By Stu Oltman - Technical Editor, Wing World Magazine
Reprinted with permission

How Long To Charge A Battery

In describing what happens as a battery charges, we found that charging is a process of reversing the chemical reaction that produced the electricity we used. Charging takes the sulfate out of the battery plates and returns it to the electrolyte, and we said that as the charging process is almost complete, hydrogen and oxygen bubbles would be seen rising from the plates. The electrical energy of the charger is converted to chemical energy stored in the battery, but the process isn’t 100 percent efficient. Some of the charger’s energy is wasted as heat and some is spent breaking down the electrolyte into its gaseous components. Therefore, if we drain 10 amp hours from a battery, we must replace those 10 amp hours plus an additional amount (usually around 10 percent). So, in this example, 11 amp hours would be required to replenish a battery that had 10 amp hours of its capacity drained.

How do you know exactly how much your battery is depleted? You don’t. Therefore, you need some other way to determine when it’s fully charged. If you’re using one of the available automatic battery maintainers, you simply plug it in, and let it do its thing. When the battery has reached 14.5 volts, the charger considers the battery fully charged, and it drops to the float mode (discussed last month). If using a trickle charger or taper charger, you have two test methods available to you: voltage or specific gravity.

Remember we said that a battery’s state of charge couldn’t be accurately determined by measuring voltage while the battery is charging. The reasons were examined last month. However, we can examine the change in voltage while charging. If the voltage fails to increase in any half-hour period during charging, the battery can be considered to be as charged as it’s going to get. Disconnect the charger, let the battery rest for at least one hour, then measure the voltage. A fully charged lead/acid battery of the flooded variety (free acid) should read 12.6 volts or slightly higher. A sealed battery (absorbed acid) should read at least 12.8 volts. Another method, and a more accurate indicator of state of charge, is measurement of the specific gravity of the electrolyte. Wait – don’t turn to that article written by the talking trunk mascot! I’ll explain!

Quick – which is heavier, oil or water? Wrong! Water is heavier than oil. That’s why oil floats on water. Liquids have different weights, and this fact makes it convenient to use weight as a measure of how much acid is in solution in the battery electrolyte. By convention, we assign the number 1.000 to the specific gravity of distilled water. As we dissolve “stuff” in the water, the solution becomes heavier. Battery electrolyte is a solution of water and sulfuric acid with a specific gravity of 1.265 for flooded batteries and around 1.310 for sealed (absorbed glass mat) batteries. This means, in the case of flooded batteries, that the electrolyte is 1.265 times heavier than distilled water. The specific gravity can be easily measured with the use of a battery hydrometer, a glass tube with a calibrated float inside. The higher the specific gravity of the solution in the tube, the higher the float will ride in the solution. This is the same principle that causes you to be more buoyant in salt water than in fresh.

Since we know that the electrolyte started life with a specific gravity of 1.265, our goal in charging is to return the electrolyte to that condition. Remember that during battery discharge, some of the sulfate from the electrolyte bonded to the battery plates. This means that it’s no longer in solution, so the specific gravity of the electrolyte must be less, and it is. If we can return that specific gravity to 1.265, we can be assured that we’ve removed all, or almost all, sulfate from the plates. As with the voltage test, and for the same reason, we can’t use specific gravity readings taken during charge to indicate the battery’s state of charge. However, as with voltage, the rate of change is useful. If the specific gravity fails to increase at all in any one-hour period, the battery is as charged as it’s going to get.

Sometimes you’ll encounter a battery with low specific gravity, 1.240 for example, and no amount of charging will cause it to rise much. This is a case, as I described last month, where the sulfate crystals have hardened to the point that charging won’t remove them. This battery is now useful for many things, but starting an engine isn’t one of them. Can’t I just pour out the old acid and pour in fresh? That would restore the specific gravity, wouldn’t it? Certainly, it would restore the specific gravity, but what of the sulfate that now clogs the plates? Since there’s little plate surface area left to conduct the chemical reaction, the specific gravity of the electrolyte now becomes immaterial. Don’t ever try this stunt, and don’t bother with “magic pills” that supposedly restore dead batteries. With your new knowledge, gained from reading these articles, your next battery will probably provide better service than your old one.

Enough About Batteries Already!

Well…just one more thing. Why is it that if you measure your battery voltage and find it to be 12.6 volts, it drops to something like 12.4 (or less) when you turn the ignition switch on? I asked this question of several people and got answers like “because the key is on,” or “because you’re draining it,” or “because the lights are using some volts.” Nice try; no cigar. Yes, the lights are using some volts. In fact, they should be using all of the volts in their circuit, but they can only use what they can get. Right now, they’re only getting 12.4 volts. Where did the other .2 volts go? Let’s answer this question as an introduction to voltage drop testing.

Take a look at the circular diagrams. To find any one thing in either diagram, simply cover that thing with your thumb, and do the indicated math on the other two things. For instance: if we want to find volts, we see that we should multiply amps by ohms (resistance) or divide watts by amps. To find amps, we divide volts by ohms or watts by volts. By substitution, we can come up with several more ways to arrive at our answers, but we don’t need to get that involved right now. Go get a calculator, take a break, and come back ready to learn something that’ll make you a big hit at cocktail parties!

The Missing Volts

You must understand a couple of things before proceeding:

  1. No current flows in the circuit unless the switch is closed (dashed arrow).
  2. When current flows, the exact same amount of current flows through ALL points in the circuit.
  3. All of the voltage rise (battery voltage) will be consumed by the voltage drop or drops in the circuit.
  4. No voltage drop will occur across any load unless current is flowing in the circuit.
  5. Internal resistance of a fully charged battery in good condition is about .010 ohms.
  6. NEVER, EVER, attempt to measure the resistance across the battery terminals. You’ll fry your meter, and there could be more serious consequences.

With these things in mind, assume that the battery voltage is 12.6 volts with the switch open. The load will draw 18 amps (normal for a GL1500) when the switch is closed (ignition switched on). The voltage drop due to the battery’s internal resistance is found by multiplying amps by ohms as the circular diagram indicates. Doing this, we find that 18 x .010 = .18 volts. Therefore, the voltage at the load will be 12.6 minus .18 = 12.42 volts. The instant the switch is turned on, the voltage would indeed be 12.42, but as the battery begins to discharge, the internal resistance increases as we discussed last month, and the voltage drop increases as a result. If the load is great, such as an electric starter, the acid depletion at the plates (check Part IV of this series) accelerates the reduction in voltage, but ignore these factors for the purposes of these examples. Consider only the drop due to resistance.

Now let’s see what happens when we use the electric starter. The starter draws about 100 amps after it gets up to speed, but the amps required to start it turning against a stationary engine will peak at around 200 for a split second. Using our little circles again, we see that .010 (battery resistance) multiplied by 100 (amps flowing) yields a voltage drop across the battery of 1 volt. This means that only 11.6 volts are available to the starter after it stabilizes at speed. To get the starter spinning, the drop would be twice as much (200 x .010), and the voltage at the starter would be only 10.6 volts. This is with a fully charged battery.

Now assume that you’ve had the bike stored in the garage for one week without being on a charger. As discussed previously, the self-discharge of the battery would have removed about 7 percent of the battery’s capacity. The draw from the keep-alive memories would remove another 18 percent. Therefore, the battery would be around 25 percent discharged with an open-circuit voltage of around 12.5 volts. The battery’s internal resistance will also have increased, perhaps to around .020 ohms. Turn on the key, and you no longer see 12.42 volts. We now have a voltage drop across the battery of 18 amps multiplied by .020 ohms = .36 volts. This leaves 12.14 for the lights rather than 12.42 as before. Operate the starter, and things really go downhill! We see that with 200 amps needed momentarily, the voltage drop at the battery would be 200 x .020 = 4 volts. This leaves 8.5 volts at the starter. If it can manage to grunt and groan and get the engine spinning, the amps flow may be reduced to 100 amps. In this case, the battery voltage would rise back up to around 10.5 volts. While this may indeed spin the motor, it may not be sufficient voltage to operate the ignition. However, I believe you can now see that removing your finger from the start button will reduce the amount of current flowing through the battery to around 18 amps again. As a result, the voltage will immediately jump back up to a level more to the ignition’s liking. If the engine is still spinning, it’ll likely fire up.

Troubleshooting A Slow Starter

As a last exercise in this installment, let’s take a look at what happens when we have a loose or dirty connection at either end of the wire running from the starter solenoid to the starter, and why voltage drop testing rather than resistance testing is necessary to find the problem.

A weak battery as previously demonstrated could cause a slow starter, or one that won’t turn the engine at all. The problem could also be the result of mechanical difficulties inside the starter or the engine itself. Quite often, however, it will be seen that the starter is not getting the voltage or current it needs to do its job even though the battery, starter, and engine are all in good condition.

We said that a fully charged battery has an internal resistance of around .010 ohms. A starter has a resistance of around .05 ohms. So, the total resistance in the starting circuit would be about .06 ohms ignoring any other factors. To find the current that will flow in the starting circuit, we divide the battery voltage by the total circuit resistance (as shown by our circular diagrams). This gives us 12.6 volts divided by .06 ohms (.010 plus .05) = 210 amps. The battery voltage drop would be 210 amps multiplied by .010 ohms = 2.1 volts, leaving us with a battery voltage of 10.5 volts. But what happens if a loose or dirty connection at either end of the starter cable causes a small resistance ahead of the starter? Consider Load 1 in the diagram to the immediate left to represent the dirty connection and Load 2 to represent the electric starter. Let’s assign a value of .05 ohms to the resistance caused by the bad connection.

You’d need an extremely accurate ohmmeter to find a resistance as small as .05 ohms. It’s likely that your equipment isn’t nearly sensitive enough. Nevertheless, let’s see what this seemingly harmless resistance does in our starter circuit. Again, referring to the circular diagrams, we figure the amps that will flow when the starter button is pressed by dividing the total voltage by the total resistance. Doing this, we see that 12.6 volts divided by .11 ohms (.05+. 010 +. 05) = 114 amps. The voltage that will be dropped by the electric starter will be 114 amps multiplied by .05 ohms (resistance of the starter) = 5.7 volts! What you’ll likely hear is “click” and nothing else! Where is the rest of the voltage getting used up? The battery is dropping 114 amps times .010 ohms = 1.14 volts, and the bad connection is dropping 114 amps times .05 ohms = 5.7 volts. The total of these numbers only equals 12.54 due to rounding off the previous answers, but you surely get the idea, and notice that the battery voltage drop is only 1.14 volts. This means that the battery will be reading almost 11.5 volts during this failed starting attempt leaving the uninitiated scratching their heads. So if you can’t measure the small resistance of this bad connection, how do you find the problem? You connect one lead of your voltmeter to the solenoid terminal at which the starter cable is attached, and connect the other lead to the terminal on the starter. Remembering that a voltage drop will occur across any resistance in a circuit when current flows, press the starter button while watching your voltmeter. You should see a reading no more than a few tenths of a volt. Any more than that, and you need to do some cleanup work on those cables to eliminate as much resistance as possible.

Next time, we’ll explore some other useful circuit problem solutions using voltage drop testing. Put some new batteries in that calculator, and don’t miss the next installment.

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