Reply to: Battery jump starters

[DoctorStrangelove]

I remember doing a practical at Uni, where you had to design an op-amp circuit. Once you'd calculated the required component values, you went and collected them from the stores technician. One guy couldn't understand the bemused look on the technician's face when he asked for a several Farad capacitor.

I had a 1 Farad capacitor in stock.

I used it to frighten people.

A bit like the 1000:1 scope probe.

Many years ago a colleague told me that he'd requested a 741 opamp from the uni stores & was puzzled that it wasn't triangular like the symbol on the circuit diagram.

[DealorNoDeal]

Deducing that the reason a battery drops voltage under load is due to internal resistance

Same for any voltage source, even the National Grid! Otherwise, you could get nearly infinite current.

I=V/R; 12.6/≈0=≈∞

[d000hg]

most if it is GCSE Resistors in series - Calculating resistance - CCEA - GCSE Physics (Single Science) Revision - CCEA - BBC Bitesize

I did it at A level as there was no electronics O' level in my school. It's great simple practical knowledge.

Knowing the formulae and understanding the principles is trivial GCSE stuff. Deducing that the reason a battery drops voltage under load is due to internal resistance very definitely isn't GCSE level... obvious once you know of course.

[DealorNoDeal]

I remember doing a practical at Uni, where you had to design an op-amp circuit. Once you'd calculated the required component values, you went and collected them from the stores technician. One guy couldn't understand the bemused look on the technician's face when he asked for a several Farad capacitor.

[WTFH]

Look on the bright side, you did your A level in SI units.

Back in the dawn of time, mine included Imperial, CGS, and MKS units.

Who'd have thunk that the CGS unit of capacitance was the centimetre.

Not something that has ever proved of any use for me to know.

You think that's bad?
Do you know how many Foals of power are required to move a Firkin along a Furlong in a Fortnight?

[DoctorStrangelove]

most if it is GCSE Resistors in series - Calculating resistance - CCEA - GCSE Physics (Single Science) Revision - CCEA - BBC Bitesize

I did it at A level as there was no electronics O' level in my school. It's great simple practical knowledge.

Look on the bright side, you did your A level in SI units.

Back in the dawn of time, mine included Imperial, CGS, and MKS units.

Who'd have thunk that the CGS unit of capacitance was the centimetre.

Not something that has ever proved of any use for me to know.

[Paralytic]

most if it is GCSE Resistors in series - Calculating resistance - CCEA - GCSE Physics (Single Science) Revision - CCEA - BBC Bitesize

I did it at A level as there was no electronics O' level in my school. It's great simple practical knowledge.

I do love a bit of humblebragging

[Lance]

Inneresting stuff. Made sense to me.

most if it is GCSE Resistors in series - Calculating resistance - CCEA - GCSE Physics (Single Science) Revision - CCEA - BBC Bitesize

I did it at A level as there was no electronics O' level in my school. It's great simple practical knowledge.

[d000hg]

Inneresting stuff. Made sense to me.

[DoctorStrangelove]

Stone me, that's reminded me of "Sideways Sid"*, an Esteemed Customer with "A First Class Degree in Physics" who, however, wouldn't agree with said diagram & proceeded to argue the hind legs off a donkey that he was right & we were wrong.

Oh, Esteemed Customers, how we loved 'em.

I forget exactly what his argument was but his maths was wrong & our maths gave the Rite Anser.

*So named since he was wall eyed & you could never be sure who he was looking it. <- see wot I did there?

He never got further than the first year IIRC.

[DealorNoDeal]

Not the best diagram but hopefully you get the gist.

[DealorNoDeal]

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?

Imagine an extreme load; a short circuit made of really heavy gauge copper wire. The voltage across the terminals will drop to nearly 0V. Other loads (starter motor, headlights) will just reduce the voltage by less. The reason this happens is, as Lance mentioned previously, the battery has internal impedance. Think of it as though there is a resistor inside the battery.

I don't know why charging leaves the voltage (initially) higher but it's probably a chemistry thing. I imagine other rechargeable batteries do this, not just lead acid.

The voltage drops more for a flat/dying battery because the internal impedance has gone up.

ps. Lance beat me to it.

[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.

[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?

[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.