Battery Isolator with Battery Drain

[DeepFrz]

Both batteries will see the same voltage when charging and be isolated when not charging. The diodes are a one way current switch but the voltage will be the same on the output of each one during the charging stage.

[OceanPlanet]

Only if the current is high enough to raise the voltage of the low battery up to the same voltage as the other one.

[DeepFrz]

Only if the current is high enough to raise the voltage of the low battery up to the same voltage as the other one.

If the alternator isn't strong enough to supply a voltage high enough to charge the battery then there are other problems to contend with. I challenge you to measure the voltage on the two legs of the diode isolator and let us know what the difference is. Measure before starting the engine and afterward.

The lowest charged battery will accept the most current during the charge, the highest charged will accept less, depending on its state of charge. But it will "see" the same voltage as the lower charged battery.

[nigelmercier]

Originally Posted by OceanPlanet

Wrong. The batteries are not "shorted" together when charging. As you say in your first sentence, the diodes are one-way valves.

Sorry, but I am not wrong. When diodes conduct (which they will as soon as there is more than 0.6V above the battery voltage), they becomes low resistance, creating an effective short at the point at which they join the charging cable.

[daddle]

Originally Posted by nigelmercier

Sorry, but I am not wrong. When diodes conduct (which they will as soon as there is more than 0.6V above the battery voltage), they becomes low resistance, creating an effective short at the point at which they join the charging cable.

You are mistaken. You've some unconventional ideas about electronics. Like the FM antenna thread earlier. Diode type isolators are problematic. They do not "create an effective short." The regulator will sense only the battery it is connected to. This may cause it's output to be low current which will put only a low voltage across the other weaker battery. Not what anybody would want. The smarter combiners or duo systems are far better. Or perhaps two alternators.

[nigelmercier]

Originally Posted by daddle

You are mistaken. You've some unconventional ideas about electronics...

Interesting, I'm a Technology Lecturer at a UK university.
When you connect a charging voltage to two batteries via two diodes, the diodes will both conduct, irrespective of the state of charge. The physical junction (as opposed to the PN junction) of the diodes is at the charging voltage, so the conducting diodes will cause some current to flow from the higher charged battery backwards through its associated diode in order to keep both batteries at the same potential. Perhaps it is this that you don't understand: diodes can pass current in both directions if they are being held in a conducting state.

I could produce a diagram and graphs, but I can't be bothered. I refer you to #16 above.

[daddle]

Originally Posted by nigelmercier

Interesting, I'm a Lecturer

Indeed.

[donradcliffe]

Nice try Nigel, but you'll have to do better than that if you want to lecture at my university. Each diode/battery connection is a separate circuit in the isolator, and there is no way you are going to get any significant current through a reverse biased diode (you are a long way from breakdown voltage), and there is no way you are going to discharge a battery through it.

[bill good]

Twin engines & twin systems with only one isolator? maybe you should have 2 isolators? I hope you kept the one you thought was faulty as you now can put it to good use. What make & model isolator do you have.

Regards Bill

[cal40john]

Originally Posted by nigelmercier

Interesting, I'm a Lecturer in Technology at a UK university.

I think we have everyone touching different parts of the elephant blindfolded.

Case 1 alternator can output voltage set by regulator, as in batteries are accepting less current than the max output of the alternator.

Let's say we measure the voltage at the alternator and it's 14.8 V. There will be a 0.4 V drop across each diode if they're schottkey, so each battery will see, or to put another way will be at a voltage of 14.4 V. If one battery is discharged more than the other, the more discharged battery will have a greater charge acceptance and will draw proportionally more of the current. From the graph in Calder's book, if the battery in each bank is a 100 amp-hr battery, and battery one is at 1/2 charge, at a battery terminal voltage of 14.4 V the battery will accept 30 amps. Battery two is at 3/4 charge, at a battery terminal voltage of 14.4 V its acceptance is 15 amps. This is 45 amps total, less than the max output of the alternator, so the regulator is throttling the output to that 45 amps to maintain this voltage setpoint.

In this case the voltage at each of the batteries is the same, it doesn't matter on which battery the sense lead is in the first order approximation. In reality the V-I curve for the diode is not vertical when forward biased and the battery that draws more current will see a lower voltage by some millivolts. So connecting the sense lead to the house bank could give a bigger output of the alternator. I never measured it, so I don't know if it is significant or matters at a practical level.

Even though you can think of this as an effective short in one way, it is only true as long as both diodes are forward biased. There is no way current can flow from one battery to the other. For the current to leave one battery it has to reverse bias its diode at which point the circuit is open.

Case 2, one battery is so deeply discharged that the alternator is putting out its maximum current. In this case the deeply discharged battery determines the voltages in the system. Let's put the sense lead on this battery, and this large bank is accepting the max current output of the alternator. At this point the battery terminal voltage is set by the physical characteristics of the battery. Let's say it is 12.0 volts with 50 amps going into it. The voltage on the alternator side of the diode is 12.4 volts. Unless the terminal voltage of the other battery is less than 12 volts, its diode is not forward biased and no current flows into this battery. And as it's reverse biased, no current leaves the battery. Unless you have a huge bank and a small alternator this case won't last long. The terminal voltage of the deeply discharged battery is going to rise, which the voltage on the alternator side of the diode will follow at 0.4 volts higher. Eventually it reaches a voltage where the diode is forward biased on the second battery. This will raise the second battery's terminal voltage by a small amount at first, and so its charge acceptance rate will be very small and a small proportion of the current flows into that battery.

In this case of the deeply discharged battery with max alternator output what happens when the sense lead is connected to the relatively charged battery? Let's say this battery is 12.4 volts. The regulator wants to get it to 14.2 volts, it turns on the alternator. The dead battery accepts all of the current, and so the voltage at the alternator is set by the dead battery since its diode is forward biased, the diode to the relatively charged battery is reversed biased and sees no current and no voltage rise, so the regulator will keep the alternator output at maximum.

John

[DeepFrz]

Thanks for the refresher in diode theory John. In my previous posts I wasn't thinking of one battery bank being so low that the diode on the other bank would be reverse biased. Although that would be for only a short time its true that the alternator would, for that short time, only be charging the lowest battery bank.

[OceanPlanet]

Originally Posted by cal40john

I think we have everyone touching different parts of the elephant blindfolded.

Case 1 alternator can output voltage set by regulator, as in batteries are accepting less current than the max output of the alternator.

Let's say we measure the voltage at the alternator and it's 14.8 V. There will be a 0.4 V drop across each diode if they're schottkey, so each battery will see, or to put another way will be at a voltage of 14.4 V. If one battery is discharged more than the other, the more discharged battery will have a greater charge acceptance and will draw proportionally more of the current. From the graph in Calder's book, if the battery in each bank is a 100 amp-hr battery, and battery one is at 1/2 charge, at a battery terminal voltage of 14.4 V the battery will accept 30 amps. Battery two is at 3/4 charge, at a battery terminal voltage of 14.4 V its acceptance is 15 amps. This is 45 amps total, less than the max output of the alternator, so the regulator is throttling the output to that 45 amps to maintain this voltage setpoint.

In this case the voltage at each of the batteries is the same, it doesn't matter on which battery the sense lead is in the first order approximation. In reality the V-I curve for the diode is not vertical when forward biased and the battery that draws more current will see a lower voltage by some millivolts. So connecting the sense lead to the house bank could give a bigger output of the alternator. I never measured it, so I don't know if it is significant or matters at a practical level.

Even though you can think of this as an effective short in one way, it is only true as long as both diodes are forward biased. There is no way current can flow from one battery to the other. For the current to leave one battery it has to reverse bias its diode at which point the circuit is open.

Case 2, one battery is so deeply discharged that the alternator is putting out its maximum current. In this case the deeply discharged battery determines the voltages in the system. Let's put the sense lead on this battery, and this large bank is accepting the max current output of the alternator. At this point the battery terminal voltage is set by the physical characteristics of the battery. Let's say it is 12.0 volts with 50 amps going into it. The voltage on the alternator side of the diode is 12.4 volts. Unless the terminal voltage of the other battery is less than 12 volts, its diode is not forward biased and no current flows into this battery. And as it's reverse biased, no current leaves the battery. Unless you have a huge bank and a small alternator this case won't last long. The terminal voltage of the deeply discharged battery is going to rise, which the voltage on the alternator side of the diode will follow at 0.4 volts higher. Eventually it reaches a voltage where the diode is forward biased on the second battery. This will raise the second battery's terminal voltage by a small amount at first, and so its charge acceptance rate will be very small and a small proportion of the current flows into that battery.

In this case of the deeply discharged battery with max alternator output what happens when the sense lead is connected to the relatively charged battery? Let's say this battery is 12.4 volts. The regulator wants to get it to 14.2 volts, it turns on the alternator. The dead battery accepts all of the current, and so the voltage at the alternator is set by the dead battery since its diode is forward biased, the diode to the relatively charged battery is reversed biased and sees no current and no voltage rise, so the regulator will keep the alternator output at maximum.

John

Bingo. We have a winner.

[daddle]

Originally Posted by OceanPlanet

Bingo. We have a winner.

Not so fast. That description is proper, as far as it goes. However I think charging performance may suffer if the regulator sense is connected to the well charged battery and the other battery is discharged. Most "smart" regulators I've met will sense the charged state of the battery and switch to low current float mode (moderate float sense voltage). Some do this by reducing the current for a moment. The discharged battery may charge very slowly, which may not be as intended. A simple charger may give better performance as it may just try to maintain some maximum sense voltage.

[OceanPlanet]

Originally Posted by daddle

Not so fast. That description is proper, as far as it goes. However I think charging performance may suffer if the regulator sense is connected to the well charged battery and the other battery is discharged. Most "smart" regulators I've met will sense the charged state of the battery and switch to low current float mode (moderate float sense voltage). Some do this by reducing the current for a moment. The discharged battery may charge very slowly, which may not be as intended. A simple charger may give better performance as it may just try to maintain some maximum sense voltage.

Connect the alt reg sense to the house bank, as it will (nearly always) require more charging than the start. It's not a complicated question.

Or (as I like to do), do away with the battery isolator, charge the house bank directly from the alt (or alts), and use a voltage-sensing charge relay to charge the start batt from the house bank. I like the Sterling Power VSR or CVSR; I'm sure there are other good ones too.

[sailorboy1]

I don't understand the sense in having more than 1 house bank connected to different terminals of the battery isolator. Unless there is a way to isolate each bank, and where doing so connecting the alternator sensor to the bank still on service for the regulator.

My boat has 2 house banks and a start battery. Each bank can supply the main DC panel alone or by together. Also each house bank can be selected to the inverter.

All my charging, either engine or with the battery charger goes though the inverter switch to the house banks. So whichever bank is selected to the inverter (1,2,or both) gets charged. Normally I have both banks on service to both the main DC panel and to the inverter (which is normally off unless I need AC for something).

So on my isolator (just got a new one that doesn't use diodes with the big voltage drop) I have the engine alternator output and the house battery charge line (to the inverter switch) connected together on the isolator input. And 1 of the 2 outlet terminals connected to the start battery. So the alternator always senses the house bank, either 1,2 or both and the regulator charges based on the house bank on line.