XPost: uk.rec.driving, alt.home.repair, uk.d-i-y
From: xenolith@optusnet.com.au
On 23/6/19 9:24 am, Commander Kinsey wrote:
> On Sat, 22 Jun 2019 11:41:15 +0100, Xeno wrote:
>>>
>>>> the *battery* is the *load*.
>>>
>>> Not when the battery is fully charges and is being charged
>>> with a bench supply that is delivering 4A to the battery.
>>
>> Take the case of an alternator charging a battery at ~4 amps. The
>> battery is the load and it also provides, as part of that function, the
>> reference *voltage* that the alternator *must have* in order to control
>> the output.
>
> I believe the reason Rod mentioned load as being seperate to the battery
> is this thread is about my car, with a fault that draws current from the
> battery, and it's also connected to a bench supply. In this situation,
> we refer to "load" as the faulty alarm system. I originally asked how
> an alternator could distinguish between the battery charging, and a load
> such as your headlamps being on.
>
>>>> In the process of being charged it is using electric current. That
>>>> makes it the load.
>>>
>>> See above.
>>
>> What happens to the charger when you disconnect the power with the
>> battery connected? It should, if designed correctly, shut down since it
>> no longer sees a load. Otherwise it may destroy itself.
Ooops, got that arse about - disconnect the *battery* with the 240V
power still connected.
>
> Bullshit. You're telling me that a bench supply with a battery on the
> output and no 240V input will blow up if it doesn't shut down? Wrong.
The battery is the load. And I'm referring to a charger and the leads to
the battery being disconnected, not the 240V input. The same as if you
open circuit the battery on a running, and charging, alternator. It can
be done but there is a high risk of a spike zapping something. The
battery acts to dampen spikes and it's removal from the circuit exposes
the regulator and any other electronic components to spikes.
> Absolutely wrong. The output end of the supply (which probably ends
> with smoothing capacitors) is just kept at the normal output voltage by
> the battery.
I used the term *charger* for a very good reason.
>
>>>> Even when it is fully charged it will still take a trickle charge
>>>
>>> 4A isnt a trickle charge.
>>
>> That depends entirely on the amp hour rating of the battery.
>> Also, my bench charger will start off at 4 amps, its maximum capacity.
>> As the battery becomes charged, that current will drop down to *1 amp*
>> and, from that point, it will maintain a *trickle charge*.
>
> How fucking big is that battery?! When I charge a 60Ah lead acid at
> 13.8V, it drops to about 150mA when full.
When I was working on trucks, it wasn't unusual to see 4 x 12 Volt, 200
AH batteries wired series-parallel, so yes, 1 amp can be a trickle.
>
>> From Wikipedia;
>> For lead-acid batteries under no load float charging (such as
>> in SLI batteries), trickle charging happens naturally at the
>> end-of-charge, when the lead-acid battery internal resistance
>> to the charging current increases enough to reduce additional
>> charging current to a trickle, hence the name. In such cases,
>> the trickle charging equals the energy expended by the
>> lead-acid battery splitting the water in the electrolyte into
>> hydrogen and oxygen gases
>
> Trouble is you might want to do a bulk charge first, at 14.4V.
>
>> The car alternator regulator is no different. It sees the battery as a
>> load, determines the voltage reference and pumps up its output.
>
> All it can do is supply a certain voltage. 13.8V is safe. If it uses
> 14.4V or higher, it has to know when to back off to 13.8. It cannot do
> that if there is an external current draw like headlamp, as it won't
> know if they're switched on, or if the battery is drawing that current.
The regulator has a Zener diode to *limit* the voltage. An external
current draw will lower the voltage.
>
>> When the
>> regulator sees the battery voltage at the peak setpoint,
>
> You can't tell a battery is full by voltage. You can only tell by it
> drawing less current. The voltage is determined by the charger.
The voltage at the battery will be determined by the battery's *internal
resistance*. As that battery's internal resistance goes higher, as it
will do as it becomes charged, the voltage will go higher and current
will reduce.
>> it too will
>> drop the current to a trickle.
>
> Actually it drops the voltage, to 13.8. I used to have a solar battery
> regulator which had very detailed instructions saying how it worked. It
> took the voltage of the solar cells and altered it up or down a bit to
> suit the battery condition.
>
>> If you add a load, say by turning
>> headlights on, that is in *parallel* to the battery and it will drop the
>> system voltage down a tad. The regulator will see that and pump up the
>> output current appropriately. The current will apportion itself to the
>> *two* loads as appropriate to their individual internal resistances.
>
> But it can't tell the difference between current going to the lights and
> to the battery.
It doesn't need to. All it needs to know is the aggregate current draw,
the total. That will be indicated by the voltage.
--
Xeno
Nothing astonishes Noddy so much as common sense and plain dealing.
(with apologies to Ralph Waldo Emerson)
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
|
