Looking for some feedback from those better versed than I in battery behavior. My custom camper's 12v power system relies on two Deka 4d agm batteries with a combined 420ah of storage. Cabin heating and cooking duties are handled by a Wallus 85DU diesel cooktop and herein lies my question. The 4-5 minute start up cycle of the Wallus draws 8 amps which, when the battery bank is fully charged (12.97v) draws the voltage down to 12.33. Once the start up cycle is completed the battery will gradually return to near "resting" voltage. Maybe I'm a bit paranoid but even after 3 years of using this system I still wonder if that amount of voltage drop is normal? I realize that the Wallus is a marine product and assumes an alternator will probably be supplying consistent voltage during start up but still, considering the size of the battery bank, should the voltage drop that much under an 8 amp draw? I've chatted with the folks from both Wallus and Deka but have yet to receive a reassuring answer. Thoughts?
Expected voltage drop?
- Thread starter Fusooka
- Start date
dwh
Tail-End Charlie
I don't know that unit specifically, but the voltage of the system will drop while there is a load on it.
I've got a cheapo 12v oscillating fan with a 10' cord and a lighter plug. It draws 10w, or 0.8a. Measuring the voltage at the fan end of the wire it might say 12.8v, but as soon as I turn it on, it'll drop (depending on how hot the ambient temp is) as low as 11.8v. As soon as I turn the fan off, the voltage jumps right back up to 12.8v.
That's measuring the voltage at the load end of a fairly long, fairly skinny wire. If I measure across the battery terminals, the voltage doesn't change when I turn that little fan on and off.
So the first question is - exactly where in the system are you getting the voltage measurement?
And the surface charge effect works both ways - when charging it builds up on the plates and then takes a while to dissipate into the electrolyte (temporarily raising the apparent voltage), when discharging it gets sucked from the plates and takes a while to build up again (temporarily lowering the apparent voltage).
I've got a cheapo 12v oscillating fan with a 10' cord and a lighter plug. It draws 10w, or 0.8a. Measuring the voltage at the fan end of the wire it might say 12.8v, but as soon as I turn it on, it'll drop (depending on how hot the ambient temp is) as low as 11.8v. As soon as I turn the fan off, the voltage jumps right back up to 12.8v.
That's measuring the voltage at the load end of a fairly long, fairly skinny wire. If I measure across the battery terminals, the voltage doesn't change when I turn that little fan on and off.
So the first question is - exactly where in the system are you getting the voltage measurement?
And the surface charge effect works both ways - when charging it builds up on the plates and then takes a while to dissipate into the electrolyte (temporarily raising the apparent voltage), when discharging it gets sucked from the plates and takes a while to build up again (temporarily lowering the apparent voltage).
dwh
Tail-End Charlie
So far, sounds normal to me.
But to proceed, the next question would be: what is the wire size/distance of the supply wire from the battery to the cooktop? Plug that, along with the known voltage and amperage load, into a voltage drop calculator and see if it is within expectations.
(Also, I was wondering about the fuel pump. Does the fuel pump keep running once the stove is though its start-up cycle? Or is the fuel pump operating intermittently? If it keeps running, that, combined with whatever is still using power (control board?) could be why it doesn't recover to full resting voltage - there is still a load pulling down the bus voltage. I'm assuming that the start-up load has something to do with ignition.)
But to proceed, the next question would be: what is the wire size/distance of the supply wire from the battery to the cooktop? Plug that, along with the known voltage and amperage load, into a voltage drop calculator and see if it is within expectations.
(Also, I was wondering about the fuel pump. Does the fuel pump keep running once the stove is though its start-up cycle? Or is the fuel pump operating intermittently? If it keeps running, that, combined with whatever is still using power (control board?) could be why it doesn't recover to full resting voltage - there is still a load pulling down the bus voltage. I'm assuming that the start-up load has something to do with ignition.)
What you are seeing is normal. 8 amps for 5 minutes is insignificant to the pair of 4d batteries. Remember, Voltage is the potential (pressure) that enables current to flow. It is not the amount (current, amps) or total amount of energy used (amp hours). The voltage will drop under any load and that drop is not linear. For example, a large starter motor, pulling 1000A will draw the voltage down to 9-10V give or take a bit. Once the load is removed the voltage will quickly return to near "resting" levels.
So, in the mean time, I took the advise offered from the folks at Wallus and re-routed my supply directly to the battery...bypassing the distribution panel. That did help with voltage dropping to 12.42v during start-up (ignition) instead of 12.3v. Once start up is completed it returned quickly to near resting (post ignition draw is .3 amps) of 12.8v. But...I'm still scratching my head a bit about the relationship between amps being drawn and voltage drop. Huffy indicates it's not linear which coincides with my experience but seems counterintuitive. I tested this with a 23 amp draw vs an 8 amp draw...both with a fully charged battery. Voltage drop was nearly the same (from 12.9 to 12.3 and 12.35 respectively) for a two minute test. Both returned to similar resting voltage. I appreciate the pressure (volts) and volume (amps) analogy but still having trouble wrapping my brain around this not being linear? Different analogy for a simple thinker?
dwh
Tail-End Charlie
Well, "voltage drop" is actually just a convenient way to describe an effect with a few different variables. Variables which themselves aren't fixed, but which can affect each other.
For instance, the amps of load can be altered by the resistance of the wire. But the resistance of the wire can also be altered by the amps of the load. Either can be altered by the supply voltage, and either can alter the supply voltage.
And then, to make it even more interesting, what readings you get depends on where you take the measurement.
An analogy...hrmmm...
How about a helicopter analogy? Helicopters are cool.
It's sort of like flying a helicopter - increasing the angle of attack of the main rotor (pulling up on the collective) then requires more throttle to maintain rotor speed. But that results in more yaw, which then requires increasing the angle of attack of the tail rotor to keep the thing headed in a straight line.
Now let's see how my off-the-cuff helicopter analogy rates when translated into electro-speak (confidence is not high...): Increasing the watts of load draws down the voltage, which in turn raises the amperage, which increases the resistance of the wire, which also increases the amperage, which draws down the voltage even more.
Eventually, it all settles down to a more or less balanced place - until something changes, like the amps load heats up the wire, which increases the resistance and now the whole system adjusts to a new state of equilibrium.
In other words, it's a balancing act where everything affects everything else. "Voltage drop" is an easy way to describe the end result, just as "headed in a straight line" is in the helicopter analogy. But the usage of the phrase "voltage drop" is a bit misleading, as it makes the whole process appear to be cut and dried, which actually it's more fluid and always in motion.
Don't ask me why, but for some reason I think I may have just made the whole even more confusing.
For instance, the amps of load can be altered by the resistance of the wire. But the resistance of the wire can also be altered by the amps of the load. Either can be altered by the supply voltage, and either can alter the supply voltage.
And then, to make it even more interesting, what readings you get depends on where you take the measurement.
An analogy...hrmmm...
How about a helicopter analogy? Helicopters are cool.
It's sort of like flying a helicopter - increasing the angle of attack of the main rotor (pulling up on the collective) then requires more throttle to maintain rotor speed. But that results in more yaw, which then requires increasing the angle of attack of the tail rotor to keep the thing headed in a straight line.
Now let's see how my off-the-cuff helicopter analogy rates when translated into electro-speak (confidence is not high...): Increasing the watts of load draws down the voltage, which in turn raises the amperage, which increases the resistance of the wire, which also increases the amperage, which draws down the voltage even more.
Eventually, it all settles down to a more or less balanced place - until something changes, like the amps load heats up the wire, which increases the resistance and now the whole system adjusts to a new state of equilibrium.
In other words, it's a balancing act where everything affects everything else. "Voltage drop" is an easy way to describe the end result, just as "headed in a straight line" is in the helicopter analogy. But the usage of the phrase "voltage drop" is a bit misleading, as it makes the whole process appear to be cut and dried, which actually it's more fluid and always in motion.
Don't ask me why, but for some reason I think I may have just made the whole even more confusing.
verdesardog
Explorer
dwh
Tail-End Charlie
The wire doesn't have to be undersized to heat up. It will heat up some whenever it is carrying power.
This is why wire has to be de-rated when there are multiple conductors in a conduit or raceway.
Properly sized wire will heat up, but not over heat. Undersized wire will over heat...
...unless there are too many wires in a small space, then even properly sized wire can and will overheat.
Thus:
Wire resistance is a variable,
based on wire temperature,
which is a variable,
based on current being carried.
(Yes, I just ignored wire length for that example.)
Variable resistance with variable current also = variable voltage.
And yes, battery resistance is also part of that whole balancing act. But I left it out since it wasn't necessary to answer the OP's generalized question about voltage drop.
4x4junkie
Explorer
To me that seems like more v-drop than should be from just 8 amps off of a 420Ah bank...
Do you have any other charge sources going to that battery bank besides your alternator? (solar, shore charger, etc.) I have to wonder if maybe it's because the batteries are not receiving proper absorptive charging and the resulting sulfation is increasing their internal resistance.
Do you have any other charge sources going to that battery bank besides your alternator? (solar, shore charger, etc.) I have to wonder if maybe it's because the batteries are not receiving proper absorptive charging and the resulting sulfation is increasing their internal resistance.
Three charging sources; 130w solar charger, shore power (Iota engineering 55amp 3 phase smart charger), and the alternator. I have relied on the solar charger because, with the exception of the stove, my power draws are minimal and have never drawn the resting voltage below 12.5v. That being said, I have wondered whether the solar charger is capable of deep charging the battery as you've described?
DiploStrat
Expedition Leader
Some Anecdotal Comments
My battery bank is larger than yours and my solar array is much larger.
I see voltage drops, and recovery, like yours, but typically only with my induction stove or microwave, not with my Dual Top which ramps up to about 10A when it is really blowing hot air.
My solar charger actually has the most aggressive profile of any of my chargers; in fact I had to lower the acceptance to float amp rate of the shore charger to get it to reach the same level of charge. Part of this is because I almost always have the background loads of the Dual Top (2-10A) and the starter batteries which present themselves as a load to the charger but which do not register on the battery meter which reads the camper batteries. (Yes, the TriMetric has the option of measuring a second bank, but it has never been worth the cost and effort.)
As long as your batteries recover immediately, i would not worry about it. I would worry more that the camper batteries are, in fact, reaching full charge.
Hope this is helpful.
My battery bank is larger than yours and my solar array is much larger.
I see voltage drops, and recovery, like yours, but typically only with my induction stove or microwave, not with my Dual Top which ramps up to about 10A when it is really blowing hot air.
My solar charger actually has the most aggressive profile of any of my chargers; in fact I had to lower the acceptance to float amp rate of the shore charger to get it to reach the same level of charge. Part of this is because I almost always have the background loads of the Dual Top (2-10A) and the starter batteries which present themselves as a load to the charger but which do not register on the battery meter which reads the camper batteries. (Yes, the TriMetric has the option of measuring a second bank, but it has never been worth the cost and effort.)
As long as your batteries recover immediately, i would not worry about it. I would worry more that the camper batteries are, in fact, reaching full charge.
Hope this is helpful.
dwh
Tail-End Charlie
Deep charging? What's that? I've heard of "deep discharge" which means to drain the battery a lot. But "deep charge"? That would mean what? Charging to 100%? More than 100%?
"Deep charge" sounds like a mythical beast to me.
It's just chemistry. The lead/electrolyte gets fully saturated with electrons and that's that. All done.
If you're seeing a resting voltage of 12.97v (in the OP you mentioned that number, though you said "full charge" rather than "resting voltage"), then I'd say that you are getting the electrolyte fully saturated.
Though actually, at that high a number, I'm thinking you were actually seeing a bit of residual surface charge that makes the reading a bit high.
dwh
Tail-End Charlie
Woops. Forgot to mention this part...
Did you actually mean, "relied on the solar charger AND the alternator"?
Did you actually mean, "relied on the solar charger AND the alternator"?