Promaster battery issue

[avanti]

Originally Posted by DCHitt

The only problem I have with your explanation that a door or key has to be pressed before it Pings is that the computers must be on in order for them to know to Ping the Key Fob (Can-Bus stuff).

So either the Ping system is running all the time to turn on the computers or the computers are running all the time to turn on the Ping. Sort of a chicken and egg problem.

The Rep could be right.

No. The CANbus spec has a special "sleep" mode that is capable of waking up various ECUs. They normally are all "asleep" until a specific device (in this case the door locks or the key lock--maybe a few others) send a "wake" message on the bus. When THAT happens, the bus wakes up and the vehicle starts pinging. This is necessary for stuff like turning on the interior lights when the door opens, and so on. It is also used for this purpose.

[DCHitt]

Originally Posted by mkguitar

the usual battery drain test with an ammeter can confirm what drain the security system places on the vehicle.

What a great idea. We could measure it.....

[mojoman]

I have a 2014 Promaster, it came with a large AGM battery. Mine is still good but I have had it on a smart trickle charger when idle and a couple of windshield solar panels when not plugged in. AGM batteries are pricey. The charger and solar panels ran me $60 which is a good investment to extend the life since the AGM battery will cost you around $200 to replace.

[DCHitt]

Originally Posted by avanti

send a "wake" message on the bus. When THAT happens, the bus wakes up and the vehicle starts pinging.

Thanks again for the information. The Can-Bus interests me. Sorry for the dumb questions. I still don't understand how you can send a "wake" message if they are powered off. Maybe not really turned off "just asleep"? Maybe the Promaster really powers them off?

[cruising7388]

Originally Posted by mkguitar

on my old chev I determined that the parked load was about 1200 mA ( 1.2 amps)...much of this extra load was corrosion on the fuse lugs causing high resistance. after cleaning up the fuses, the parked load ( alarm and radio) is closer to 600 mA)

Assuming a fixed applied voltage, how does decreasing the resistance in a circuit feeding loads result in a decreasing current flow?

[mkguitar]

It takes more power to push through a dirty connection than a clean connection- electrons flow more easily through a clean large cool conductor than a dirty, smaller or hot conductor.

Let's say the radio memory itself draws 100mA, if there are dirty wire connections then that will increase the current draw on that circuit


as the snake eats it's tail...a dirty connection can cause heat to build in that area...the heat itself can cause more resistance which results in even more heat.
eventually the wires can burn- unless a fuse pops

Mike

[avanti]

Originally Posted by DCHitt

Thanks again for the information. The Can-Bus interests me. Sorry for the dumb questions. I still don't understand how you can send a "wake" message if they are powered off. Maybe not really turned off "just asleep"? Maybe the Promaster really powers them off?

You are correct that they aren't really off, but they are very close. Almost all modern microprocessors have a facility for sleeping and waking up on various events. It's like when your TV turns on via an IR command from your remote control, or a calculator that turns on when you press any key. The CANbus controller chips have a specific facility to wake up themselves and the rest of the ECU when they see any message on the bus. There are dozens of devices in your van that are sleeping in this mode when you are parked. So, yes, there is a significant parasitic load. It's just that I doubt that it has anything to do with your key fob.

[Bruceper]

Originally Posted by cruising7388

Assuming a fixed applied voltage, how does decreasing the resistance in a circuit feeding loads result in a decreasing current flow?

Because I=V/R

Yes a smart alec answer, but it is a correct answer. When you reduce resistance and have a static voltage, the current will drop.

[avanti]

Originally Posted by Bruceper

Because I=V/R

Yes a smart alec answer, but it is a correct answer. When you reduce resistance and have a static voltage, the current will drop.

Uh... Might want to check your math.

[DCHitt]

Originally Posted by avanti

There are dozens of devices in your van that are sleeping in this mode when you are parked. So, yes, there is a significant parasitic load. It's just that I doubt that it has anything to do with your key fob.

Which brings us back to the start. If the Rep is right the Promaster is getting rid of a great deal of the parasitic load by powering the computers off until the key fob is close? A ping once is a while might be a lot less power than having all the computers powered (in sleep mode) all the time.

That could be a good idea. I guess we could test this theory with an amp meter and a key fob.....

[avanti]

Originally Posted by DCHitt

Which brings us back to the start. If the Rep is right the Promaster is getting rid of a great deal of the parasitic load by powering them off until the key fob is close? A ping once is a while might be a lot less power than having all the computers powered (in sleep mode) all the time.

That could be a good idea. I guess we could test this theory with an amp meter and a key fob.....

Read my original explanation. If it is "off" it can't ping. If it can ping, it can listen with less current. Your tech's theory is almost certainly nonsense.

[cruising7388]

Originally Posted by mkguitar

It takes more power to push through a dirty connection than a clean connection- electrons flow more easily through a clean large cool conductor than a dirty, smaller or hot conductor.

Let's say the radio memory itself draws 100mA, if there are dirty wire connections then that will increase the current draw on that circuit

Not always. I think it's somewhat more complicated than this. If the load impedance is resistive (devoid of any j reactance component), an intervening resistance like a flaky connector will not result in any increase in the source voltage and will result in a decrease in the amperage flowing to the load. Under these circumstances, Ohms law dictates that voltage and power dissipated is divided between the connector and the load.

If the load impedance is purely resistive, the loss of some load driving voltage and reduced current may impair the load function but doesn't damage it.. But loads like motors that have a reactance component are subject to damage from the same condition. To address this, designers craftily dodge this consequence by employing a winding array that compensates for lower voltage with a higher current demand to effectively permit continued operation at the design power level. In this scenario, a flaky connector would result in a higher current demand that would initially satisfy the motor but which, as you point out, produces an overheating condition that invariably threatens to burn up the connector and if it advances beyond the voltage compensation capability of the motor would probably damage the motor too if it doesn't have over-currrent protection.

I have no idea of what impedance a volatile memory circuit expresses but I wonder whether a voltage reduction seen by the memory would double the memory current demand you observed. Does the fuse you serviced support only memory and no other circuits?

[avanti]

Originally Posted by cruising7388

Assuming a fixed applied voltage, how does decreasing the resistance in a circuit feeding loads result in a decreasing current flow?

I don't think we should drop this [sub]topic without fully-clarifying the above question. No offense to anyone, but the various answers are confusing and in some cases just wrong, and I think we should leave the record clear, lest some future reader burn down his or her van due to faulty math...

Here is my attempt:

Originally Posted by mkguitar

It takes more power to push through a dirty connection than a clean connection- electrons flow more easily through a clean large cool conductor than a dirty, smaller or hot conductor.

This is true. BUT "power" and "current" are not the same thing.

Let's say the radio memory itself draws 100mA, if there are dirty wire connections then that will increase the current draw on that circuit

As @crusing explains below, this is not in general true. According to ohm's law, adding a resistor in series with a load will decrease current. Think of it this way: If you open a switch in a circuit, the resistance goes to infinity, so the current goes to zero.

as the snake eats it's tail...a dirty connection can cause heat to build in that area...the heat itself can cause more resistance which results in even more heat.
eventually the wires can burn- unless a fuse pops

The phenomenon described certainly happens, but it is NOT due to runaway current caused by runaway resistance. The vicious cycle is as follows: increased resistance at the bad joint causes a greater voltage drop across the joint (at the expense of voltage at the load). This, in turn generates heat at the joint, which further damages the bad joint, which causes further voltage drop which causes increased heat... It is voltage drop over the increased resistance that causes the death spiral, not increased total current.

Now, as @crusing explains, things are different in the case of an inductive load. Things with coils act like auto-transformers which can actually generate currents in certain parts of their cycles. In these cases we talk about impedance, not just resistance. This is why you particularly see those kinds of failures when electric motors (e.g., starter motors) are involved. It is very unlikely that a modern piece of electronics will present any significant inductive load.

Originally Posted by Bruceper

Because I=V/R
Yes a smart alec answer, but it is a correct answer. When you reduce resistance and have a static voltage, the current will drop.

As I said, this statement includes a math error:
I=V/R (i.e. Ohm's Law: current=voltage/resistance) is certainly correct and relevant. BUT, "R" is in the denominator, so if you increase "R" and keep "V" constant, then "I" goes down, not up.

Originally Posted by cruising7388

Not always. I think it's somewhat more complicated than this. If the load impedance is resistive (devoid of any j reactance component), an intervening resistance like a flaky connector will not result in any increase in the source voltage and will result in a decrease in the amperage flowing to the load. Under these circumstances, Ohms law dictates that voltage and power dissipated is divided between the connector and the load.

If the load impedance is purely resistive, the loss of some load driving voltage and reduced current may impair the load function but doesn't damage it.. But loads like motors that have a reactance component are subject to damage from the same condition. To address this, designers craftily dodge this consequence by employing a winding array that compensates for lower voltage with a higher current demand to effectively permit continued operation at the design power level. In this scenario, a flaky connector would result in a higher current demand that would initially satisfy the motor but which, as you point out, produces an overheating condition that invariably threatens to burn up the connector and if it advances beyond the voltage compensation capability of the motor would probably damage the motor too if it doesn't have over-currrent protection.

I have no idea of what impedance a volatile memory circuit expresses but I wonder whether a voltage reduction seen by the memory would double the memory current demand you observed. Does the fuse you serviced support only memory and no other circuits?

This analysis is correct. As I said, though, it is very unlikely that a piecer of electronics would present a significant inductive load. Assuming this, a flaky connector will DECREASE total current, stealing voltage from the load. If it gets bad enough, the voltage on the load will drop until it no longer functions properly. If total current is high enough, you MIGHT get runaway heat damage (also unlikely with electronics), but you will never get increased current draw.

Once again, no criticism intended of anyone--just trying to keep the record straight.

[cruisefx]

To the poster who started this thread, did you find a resolution to your chassis battery issue? I have read elsewhere of someone having a similar issue on a new Promaster. The Promaster dealer told the owner that it is possible for the battery to drain after a few days. No explanation given so it sounds crazy

[gerrym51]

Originally Posted by cruisefx

To the poster who started this thread, did you find a resolution to your chassis battery issue? I have read elsewhere of someone having a similar issue on a new Promaster. The Promaster dealer told the owner that it is possible for the battery to drain after a few days. No explanation given so it sounds crazy

i have a 2015 zion 2015 promaster. the manual says it will go 21 days without needing starting.

I have gone as long as 14 days twice and it has started.

Look in your promaster manual. i forget the catagory but it was in there

what van do you have. i have read of roadtrek owners with voltstart causing yhis

[cruisefx]

I don't have the issue but there is a new Aktiv owner with the issue...just trying to assist. She called a Dodge dealer and she posted the following.

"I called Promaster about my chassis battery going dead after several days. Was told that the chassis battery WILL discharge if engine not started within 5 days or so. The tech spoke very confidently that this is the case on all Promasters and stated that many modern vehicles have the same issue. Is this true or is the tech brushing this issue aside?"

Does not sound right to me but then I have a Chev RT.

[avanti]

I don't know anything about Promasters, but all modern vehicles have significant vampire loads when parked. If you are going into storage, you have three options:

1) Start and "exercise" the vehicle regularly. This is generally not good for the engine (unless done very thouroughly) and really should be avoided if possible.

2) Disconnect the battery when in storage. This is effective--a truly disconnected battery will last for many months. Some vehicles (e.g., Sprinters) have a provision for this, and it is possible to add an aftermarket battery switch. Some modern vehicles have a protocol for doing this--often involving a wait time before disconnecting the battery.

3) Arrange for some kind of a battery maintainer. If you keep your chassis plugged into shore power, or if you have effective solar during storage, the best way to do this is a DC-DC trickle-charger such as the Trik-L-Start.

[Bud]

"I don't know anything about Promasters, but all modern vehicles have significant vampire loads when parked."

2013 Honda Accord EX is happy starting after 4 weeks, but maybe it is not all modern.

Bud

[cruisefx]

The Aktiv owner who has the issue indicated that she needed a boost after the rig sat for 3 days. She drove it halfway across the country to get home and then it would not start. It was boosted once and then a few days later, the chassis battery was dead again. I have a hard time believing that a battery cannot sit for 3 days without needing it to be started up daily. My Chev RT sits in my driveway which is located in cold wintery Ontario and it will start even when unused for 2 -3 weeks. She has indicated that everything is off but I have not asked about her key fob.

[avanti]

I agree that 3 days (or even 2 weeks) is unreasonable, and probably not normal. I think that maybe 6 weeks is typical. But, few current vehicles will last through a winter's storage with the battery connected.