12 volt charging

Good evening to all. Picked up the new jayco van and found out that l dont have a hot wire to trickel charge the battery in the van while driving van has a ST20-II power supply can l run a cable from van battery +/- to anderson plug on car with out damageing STpower supply. thanks to all.
Back Expand Un-Read 0 Moderator

Reply By: Battery Value Pty Ltd - Saturday, Mar 27, 2010 at 19:33

Saturday, Mar 27, 2010 at 19:33
Ron,

if I understand this right, the output of the power supply is hard wired to the second battery.
If you intend to charge this second battery by alternator, that's ok, you're not going to damage the power supply that way.

Best regards, Peter
AnswerID: 410695

Follow Up By: Member - Ron O (VIC) - Saturday, Mar 27, 2010 at 20:26

Saturday, Mar 27, 2010 at 20:26
At the moment only 1 battery in the van and is wired to ST20 unit but will only trickle charge through 240. So if l add second battery conected in parellel and charged through car will it damage STunit just making sure so l dont bugger the unit.thanks Peter
0
FollowupID: 680573

Follow Up By: Battery Value Pty Ltd - Saturday, Mar 27, 2010 at 20:53

Saturday, Mar 27, 2010 at 20:53
no worries Ron,

no probs, you can have your second battery connected in parallel to your car battery and in parallel to your charger.

Just make sure there is an isolation switch, or a diode to prevent the fuse on the second battery from blowing, when cranking.

Best regards, Peter
0
FollowupID: 680582

Follow Up By: Member - Ron O (VIC) - Saturday, Mar 27, 2010 at 21:36

Saturday, Mar 27, 2010 at 21:36
Peter could you give me some guidance on isolation switch or diode never had to do this before
0
FollowupID: 680591

Follow Up By: Battery Value Pty Ltd - Saturday, Mar 27, 2010 at 22:25

Saturday, Mar 27, 2010 at 22:25
Ron,

instead of running the 12V +/- from your starter battery to the Anderson plug directly, put this isolation switch in series with the positive wire.
If you go the more elegant way, you can replace this switch with a solenoid contact. The solenoid is basically a relay with a high contact current rating. Whenever the motor runs, the solenoid is energized and the contact closed. Now you have a direct connection between your starter battery and your second battery via the Anderson plug.
If you turn off the motor, the solenoid also gets turned off and the contact opens. Thus, the next time you start the motor, no (fuse blowing) current can flow from your second battery into the first one and through the starter motor.
Better solenoids have some electronic circuitry in them which delays the power to the solenoid for a few seconds after motor start. This ensures the starter battery receives all available alternator current for a little while before part of it is allowed through the second battery.

Instead of the isolation switch, you could also use an isolation diode which is basically a one way valve for current. If it's wired correctly, it only allows current from the starter battery to the second battery, but not the other way around.
Some diodes add excessive voltage drop which make them a less than ideal solution.

Hope this little primer is of some help.

Best regards, Peter
0
FollowupID: 680603

Follow Up By: Member - Jeremy W (SA) - Sunday, Mar 28, 2010 at 09:12

Sunday, Mar 28, 2010 at 09:12
I use diodes (then switch them out of circuit when charging) but the catch is that you loose 0.7V across the diode. This diode will dissipate power ... if 20A is in circuit then the diode will dissipate 14W and get very hot ( and ultimately fail) if one does not use a heat sink.

Source of cheap high current diodes: I use the diode in a power mosfet.

Part number: IRLB5814PbF ($ 4.72 each). This is the device I use in my battery management switches.

Hope this helps.

Jeremy.



0
FollowupID: 680654

Follow Up By: Battery Value Pty Ltd - Sunday, Mar 28, 2010 at 09:41

Sunday, Mar 28, 2010 at 09:41
Jeremy,

brilliant idea to bypass the blocking diode with a switch during alternator charging - that way you don't have to worry about them dropping voltage at all.
I think the free wheeling diodes in power mosfets are majority carriers with minimal forward voltage - so even if you don't do this bypass trick, their heat dissipation is the least you can have (double up for even less heating).


Best regards, Peter
0
FollowupID: 680665

Follow Up By: Member - Jeremy W (SA) - Sunday, Mar 28, 2010 at 11:44

Sunday, Mar 28, 2010 at 11:44
Hello Peter,

Judging by the characteristics they may be normal PN junctions. PIN diodes will drop a lower voltage but they cost a veritable fortune!! Don't even know if they are available in the currents we routinely require.

Without getting into an arguments with everybody what I do - current design project - is to use multiple batteries of lower capacity and bring them online as the current demand increases, in this way if a battery should fail then I do not loose most of my capacity also I mix and match deep cycle and "normal" batteries.
Those not supplying current to the load are charge managed individually.

Jeremy

0
FollowupID: 680692

Follow Up By: Mike DiD - Sunday, Mar 28, 2010 at 20:16

Sunday, Mar 28, 2010 at 20:16
"the catch is that you loose 0.7V across the diode. "

If you use Schottky Diodes the voltage drop will be closer to 0.4 volts.
0
FollowupID: 680803

Follow Up By: Member - Jeremy W (SA) - Sunday, Mar 28, 2010 at 21:03

Sunday, Mar 28, 2010 at 21:03
Yes, definitely - A Schottky Barrier diode also called a PIN diode because of its construction (P-type Intrinsic N-type).
0
FollowupID: 680817

Reply By: DIO - Sunday, Mar 28, 2010 at 10:03

Sunday, Mar 28, 2010 at 10:03
Check this out LINK
AnswerID: 410744

Sponsored Links