Need to disconnect battery before charging?1

The message from J0EY contains these words:

No. Just don't leave it on charge for ever.

Read the small print on the label. 11 amps in large letters, but 7.5 actual. And you'll be lucky to see more than 5 in reality - and that only with a good but discharged battery.

They talk about RMS amps - as a DC output. Wankers. They should be taken to the trades description people.

It's fine for any car battery.

The message from "Dave Plowman (News)" contains these words:

Well, to be fair, you can measure DC with RMS if you really want to - it's just not really worth it!

Maybe it's unsmoothed output ? Google suggests opinions vary on whether lack of a decent DC supply is good or bad for recharging lead-acids.

Apart from leisure batteries, most car chargers are emergency-use only. I would guess normal battery life is affected about 99.999% by the performance of the car's charging circuit.

You mean the ripple still present?

Well, there's no smoothing on a car alternator either. Higher frequency ripple, though.

It doesn't matter, the average amperage (x time) tells you how quickly the battery will charge, the rms just tells you how hot it will get.

RMS (root mean square) is lower than a peak value - obviously. But the Halford charger quotes the higher figure as RMS - and it's the large one on the front. Only the small print says the actual output to the battery is considerably less. So purely there to mislead.

Also, claiming a 7.5 amp charger is only suitable for large cars is rubbish - which car has an alternator with a peak output that low? Even dynamo equipped cars could manage 30 amps or so - and with much cruder regulation.

The Halfords charger in question is actually pretty good and gentle with batteries - it's just the misleading spec that annoys me.

Me no comprende. I meant perhaps the 14V or whatever has just been full-wave rectified (hope that's the right term, my electronics is very rusty). This, I think, inverts the negative-going wave so you get bump-bump-bump rather than bump-flat-bump (never mind rusty, it may be non-existent). Calculating the RMS of this seems reasonable.

I might guess that a hefty lead-acid cell acts as a fairly decent capacitor, removing the need for smoothing...

... which would seem to be the case. Frequency will be variable but not far from 50Hz I would have thought. (And probably starts at a lower value.)

All you ever needed to know about car batteries can be found here:

in news: snipped-for-privacy@davenoise.co.uk, "Dave Plowman (News)" slurred :

Why? The regulation will undoubtedly be terrible, and there will be significant ripple, so RMS is a perfectly valid measure for the output. There is anyway no fundamental reason, other than it being redundant, why you shouldn't quote RMS for DC.

Agreed.

The message from "Dave Plowman (News)" contains these words:

Bit like the Lottery claiming that your chances of winning are increased by 100% if you actually buy a ticket.

Duh.

The message from John Laird contains these words:

Though unecessary, just a plain average would do. The RMS is only done to cope with AC that crosses zero which would otherwise average to zero.

Please explain how RMS has got anything to do with DC?

You may be right. Doesn't using RMS give you the power properly though, if you then use that value (as current or voltage as appropriate) and do the I*I*R or V*V/R calculation naively ? A simple voltage or current average won't yield the same (correct) answer.

The message from John Laird contains these words:

A simple average of DC with ripple is is the same as the RMS of DC with ripple. The only reason for taking the root of the mean of the square is to sidestep the problem of the average voltage of AC being zero. In effect, it rectifies the signal mathematically instead of with diodes, but that's /all/ it does.

You could just quote mean magnitude then. I maintain it is useful to compute RMS voltage as this will yield the correct power into resistive loads if plugged into the usual V*V/R equation. For AC, DC, DC with a ripple, whatever.

I haven't quite worked out in my head whether it has the same relevance in charging circuits, as I suspect you wish to compute the integral of current over time instead. Maybe mean voltage is better in that case.

The message from John Laird contains these words:

RMS /is/ mean voltage. It's just that for AC you have to use a mathematical trick to get round the mean voltage of AC being zero.

IIRC, RMS on a standard sine wave AC signal equates to DC in terms of heating effect.

'taint. Consider a simple case - a square wave alternating 0.5v and 1.5v for equal time periods. Arithmetic mean is 1.0v. RMS is 1.12v.

(and as others have pointed out, if you're trying to work out eg heating power, RMS is the correct thing to use).

cheers, clive