

Alternators produce Wattage. That's a combination of both Voltage and Current, and the two are interchangeable because Wattage = Voltage x Current. If your alternator is putting out 110W, it could be 11A at one Volt or it could be 11V at one Amp, or any combination that adds up to 110W. The actual combination you see will be determined by Mr. Ohm and his famous law.

Let's say you hook up your alternator to a circuit with four Ohms of resistance. We know that Ohm's Law is as follows:
Current = Voltage / Resistance and we also know that Watts = Current x Voltage. Now we just need to figure out the numbers that satisfy both of those equations.

Current = Voltage / 4
and
110 = Current * Voltage

From the first equation, we already know that Current = Voltage / 4, so let's sub that into the second equation (Algebra is fun!) and we get:
110 = Voltage^2 / 4

Simplifying gives us 440 = Voltage^2 or approximately 21 Volts.

Plug it back into the top to get Amps and we have 5.25. Double check ourselves: 21 * 5.25 = 110? Check. 5.25 = 21 / 4? Check.

And to make things even more complicated, in a permanent magnet alternator system, the Wattage output of the alternator is changing along with the RPMs of the engine. This is where the voltage regulator comes into play and also why we don't usually care too much about the AC voltage readings. Most voltage regulators for permanent magnet alternators are a "shunt type". They work by monitoring DC voltage levels and consuming the additional power that's being generated by the alternator. So for the purposes of our above calculations, we can fix the voltage at a suitable level (14V is fine) and our resistance becomes variable (due to the electronics within the regulator).

Anyway... not sure if that answers your question or not... I sort of got off on a tangent, I guess.
 
Texasstar said:
So a 100w system / 35 VAC=2.87 amps to sustain a current charged battery

For lead acid, at least, most batteries like to see at least an additional Amp of charging capacity within the system. Batteries will have their own internal resistance that changes with charge. As the battery gets topped up, its resistance will increase, which will lower the amount of power the battery is taking out of the system. A battery with a low enough charge (and high enough capacity), may actually pull more Wattage than your alternator can supply, leading to a situation where your bike doesn't seem like it's charging because a dead battery is sucking all the juice instead of the components you're trying to actually power.
 
Sonreir said:
Not that simple, unfortunately.

Alternators produce Wattage. That's a combination of both Voltage and Current, and the two are interchangeable because Wattage = Voltage x Current. If your alternator is putting out 110W, it could be 11A at one Volt or it could be 11V at one Amp, or any combination that adds up to 110W. The actual combination you see will be determined by Mr. Ohm and his famous law.

Let's say you hook up your alternator to a circuit with four Ohms of resistance. We know that Ohm's Law is as follows:
Current = Voltage / Resistance and we also know that Watts = Current x Voltage. Now we just need to figure out the numbers that satisfy both of those equations.

Current = Voltage / 4
and
110 = Current * Voltage

From the first equation, we already know that Current = Voltage / 4, so let's sub that into the second equation (Algebra is fun!) and we get:
110 = Voltage^2 / 4

Simplifying gives us 440 = Voltage^2 or approximately 21 Volts.

Plug it back into the top to get Amps and we have 5.25. Double check ourselves: 21 * 5.25 = 110? Check. 5.25 = 21 / 4? Check.

And to make things even more complicated, in a permanent magnet alternator system, the Wattage output of the alternator is changing along with the RPMs of the engine. This is where the voltage regulator comes into play and also why we don't usually care too much about the AC voltage readings. Most voltage regulators for permanent magnet alternators are a "shunt type". They work by monitoring DC voltage levels and consuming the additional power that's being generated by the alternator. So for the purposes of our above calculations, we can fix the voltage at a suitable level (14V is fine) and our resistance becomes variable (due to the electronics within the regulator).

Anyway... not sure if that answers your question or not... I sort of got off on a tangent, I guess.
I get it. It is like water flow. For example when designing an irrigation system we use 5 Feet per second as a maximum speed limit through the pipe. You can go faster but your friction loss increases and you may have water hammer in your toilet.


I am over simplifying things down to understand the Honda family joules. Fixing the VAC to the max we produce and working the equation backwards from there to arrive at how many amps are truly available on a stock cb175 to keep a battery charged. Weren't they worried about overcharging the lead battery and therefore split one generator leg just to supply the headlight? Matt should Zeke go into Mechanical or Electrical Engineering?


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Texasstar said:
I get it. It is like water flow. For example when designing an irrigation system we use 5 Feet per second as a maximum speed limit through the pipe. You can go faster but your friction loss increases and you may have water hammer in your toilet.


I am over simplifying things down to understand the Honda family joules. Fixing the VAC to the max we produce and working the equation backwards from there to arrive at how many amps are truly available on a stock cb175 to keep a battery charged. Weren't they worried about overcharging the lead battery and therefore split one generator leg just to supply the headlight? Matt should Zeke go into Mechanical or Electrical Engineering?

Yup. When I was first learning electrical stuff, comparing it to plumbing was exactly what I did. The alternator is your water pump. The resistance is the diameter of your pipes (low resistance is big pipes). The amount of water flowing through the system is current. The system pressure is voltage. Your battery is the water tower that the pump feeds. Your pump is only good for so much water output and this can be used as pressure, flow, or a certain related combination. If the pipes are too big (resistance is low and too much wattage draw on the system), then your pressure and your flow will be augmented by the water tower until it runs dry, then you're on whatever the pump can supply directly. But on the other hand, if your pump is too powerful for the system you might break some pipes or even your water tower.

For your testing, I would skip the VAC side of things. AC checks are only useful to see if your alternator is functioning and, on their own, they are not useful for determining alternator output. The reason for this is because the internal resistance of your multimeter is very high and you probably don't know the exact value. You might be able to use another multimeter to figure it out, but then we start getting into Inception-like scenarios. Measuring current directly from your alternator may cook it, so I don't recommend that choice either. And, ultimately, it's the DC side of the equation that's running the bike and charging the battery, so we should be measuring DC. :p

For the battery, overcharging was not the same consideration in the 70s as it is now. The battery was actually specifically chosen because of the beneficial effect it has on helping to regulate power within the overall system. You may notice that some of the early twins and singles don't even use a voltage regulator and that's because of how the battery acts. A battery with an overly large capacity is more resistant to overcharging than one with a small capacity and so you'll often see batteries with a 12Ah capacity put on relatively small motorcycles, even those without electric starters. With a flooded cell lead acid battery, there will reach a point when the battery is fully charged and the voltage differential between your circuit and the battery exceeds about 1.5V. For instance, lets say your system voltage is around 15V and your battery is floating a full charge at 13.5V. When the voltage differential between the plates in your battery exceeds that threshold, the additional power your system is generating goes into electrolysis. You start splitting the water molecules in your electrolyte into Hydrogen and Oxygen atoms and additional power also gets lost as heat. You had to periodically top them up with water. So the batteries were specifically chosen to compliment the charging system so that they could help ease the burden on the voltage regulators, which still had a ways to go before they could handle the task on their own. A modern voltage regulator will almost always permit the use of a smaller battery on an older bike because the battery is no longer serving in the capacity of an auxiliary voltage regulator. The inverse is also true: Modern batteries don't play nicely with older voltage regulators.

For Zeke, I would pick whatever interests him the most. I work for the College of Engineering at Oregon State and both are growing programs with good placement after graduation. It's purely anecdotal, but it seems the EECS (Electronic Engineering and Computer Science) students seem to do better in our neck of the woods because of the presence of Intel and Microsoft. I'm not sure there's a"bad" engineering choice at the moment. Our economy seems to be in a transition away from manufacturing, but we still do a heck of a lot of design. All the stuff you see manufactured overseas is still being designed and engineered over here.
 
Sonreir said:
Yup. When I was first learning electrical stuff, comparing it to plumbing was exactly what I did. The alternator is your water pump. The resistance is the diameter of your pipes (low resistance is big pipes). The amount of water flowing through the system is current. The system pressure is voltage. Your battery is the water tower that the pump feeds. Your pump is only good for so much water output and this can be used as pressure, flow, or a certain related combination. If the pipes are too big (resistance is low and too much wattage draw on the system), then your pressure and your flow will be augmented by the water tower until it runs dry, then you're on whatever the pump can supply directly. But on the other hand, if your pump is too powerful for the system you might break some pipes or even your water tower.

For your testing, I would skip the VAC side of things. AC checks are only useful to see if your alternator is functioning and, on their own, they are not useful for determining alternator output. The reason for this is because the internal resistance of your multimeter is very high and you probably don't know the exact value. You might be able to use another multimeter to figure it out, but then we start getting into Inception-like scenarios. Measuring current directly from your alternator may cook it, so I don't recommend that choice either. And, ultimately, it's the DC side of the equation that's running the bike and charging the battery, so we should be measuring DC. :p

For the battery, overcharging was not the same consideration in the 70s as it is now. The battery was actually specifically chosen because of the beneficial effect it has on helping to regulate power within the overall system. You may notice that some of the early twins and singles don't even use a voltage regulator and that's because of how the battery acts. A battery with an overly large capacity is more resistant to overcharging than one with a small capacity and so you'll often see batteries with a 12Ah capacity put on relatively small motorcycles, even those without electric starters. With a flooded cell lead acid battery, there will reach a point when the battery is fully charged and the voltage differential between your circuit and the battery exceeds about 1.5V. For instance, lets say your system voltage is around 15V and your battery is floating a full charge at 13.5V. When the voltage differential between the plates in your battery exceeds that threshold, the additional power your system is generating goes into electrolysis. You start splitting the water molecules in your electrolyte into Hydrogen and Oxygen atoms and additional power also gets lost as heat. You had to periodically top them up with water. So the batteries were specifically chosen to compliment the charging system so that they could help ease the burden on the voltage regulators, which still had a ways to go before they could handle the task on their own. A modern voltage regulator will almost always permit the use of a smaller battery on an older bike because the battery is no longer serving in the capacity of an auxiliary voltage regulator. The inverse is also true: Modern batteries don't play nicely with older voltage regulators.

For Zeke, I would pick whatever interests him the most. I work for the College of Engineering at Oregon State and both are growing programs with good placement after graduation. It's purely anecdotal, but it seems the EECS (Electronic Engineering and Computer Science) students seem to do better in our neck of the woods because of the presence of Intel and Microsoft. I'm not sure there's a"bad" engineering choice at the moment. Our economy seems to be in a transition away from manufacturing, but we still do a heck of a lot of design. All the stuff you see manufactured overseas is still being designed and engineered over here.
Printing this...


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Is there any difference between a 6v reg rectifier and a 12v reg rectifier like for a 70 watt ducati 6v system vs 100 watt Honda 12v system? Is there an operating range like 6-12 volts just like a test light? Both are changing ac to dc. It would seem to me that they would be compatible?


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