Re: Charging a battery

"hls" <hls@xxxxxxxxxx> writes:

"Thomas Tornblom" <thomas@xxxxxx> wrote in message
It is still in effect. The secondary winding has a very low
DC-resistance and you will feed the voltage in the battery through one
or two diodes (half/full wave rectifier) and through the secondary
winding, which will be almost a dead short.

Having the charger plugged in will theoretically double the voltage as
you will then have both the charger output voltage and the battery
voltage in the circuit. The charger output current will most likely be
much less than the current from the battery, so it will not make much
difference.

Let's forget about the charger being plugged in.. This was a stipulation in
the original post, and perhaps shouldnt even be a part of this analysis.

If the battery is hooked up INCORRECTLY (positive terminal to charger
negative)
then the maximum current is a function of

(transformer secondary resistance) plus (diode impedance at the
voltage employed) plus
any other resistances in the circuit.

We did not stipulate that the battery had to be hooked up correctly.
If the impedance is
only 12 ohms, effectively, we have limited the discharge current to
about one amp.

The DC resistance in the secondary winding is much less than 12
ohms. I can't easily check the actual DC resistance of a charger
transformer, but I just checked the DC resistance of a 42V 1.5A
secondary winding and it measures about 1.4 ohms. Interpolating this
to 12V 1.5A gives a resistance of around 0.4 ohms, and if you step
this up to say 4.5A, then the DC resistance becomes around 0.13 ohms.

Say you have a full wave rectifier and 12V battery, then the current
theoretically becomes (12 - (2*0.7))/0.13 = 81A, hardly "a small bit
of current". The leads has resistance and the diodes will have a higher
forward voltage at high current, so the current will be lower, but it
will still be on the order of tens of amps.


In my original post, I specified that :
No, it will not normally cause a battery to discharge quickly, especially IF
you have red to positive, and black to negative

In your original post you also stipulated:
---
If you had those reversed, then you might draw a small bit of current.
---

Which is clearly wrong, it will draw a fair amount of current


Now, if you want to continue to speculate about odd combinations, you could
also add (1) diode is shorted, (2) capacitors, if they exist, are
shorted, etc
ad nauseum.

I dont doubt you abilities with electronics, nor mine. We could teach
a f***ing
course with lesser problems.

.

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