Basic electrical

[Normk]

Here's an important concept, going back to the child standing on the desk, again. How high is the child? 6' or 3' tall. The desk is 3' so the child's head is 6' above the floor. Is the child 3' tall or 6'?

Obviously there is a context issue here and important one! If your neighbor threatens to send his 6' 6" grandson over to deal with you, it might be important, before sassing him back, to know if the grandson was measured while standing on a 3' desk....

When we discuss a "12 volt" battery, what does that tell us about the battery> How large it is? How much electricity it can deliver?

A "12 volt" lead acid battery isn't 12 volt unless it is nearly dead. Fully charged it has 12.8 to 13.2 volts difference when measured between the positive and negative posts. Old timers will recall the batteries of the 1940's and up into the 1960's as having exposed connectors and cell caps which were removed in order to check the eletrolyte level. One could tell a 12 volt battery because it has 6 caps while a 6 volt has three. Spot a helicopter battery right away because it has 12 caps and the stand-by generator starting batteries used by some telephone companies have 4 caps. Wierd, eh?

These lead-acid batteries are, in fact, batteries. A battery is a group of things which work together as in " a battery of guns" or a "battery of storage cells". A 12 volt battery has six storage cells, each of which has about 2.1 volts across the cell. These cells are connected in series, one after the other to produce the 6 volt, 8 volt, 12 volt or 24 volt potential across the total battery. Modern batteries have the cell connectors hidden inside the battery case in order to reduce parasitic drain losses, etc. but the old ones had exposed connectors between the cells. In fact there are manufacturers who make "look alike" batteries intended to resemble the old ones for use in classic and antique cars.

In this type of battery, one could use a voltmeter to test the potential of each cell in order to compare cell condition. In the modern ones, we cannot do this easily although testers were made a couple of decades ago which had electodes intended to reach down into the battery electrolye to contact the two plate groups within the cell to allow voltage to be read.

So what makes a 12 volt battery as compared to a 6 volt? If you own a motor home or travel trailer, you will have a "house battery" which provides power for lights, water pump and so on during times in which the rig is not plugged into shore power (from the boat term) or doesn't have a power generator in service. It is common practice to use two, 6 volt batteries connected into series in order to create 12 volts. The negative of one battery is the negative of the set, while the positive of that battery is connected to the negative of the other in the pair. The positive of the other battery becomes the positive of the battery set.

6 volts + 6 volts = 12 volts as this is a series set. I will post a diagram later but just to conclude as I need to get back to installing an electric fuel pump and pressure regulator onto a CBR400 Honda.

If the two batteries were connected positive to positive, and negative to negative, which we call a "parallel" connection, the voltage would be 6 volts. What's the advantage, disadvantage of these two configutations?

More on that later. You rig may have two 12 volt batteries in the house battery box, so how are they connected?

Of to the Honda,

Norm

[Normk]

Check out the thread in which someone is intending to use 24 volts for starting. This is an old strategy used on heavy equipment and other large engines in order to improve cranking speed for starting purposes. Like older 24 volt starting systems, it is more desireable to run the electrical system on 12 volts because lighting and other components are more readily avialable. He is exploring means of operating two batteries in parallel for charging and running purposes but connecting into series to operate the starting motor. The remainder of the electrical system remains operating at 12 volts off one battery during cranking. Have a look at some of the proposed wiring diagrams as this is simply an exercise in logic which seems well in hand so doesn't need more comments from such as I.

Everyone who wishes to work with electrical systems is well advised to understand Ohm's Law which relates the relationship between voltage, current and resistance. In mathematical formula volts = current multiplied by resistance or E = IxR. We usually teach the formula as a triangle having E (voltage) at the top with I (current or amps) and R (resistance) below.

_E_
I | R

If one needs to calculate one quantity, simply cover the unknown and the formula remains. Example: If one knows voltage (E) and Current/Amps (I) then cover R (resistance/ohms) which leaves the formula R = E/I

Without making calculations, one can see the relationship clearly by considering the formula. Increasing E while R remains the same must cause I to increase. Increasing I while E remains the same must require a reduction in R.

This should be obvious as pushing harder against the same resistance or friction will result in more progress, more current flow. Pushing with the same effort against less resistance will have the same effect.

Another principle to recognize in electrical work is that R (Ohms or Resistance) and E (Volts or electromotive force) are causal factors while I (current or amps.) are not. By this I mean that one can increase or decrease voltage by increasing alternator speed, or adding batterys in series which will result in more current flow in proportion to the increase in voltage. One can also reduce or increase resistance (Ohms) by adding or subtracting to the circuit which will change current inversely.

Current flow is proportional to voltage. Increase voltage and current (amps) goes up. Double voltage and current doubles, ditto decreasing voltage. Current flow is inversely proportional to resistance which means that more resistance means less current. Double R = 1/2 I.

This is all fine and good so long as we remain within the bounds of Ohm's Law but many circuits such as those containing a DC motor (recall, DC = Direct Current?) do not obey the Ohm's Law equation in an obvious manner but more on that later.

So Volts are how much "push" is acting on the electrons in a circuit. Amps are the measure of how many amps are flowing (by memory one ampere is the flow of 6.23 billion billion electrons per second past a given point) but it could be a few less as my eyesight isn't as good these days.

The point here is that the amps (short for Amperes) are the measure of the physical stuff which is acted apon and which then acts apon the other aspects of the circuit.

Let's say that we have a one volt battery connected to a one ohm resistance....OK, so the current flow will be E over I R, One volt divided by one Ohm = 1 Amp of current. The same would be true if we had 100 volts pushing current through a 100 Ohm resistor = 1 Amp. So these two circuits are the same, right?

Logically, we can see that it must take more effort to push that one amp of electrons through a resistance which is 100 times as great and this is reflected in the requirement for 100 volts to do so.

If we applied 100 volts to the 1 Ohm resistor, clearly we would have a current flow of 100 amps. much more work being done.

The Watt is the unit of electrical work and is used to express such measures as the amount of heat or light delivered by a bulb. A 24 Watt 12 Volt bulb requires 2 amps. of current flow in order to light to rated output.

Watts, then equal Amps. times Volts.

1 Amp. being pushed through a 1 Ohm resistance requires 1 Volt and 1 Watt of work is done by this effort.

A 60 Watt headlight bulb (assuming it were rated at 12 volts) requires 60 Watt/12 Volts = 5 Amps. But, a 60 Watt trouble light bulb in a shop light requires only 0.5 Amps so how can that be? Note that the trouble light bulb plugs into a wall socket which applies 120 Volts. 60 Watts divided by 120 Volts = 0.5 Amps.

Now. let's spin those around and look at some other effects: How much resistance does that 60 Watt headlight bulb have?

Plug the two known quantities in to the Ohm's Law equation.

Do the same for the 60 Watt trouble light bulb.

Now, here's where things can be interesting! What will be the Wattage of the headlight bulb if we connect it to the wall socket?

How about the trouble light bulb connected to 12 volts?

I'll leave you there with your figuring.

HIH

Norm

Let's consider a circuit containing a battery, and one light bulb:

[Steve]

And do any of the exerienced electricals here have any reservations or wish to add clarifications?

Just one little detail, if you don't mind.

If you see 12.8 then you are seeing a fully charged 12 volt lead acid battery. But when you remove the leads, maybe the meter says 134, so what's going on? Have a close look at the scale and you will likely see a symbol such as mV, or milli, something like that.

What is happening is that the meter is auto ranging. It has automatically switched to a lower voltage scale in order to display the voltage which it sees in whole numbers. It is sensing, maybe, 134 millionths of a volt from magnetic effects out of the air. That's the 134!

Reading 134mV is not all that uncommon, but that is 134 thousandths of a volt, not millionths. Millionths would have the prefix "micro-".
Still a rather small number, but off by a factor of 1000.

How about if I ignore the advise regarding not having the Nissan's engine running? The Nissan's voltage regulator has a 14.8 volt set point and a 120 amp. alternator. With the Nissan engine running, the bike will crank even faster because the Nissan alternator is more than capable of providing enough current at 14.8 volts to crank the bike engine. That's good, right? Faster cranking and higher ignition voltage means faster starting?

Question: because our bikes volts output is similar but the vrr amp. ratings are different this is the reason why we would over load our system and fry our electrical system? Our bikes vrr can vary but for simplicity my Honda vrr is around 40 amps versus a cars amps of 120 therefore, poof???

Can you explain a bikes amp output and how do we know what and how many accessories we can add to our electrical system without discharging our battery? Like if I want to plug in a radio, GPS, phone charger, and/or outlets for my heated vests and pants.

Thanks for your tutorials.

These questions might be better answered by using one of my favorite electrical analogies, water.

Electrical volts can be expressed as water pressure (psi).
Electrical amps can be expressed as gallons (liters) per second (or minute).

Your R/R is designed to maintain a certain pressure (14.x volts) up to a maximum flow of 15 amps (gallons per minute). The stator is a pump that is driven by the engine. The faster you turn the pump, the more current you generate. Since you are tryng to push a lot of current through a small passage (a wire), there is some resistance that shows up as increased pressure (volts). The regulator will allow whatever comes through, until it reaches 14.x volts, then holds that pressure by opening a bypass. (It actually diverts the flow back and forth very quickly, but the bypass analogy works, too.) The whole assembly will only handle a flow of 15 gallons per minute, anything more than that will damage it.

Simply connecting jumper cables to another battery just lets current come from another source at the same pressure. However, if the donor's engine is running, its "pressure" is also higher. If that "pressure" is higher than your regulator, your regulator will try to control it. Since the donor's "pump" can supply more current than your regulator can handle, bad things happen.


Now, how much can you add? That depends on how fast your engine is turning. Let's continue with the water analogy. Your pump (engine/stator assembly) will provide more current as it is turned faster. Anything that requires water will reduce that pressure just a bit, because it has someplace else to go. If the pump is pumping fast enough to handle the extra flow, it will get to the regulator's set pressure. If you add aditional loads (lights, GPS, etc.) you will require a bit more current. If the pump is still able to maintain enough current, it will still hit the regulated pressure, and the excess will be bled off.

Fortunately, electrical "pressure" is very easily measured with a voltmeter. If you so some simple tests, you will find what your bike is capable of "pumping", and use that for a reference. Anything you add that draws more current will reduce what the regulator has to bypass. As long as it is bypassing, it is wasting current. If you could somehow adjust your load to perfectly balance the output of the stator (pump), you would still have adequate pressure (voltage). The problem there is that, as you slow down, you will pump less, and will have to reduce your loads to maintain balance. If you can remember to turn off lights, heated jacket, heated hand grips, etc, every time you slow down to a stop, great. For the rest of us, we can just watch the voltmeter to see if we are maintaing our electrical "pressure" somewhere near the regulator's set point.

I second the request about how to know what we can add without killing our charging system. Or, how do we know how much more space we have or if we've gone over already?

Again, a voltmeter is the answer. As the engine speed increases, it will push the voltage up to the regulator's set point, then drop one or two tenths of a volt. On my bikes, that happens about 2500 rpm. The bike might idle at 13 volts, rise to 14.4, then, at 2500 rpm, drop to about 14.2 and stay there. As I rev the engine faster, I am generating more current, but wasting it in the regulation process. If I turn on some driving lights, I might not reach the regulator set point until about 3500 rpm, meaing the voltage will be lower until then, too. Anything above 3500 will be wasted. Now, let's run at 4000 rpm (wasting some power by regulating), then turn on the heated jacket. Since the load is a bit too much, the stator will not be able to keep up and the voltage will drop a bit. Simply knowing the set point for your bike and your present voltage reading will tell you your status.

.

[Normk]

Steve posted: "Reading 134mV is not all that uncommon, but that is 134 thousandths of a volt, not millionths. Millionths would have the prefix "micro-".
Still a rather small number, but off by a factor of 1000."

I was exaggerating the reading to illustrate that the 134 could be almost anything and to far out of whack as to be meaningless. It's quite strange, though how often a displayed number, due maybe to the other meter lead coming disconnected, can be in the expected range. If one isn't looking closely, one can be lead off into tangents.

My point was that people, even experienced ones, sometimes don't see the mV, etc. Even worse if one does not understand the significance of the label.


Another perspective is something I was hoping to see as a different analogy with different description may clarify points which were not clear from my attempt.

One point regarding how much one can add to the system:

Whether the alternator runs full output all the time with the excess shunted to ground as you state, depends on whether the regulator is truly a load type such as a Zenor. Try measuring the current in the VRR's ground wire. In some systems, the output current is near to maximum at all times. These are systems which are loading in order to control maximum voltage.

In these systems the VRR ground circuit flows nearly maximum output at all times with some current delivered to the vehicle loads and the remainder is load to ground for voltage control by shunt loading.

In other VRR systems such as my 1979 GS850G and many others I have checked, the VRR ground circuit current matches the load current to the electical system. This proves that this VRR is not shunt loading because the voltage remains at set point without additional current flow in the VRR ground circuit.

The debate rolls on as to whether loading the alternator to 100% shortens stator life or is a causal factor to stator burn-out but it seems clear that higher stator currents are present only at higher electrical loads in systems which use non-loading regulation.

Up early for son's soccer game so off to bed.

HIH

Norm

[Steve]

The debate rolls on as to whether loading the alternator to 100% shortens stator life or is a causal factor to stator burn-out but it seems clear that higher stator currents are present only at higher electrical loads in systems which use non-loading regulation.

I was only aware of "shunting" regulation, not "loading" regulation happening in our GS bikes.

With a "shunting" setup, the stator is putting out 100% all the time anyway, so adding additional load only diverts the current from the regulator, without changing the load on the stator.

.

[koolaid_kid]

I would also like to see a discussion on the following:
What is the source of electrical power while the engine is running? Is it the battery or the charging system? (Assuming a healthy, fully functioning charging system.) What is the function of each component (battery, charging system) during this period?
Thanks.
Please note, there is confusion all across the web to this issue, it is not limited to GSR.

[850 Combat]

i would like to know if a higher amperage rating on a regulator/rectifier set up is automatically better. I was thinking about upgrading to high amperage R/R's to help keep from burning them out. With a fuse or a breaker, for instance, this is not that great of an idea. my feeling is that on an R/R, it is a good idea, but I would like to confirm it.

[Normk]

Very late and need to sleep so a quick attempt at the last two posts.

A fuse or breaker is intended to open, or stop the current flow in a circuit if the current flow reaches the set point of the fuse or breaker. This would typically be because of a decrease in circuit resistance rather than an increase in voltage so same Voltage/smaller Restance = more Amps. Higher current (more amps) times same Voltage = more Watts, so higher power factor than that for which the circuit was intended and smoke.

Installing a larger fuse would allow a larger current flow before the fuse controlled the circuit so smoke for sure if something went wrong.

When considering a VRR of larger capacity (designed to service a larger capacity alternator output) the rectifier bridge will be capable of carrying a larger current so will be less challenged by the smaller load to which it is actually attached.

In addition, regardless of the regulator type (assuming a permanent magnet alternator) the regulator's capacity to shunt current to ground will be greater than for the smaller alternator so, again a greater margin. If one of the SCR or other circuits, the regulator will be at least as capable as needed.

An additional consideration which seems to be largely ignored is that of the forward bias voltage drop in semi-conductors. Diodes, for example, which are used for rectification in most alternators as electrical one-way valves allow current flow in only one polarity as everyone knows. What few seem to appreciate is that common silicone diodes have a no-load forward bias of about 0.7 volts which rises to over 1 volt at full load.

Yep, put a diode in series and you must expect a voltage drop (loss) of a volt or so. We had not end of problems with travel trailers, motor homes, service trucks and boats in which electronic battery isolators were installed. Seems like a great idea as there are no moving parts but knock one volt off your 14.6 volt charging voltage and see what happens to the time required to re-charge a battery! You can time the charging with a calendar!

Put 20 amps through a pair of diodes (one postive and one negative) and that requires a voltage drop of 2 volts. No big deal, you say as the regulator simply dials up the alternator voltage to compensate. Yes, that's correct but you also have at least 40 Watts of heat from the rectifier alone in that VRR box. The voltage regulator will contribute more.

The VRR intended for the larger alternator will be designed to dissapate the heat from a higher rectification current so will be happier when dealing with a smaller output.

Hope that explains.....

Norm

[Normk]

I installed Ispell during writing of a reply. It cleared my post so will have to start over in the morning.

Norm

[Normk]

It seems that a post which required more than 2 hours to craft has been lost because the forums were down for cleaning. At least, I don't see it now and that was the message I received when the post was submitted.

Let's see if this one posts...

Norm

[Normk]

What can I say?

Certainly not going to go to that trouble again.....

Norm

[Normk]

Fortunately, I had copied the post which is unusual so was able to recover. Hope it is the copy I intended.

This would not be the ideal time for someone to instruct me as to how I might better arrange my postings.....


Here it is despite that I'm in a very foul mood:

Koolaide referered to a subject often wondered about but seldom discussed. It speaks more to basic understanding of electrical system dynamics than having direct application.

In short, when the engine is turning at suffient speed for the alternator voltage to rise above that of battery voltage, the alternator will provide the current. In other words, the electrical flow will be from the alternator rather than "out of the battery".

Various concepts include a picture of current flowing from the alternator, into the battery and then back to the electrical system and so on. These are easily clarified by placing an ammeter into the circuit between the VRR output wire and its connection to the electrical system. Place another ammeter between battery and its "output" circuit.

Run the bike as per normal with lights, etc. and monitor the current flows and, most importantly, the current direction. I am going to over simplify here and clarify in more detail later so please pardon the license.

When the switch is turned on and before the engine is started, the lights, ignition and other normal loads are connected, through the ignition switch (and/or relays controlled by the ignition switch) to the fuse box from where power is distributed to the various circuits. In this case the ammeter connected in the battery circuit can be seen to show current flowing from the battery. The current flow for my GS850G is about 8 amps. although a stock one would be higher. The current flow indicated by the ammeter in the alternator output (the circuit leading from the VRR to the wiring harness) is zero. If it is not virtually zero, it would indicate a problem with the VRR since the VRR would be allowing a drain to ground but that is aside.

Hit the starter button and current flow from the battery will rise (assuming we are measuring the battery cable to starter relay circuit as well as the lead to the main fuse) exponentially to (guess for the 850 as haven't measured) 60 - 90 amps. as the starter operates. If we are not including the starter cable circuit, the current will still likely rise a bit to include the starter relay's magnetic winding. Engine is cranking but has not begin to run and alternator ammeter still shows zero.

The engine begins to run and increases in speed from around 150 RPM to 3,000 RPM (Suzuki liked fast idle on the choke). Monitoring the two ammeters indicates a shift in current flows. Now the current flow in the battery circuit ammeter is about the same (6 amps) but the current in the VRR circuit is near 14 amps. How does that make sense???

Oh, just noticed: the ammeter is indicating the opposite polarity! Current is not flowing out of the battery, it is flowing into the battery. The alternator is producing its maximum output of 14 amps., 8 amps of which are used in operating the bike's electrical loads and the remainder flowing back into the battery in order to re-charge the battery. I am just grabbing numbers here of about the correct magnitude to fit conditions.

As the engine begins to run, the battery ammeter reading drops off gradually as the battery is re-charged and its counter-voltage returns to normal. The VRR/alternator ammeter continues to indicate 8 amps. (or there abouts) because this is the current flow required to supply the normal loads which are switched on. We turn the headlight switch off and the VRR ammeter drops from 8 amps. to 4.8 amps. because we have removed the headlight's 3.2 amp. load. Turn on the turn signals and the ammeter will alternate between 4.8 amps. when the signal bulbs are not lit and (depending) 6.5 amps. when the lights are on. Press the brake lever and the ammeter's reading indicates another load.

If we turn everything on we return to an 8 amp. current reading from the VRR ammeter. Let's assume that the battery has had sufficient time to fully re-charge.

We connect an additional headlight bulb by jumpers to the battery, the VRR ammeter will indicate an increased current flow of 4 amps. The battery ammeter still shows near zero. The VRR shows 12 amps.

Clip on another headlight bulb and what happens? The second additional headlight illuminates so we know that it has current flow. As it is identical with the first additional headlight, we expect that it will also increase VRR current reading by 4 amps, correct?

The VRR ammeter reads 14 amps. but that's only 2 amps. more but not the 4 addional amps. expected.....

Recall that my GS850G has the early, small alternator which delivers a maximum of 14 amps. so that all we get. So what happens?

Let's keep the engine running and disconnect the battery.

The additional load will be reflected by a drop in voltage so that the Watts delivered from the alternator. If we clip in another additional headlight, current flow from the voltage will drop further. This may sound to be very complex but in fact is simple.

The alternator delivers a maximum current flow of 14 amps. into the electrical system. This much current flow can maintain a pressure of 14.5 volts but if we attempt to increase current flow beyond that 14 amps. the voltage will fall to the point where the voltage applied is just sufficient to cause a flow of 14 amps. through the total circuit resistance.

Let's back up and look at the condition existing when the engine is running (battery disconnected as it will simply confuse things right now) with only the headlight and other normal loads. We will use the value of 14.5 volts as the VRR set point so we obviously have a specific circuit resistance for all electrical loads which we know how to calculate using Ohm's Law equation.

E=IR or Volts = Amps. times Ohms

Realigning the equation to calculate the value of Ohms = Volts/Amps.

Ohms = 14.5 Volts/8 amps. = 1.8 Ohms (1.8125)

So the electrical system resistance under normal conditions = 1.8 Ohms

When we clip on an additional headlight, the headlight is in parallel so the total resistance of the electrical system falls. We know this has to be true because the total current flow (indicated by the VRR ammeter increased) the only way to increase current is to increase voltage (which didn't happen) or by decreasing resistance. Decreasing resistance is the only possible effect.

So, clip on the additional headlight and VRR ammeter indicates an additional 4 amps. (8 amps. + 4 amps) to 12 amps. but still the same voltage of 14.5 volts.

Ohms = 14.5 volts/12 amps. = 1.2 Ohms (1.208).

So there is a crucial concept for you to recognize. When we turn on additional circuits, we decrease total circuit resistance so total current rises. If you don't understand this, please post and I will try to offer a better explaination.

That all follows so let's add the second additional headlight. The headlights have how much resistance? 14.5 volts/4 amps. = 3.6 Ohms (3.625), so why doesn't the current from the VRR go up the same, 4 amps.?

In fact, the current remains the same but the Voltage falls. Let's pick a number and agree that the Voltage drops from 14.5 to 13 volts and see if the calculations match up. If not it will just show how well we guess.

We determined that this was because the alternator could not provide more than 2 additional amps. The VRR ammeter indicated 14 amps. a rise of 2 amps. so let's look at what happened.

We have a total resistance of 1.2 Ohms and then add the additional headlight's resistance of 3.6 Ohms. We don't add 3.6 Ohms + 1.2 Ohms to equal 4.8 Ohms and can prove this because we know that more resistance equals less current.

In order to calculate the values of resistances in parallel, that is to say resistances which are connected beside one another to the same source, we use the equation:

1/Resistance Total = 1/R1 + 1/R2

1/Resistance Total = 1/1.2 Ohms + 1/3.6 Ohms = 4/3.6 Ohms = 0.9 Ohms.

The first logical test shows that this value is in the correct range as it is less that either resistor value which we know must be true as current goes up in this circuit and not down when the second resistance is connected. A "guesstimate" shows the change in degree of value is in the correct magnitiude, not a big change.

OK, we have 0.9 Ohms with a 12 amp. current flow.

0.9 Ohms x 12 amps. = 10.8 volts.

Add the second additional headlight 3.6 Ohms and what happens to the voltage?

Yes, it drops.

1/Resistance Total = 1/R1 + 1/R2 + 1/R3

1/Resistance Total = 1/1.2 Ohms + 1/3.6 Ohms + 1/3.6 Ohms = 5/3.6 Ohms = 0.72 Ohms.

0.72 Ohms x 12 amps = 8.64 volts.

In fact things won't happen exactly as these values indicate as the resistances will also change due to the fact that they will not heat to as high a temperature. The effect will follow the same progression.

Now, let's reconnect the battery. When the system voltage drops below that of the battery's 12.8 volts, the battery voltage will cause a current flow out of the battery into the electical system which will help to maintain the system voltage higher than it would otherwise as the battery has sufficient capacity to withstand the load. The voltage will gradually drop and this will be reflected by the current indicated by the battery's ammeter.

I'm going to have to end here as life demands attention. Hoping this was useful to someone. Again, apologies for typos and grammatical errors as haven't proof read as due to time.

Norm

[Normk]

I'm thinking that it might be time to close this thread as it is becoming quite extended. What is your pleasure? Continue in this thread or begin another to carry on into another direction?

[cowboyup3371]

I think keeping this thread going would work fine. Maybe talk to Frank about making it a sticky though. Heck, I'm only on the second page here (40-post settings)

[Smellii]

Thread length is irreverent as long as it's productive.