Charging a battery

I've always wondered about this. Will leaving a car battery connected to a
charger that is not plugged in discharge the battery?

Thanks,
Dennis

"Dennis" > wrote in message
. ..
> I've always wondered about this. Will leaving a car battery connected to a
> charger that is not plugged in discharge the battery?
>
> Thanks,
> Dennis

No, it will not normally cause a battery to discharge quickly, especially IF
you
have red to positive, and black to negative.

If you had those reversed, then you might draw a small bit of current.

I tested my charger and found greater than 20 megohms when red to positive
was connected.

In the reverse direction, varying apparent resistance was seen, depending
upon the
range of the ohmmeter. (And I will suggest you figure out why this would
be.)

If the lowest observed resistance was correct (and it isn't, but lets say
about 500
ohms), then you would drain the battery at a rate of about 24
milliamps...not much.

Other chargers may be different. The capacitor drain resistors, if they
exist, and
the forward breakdown voltage of the diode, could vary somewhat.

"hls" > writes:

> "Dennis" > wrote in message
> . ..
>> I've always wondered about this. Will leaving a car battery
>> connected to a charger that is not plugged in discharge the battery?
>>
>> Thanks,
>> Dennis
>
> No, it will not normally cause a battery to discharge quickly,
> especially IF you
> have red to positive, and black to negative.

Correct.

>
> If you had those reversed, then you might draw a small bit of current.

You will draw a lot of current, and if the charger has a fuse on the
secondary side, it will likely trip, at least if it is a simple charger.

>
> I tested my charger and found greater than 20 megohms when red to positive
> was connected.
>
> In the reverse direction, varying apparent resistance was seen,
> depending upon the
> range of the ohmmeter. (And I will suggest you figure out why this
> would be.)

You can't reliably use an ohmmeter to measure the resistance across a
rectifier, which is what you are doing here. The rectifier/diodes has
a forward voltage drop, on the order of 0.7V per diode (silicon
diodes) and the ohm meter puts out a voltage/current on the leads, and
measures the voltage drop across the leads, and calculates the
resistance from that. Depending on the voltage the meter uses, you can
get about any resistance reading, and the higher the voltage the meter
uses, the lower the resistance reading will be, as the diodes starts
conducting when the voltage reaches above the forward drop.

The varying readings you got was for this reason. The meter uses
different voltage and/or current for the different ranges.

>
> If the lowest observed resistance was correct (and it isn't, but lets
> say about 500
> ohms), then you would drain the battery at a rate of about 24
> milliamps...not much.
>
> Other chargers may be different. The capacitor drain resistors, if
> they exist, and
> the forward breakdown voltage of the diode, could vary somewhat.

There are normally no capacitors in cheap chargers. There is no need
for smooth voltage/current to charge a battery. The battery will do
that nicely for you.

In article >, Thomas Tornblom >
wrote:

> "hls" > writes:
>
> > "Dennis" > wrote in message
> > . ..
> >> I've always wondered about this. Will leaving a car battery
> >> connected to a charger that is not plugged in discharge the battery?
> >>
> >> Thanks,
> >> Dennis
> >
> > No, it will not normally cause a battery to discharge quickly,
> > especially IF you
> > have red to positive, and black to negative.
>
> Correct.
>
> >
> > If you had those reversed, then you might draw a small bit of current.
>
> You will draw a lot of current, and if the charger has a fuse on the
> secondary side, it will likely trip, at least if it is a simple charger.
>
> >
> > I tested my charger and found greater than 20 megohms when red to positive
> > was connected.
> >
> > In the reverse direction, varying apparent resistance was seen,
> > depending upon the
> > range of the ohmmeter. (And I will suggest you figure out why this
> > would be.)
>
> You can't reliably use an ohmmeter to measure the resistance across a
> rectifier, which is what you are doing here. The rectifier/diodes has
> a forward voltage drop, on the order of 0.7V per diode (silicon
> diodes) and the ohm meter puts out a voltage/current on the leads, and
> measures the voltage drop across the leads, and calculates the
> resistance from that. Depending on the voltage the meter uses, you can
> get about any resistance reading, and the higher the voltage the meter
> uses, the lower the resistance reading will be, as the diodes starts
> conducting when the voltage reaches above the forward drop.
>
> The varying readings you got was for this reason. The meter uses
> different voltage and/or current for the different ranges.
>
> >
> > If the lowest observed resistance was correct (and it isn't, but lets
> > say about 500
> > ohms), then you would drain the battery at a rate of about 24
> > milliamps...not much.
> >
> > Other chargers may be different. The capacitor drain resistors, if
> > they exist, and
> > the forward breakdown voltage of the diode, could vary somewhat.
>
> There are normally no capacitors in cheap chargers. There is no need
> for smooth voltage/current to charge a battery. The battery will do
> that nicely for you.


Yep... Lots of old motorbikes (and no doubt, at least some current ones)
just pumped pulsed DC straight off the half-wave rectifiers into the
system, and relied on the battery to smooth it out. And smooth it would
be.

One thing I've often wondered in connection with
batteries/chargers/alternators: What kind and quantity load does the
battery present to the alternator? I'm sure it varies, both from battery
to battery, and depending on the current state of charge, but *IN
GENERAL*, does a good battery look like, for instance, a 500 ohm
resistance, as far as the alternator is concerned?

I've wanted to load-test an alternator, but don't know how I should load
it to get something like an accurate result.

Obviously, they're loaded somehow when you put them on the test machine
at Autozone or whatever, but the "somehow" is a mystery to me.

--
Don Bruder - - If your "From:" address isn't on my whitelist,
or the subject of the message doesn't contain the exact text "PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without my
ever knowing it arrived. Sorry... <http://www.sonic.net/~dakidd> for more info

"Thomas Tornblom" > wrote in message
...

>> If you had those reversed, then you might draw a small bit of current.
>
> You will draw a lot of current, and if the charger has a fuse on the
> secondary side, it will likely trip, at least if it is a simple charger.

He specifically qualified this scenario by saying that the battery charger
was unplugged... If you reverse the connections, then you would have a
current discharge related to the the effective impedance of the diode, any
current limiting resistor, and the transformer secondary impedance.

Now, if it were plugged in (which is NOT the case) your assessment would
certainly be in effect.



>> I tested my charger and found greater than 20 megohms when red to
>> positive
>> was connected.
>>
>> In the reverse direction, varying apparent resistance was seen,
>> depending upon the
>> range of the ohmmeter. (And I will suggest you figure out why this
>> would be.)
>
> You can't reliably use an ohmmeter to measure the resistance across a
> rectifier, which is what you are doing here. The rectifier/diodes has
> a forward voltage drop, on the order of 0.7V per diode (silicon
> diodes) and the ohm meter puts out a voltage/current on the leads, and
> measures the voltage drop across the leads, and calculates the
> resistance from that. Depending on the voltage the meter uses, you can
> get about any resistance reading, and the higher the voltage the meter
> uses, the lower the resistance reading will be, as the diodes starts
> conducting when the voltage reaches above the forward drop.
>
> The varying readings you got was for this reason. The meter uses
> different voltage and/or current for the different ranges.

Not news to me. That is why I suggested that the OP look into this.



>> If the lowest observed resistance was correct (and it isn't, but lets
>> say about 500
>> ohms), then you would drain the battery at a rate of about 24
>> milliamps...not much.
>>
>> Other chargers may be different. The capacitor drain resistors, if
>> they exist, and
>> the forward breakdown voltage of the diode, could vary somewhat.
>
> There are normally no capacitors in cheap chargers. There is no need
> for smooth voltage/current to charge a battery. The battery will do
> that nicely for you.

True, there are often not capacitors in cheap chargers, or drain resistors,
or current limiting resistors. In fact, you may not even know what sort of
rectifier you have. A lot of the older and cheaper ones had seleniums..
Not so much now.

You and I dont disagree. In fact, we agree totally.

His question was a bit too open.. Look back at it, and consider the worst
case
scenarios...he didnt even say the charger was operating properly.. Shorted
rectifiers,
wrong connections, and other things could enter into his very open scenario.

"Don Bruder" > wrote in message news:46d0a68f$0$14137
> One thing I've often wondered in connection with
> batteries/chargers/alternators: What kind and quantity load does the
> battery present to the alternator? I'm sure it varies, both from battery
> to battery, and depending on the current state of charge, but *IN
> GENERAL*, does a good battery look like, for instance, a 500 ohm
> resistance, as far as the alternator is concerned?
>
> I've wanted to load-test an alternator, but don't know how I should load
> it to get something like an accurate result.
>
> Obviously, they're loaded somehow when you put them on the test machine
> at Autozone or whatever, but the "somehow" is a mystery to me.

It appears as a varying load. Consider that you have a varying electrolyte,
varying polarization, varying resistive component.

The test rigs to show battery function under load make some assumptions.
In my experience, the (cheap) tests are not always correct.

On Aug 25, 5:00 pm, Don Bruder > wrote:

>
> One thing I've often wondered in connection with
> batteries/chargers/alternators: What kind and quantity load does the
> battery present to the alternator? I'm sure it varies, both from battery
> to battery, and depending on the current state of charge, but *IN
> GENERAL*, does a good battery look like, for instance, a 500 ohm
> resistance, as far as the alternator is concerned?
>
> --
> Don Bruder -

A battery appears to the charger as two resistors and a battery. The
"battery" is in series with one resistor, and in parallel with the
other. The value of all three circuit elements vary during the
charge.

"hls" > writes:

> "Thomas Tornblom" > wrote in message
> ...
>
>>> If you had those reversed, then you might draw a small bit of current.
>>
>> You will draw a lot of current, and if the charger has a fuse on the
>> secondary side, it will likely trip, at least if it is a simple charger.
>
> He specifically qualified this scenario by saying that the battery charger
> was unplugged... If you reverse the connections, then you would have a
> current discharge related to the the effective impedance of the diode, any
> current limiting resistor, and the transformer secondary impedance.
>
> Now, if it were plugged in (which is NOT the case) your assessment would
> certainly be in effect.

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.

"Thomas Tornblom" > 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.

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

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.

"hls" > writes:

> "Thomas Tornblom" > 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.
>