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Car Forum / Driving, Maintenance, Tuning / Maintenance and Repair / August 2007Charging 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 > 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. >> I've always wondered about this. Will leaving a car battery >> connected to a charger that is not plugged in discharge the battery? [quoted text clipped - 5 lines] > 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. [quoted text clipped - 3 lines] > 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 [quoted text clipped - 4 lines] > 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. > >> I've always wondered about this. Will leaving a car battery > >> connected to a charger that is not plugged in discharge the battery? [quoted text clipped - 46 lines] > 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.  SignatureDon Bruder - dakidd@sonic.net - 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 "Don Bruder" <dakidd@sonic.net> 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 [quoted text clipped - 8 lines] > 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. > One thing I've often wondered in connection with > batteries/chargers/alternators: What kind and quantity load does the [quoted text clipped - 5 lines] > -- > Don Bruder - dak...@sonic.net 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. On Aug 26, 9:44 am, Don Stauffer in Minnesota <stauf...@usfamily.net> wrote: > > One thing I've often wondered in connection with > > batteries/chargers/alternators: What kind and quantity load does the [quoted text clipped - 10 lines] > other. The value of all three circuit elements vary during the > charge. A battery has an internal resistance of considerably less than one ohm, and if we work this into a formula using the voltage differential between the alternator's output and the battery's voltage, we can come up with a current flow. This site http://www.flex.com/~kalepa/desulf.htm specifies a 0.1 ohm internal resistance for a fully-charged battery, and this resistance will increase somewhat as the battery discharges. Supposing an internal resistance of 0.15 ohms after the engine starts (probably too high an assumption), and a battery voltage of 12 volts (somewhat discharged from its 12.6 charged state) we could see a differential of 2.5 volts between the battery and the alternator charging voltage of 14, which would give us a charging current flow of 16.7 amps. If the internal resistance increased less than I have assumed, the current would be higher. Another site: http://www.buchmann.ca/article25-page1.asp Dan >> 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 [quoted text clipped - 17 lines] > 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 [quoted text clipped - 8 lines] > 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. >>> If you had those reversed, then you might draw a small bit of current. >> [quoted text clipped - 8 lines] > 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" <thomas@Hax.SE> 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 [quoted text clipped - 6 lines] > 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. > "Thomas Tornblom" <thomas@Hax.SE> wrote in message >> It is still in effect. The secondary winding has a very low [quoted text clipped - 23 lines] > 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 [quoted text clipped - 4 lines] > a f***ing > course with lesser problems. > Which is clearly wrong, it will draw a fair amount of current Easy enough to find out. In the course of the day, I will hook mine up backwards, and unplugged, and find out what actually happens. No need to play Ohms laws games when we can measure a typical application. > Which is clearly wrong, it will draw a fair amount of current In fact, I connected an old (expendable) VOM in series with the unpowered battery charger and hooked the charger up + to -, - to +. When I touched the probes to the terminals, I got an instantaneous reading of 16 amps, which is indeed a fair amount of current. More that I would have thought. Oddly, there was no sparking at the terminals, no visual observation of that much current draw. In any case, that much current would clearly draw your battery down in short order. When connected in the conventional and correct way, current draw is negligible. > > Which is clearly wrong, it will draw a fair amount of current > [quoted text clipped - 7 lines] > Oddly, there was no sparking at the terminals, no visual observation of that > much current draw. It didn't trip a circuit breaker inside the charger? > In any case, that much current would clearly draw your battery down in short > order. Many chargers come with a circuit beaker that will stop the current after a few seconds. It may take a minute or so for it to reset and then should trip again after a second or 2 (the second time should be faster than the first because it is already warm). At that rate it could take a while to discharge the battery. -jim > When connected in the conventional and correct way, current draw is > negligible. >> > Which is clearly wrong, it will draw a fair amount of current >> [quoted text clipped - 23 lines] > > -jim I only left it connected for seconds. I suspect it would have tripped the breaker in short order. My observations were, of course, only on this one unit, and I suspect that designs vary with price and complexity of the charger. I was a little surprised that I didnt draw sparks. Maybe I should have already discarded that VOM :>) |
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