Electrical 101 - Read This First
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And this is why I am a part of this forum. Subscribed!
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frankxedge
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This rules!
mydlyfkryzis
当有疑问时踢你的敌人在生殖器上，你可以道歉后
I'd like to further add to this...You may wonder why I rated the 14.5 volt condition lower in current on the alternator :
But then raised the current on the coil amp when voltage went up.
There is something many do not realize about these permanent magnet (PM) type alternators. They do NOT generate voltage at all. They generate CURRENT (AMPS). At a given RPM, the PM alternator generators a specific, fixed amount of current.
Let's figure out the capacity of the alternator from the manual.
The last of the CB350 Manuals list the loads:
For Nightime, is has 1 35W lamp, 3 3W lamps, 2 ignition coils and the battery. For this instance, let's consider the battery fully charged.
At 5000 RPM, the Alternator is rated 14.5 Volts, with 1.5 to 2.5 amps left over for charging.
Lets convert the coils to amps at 14.5 V : V=IR, so I=V/R... 14.5/5 = 2.9 Amps, there are 2 coils, so 5.8 amps for coils. The coils are only activated for about 25% of the time (when the points are closed for that coil) so the ignition requires 1.45 amps. Let's say 1.5 to keep it simple.
Lights are 35 + 9 = 44 watts. Watts = Volts *Amps so Amps = watts/volts = 44/14.5 = 3 Amps....
Finally, the book says 1.5 to2.5 available, assuming the alternator system is "optimum", lets assume 2.5 A.
So, we get 1.5 + 3 + 2.5 = 7 amps. Lets double check our work: 7 A x 14.5 V = 101 watts. The Alternator is rated 110 watts, and I bet my coils use that wee bit more. So say 2 Amps for the coils and the math checks out. That makes the system 7.5 Amps at 5000 RPM.
that's the "rating".
Let's go back to the rating at 14.5 volts, 7.5 amps. We are missing Resistance... R = V/I so 14.5/7.5 = 1.933 Ohms. The Bikes electrical system, with all the lights in parallel, the battery, and coil resistance in the alternator, total 1.933 ohms.
So back to the original idea, the Alternator produces CURRENT. the volts in the system are dependent on the resistance in the system. Add 2 more lights and the reisistance drops. Lower resistance means lower voltage. the amps may still be 7.5, but you notice the voltage is now 14.0 volts. do the math, the resistance of the system is now 1.866 Ohms. This is how the voltage regulator works. when the regulator senses the voltage above 15 volts, it starts to decrease the resistance in it's circuit to ground, lowering the overall resistance, causing the voltage to drop. the alternator is still putting out the SAME current, but now the system is lower resistance, so the voltage is too.
If you grounded the output of the stator at 5000 RPM, on this particular bike, through an ammeter, the voltage would drop to zero (shorted voltage is zero) but the current would still be 7.5 amps.
The capacity (or ampacity) of the alternator goes up as the RPM climbs. However, since the resistance stays the same until the voltag regulator kicks in, the voltage rises with the current,and then the regulator starts reducing the resistance to maintian the maximum allowed voltage in the system
this may clear up how the system works for some, confuse it for others. It is unusual to think of the system as current based, but it is. The rest of the system defines the voltage.
Remember though, there are losses. The magnets get weaker over time, the internal resistance of the alternator coils can change, other items can cause the current you want to go somewhere besides the lights, coils and battery.
This outline for a permanent magnet alternator does not apply directly to field excited alternators. While they output current alos, their output is not tied to the RPM like a PM alternator is. That's for a different post and a little more complicated.
But then raised the current on the coil amp when voltage went up.
There is something many do not realize about these permanent magnet (PM) type alternators. They do NOT generate voltage at all. They generate CURRENT (AMPS). At a given RPM, the PM alternator generators a specific, fixed amount of current.
Let's figure out the capacity of the alternator from the manual.
The last of the CB350 Manuals list the loads:
For Nightime, is has 1 35W lamp, 3 3W lamps, 2 ignition coils and the battery. For this instance, let's consider the battery fully charged.
At 5000 RPM, the Alternator is rated 14.5 Volts, with 1.5 to 2.5 amps left over for charging.
Lets convert the coils to amps at 14.5 V : V=IR, so I=V/R... 14.5/5 = 2.9 Amps, there are 2 coils, so 5.8 amps for coils. The coils are only activated for about 25% of the time (when the points are closed for that coil) so the ignition requires 1.45 amps. Let's say 1.5 to keep it simple.
Lights are 35 + 9 = 44 watts. Watts = Volts *Amps so Amps = watts/volts = 44/14.5 = 3 Amps....
Finally, the book says 1.5 to2.5 available, assuming the alternator system is "optimum", lets assume 2.5 A.
So, we get 1.5 + 3 + 2.5 = 7 amps. Lets double check our work: 7 A x 14.5 V = 101 watts. The Alternator is rated 110 watts, and I bet my coils use that wee bit more. So say 2 Amps for the coils and the math checks out. That makes the system 7.5 Amps at 5000 RPM.
that's the "rating".
Let's go back to the rating at 14.5 volts, 7.5 amps. We are missing Resistance... R = V/I so 14.5/7.5 = 1.933 Ohms. The Bikes electrical system, with all the lights in parallel, the battery, and coil resistance in the alternator, total 1.933 ohms.
So back to the original idea, the Alternator produces CURRENT. the volts in the system are dependent on the resistance in the system. Add 2 more lights and the reisistance drops. Lower resistance means lower voltage. the amps may still be 7.5, but you notice the voltage is now 14.0 volts. do the math, the resistance of the system is now 1.866 Ohms. This is how the voltage regulator works. when the regulator senses the voltage above 15 volts, it starts to decrease the resistance in it's circuit to ground, lowering the overall resistance, causing the voltage to drop. the alternator is still putting out the SAME current, but now the system is lower resistance, so the voltage is too.
If you grounded the output of the stator at 5000 RPM, on this particular bike, through an ammeter, the voltage would drop to zero (shorted voltage is zero) but the current would still be 7.5 amps.
The capacity (or ampacity) of the alternator goes up as the RPM climbs. However, since the resistance stays the same until the voltag regulator kicks in, the voltage rises with the current,and then the regulator starts reducing the resistance to maintian the maximum allowed voltage in the system
this may clear up how the system works for some, confuse it for others. It is unusual to think of the system as current based, but it is. The rest of the system defines the voltage.
Remember though, there are losses. The magnets get weaker over time, the internal resistance of the alternator coils can change, other items can cause the current you want to go somewhere besides the lights, coils and battery.
This outline for a permanent magnet alternator does not apply directly to field excited alternators. While they output current alos, their output is not tied to the RPM like a PM alternator is. That's for a different post and a little more complicated.
A lot of good info there, but it might help to clarify a few aspects.
Alternators and generators produce wattage, not just current or voltage. As an alternator's speed increases, its wattage increases as well. This manifests as both an increase in voltage and in amperage. Amperage and voltage are relational outputs from the alternator. Change one and the other will follow suit. This relationship is governed by Faraday's Law (amount of voltage generated by magnetic fields) and the characteristics of the alternator (specifically: number of windings, rotational speed of the rotor, strength of the magnets used, and the resistance of the wire in the windings).
This relationship can be seen in effect when you run the math through on a theoretical resistor and generator. Lets assume as have a 35W light bulb within a 12V circuit. We already know that voltage times amperage is wattage, so we know that the 35W bulb is drawing 2.9 amps. Following through with Ohm's law means the bulb has a resistance of 4.1 ohms. So everything is working fine at 35W, but what happens when we're only generating 30W? From experience, most of us know that the bulb will dim, but the relationship governing the dimming of the bulb could probably use more explanation.
We already know that wattage is voltage times amperage, but there's more to it than that. Wattage is also voltage squared, divided by resistance which means we can also specify wattage as amperage squared, times resistance.
Resistance of our light bulb doesn't change, so we can figure out the voltage in the circuit when 30 watts are supplied instead of 35. 30W = V²/4.1. V = 11.09. The rest of Ohm's Law gives us 2.7 amps as the current draw. So when we have a 4.1 Ohm light bulb that's meant to operate at 35W, but is only supplied 30W, we have a voltage drop of nearly a volt and the current is reduced by .2A as well.
Alternators produce wattage and it's the resistance of your system that determines the current drawn from the alternator and the voltage at which it is delivered (to an extent).
That's only half of the story though. What happens when you're generating too much wattage for your system?
That's why you have a voltage regulator. If you have an electromagnetic alternator, the regulator will change the strength of the magnetic field inside your alternator by regulating the voltage going to the electromagnets. A weaker electromagnetic field means less wattage is being generated for a given RPM. For a permanent magnet alternators, there is no way to regulate the strength of the magnets and changing the RPMs wouldn't be productive, so instead the system needs to be designed to use up more wattage. Your voltage regulator will dynamically change it's own resistance to attempt to keep the voltage of the system at a set level. The regulated voltage amount is a static value (let's say 14V) and so it's the Wattage being generated that determines the resistance. Since W = V²/Ω, then Ω = V²/W, then we can say the amount of Wattage being used by the regulator is 196/Ω or the resistance of the regulator at this moment in time is 196/W.
Quick recap: Alternators produce wattage. Wattage is both voltage and current. As wattage increases, there will be an increase in both voltage and current because voltage is a measurement of how effective electricity is at overcoming resistance in a circuit. If you regulate voltage, then you're also preventing amperage from rising because of the relationship defined in Ohm's Law.
Finally, just to clarify the "adding a headlight drops resistance" statement made by Rich. That's true, but only if the headlight is added in a parallel circuit. If you add it in series, the resistance will go up and the wattage consumed by your circuit will drop (as will the intensity of the lights).
Alternators and generators produce wattage, not just current or voltage. As an alternator's speed increases, its wattage increases as well. This manifests as both an increase in voltage and in amperage. Amperage and voltage are relational outputs from the alternator. Change one and the other will follow suit. This relationship is governed by Faraday's Law (amount of voltage generated by magnetic fields) and the characteristics of the alternator (specifically: number of windings, rotational speed of the rotor, strength of the magnets used, and the resistance of the wire in the windings).
This relationship can be seen in effect when you run the math through on a theoretical resistor and generator. Lets assume as have a 35W light bulb within a 12V circuit. We already know that voltage times amperage is wattage, so we know that the 35W bulb is drawing 2.9 amps. Following through with Ohm's law means the bulb has a resistance of 4.1 ohms. So everything is working fine at 35W, but what happens when we're only generating 30W? From experience, most of us know that the bulb will dim, but the relationship governing the dimming of the bulb could probably use more explanation.
We already know that wattage is voltage times amperage, but there's more to it than that. Wattage is also voltage squared, divided by resistance which means we can also specify wattage as amperage squared, times resistance.
Resistance of our light bulb doesn't change, so we can figure out the voltage in the circuit when 30 watts are supplied instead of 35. 30W = V²/4.1. V = 11.09. The rest of Ohm's Law gives us 2.7 amps as the current draw. So when we have a 4.1 Ohm light bulb that's meant to operate at 35W, but is only supplied 30W, we have a voltage drop of nearly a volt and the current is reduced by .2A as well.
Alternators produce wattage and it's the resistance of your system that determines the current drawn from the alternator and the voltage at which it is delivered (to an extent).
That's only half of the story though. What happens when you're generating too much wattage for your system?
That's why you have a voltage regulator. If you have an electromagnetic alternator, the regulator will change the strength of the magnetic field inside your alternator by regulating the voltage going to the electromagnets. A weaker electromagnetic field means less wattage is being generated for a given RPM. For a permanent magnet alternators, there is no way to regulate the strength of the magnets and changing the RPMs wouldn't be productive, so instead the system needs to be designed to use up more wattage. Your voltage regulator will dynamically change it's own resistance to attempt to keep the voltage of the system at a set level. The regulated voltage amount is a static value (let's say 14V) and so it's the Wattage being generated that determines the resistance. Since W = V²/Ω, then Ω = V²/W, then we can say the amount of Wattage being used by the regulator is 196/Ω or the resistance of the regulator at this moment in time is 196/W.
Quick recap: Alternators produce wattage. Wattage is both voltage and current. As wattage increases, there will be an increase in both voltage and current because voltage is a measurement of how effective electricity is at overcoming resistance in a circuit. If you regulate voltage, then you're also preventing amperage from rising because of the relationship defined in Ohm's Law.
Finally, just to clarify the "adding a headlight drops resistance" statement made by Rich. That's true, but only if the headlight is added in a parallel circuit. If you add it in series, the resistance will go up and the wattage consumed by your circuit will drop (as will the intensity of the lights).
mydlyfkryzis
当有疑问时踢你的敌人在生殖器上，你可以道歉后
I might disagree slightly on the wattage. You can produce current from the alternator/generator at Zero Volts. In addition, the voltage is the variable for a given RPM, not the current. At a fixed RPM, the Current is fixed. Depending on the resistance, the voltage will vary. The wattage is not produced, but is the measurement of what is being used...
I do work with a few current based sources. When the current source is 2 amps, it's 2 amps. Depending on the resistance, the voltage will vary, but not the amps.
4-20 milliamp circuits work on this principle. The voltage is often zero, yet the current remains fixed. From a pure theoretical sense, there is a resistance in the circuit, but practically, the voltage is zero. Which implies no wattage.... However, if I put a resistor in the circuit, the volts will climb, but the current remains fixed. No additional wattage (power) being used. Wattage calculations kind of fall apart on these type systems. Power is being used, but conventional wattage calculations don't do it well. The engine itself, if you will, also is a Wattage generator. 746 watts = 1 HP. A CB30 is 34HP Nominal, so you could also say it is 25 Kilowatts. 3.412 BTU = 1 watt. So in a case where voltage goes to zero, other measurements of Power work too...if you measure the heat generated in a circuit, you can also calculate watts despite zero voltage...
Another thing, Unlike a battery, which will supply a varying amount of amps, as well as volts, amperage sources are fixed capacity.
Of course, the system has varying resistance, the RPM (amp output) varies with RPM, and the regulator is voltage only regulation, not current.....so it is a complex system for calculation.
We need to find a nice pub and a couple of pints and have a good discussion....lol
I spell everything out for the general public....not implying you don't know this Matt...
I do work with a few current based sources. When the current source is 2 amps, it's 2 amps. Depending on the resistance, the voltage will vary, but not the amps.
4-20 milliamp circuits work on this principle. The voltage is often zero, yet the current remains fixed. From a pure theoretical sense, there is a resistance in the circuit, but practically, the voltage is zero. Which implies no wattage.... However, if I put a resistor in the circuit, the volts will climb, but the current remains fixed. No additional wattage (power) being used. Wattage calculations kind of fall apart on these type systems. Power is being used, but conventional wattage calculations don't do it well. The engine itself, if you will, also is a Wattage generator. 746 watts = 1 HP. A CB30 is 34HP Nominal, so you could also say it is 25 Kilowatts. 3.412 BTU = 1 watt. So in a case where voltage goes to zero, other measurements of Power work too...if you measure the heat generated in a circuit, you can also calculate watts despite zero voltage...
Another thing, Unlike a battery, which will supply a varying amount of amps, as well as volts, amperage sources are fixed capacity.
Of course, the system has varying resistance, the RPM (amp output) varies with RPM, and the regulator is voltage only regulation, not current.....so it is a complex system for calculation.
We need to find a nice pub and a couple of pints and have a good discussion....lol
I spell everything out for the general public....not implying you don't know this Matt...
I could use a little more background on the details behind current generated at zero volts... Unless we're talking about voltage differential and not actual volts?
I understand the mechanics of generating voltage without current, but I didn't know the opposite was possible.
I understand the mechanics of generating voltage without current, but I didn't know the opposite was possible.
mydlyfkryzis
当有疑问时踢你的敌人在生殖器上，你可以道歉后
It's a shame you live so far away....We would have a great time filling in the blanks....I am partial to Guinness Stout.
Oddly, with current sources, as resistance goes to zero, voltage goes to zero. As resistance goes to infinity, voltage goes to infinity. Except in real life....
We use current transformers in a lot of equipment. You may be familiar or not. Basically the working part of a clamp on ammeter. They supply 0-5 amps in application, with the ratio of windings set up for the current level they are measuring. So a a 600 amp circuit would have a 600:5 ratio transformer.
One problem with this regards open circuit voltage. since insulation doesn't handle "infinity" volts, we never reach it. But an open circuit can have some very high 5 amp, HIGH voltage on it that likes to jump around. The cure for this is to make sure any unused outputs are shorted with heavy wire. If you put another amprobe on, or a meter, you can see that the current at 600 amps load is 5 amps, even though the voltage across an inch of 10 gauge wiring is still zero and since one side is always grounded, zero potential to ground. Yet the current is still going round and round.
With sensitive enough equipment, you could measure the resistance in a 1 inch long 10 gauge wire, but it would be very very low.
Even so, the watts would still be infinitesimal using the volt x amp formula.
I find electricity very interesting as the rules that govern how we deal with it under "normal" circumstances go away when either the numbers are really low or very high. 500,000 Volts does not follow a lot of the rules we know like 5 volts will. As we all learn on our ignition systems, when the voltages start getting the letter "K" in the description (as in 15K volts) things that insulate (like Air) suddenly are conductive....
Oddly, with current sources, as resistance goes to zero, voltage goes to zero. As resistance goes to infinity, voltage goes to infinity. Except in real life....
We use current transformers in a lot of equipment. You may be familiar or not. Basically the working part of a clamp on ammeter. They supply 0-5 amps in application, with the ratio of windings set up for the current level they are measuring. So a a 600 amp circuit would have a 600:5 ratio transformer.
One problem with this regards open circuit voltage. since insulation doesn't handle "infinity" volts, we never reach it. But an open circuit can have some very high 5 amp, HIGH voltage on it that likes to jump around. The cure for this is to make sure any unused outputs are shorted with heavy wire. If you put another amprobe on, or a meter, you can see that the current at 600 amps load is 5 amps, even though the voltage across an inch of 10 gauge wiring is still zero and since one side is always grounded, zero potential to ground. Yet the current is still going round and round.
With sensitive enough equipment, you could measure the resistance in a 1 inch long 10 gauge wire, but it would be very very low.
Even so, the watts would still be infinitesimal using the volt x amp formula.
I find electricity very interesting as the rules that govern how we deal with it under "normal" circumstances go away when either the numbers are really low or very high. 500,000 Volts does not follow a lot of the rules we know like 5 volts will. As we all learn on our ignition systems, when the voltages start getting the letter "K" in the description (as in 15K volts) things that insulate (like Air) suddenly are conductive....
frankxedge
New Member
Ok one question. If I were to remove a wiring run, such as in the DIY FAQ, would I leave the ground alone that isn't outlined in red?
frankxedge
New Member
Ok cool thanks sonreir!
Is there a sticky on how to start trouble shooting? My 1972 CB 350 k4 has 0 electrical power in the middle of my ride. Nothing will turn on. Battery holds at 12.8 right after it happened, Main fuse in front of battery is intact, i need to check grounds. But does anybody have a protocol on what to check for?
- Battery
- Fuse
- Grounds clean?
-?
THanks!
- Battery
- Fuse
- Grounds clean?
-?
THanks!
Time to get out your multimeter and start measuring.
With the ignition switch on and the kill switch off, check for power at the following points:
1.) Battery positive terminal
2.) Right after the main fuse
3.) Right before the ignition switch (red wire)
4.) Right after the ignition switch (black wire)
If there really is zero power to the bike, then you should see zero power at one (or many/all) of those points.
With the ignition switch on and the kill switch off, check for power at the following points:
1.) Battery positive terminal
2.) Right after the main fuse
3.) Right before the ignition switch (red wire)
4.) Right after the ignition switch (black wire)
If there really is zero power to the bike, then you should see zero power at one (or many/all) of those points.
I did what you had told, and saw no power after the fuse. So i replaced the fuse, and the bike started up :-X I apparently put in a 20 amp fuse instead of a 15 amp prior. The fuse looked untouched which was why I thought it was still working, but after the power reading, i replaced it anyway, and bike started up. Many hours of tinkering around and learning a alot, only to miss an obvious one!
Thanks much
Thanks much
Mnguyendc said:I did what you had told, and saw no power after the fuse. So i replaced the fuse, and the bike started up :-X I apparently put in a 20 amp fuse instead of a 15 amp prior. The fuse looked untouched which was why I thought it was still working, but after the power reading, i replaced it anyway, and bike started up. Many hours of tinkering around and learning a alot, only to miss an obvious one!
Thanks much
I had the same issue the first week after I got my first bike.
I'm not a fan of the old glass fuses. I replace with modern ATC fuses every chance I get, now.
