Post by pamcopete on Feb 18, 2008 15:10:05 GMT -5
The forum for this thread has been changed. Rather than a discussion style, this thread will be a "text book" resource. You may post your questions and comments in this thread instead:xs650temp.proboards29.com/index.cgi?board=ElectricOther&action=display&thread=1204727171&page=1#1204727171
My thanks to those who did contribute earlier and I welcome your continued participation.
Many members have from time to time expressed a desire to know more about the basics of electricity to assist them in understanding how the electrical system on the XS650 works and to be better able to find and fix problems.
I invite your participation and feedback, but I would like to keep everything on topic and follow the list, so please don't be offended if I delete your post in the interest of keeping the thread brief and on topic.
This is not intended as a trouble shooting course, as there are probably an infinite number of possible problems that can affect the electrical system, so you should continue to post your requests for help with your particular problem in the usual sections.
Post by pamcopete on Feb 19, 2008 10:51:45 GMT -5
George Simon Ohm was a German physicist, 1787 to 1854, who came up with the law:
" for any electrical circuit, the current is directly proportional to the voltage and is inversely proportional to the resistance"
The three formulas that express this law mathematically are:
I = E/R
E = IR
R = E/I
Where I is current in amps, E is voltage in volts and R is resistance in Ohms.
Resistance is expressed in Ohms where an Ohm is the resistance in a circuit with a 1 Volt source and a current flow of 1 Amp.
Here is a drawing that I will use to illustrate Ohms law.
Voltmeter V1 reads 2 Volts from the one cell battery. The switch is open, so there is no current flow.
Voltmeter V2 reads 0 volts because there is no current flow with the switch open, so using Ohms law:
E (Volts) = 0 Amps X 1 Ohm = 0 Volts
Amp meters A1 and A2 read zero because the switch is open and there is no current flow.
Here is the same circuit with the switch closed:
With the switch closed (on), current will flow from the cell through the internal resistance of the "battery" and through the wiring and resistor R1. So, the total resistance in the circuit is 1.5 Ohms, the source voltage is 2 volts, so the current is:
I = 2 Volts/1.5 Ohms = 1.33 Amps
Voltmeter V1 will now read the cell voltage minus the drop across the internal resistance of the "battery"
Voltage drop across internal resistance:
E = 1.33 Amps X .5 Ohms = .66 Volts
V1 = 2 Volts - .66 Volts = 1.34 Volts
V2 will now read the product of the resistance (1 Ohm) times the current (1.33 Amps)
E (V2) = 1.33 Amps X 1 Ohm = 1.33 Volts
I introduced the concept of internal resistance of the battery because we always encounter this phenomena in the circuits of the XS650, especially when you start up. The .66 voltage drop in this illustration of a one cell "battery" would be multiplied by 6 in the 6 cell battery for a total drop of 3.96 leaving:
12.7 - 3.96 = 8.74 volts for the starter motor and ignition system during cranking.
The other point of this little drawing is to show that voltage readings are location sensitive. Meaning, your voltage reading can be different in different parts of the same circuit, but the current reading is always the same throughout the circuit.
An example of this is when you are checking the circuit for the tail light. Starting at the battery, you may read something like 12.5 volts with the taillight on and everything else off, like in the "park position of the ign switch. As you trace the red wire from the battery to the ignition switch, you may read something like 12 volts on the blue wire coming from the ignition switch. This means that you have "dropped" a half a volt across the contacts of the ign switch. However, if you had put an Amp meter in place of the main fuse, you may have read something like .66 Amps with the tail light on. If you went all the way to the blue wire on the taillight socket, you would still read .66 Amps, but the voltage might be something like 11 volts because of the drops encountered along the way.
If you have any questions, feel free to post them, but please stay with the current topic. Specific troubleshooting questions should be posted in the usual sections.
Post by pamcopete on Feb 20, 2008 8:54:37 GMT -5
The total resistance in the circuit:
12.5/.66 = 18.94 Ohms
In the tail light example, we start out with 12.5 volts at the battery, but end up with 11 volts at the taillight. That's a drop of 1.5 volts with a current of .66 A, so the circuit resistance, including the ign switch, is 1.5 V / .66A = 2.27 Ohms. So, your question is, where is the rest? The rest, 16.67 Ohms, is the resistance of the tail light filament.
Now, we have all measured the resistance on the blue wire to ground, and came up with something around 3 Ohms. The brake light is even lower, like 1 or 2 Ohms, so what gives? Why does Ohms law show a resistance of 16 Ohms for a bulb that we measure at 2 or 3 Ohms?
The answer is heat. The resistance of wire, including the tungsten wire for the filament in a light bulb, increases with an increase in temperature. So, the effective resistance of the tail light bulb is about 16.6 Ohms when it gets hot.
So, the total effective resistance is:
2.34 Ohms Circuit resistance
16.6 Ohms Effective Filament Resistance
18.94 Ohms Total Effective resistance
So, what would the measured resistance be?
2.34 Ohms circuit Resistance
2.0 Ohm Cold filament Resistance.
4.34 Ohms total circuit measured resistance.
The other connection to the tail light is ground (negative battery). So, if you were to measure the blue wire resistance to ground when trouble shooting the taillight, and there was nothing wrong with the tail light, you would measure about 2 Ohms because the blue wire circuit to the taillight comes after the ign switch, so you would not be including the resistance of the ign switch, so you might come up with a resistance of 2 Ohms to ground and conclude that there is a partial short to ground. What to do? whenever you are measuring the blue wire resistance to ground, looking for a short to ground, remove the taillight to eliminate that false reading of a partial (2 Ohms) short to ground.
Post by pamcopete on Feb 20, 2008 9:33:03 GMT -5
In the tail light example above, I used an effective resistance value for the tail light of 16.67 Ohms. You can't measure the resistance when the bulb is on, so how do you come up with the effective resistance of the filament in a light bulb when it's on?
Fortunately, the manufacturer of the motorcycle and the taillight spec the bulb in Watts at a certain operating voltage, so we can use Ohms law for power to determine the effective resistance and current.
Ohms law for power is expressed mathematically as:
P (Watts) = I (Amps) X E (Volts)
So, in our example, we are given the wattage of the tail light (8 Watts, Haynes manual) and the operating voltage, 12.5 V.
So, if P = I X E, then I = P / E and E = P / I
In our example:
I = 8 Watts / 12.5 Volts = .64 Amp
So, the effective resistance would be:
R = 12.5 Volts / .64 Amps = 19.53Ohms
However, in our circuit, we measured the actual voltage at the taillight and came up with 11 Volts instead of the ideal 12.5, so we know the bulb will not be able to produce it's full rated wattage, and that the resistance will be less because the bulb will be cooler, so I derated the bulb to 7.25 Watts ( just a guess).
I = 7.25 Watts / 11 Volts = .66 Amps
R = 11 Volts / .66 Amps = 16.6 Ohms
So, with this example perhaps we can see that even with the simplest circuit on the bike, the taillight, complications can develop. This means that you have to be more methodical when you trouble shoot and stop and think about the meaning of your test results and procedures.
Post by pamcopete on Feb 20, 2008 11:00:12 GMT -5
Here is a partial wiring diagram for a typical (1978/E) XS650 when you first turn on the key but before you have started the engine:
After you start the engine and the battery has recovered, at 3,000 RPM, using Ohms law, what would the following values be?
1. Battery Voltage (from the specs)
2. Current through the main fuse.
3. Voltage drop across the main fuse.
4. Voltage drop across the Ign switch.
5. Voltage on the coil
6. Voltage for the headlight (do not derate, just use the spec to calculate the effective resistance)
A clue here is to calculate the resistance of the main fuse, ign switch and kill switch before you start the engine, because the resistance will be the same after you start the engine. Calculating the total resistance in the circuit is the key to figuring out the current through the main fuse.
A clue here is something that we have not discussed yet, and that is the resistance of resisters in series and parallel. After the engine starts and the safety relay has turned on the headlight, the circuit for the coil and kill switch in series and the headlight are connected in parallel, so calculate the resistance for each and use these formula:
R1 = Coil Resistance + Kill Switch Resistance
R2 = Headlight Resistance X R1/ Headlight Resistance + R1
R2 is the total resistance after the ign switch and includes the kill switch resistance in series with the coil resistance and in parallel with the headlight. The safety relay contacts are considered to be Zero Ohms for this little demo, but in fact, they have some resistance too. The main purpose in this demo is to see the effect the voltage drops have on available voltage and current to the all important coil, with the head light on.
I'm jumping ahead here a little bit, but I will supply the answers in stages to help.
Post by pamcopete on Feb 21, 2008 8:30:33 GMT -5
Well, lets get started on solving our little riddle:
We measured the voltage and resistance at the coil, so we can determine the circuit current with the formula I = E/R.
I Current = 11.25 Volts / 4 Ohms = 2.8 Amps.
This current goes through all of the other components (except the headlight), so we can calculate the resistance of the fuse, ign switch and kill switch using the formula R = E/I for each.
R = .25 V / 2.8 A = .09 Ohms
R = .5 V / 2.8 A = .18 Ohms
R = .5 V / 2.8 A = .18 Ohms
Just before we start the engine, we could go ahead and calculate the effective (hot) resistance of the head light on low beam.
From the Haynes manual we find the Wattage of the headlight on low beam as 40Watts. Assuming the spec is at 12.5 volts, then use the formula I (amps) = P (Watts) / E (Volts)
I = 40 W / 12.5 V = 3.2 Amps.
So, without going any further, we can see that the headlight draws more current than the ignition system. In fact, the headlight is the greatest user of current and power in the whole electrical system, one of the reasons why it should only be turned on when the engine and alternator are running, which is one of the reasons for the Safety Relay.
The effective (hot) resistance of the headlight on low beam would be:
R = E /I
R = 12.5 V / 3.2 A = 3.9 Ohms
Almost ready to start the engine.
Post by pamcopete on Feb 21, 2008 23:07:07 GMT -5
The other components use the same current going through their resistance to create power, expressed in Watts. The battery is "putting out" 12.5 volts at 2.8 Amps. P = I X E, so the battery is "providing" 31.25 Watts of power. The coil is "using" 2.8 Amps at 11 volts, so it is consuming P = 11 V X 2.8 A = 30.8 Watts of power. The remaining power is wasted in the fuse and switch resistance.
Think in terms of a hydraulic pump providing a certain flow of oil through a series of hydraulic motors. The pressure at the outlet of each motor is less than the input, but the flow of oil is constant. Voltage is like pressure, Amps are like the flow of oil. What goes in has to come out.
The difference in pressure between the inlet of the pump and the outlet is converted to power that turns the shaft of the motor, which can be measured in Watts. So, you can experience a drop in pressure from the inlet of a pump to the outlet, just as you can experience a drop in voltage between the in and out contacts of a switch, but the volume of electrons (Amps) is the same, just as the volume of oil is the same from the inlet to the outlet of a hydraulic pump.
Post by pamcopete on Feb 26, 2008 14:16:56 GMT -5
Well, time to start the engine:
After the engine has started and the battery has "recovered" from the effect and the safety relay has operated to turn on the headlight, we would have the following situation:
The battery voltage is now 14.5 because of the energy supplied from the alternator and the headlight is on. Before the engine started, we had measured the coil resistance (4 Ohms) and calculated the "hot" headlight resistance (3.9 Ohms) from the Haynes manual spec of 40 Watts @ 12.5 volts for the low beam. We had also calculated the resistance of the other components:
Main fuse .09 Ohms
Ign Switch .18 Ohms
Kill Switch .18 Ohms
Now we need to calculate the total resistance (load) in the circuit so we can also calculate the total current in the circuit.
The circuit splits into two paths after the Ignition switch. One path feeds the coil and the other path feeds the headlight, so these paths are in parallel. To determine their combined resistance, we use the formula for resistors in parallel:
R = R1 X R2 / (R1 + R2)
Where R1 is the headlight resistance and R2 is the coil resistance and kill switch resistance in series.
R = 3.9 X (.18 + 4) / (3.9 + (.18 + 4)) = 2.02 Ohms
Add up all the resistances in the circuit:
Main fuse .09
Ign Switch .18
Headlight and coil in parallel 2.02
Total circuit resistance 2.29 Ohms
This is the resistance that the battery and alternator "sees" and is referred to as the load.
From that you can calculate the load current:
14.5 (battery voltage) / 2.29 (load) = 6.33Amps
Next we will calculate the voltage that feeds both the headlight and the coil:
Using the fuse and ign switch resistance of .09 + .18 for a total of .27 Ohms, we can calculate a drop from the battery to the out side of the ign switch:
E = 6.33 X .27 = 1.7 Volts
So the voltage to the headlight and coil is 14.5 - 1.7 = 12.8 V
The combined resistance of the kill switch and coil is 4 + .18 = 4.18 Ohms. The current for the coil would be:
12.8 V / 4.18 Ohms = 3.1 Amps
And the voltage across the coil terminals would be:
3.1 X 4 = 12.4 Volts
I was kinda in a hurry when I did this this afternoon, so here's your chance to catch a mistake! Check it out and post the errors. I'll review it later.
Post by pamcopete on Feb 27, 2008 10:38:18 GMT -5
Normally, you would probably use a voltmeter to check the drops across the various components, so this exercise is just to show Ohms law at work and perhaps provide a better insight as to the significance of your readings, especially the drops across the ign switch and kill switch, which in the example are close to real world on many bikes out there. For example: The battery has 14,5 volts available, but only 12.4 volts end up at the coil due to the drops across the fuse, ign sw, and kill switch. The drops will be roughly in proportion when you start the bike which causes the battery voltage to go down as low as 10 volts, and that's with a good battery, so lets do some more fun math and calculate the available voltage and current at the coil during starting when the battery can drop as low as 10 volts. The headlight will be out of the circuit because the safety relay has not activated yet, but it does activated immediately after start when the battery has not fully recovered, and that explains what I call a "partial" start. As soon as the safety relay turns on the headlight, there would be an immediate additional load (less resistance) on the battery. Remember also that when current is first applied to the headlight, the filament is cold, and it's effective resistance would be very low, perhaps only 1 or 2 Ohms, so it can actually drop the available voltage and current to the coil so low for just an instant that the coil cannot produce a spark and the engine quits after one or two pops. This effect is even more pronounced in a TCI system because the electronics in the TCI box require at least 8 volts to function properly. So, the fact that the ALU keeps the headlight on after a "partial start" is actually a good thing because the headlight would be drawing less current due to it's higher effective resistance. Of course, it would be preferable to be able to keep the headlight off during the start sequence, but that's not the case with stock wiring.
So, here's a quick calculation for the available voltage and current to the coil while the starter motor is turning but before the safety relay has turned on the headlight.
Total circuit resistance:
Fuse .09 Ohm
Ign Sw .18 Ohm
Kill Sw .18 Ohm
Coil 4 Ohm
Battery 10 Volts
I = 10 / 4.45 = 2.25 A
Voltage at the coil = 4 X 2.25 = 9 Volts
Well, that's pretty close to the minimum operating voltage for the TCI.
Right after the engine starts and the safety relay has turned on the headlight, for just an instant, the headlight cold resistance is maybe 2 ohms, so here's a quick look at the available voltage and current for the coil.
Effective resistance of the headlight and coil in parallel:
2 X 4.18 / 6.18 = 1.35 Ohms
Total circuit resistance:
.09 + .18 + 1.35 = 1.62
Circuit current = 12 / 1.62 = 7.4 Amps
Available voltage at the output of the Ign switch, assuming that the battery bounces back to 12V when the starter disengages:
12 - (.27 X 7.4) = 10 Volts
Current to coil:
10/ 4.18 = 2.4 Amps
Voltage at the coil = 2.4 X 4 = 9.6
Again, very close to the minimum for the TCI.
A slightly lower battery voltage to start and you can see that the engine either would not start, or you would have a succession of "partial" starts, which we have all experienced.
Another math check would be appreciated!
Post by pamcopete on Feb 29, 2008 17:20:11 GMT -5
Well, that's about it for Ohms Law. When working on your bike's electrics, you probably won't need to really do a lot of math, because you will be using a meter that just gives you the results, but I think it helps to have an understanding of the basics, like Ohms law, to better interpret what the meter reading means.
The next section will cover meters, which are nothing but Ohms law in action, so I will have to reintroduce some of the math just so we can have a better understanding of what the meter reading means.
Next up: Voltmeters, ohmmeters and amp meters, or VOM's.
Post by pamcopete on Mar 5, 2008 10:30:58 GMT -5
The Voltmeter is the simplest of the Volt, Ohm and Miliampere meters that come packaged together in a Multimeter (VOM). It's the meter function that we use most often to measure the battery voltage. Here is a simple little depiction of an analog (mechanical) voltmeter: (Digital meters are next)
The meter itself is basically a tiny DC motor that has a spring on its shaft so it can't rotate more than about 90 deg.. The spring is calibrated to limit the movement of the motor's shaft in proportion to the current that is applied to the "armature". Wait, isn't this post about voltmeters? Why are we talking about current?
From the Ohms law discussion, we saw that nothing happens without current flow.
The "motor" part of the voltmeter has to have current flow in order to move the needle attached to the "armature". A typical meter movement will need 50 microamps to move to the full scale deflection. A microamp is .0000001 amp so 50 microamps would be .00005 Amps. As small as it is, some power has to be developed in order to move the meter pointer and the meter has to have some resistance in order to cause a current flow. A typical meter movement is rated at 0.1 Volt full scale deflection. So, a quick visit back to Ohms law would give us the resistance of the small winding of wire on the armature of the meter:
R (Ohms) = E (Volts) / I (Amps)
R = .1 / .000005 = 2000 Ohms
If we tried to measure the battery voltage by connecting the meter directly across the battery terminals, how much current would flow through those tiny wires on the armature of the meter?
I (Amps) = E(Volts) / R (Ohms)
I = 12 / 2000 = .006 A
That's about 120 times the current required to move the needle to it's full scale deflection, so something would have to give, and that would be either the mechanical destruction of the meter movement and / or the heat created in the wire of the "armature" would burn up the tiny wires.
So, the answer is to limit the current by inserting more resistance in the circuit. I = E / R. As we increase the resistance, the current goes down.
In the example meter, for a 10 volt range, we would want a total resistance in the circuit of:
R = 10 / .00005 = 200,000 Ohms
R1 = 200,000 - 2,000 = 198,000 Ohms
The meter itself has a resistance of 2000 Ohms, so if R1 was 198,000 Ohms we would have a 10 volt full scale deflection meter.
Doing the same thing with R2 to make a 100 volt meter:
R = 100/.00005 = 2,000,000 Ohms (2 Meg Ohms)
R2 = 2,000,000 - 2,000 = 1,998,000 Ohms
Now, install a switch to select either R1 or R2 and you have a little 2 range voltmeter.
You always want to use a meter with a high overall resistance so that the meter does not affect the voltage you are trying to measure. This is usually expressed as "Ohms per Volt" sensitivity. In our example meter, the total resistance in the 10 volt range is:
200,000 Ohms / 10 Volts = 20,000 Ohms per volt.
You may see this on the face of the meter or in the literature that came with your meter.
Up next: How to make an Ohmeter from a voltmeter.
Post by pamcopete on Apr 5, 2008 17:28:53 GMT -5
The Ohm meter:
In the last post, I said that we would make an Ohmmeter from a voltmeter, and as you probably guessed, all we have to do is add a battery to the basic voltmeter and it will then measure Ohms:
By inserting a 9 volt battery in our voltmeter circuit, along with a variable resistor, we can cause the meter to go to full scale if we adjust the resistor to produce .00005 Amps, or 50 micro amps of current when the test probes are connected together. When the probes are not touching, there is no meter movement because there is no current, so the resistance is infinite. A lot of members make the mistake of saying that they do not measure any Ohms, when they really mean that the resistance is infinite, an open circuit, in other words.
For each different value of resistance, the current through the meter will be somewhere between 0 and infinity. The meter is just an amp meter calibrated in Ohms.
Post by pamcopete on Sept 21, 2008 19:30:41 GMT -5
Well, finally I'm back to complete just the basics, finishing with a brief description of Digital Voltmeters.
A digital voltmeter displays the measured voltage in a numeric display (digital). It does this by converting the analog voltage from the probes to a digital set of values that are fed into a display, such as a LED or Liquid Crystal display.
All values in nature are analog, meaning that they have an infinite number of possible values. Temperature as read on an ordinary thermometer can be any of an infinity of values from the lowest to the highest values depicted on the thermometer. For all practical purposes, we do not need that level of accuracy nor can we read the difference between,say 70 F and 70.0000002 F. The same with battery voltage or any other voltage we might be interested in on the XS650.
So, the digital voltmeter samples the measured voltage several times a second. The sample is fed into a chip called an Analog to Digital converter, which has a discrete set of transistors that turn on or off depending on the value of the sample voltage. The on or off condition of the set of transistors is in turn fed to a display driver that then turns on the appropriate segments of the individual display digits. The number of on or off transistors available determines the resolution of the final display.
Because a digital voltmeter samples the voltages several times a second, it only displays the voltage that was true at the exact moment that it was sampled. This is evident when you read the battery voltage with a digital meter. You may see the tenths of a volt display change constantly because you are seeing the voltage at the exact moment of the sample. The battery voltage in turn increases when the regulator turns on and decreases when the regulator turns off.
The Ohmmeter and amp scales on a digital VOM work the same as the analog meter except that the resultant voltage levels are converted to the digital display.