Dual Battery setups

[nsxxtreme]

I'm in a mood, so I figured I would jump in and stir things up.....

If voltage drops, often current increases.

The rate at which a battery can charge is dependent on voltage. The battery has a certain resistance at a specific charge level. Usually resistance will go up as the battery becomes more charged. Nothing is rarely ever strictly DC. A staright piece of wire will have some impedence to it. So really you need that AC and DC component to get the ESR (Equivalent Series resistance). In order to get the ESR of a battery you need to use an impedence meter.

assume a batteries ESR is .02 ohms at 14.7 volts you have 735 Amps available. Dropping the voltage to 14V the current goes to 700 amps. The ESR of the battery doesn't change because you dropped the voltage.

Now to take this a step further what does that mean? It means you battery will now also deliver less current because it can only charge to 14V instead of 14.7 volts. Let's give an example so its clear.
14/.02=700 amps vs 14.7/.02=735 amps.

As long as W=V x I and W is constant. This is why you can burn a motor up with low voltage.

In order for W to be constant you are saying R changes in relation to the change in voltage. Usually that occurs with some kind of regulated power supply. W is rarely constant in a car,especially in old cars like ours. That's why head lights dim when voltage drops.

Current flows, voltage drops. Always remember that. Also R equals resistance in the ohms law you are referring to. The current in the whole circuit would be slightly higher without the diode, but not enough to effect anything. Voltage will drop by .7 volts across the diode, but current will never drop across any component. Thats basic DC circuits. Im an electrical engineering student, had those classes last spring.

mmmm.....study a little more in class I guess?

assume you have a head light and you apply 12V to it. How much voltage did you drop accross the bulb?
(V2-V1)/R=I assume .5 ohms for the bulb resistance. (12-0)/.5= 24 amps
Now assume you put a diode in series with the bulb. How much voltage did you drop accross the load? (11.3-0)/.5 =22.6

It matters where you take your measurement.

Another way to look at it is V=IR is a linear equation. Both sides MUST be equal. A change in one side of the equation requires an equal change on the other side. The only way for current to go up is if R goes down.
The purpose of my post wasn't to argue if the loss was significant or not just to confirm that a drop in voltage equals a drop in current.

Some may wonder where that extra power goes. A battery isolator is made out of a LARGE heat sink there is a reason for that. The loss in power is because of the voltage drop across the diode which is turned into heat and dumped into the heat sink. Which P=IV where P = watts..7V * (however much current you pull) assume 100 amps. (.7v*100amps)=70 watts wasted energy dumped into the battery isolater heat sink as heat. That is just for one battery. Usually no one runs a single battery on an isolater.
Add a second battery and you have a second voltage drop and more wated energy wasted into the heat sink.

 

[bellbrandon83]

mmmm.....study a little more in class I guess?

assume you have a head light and you apply 12V to it. How much voltage did you drop accross the bulb?
(V2-V1)/R=I assume .5 ohms for the bulb resistance. (12-0)/.5= 24 amps
Now assume you put a diode in series with the bulb. How much voltage did you drop accross the load? (11.3-0)/.5 =22.6

It matters where you take your measurement.

I dont understand what you are saying here. No matter how you word it, or what components you add, a 12 volt circuit will always drop 12 volts across its components. Yes, .7 volts will drop across the diode, and yes, the current changes as a result of that, but current will never "drop" across a component. The current in the circuit between the diode and the battery and the current between the diode and the bulb would be equal. Kirchhoffs law. I understand the current will change with the addition of components, thats ohms law. Maybe you missunderstood me, but I assure you, I completely understand DC and AC circuits, as well as active devices such as diodes.

 

[Fordum]

I'm not going to quote your last, the darn things get too big. Lets see if I can wing it.

Ok, true, even a straight piece of wire has inductance. However its impedance is 0 and can be disregarded as long as the voltage is constant.
Which DC is.
That was why I threw in the part about no impedance after the circuit reaches full voltage.
When a circuit is first turned on, there is a voltage rise, which is resisted by any impedances in the circuit.

If there is no AC component to a voltage, which is pretty much the case with a battery powered circuit, then it can be disregarded.
Even the charging voltage from an alternator, which does have some pulsing DC on it, would work just as well if all the ripple was filtered out.

As for current increases with voltage drop, resistance often changes with voltage. I threw the motor example in to make it clearer.
A DC motor, such as a starter motor is trying to do a certain amount of work. Such as cranking an engine.

If it gets full voltage, it spins at a certain speed and draws a certain amount of current which is the amount needed to do the set amount of work.
If the voltage is low, the motor slows down, and begins to draw more current. I have seen more motors burned out by low voltage than anything else.
They draw too much current and run hotter.

And, just to be picky, your headlight example is another place where current will often increase with voltage drop.
The hot resistance of the filament is much higher than the cold resistance. As it cools off due to lower voltage, the resistance goes down.

How much? Not a clue. I just did a very depressing calculation, and realized that is has been almost exactly 40 years since I did that chart of filament resistance VS voltage in Dr. Jone's Electronics Lab.

That is depressing. WHERE THE HECK IS MY FLYING CAR???
I was promised flying cars by now.

My leg hurts, and most of my friends are dead. I'm going to go over and sit in the corner in my rocking chair and pet my cat........

I'm OLD.........


Screw that. Got too much stuff to do. I'll be old later...

 

[nsxxtreme]

I dont understand what you are saying here. No matter how you word it, or what components you add, a 12 volt circuit will always drop 12 volts across its components. Yes, .7 volts will drop across the diode, and yes, the current changes as a result of that, but current will never "drop" across a component. The current in the circuit between the diode and the battery and the current between the diode and the bulb would be equal. Kirchhoffs law. I understand the current will change with the addition of components, thats ohms law. Maybe you missunderstood me, but I assure you, I completely understand DC and AC circuits, as well as active devices such as diodes.

*edit* Yes I think we are saying the same thing. The current will reduce for a given load because of the voltage drop.

Ok, true, even a straight piece of wire has inductance. However its impedance is 0 and can be disregarded as long as the voltage is constant.
Which DC is.
That was why I threw in the part about no impedance after the circuit reaches full voltage.
When a circuit is first turned on, there is a voltage rise, which is resisted by any impedances in the circuit.

If there is no AC component to a voltage, which is pretty much the case with a battery powered circuit, then it can be disregarded.
Even the charging voltage from an alternator, which does have some pulsing DC on it, would work just as well if all the ripple was filtered out.

there is inductance AND capacitance to a wire. There is also capacitance internally to the battery. Is it negligable? The wire at 12V yes I agree....negligable. But it really depends on how long the wire is, what guage it is an how much current you pull through it. Pull enough currecn through a small long wire and the voltage drop accross the wire is no longer negligable. Which is why I said upgrade the wire AND the alternator. The capacitance of the battery (I don't believe its negligable) I'm no lead acid battery expert though. ESR of batteries are typically measured with an impedence meter (AC).

If it gets full voltage, it spins at a certain speed and draws a certain amount of current which is the amount needed to do the set amount of work.
If the voltage is low, the motor slows down, and begins to draw more current. I have seen more motors burned out by low voltage than anything else.
They draw too much current and run hotter.

The explanation of this is complicated. There are a couple reasons for this.
1. the faster a motor spins the more cooling the fan at the front of the motor provides. A slower moving motor the less cooling provided.
2. A DC motor is wires wraped around an armature. There are contacts on the armature that turn a segment of wire on and off. This is essentially an AC signal through the wire which then inductance and impedence applies.
As long as this is spinning aggainst an opposite magnetic field current is controlled. If it gets slow enough then it just becomes a wire. Take 20 feet of wire wrap around a spool connect it directly to a 12 volt battery and see how fast it becomes a heating coil. Its a little more complicated then that but thats the best way I know how to explain for then non technical.

We aren't talking about how much current the starter will pull though. Keep in mind that the battery isolator becomes parts of the system once you add it to the system. There is no loss in power if you take your measurement before the isolator. The loss in power is because of the voltage drop accross the diode. Nothing is free in life. Anytime there is a voltage drop there is going to be a net loss in power available.

Again this is why battery isolaters are made on large heat sinks. You can't have heat without voltage being consumed. P=I*(V2-V1)

(14.7-14)*100=70 watts lost as heat into the battery isolator for a single battery. Two batteries consuming 100 amps is 140watts.

So I did an internet search after I wrote this to see if anyone explained it better then I could. This website has some examples that might help explain what I have been trying to explain. Try those if it still doesn't make sense feel free to ask questions. But I think we have drifted from the original intent of the OP.
http://sterling-power-usa.com/prosplit-rzerovoltdropmarinebatteryisolator-2.aspx

 

[StreetshocK]

WHEW!

All I know is red(+) and black(-)! But you guys could sure convince me otherwise

My head hurts, but thanks for the detail

 

[big pappa b]

I just need pictures of how it's connected, nothing more

 

[Blue85]

This is silly. Your light bulbs do not draw more current when the voltage is reduced. The V-I relationship is not linear, but the resistance is not negative. If it was, applying 0V to the bulb would be infinite current! Then if I just made the voltage high enough, current would eventually go to zero! (Actually, the second part is true - once the filament is detroyed the current goes to zero until I hit a high enough voltage to arc across the filament holders.)

As for electric motors (especially DC), they draw high current at lower speeds because they are not generating EMF. This is why you can't create as much torque at higher speeds without advancing the phase of the applied voltage.

 

[Fordum]

I've got people in the office, so I got to be brief.

The filament in a light bulb has different resistance depending on its temperature like most conductors.
It can be measured either directly or indirectly. Back when I was in college, I did it both.
Rigged up a double pole switch to switch the filament between power and an ohm meter.
Hook the meter up first, measure the resistance. Then switch to power and let the bulb come to full brightness and wait a couple of minutes.
Then switch from power to the meter, and you will read a much higher resistance than when cold.

Plus you can watch it change as it cools.

Of course it does not go to zero unless you cool it to superconductive temps, or go to infinite unless, like you say, you melt it.

As for the motor part, I'm going to have to reread it when I have more time. It just does not make any sense right off hand.
Not sure what Electro Motive Force has to do with it, nor how you adjust the phase angle of a DC current.......

 

[Blue85]

Yes, obviously the filament resistance varies with temperature. This is what makes the I-V curve non-linear. But what you said is that current goes up when voltage goes down. That part is not true - that would be negative resistance (by definition), which can be created with active circuits, but generally doesn't apply to simple physical devices.

EMF is the motor acting like a generator, which it always does when spinning. As the EMF voltage increases (with speed), the difference between it and the applied voltage decreases until you reach the steady-state speed. As you load the motor down and slow it down, this voltage difference increases - the current and torque go up.

I shouldn't have brought phase angle into the discussion, I suppose. But a DC motor is only DC on the outside. On the inside the current is chopped up ("commutated") between the different windings to make it spin. The phase advance term applies most directly to sinusoidal drive DC brushless and AC motors, but also to motors in general. But this isn't really the point of the discussion.

We're talking about dual batteries - I apologize for contributing to the derail.

 

[Fordum]

We're talking about dual batteries - I apologize for contributing to the derail.

Yeah, i think i started it, too, my bad.

last post for little while, i just lost the use of my right arm.....

typing this one handed with my left. not bad trouble, stroke or anything like that, got something locked up in the shoulder

 

[colbystephens]

 

[Fordum]

Turtles be coool......

 

[nsxxtreme]

Yes, obviously the filament resistance varies with temperature. This is what makes the I-V curve non-linear. But what you said is that current goes up when voltage goes down. That part is not true - that would be negative resistance (by definition), which can be created with active circuits, but generally doesn't apply to simple physical devices.

EMF is the motor acting like a generator, which it always does when spinning. As the EMF voltage increases (with speed), the difference between it and the applied voltage decreases until you reach the steady-state speed. As you load the motor down and slow it down, this voltage difference increases - the current and torque go up.

I shouldn't have brought phase angle into the discussion, I suppose. But a DC motor is only DC on the outside. On the inside the current is chopped up ("commutated") between the different windings to make it spin. The phase advance term applies most directly to sinusoidal drive DC brushless and AC motors, but also to motors in general. But this isn't really the point of the discussion.

We're talking about dual batteries - I apologize for contributing to the derail.

Someone else that knows what they are talking about ^^^

The problem becomes trying to explain something technical to the non technical. I thought that displaying the linear equation would have helped show the point I was making but I guess not. If you speak in complete technical terms it goes right over their head and eye's just roll. If you try to speak in non technical terms you will get ripped to shreds for not saying things 100% technically correct. Damned if you do damned if you dont.

At anypoint I think the point of this discussion was just to say that battery isolators consume power. I would rather use a solenoid or a low ESR battery isolator. Also a second battery is an additional load. If you need more power while the car is OFF then a second battery is the way to go. If you need more power while the car is running a larger alternator is the way to go.

A winch would be a good example of a need for a second battery. The pulling time is short and the current is high.