Re: How to calculate (predict) voltage drop? Can somebody check my math?

I thought cells in a battery were 2.3-2.4V each... or just under... IDK I am retarded. lol.

But to add amperage and voltage are directly related, more amperage=less voltage hence why we have voltage drop when there are large amperage loads.

04-11-2013 01:45 PM

cxa0897

Re: How to calculate (predict) voltage drop? Can somebody check my math?

Quote:

Originally Posted by profundus-sanus

output voltage of the alternator does not drop until after the draw exceeds the max amperage that the alt can produce.

There's a bit of steep curve there where the draw is exceeded. . Like a 270a unit will put out 0-270a @ 14.4v But it might eek out 280 @ 10 volts.

cool. that basically confirms what I believe happens with that second pic. The voltage of the battery will dominate output voltage while producing that extra bit of current.

04-11-2013 01:18 PM

profundus-sanus

Re: How to calculate (predict) voltage drop? Can somebody check my math?

output voltage of the alternator does not drop until after the draw exceeds the max amperage that the alt can produce.

There's a bit of steep curve there where the draw is exceeded. . Like a 270a unit will put out 0-270a @ 14.4v But it might eek out 280 @ 10 volts.

04-11-2013 01:03 PM

cxa0897

Re: How to calculate (predict) voltage drop? Can somebody check my math?

I've done a bit of thinking about this and I think I have it semi figured out.

We can treat the battery and the alternator as both a voltage source, and a current source, which is shown in the pic above.

Basically, we have two scenarios. When load is less than the alternators production, and when the load is greater than it.

When load is less than the alternators production, the alternators voltage dominates the circuit since it is higher. The current will flow from the alternator up and and right. It will then split to meet the load requirement(the resistor in this circuit, or the sub in real life). Excess current will flow through the battery charging it. The battery here can actually be assumed to be a load.

When the alternator cannot supply enough current, the battery will also become a current source. Since there is a general rule that voltage cannot be different at a single point, I assume the alternator's voltage must drop. This will allow both the alt and battery to produce the current being sent to the load, in picture 2. In this scenario, we can treat both sources as having the voltage of the battery.

I'm not too knowledgeable in alternator voltage-output behavior, but I have to assume output voltage starts to suffer as current really starts to increase. Maybe somebody will be able to toss a few words in on that topic, and I can revise what I've written

04-08-2013 11:09 AM

recon440

Re: How to calculate (predict) voltage drop? Can somebody check my math?

Thanks. There is definitely some oversimplifications and assumptions I'm making but the numbers I'm arriving at are quite reasonable.

ie.
-I'm assuming that the alternator is a constant voltage source which is obviously not the case when you are sucking more than it can provide. Net voltage is a result of equilibrium between two sources (which means at after the alternator runs out of juices there is a region where it tapers off and is actually supplying less than 14.4V) and plays a balancing act with the battery.
-The other assumption is in the battery voltage drop. I've seen 30s 500A load tests where the battery is getting pulled down to 10V but I would imagine that if I were to draw 350W for a second or two it it would probably sag to 12.5V or so (another assumption).
-I'm sure there would be a different way of figuring it out if we know the true wire resistance, true internal resistance of the battery, real output curves of the alternator, etc...

I'm not sure about the sag on a battery (I don't put 500A constant loads on a battery (where essentially the deciding factor in these tests is the battery's internal resistance and capacity)). Would it be reasonable to assume that a quality AGM battery could maintain 12.5V for a second or two while pulling 100-150A?

04-08-2013 10:51 AM

cxa0897

Re: How to calculate (predict) voltage drop? Can somebody check my math?

Alright then. This is interesting, and I'm not entirely sure how the systems interact.

Mathematically speaking, there is nothing wrong with the equations. You are basically using superposition to get a net effect. We use this to solve some basic circuits, but it assumes the sources do not interact directly, since the node in between them would have two different voltages, a physical impossibility.

How much this assumption matters I do not know. It may be negligible, and you may be spot on. It may also be more complex, and this would only hold up during certain circumstances. I'll spend some more thought on it after my test later today, and maybe Jud will pop in. He is the other resident engineer

04-08-2013 10:40 AM

recon440

Re: How to calculate (predict) voltage drop? Can somebody check my math?

I'm pretty familiar with that (wire has resistance depending on length and in turn voltage drop is inversely related to current running through it).

I'm more interested in the math behind how two voltages sources add up (if your alternator is not pushing enough amperage it's going to have to come from the battery). So your net voltage is going to have be somewhere between 14.4V and 12.86V.

So I'm just wondering how the calculations are done cause I'm seeing people making claims that that 2k+ RMS amplifiers on stock alternators (80-90A) using just high current AGM batteries their voltage stays in the high 13s (which I think is an exaggeration).

04-08-2013 10:33 AM

cxa0897

Re: How to calculate (predict) voltage drop? Can somebody check my math?

you would need to find the resistance of the power wire you are running. you can look up awg charts for a decent approximation (all wires are not true gauge)

Take that resistance, and multiply it by the current draw from your amplifier. a good estimate is rated power, and with a good amp you can expect 60/65% efficiency at 1 ohm, 75-80% at 4 ohm

this will give you the voltage drop due to resistance in the line, subtract that from the resting voltage at the front, and you have the voltage at the amp.

04-08-2013 10:30 AM

recon440

How to calculate (predict) voltage drop? Can somebody check my math?

Does anybody know how to calculate the voltage drop given two different voltage sources (a 14.4V alternator and a 12.86V battery)?

Say for instance I want to run 1200W RMS and have a 14.4V alternator that puts out 100A at idle. 40A are required to run the vehicle leaving 60A free.
The power from the alternator is coming at 14.4V, so 14.4V*60A = 864W. Now I require the other 336W which the alternator cannot supply so it must be provided by the battery. I'm assuming that a fully charged (12.86V) battery will sag to about 12.5V when it's called upon. Under this underlying assumption I'm calculating the total voltage as follows...

This seems right in line with what I'm seeing on my system (130A Denso SC 6-phase alternator (100A @ 800RPM, 130A @ 1300RPM), 44Ah Optima Redtop, big 3, Alpine PDX-M12 on dual Sundown Zv3s D2 wired to 2 ohms nominal).

Several questions I have...
I plan to add another PDX-M12 so each sub will be wired up to 4ohms (so it should run a little more efficient). In any case, at idle I'm going to be able to give 60A from the alternator and I'll be requiring the rest of the current from the battery.

What kind of a sag do batteries produce under, say, 150A of current draw for 1 second?
I plan to go with a single Odyssey PC1500 under the hood (no secondary)? Would this be sufficient to run two PDX-M12s @ 4 ohms each?

It seems to me, no matter what, there will always be voltage drop since the batteries can only give 12V power instead of 14V, correct? Even with a PC1500 and a PC1700 in the back.

P.S.
I can't afford the XS Power in this country (Japan). That's why I've chosen Odyssey. Even though Odysseys are relatively expensive I will not be extorted into paying $700 for a D5100R (which is what these greedy dealers sell them for here).

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