Author |
Message |
Matt
Rating: N/A Votes: 0 (Vote!) | Posted on Sunday, July 31, 2005 - 2:03 pm: | |
I have a 3000 W. Heart inverter with the 3 stage battery charger and a 130 amp alt to charge the house batteries. Question if Im running the gen alt the inverter is charging and also the alt on the eng. is there a chance that I might do damage to the alt or inverter since they both would be charging the house batteries. |
Jim (Jim_in_california)
Rating: N/A Votes: 0 (Vote!) | Posted on Sunday, July 31, 2005 - 2:12 pm: | |
I'm pretty sure you CAN run multiple chargers at once like that but I'm not certain in this case. Anybody with a solar array is always charging from that source and occasionally pumping in from either alternator or genset/charger (or both). Even if it's OK in your case (and more knowledgable people should chime in here) you might however want to monitor just how much juice you're pumping into the battery bank via a good amp meter like the Bogart Trimetric, esp. if your battery investment is substancial. |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Sunday, July 31, 2005 - 2:53 pm: | |
If I understand your question, you are running a gen set supplying 120 volts to the inverter charger and at the same time your main engine is running with its alternator conected to the batteries. I am not clear if you also have the genset alternator connected to the same batteries. The source of charging current will be from the highest voltage. If your main engine alternator is set to a higher voltage than the inverter three stage charger the inverter will not do anything. Coversely, if the three stage charger is set to a higher voltage it will do the charging. Since the three stage charger supplies three different voltages it will be difficult to get a proper charge out of it unless you isolate the batteries from the main engine alternator when you have the genset running. Another alternative would be to put a three stage regulator on the main engine alternator and turn off the inverter when you are driving and running the genset. |
James Maxwell (Jmaxwell)
Rating: N/A Votes: 0 (Vote!) | Posted on Sunday, July 31, 2005 - 7:18 pm: | |
Multiple simultaneous charging sources into the same batteries, using different regulators for each source, tends to confuse all the regulators and cause erratic charging; the regulators tend to fight with one another. I would install some means of disconnecting the 130w eng. alt from the house batteries while operating the genset. Let the engine alt take care of the chassis batts and the genset take care of the house batts when the engine is running. You can accomplish this by various means, however, I prefer a power solenoid on a switch that disconnects the 2 battery systems when both charging sources are outputting current. |
Jerry Liebler (Jerry_liebler)
Rating: N/A Votes: 0 (Vote!) | Posted on Sunday, July 31, 2005 - 10:54 pm: | |
James, Stan had it right, almost. Respectfully, you are simply wrong! The regulators sense the battery voltage and control their respective charging source, they do not get confused. Multiple charging sources are fine, just don't count on all of them working at once as the batterys near fully charged only the one with the highest voltage setting will be doing any charging. Below each regulator's cutoff voltage it's charging source will be delivering up to it's rated current. This means that significantly discharged batterys will be charged by all connected sources, this is the part Stan didn't quite have right. Absolutely nothing erratic about it. I say this as a BSEE with over 30 years in power electronics. Regards Jerry 4107 1120 |
Tom Caffrey (Pvcces)
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 1:39 am: | |
Matt, sometimes, mostly a long time ago, regulators would behave erratically, but mostly, that's a thing of the past. If it happens that you get some misbehavior, just turn off the inverter's charger from the remote panel. If you don't get any misbehavior, don't worry about it. You DO have a remote on your inverter, dont you? Tom Caffrey PD4106-2576 Suncatcher |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 8:29 am: | |
Jerry: I would appreciate a better explanation of how an automotive regulator works. My experience with the Delco electronic regulator on the DN 50 alternator is that it is only a current regulator. Once the output voltage is preset it varies the field current on the alternator (which varies the output voltage) according to the voltage difference between the batteries and the setpoint. If in Matt's example he has discharged batteries, then three stage regualtor on the inverter will go into high charge (high voltage) and the main engine alternator would shut down field current because it is seeing a battery voltage above the setpoint. When the three stage regulator drops down to the middle rate it will be somewhere close to the amin engine alternator set point and it may or may not charge depending on the difference between the inverter charge voltage and the alternator set point. I would really appreciate if you could clear up my missunderstanding on the regulator. A really good read is MCI's service manual on how to test the alternator and regulator. |
Jerry Liebler (Jerry_liebler)
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 11:18 am: | |
Stan, The regulator on my 4107's DN50 (stock) senses the battery voltage and controls the alternator's field current to produce an output voltage at the 'set point'. If the load is great enough (low enough resistance) The field current will be all it can be and the alternator will produce as much curent as it can (300 amps if the engine speed is at or above fast idle). As the load voltage rises, the regulator will reduce the field curent maintaining the output voltage at the set point. In Matt's example, when he has discharged batterys the three stage regulator will go into high charge but the batterys will not initially let the voltage rise, they will take it's full rated current and the voltage rise will be small, this is the 'bulk' stage of the three, during this stage the alternator will contribute current (up to 300 amps) to the charging process. As the batterys become charged the voltage will rise once their voltage exceeds the alternator regulator's set point it's contribution to charging will stop and the three stage charger will continue charging through the absorption stage. Once the three stage charger thinks the batterys are charged (determined either by time or curent drop)it'll drop it's voltage to the float level, now either it or the alternator will hold them at about this level. I believe some of the confusion is due to the fact that both of these carging sources are inherently current limited and will deliver their limit current to well discharged batterys. The reason for three stage chargers and the 'absorbtion' stage is to shorten the time to fully charge the batterys, at the beginning of the absorbtion stage (when the voltage rises above the float level) the batterys are only about 80% charged without the voltage boost above float it would take about 24 hours at the float voltage to get them to 95% charged and theoretically forever to fully charge them. Hope this is clear if not I'll gladly try some more. Regards Jerry 4107 1120 |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 12:41 pm: | |
Jerry: Thank you. Just one further question. If a charging system (regardless of type) is putting out 15 volts into discharged batteries (10.5 volts), where does the voltage drop occur? Is it not inside the battery? |
Richard Bowyer (Drivingmisslazy)
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 1:31 pm: | |
Stan, my opinion is that no charger will be putting out that voltage (15) into a fully discharged battery. The open circuit voltage, before you connect the charger may well be 15 volts or higher. As soon as you connect it to the battery, the voltage you measure on the terminals will be the just above the battery voltage. The internal impedance in the battery charger will reduce the voltage to the point that you do not overload the charger. At least that is what I have found on most chargers. That is the long answer. The short answer, in my opinion, is that the charger will only put out enough voltage to maintain the rated output current into the impedance of the battery. As the battery comes up to charge, the internal impedance increases, and the charger voltage will increase proportionally to maintain full current charging, up to the capacity of the charger. Richard |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 5:41 pm: | |
Richard: You are correct that the internal impedance will keep a battery charger from self destructing if you exceed its maximum rating. On a DN50 you can put out about 300 amps on a 12 volt system before that internal impedance kicks in. The state of charge of the batteries is not involved when you adjust the regulator. As long as the batteries are not absorbing more than 300 amps you just adjust the regulator for the desired voltage and it will put out that voltage. On a DN50 with its Delco regulaotr, if you set if for 14 volts it should never vary once the regulator is temperature stabilized. A regulated charger (such as an inverter with three stage charger) will operate the same way. It will put out its regulated voltage until you exceed its capacity. This does not apply to low cost chargers sold at the auto parts store. They are just cheap transformers with a rectifier. The cheap transformer is self protected by not putting enough iron in the core. When you approach maximum current the core saturates and the voltage drops. |
Jerry Liebler (Jerry_liebler)
Rating: N/A Votes: 0 (Vote!) | Posted on Monday, August 01, 2005 - 10:27 pm: | |
Stan, If a charger is puting out 15 volts when connected to a discharged battery the battery is seriously sulphated and probably scrap. If you mean the open circuit voltage of the charger is 15 volts and when it's connected to the battery the battery shows 10.5 and the question is where did the 4.5 volt difference go. The answer is the difference is being 'dropped' accross the effective resistance of the charger. The charger's effective resistance can be determined by such a test and measuring the current going into the battery. Dividing the volts by the amps gives the ohms. Regards Jerry 4107 1120 |
Tom Caffrey (Pvcces)
Rating: N/A Votes: 0 (Vote!) | Posted on Tuesday, August 02, 2005 - 1:55 am: | |
Stan, it is not usual in automotive applications to use any current control on an alternator. They are voltage controlled only. One reason for this is they are self limiting; no matter how fast you spin them, they will not put out more current than their maximum rating. Even if you short the output, as far as I know. For what it's worth. Tom Caffrey PD4106-2576 Suncatcher |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Tuesday, August 02, 2005 - 8:18 am: | |
Jerry: I think you are thinking about low cost unregulated chargers. A DN50 puts out the exact voltage that you set it for, on a continuous basis. Think of the battery as a series string of resistors. Some voltage drop occurs across each plate while it is being charged. The full output voltage of the alternator appears at the battery posts. See my above post to Richard. You can adjust the regulator on a DN50 with a battery in any state of charge as long as it is not absorbing more than the maximum capacity of the alternator. The common electrical theory is that voltage drop occurs in the load, which is the battery in this case. If a 15 volt charger and 10.5 volt battery gives you a problem, then think of a 13.8 volt alternator and a 11.5 volt battery. Are you seriously suggesting that the alternator output drops to 11.5 volts when it is charging the battery? |
Jerry Liebler (Jerry_liebler)
Rating: N/A Votes: 0 (Vote!) | Posted on Tuesday, August 02, 2005 - 5:16 pm: | |
Stan, Absolutely yes a good 12 volt pair of 8D batterys will have an equivalent resistance of about 5/1000 ohms so 300 amps will raise their output about 1.5 volts. If no load or charge they produce 11.5 and the alternator starts charging them and is delivering 300 amps the battery voltage will rise to 13 volts and the regulator is maxed out. My 4 L16s have an equivalent resistance of about 2/1000 ohm so 300 amps into them will raise their terminal voltage all of 0.6 volts. Your statment about adjusting the regulator is correct BUT how do you know the alternator isn't delivering all it can. much better to adjust the regulator with fully charged batterys then you know it's not delivering all it can. Regards Jerry 4107 1120 |
Stan
Rating: N/A Votes: 0 (Vote!) | Posted on Tuesday, August 02, 2005 - 8:23 pm: | |
Those are interesting numbers Jerry. Do you have an ammeter on your bus? Even after cranking until batteries were unable to crank (theoretically this is 10.5 volts per battery) I have never seen the ammeter go anywhere near 250 amps (24 volt bus) when I started the engine. I think the internal impedance of a battery at 11.5 volts is high enough that the 2.5 volt differential will not push 300 amps through it. I don't have the chart in front of me, but I think a DN50 (on 24 volts) reaches full output at 9 amps field current. It is part of the test procedure on the alternator. |
Jerry Liebler (Jerry_liebler)
Rating: N/A Votes: 0 (Vote!) | Posted on Tuesday, August 02, 2005 - 9:42 pm: | |
Stan, No I do not have a conventional ammeter in my bus. I do have an electronic ammeter on my house battery (12 volt). The house battery is inter-tied to the bus alternator through a 200 amp circuit breaker, so I've never seen a 300 amp charge either, I have seen 200 going into the house battery. With a 24 volt system the equivalant resistance of the pair of 8Ds will be 20/1000 ohm (10/1000) each. So with 11.5 volt battery about 200 amps sounds right. There is also some resistance in wires and connections. Did you know that the only difference between a 12 volt and 24 volt DN50 is the field. The stators, rotors, diodes are all the same. 9 amps sounds like the 12 volt field current, I think the 24 is more like 5. Regards Jerry |