🚀 go-pugleaf

In article <1992Dec22.190416.26827@ucsu.Colorado.EDU>, barr@tramp.Colorado.EDU (BARR DOUG) writes:
|> I notice when charging my car battery that the amps goes up to about
|> 5 when battery is dead and gradually goes down to nothing as the battery
|> is charged. How does a battery charger sense this? The internal resistance
|> of the battery? (but no, becuase that goes down as the charge goes up,
|> doesn't it?). What sort of sensing circuit is used to reduce the amperage?
|> Anyone know?

None at all.

What you basically have is:

	---[int R1]--o---[int R2]--
        |                         |
        |                         |
       (v) Charger               ---
        |     14v               ----- Battery
        |                         |
        |                         |
        -------------o-------------

Your charger, with no load applied, outputs about 14 volts,
it has some internal resistance (int R1), so as you draw current from
it, some voltage is dropped across that resistance.

A fully charged battery is about 13.2 volts, and if slightly
overcharged, about 14 volts. The battery also has some
internal resistance (int R2), but usualy negligable compared to the
internal resistance of the charger.

An uncharged battery has a voltage of about 10.8 volts.

So, we can get an idea of the value of R1 - since you say that
an uncharged battery draws about 5 amps:

	14 - 10.8 = 3.2;  3.2/5 = 0.64 ohms

So, your charger internal resistance is about 0.64 ohms.
When you draw 5 amps from it, there will be 3.2 volts
dropped across it, so the voltage across the battery will
be 10.8 volts.

As the battery charges, the voltage across it increases.
When it is up to 12 volts, the charging current will
be:

	14 - 12 = 2; 2/0.64 = 3.125 amps

When the battery is fully charged, the current will be:

	14 - 13.2 = 0.8; 0.8/0.64 = 1.25 amps.

Once fully charged, the voltage across the battery rises
quickly, and once it is at 14 volts, the current is:

	14 - 14 = 0; 0/0.64 = 0 amps

Philip