Battery jump starters

[Lance]

The first few production models of the DB9 had a great issue - the OLED instrument cluster not only displayed what gear you were in, but also controlled it. Any gear change went via the display to the gearbox.
If the display had an issue, then the gearbox got stuck.
One of the connectors on the display had a tendency to have issues. Guess what went through that connector.

[DealorNoDeal]

If you know the internal impedance of the battery, and the current that the starter draws you can calculate the voltage drop.
But measuring the voltage drop is easier (and safer) than the current draw.

An older battery will drop more than a new battery. A screwed battery could well read 12.6V but drop to under 10V when trying to crank, and likely won't drive the starter (depending on engine size, temperature, etc.)

According to google, the voltage will drop to between 10 & 11 volts during cranking (for a good battery).

If a battery is very low on charge, I expect cranking would drag the voltage way down. FWIW, I've had that happen a few times and it's never caused any problem with the electronics.

[d000hg]

If you attach a fully charged battery to a totally dead battery whilst still in situ there will be a surge of current but the volts aren't going to jump above what the charged battery can provide so I don't see an issue.

But you don't normally jump-start unless the donor car has the engine running, AFAIK - wouldn't that mean the voltage is higher since the battery is charged at more than 12V (14V or so from memory?) I never really understood why a 12V battery can be higher or lower than 12V depending on charge though.

[WTFH]

But you don't normally jump-start unless the donor car has the engine running.

I'm happy to be corrected on this, but I believe the reason for it is to stop the donor battery from draining, and ending up with 2 dead cars.

[Lance]

But you don't normally jump-start unless the donor car has the engine running,

it depends how flat the battery is and how good the donor battery is. A battery will give a bigger boost than an alternator anyway. Alternators can't give 100+A.
In any case the topic of this thread is about jump starting from a battery pack so not really the same.

AFAIK - wouldn't that mean the voltage is higher since the battery is charged at more than 12V (14V or so from memory?) I never really understood why a 12V battery can be higher or lower than 12V depending on charge though.

voltage isn't fixed. 12V is an approximation. And sticking multiple 12V batteries in parallele will not change the voltage anyway.

[DealorNoDeal]

I never really understood why a 12V battery can be higher or lower than 12V depending on charge though.

When I was having battery problems many years ago, I kept an eye on the voltage. While the engine was running, it was about 13.5V. (Ditto when on a charger.) After turning the engine off, it starts dropping and then takes a couple of hours or so to stabilise.

If you ever want to check your battery's condition, wait a few hours after turning the engine off before measuring the voltage.

If you leave a car standing, you'll notice the voltage drops a tiny amount every day. That wouldn't happen if the battery was on the bench; it's the car's electronics which are putting a small drain on the battery.

The voltage is a kind of crude indicator of the charge level (and the condition of the battery). 12.5V is good. 12V means it's lost quite a lot of charge. In my experience, 11.5V is where it starts to be touch and go whether it will crank the engine.

[DoctorStrangelove]

The first few production models of the DB9 had a great issue - the OLED instrument cluster not only displayed what gear you were in, but also controlled it. Any gear change went via the display to the gearbox.
If the display had an issue, then the gearbox got stuck.
One of the connectors on the display had a tendency to have issues. Guess what went through that connector.

I see that Xogg's hero, Jeremy Clarkson, had some fun with a car he had on test: got to the car park & couldn't get out of the damn thing because he & his passenger couldn't locate The Magic Button that opened the doors.

Similarly, once they'd found said Magic Button & got out, car starts rolling away because lo! in some demented software engineer's brain you had to do something equally Magick to a) put it in Park and b) put the fecking handbrake on.

One really misses the days of standard controls.

https://uk.style.yahoo.com/jeremy-cl...103547109.html

Mate had a car where the alternator, some black box, and the battery talked to each other & it was a major pain in the arse to change the battery.

All this tulipe seems to be designed by The Sirius Cybernetics Corporation in an attempt to kill us all.

Just because you can put microcontrollers in everything doesn't mean It's A Good Idea To Do So.

[DealorNoDeal]

All this tulipe seems to be designed by The Sirius Cybernetics Corporation in an attempt to kill us all.

Just because you can put microcontrollers in everything doesn't mean It's A Good Idea To Do So.

We've gone from cars which rust to bits as they age, to cars which die due to electrical gremlins.

One of the reasons I wouldn't want a newer car is it will have even more shyte to go wrong.

[d000hg]

voltage isn't fixed. 12V is an approximation. And sticking multiple 12V batteries in parallele will not change the voltage anyway.

I know all those things, I said I never understood why it's not fixed. Why does putting load on a nominal 12V battery cause the voltage to be a bit lower and charging it cause it to be higher, and when the battery is flat/dying why does the voltage drop much more rather than just the ampage?

[Lance]

I know all those things, I said I never understood why it's not fixed. Why does putting load on a nominal 12V battery cause the voltage to be a bit lower and charging it cause it to be higher, and when the battery is flat/dying why does the voltage drop much more rather than just the ampage?

in summary... ohms law..... and the fact that the battery isn't perfect (it has an internal impedance)


>>>>>>>>>>>>>>>>>
at greater length..... The battery has an internal impedance (resistance). As per all electrical circuits, whenever current flows through any device with an impedance there is a potential difference between either end of the resistor. This is the same for the internal impedance of the battery.
A battery is in effect a power source, and a resistor.

When not connected, no current is flowing. the potential difference between the poles of the battery is 12 V (let's call it 12V for ease)

When connected there is a current draw that results in a potential difference between the 2 connectors of the load - lets' say that it is a 0.1 Ohm load for the starter motor. That would draw 120A from the battery. And if the battery was perfect, potential difference across the start would be 12V. No voltage drop.

So let's say the internal impedance of the battery is .01 ohms.

You have a circuit with a 'perfect' 12V source, and 2 resistors in series. It's not drawing 120A as the load is 0.11 Ohms (resistors in series are summed).
Using ohms law (V=IR) the current draw is 12/0.11 = 109.09A

The potential difference across the starter is not 12V as the actual current is 109.09A so using ohms law again, 109.09*0.1=10.9Volts ( a drop of 1.1V due to the internal impedance of 0.1 ohms). If you do 109.09A * 0.01 ohms you get 1.1V which is the potential difference across the internal impedance of the battery.

I've probably not explained it very well. It's been 30 years since I last worked this out.