David, Lee sent me also a direct mail and there is a difference - he indicated that relay 1 needs *TWO* small "NO" contacts, because indeed with only 1 contact the relay 2 will keep its own coil powered from the battery via its own contact. So, the mail from Lee indicates that relay coil 2 gets power from the output of the charger *through* the second NO contact of relay 1. When you pull the AC power to the charger, relay 1 will lose coil power and disconnect both its contacts: the charger is still connected to the battery via relay 2 contact, but relay 2 coil is interrupted from the charger output by relay 1 second contact, so relay 2 also drops and the battery is totally isolated (both relay 1 and 2 contacts open, so no leakage current) Cor.
On Tue, Mar 22, 2022 at 11:40 AM David Nelson via EV <ev@lists.evdl.org> wrote: > > On Sun, Mar 20, 2022 at 3:34 PM EV List Lackey via EV <ev@lists.evdl.org> > wrote: > > > > It sounds like your charger has an electrolytic capacitor across its output. > > > > Could you add a precharging circuit across the contactor? > > I thought of this but wanted something as simple as possible. Of > course, a diode has heat dissipation to consider and who knows how > long it will last. I have more responses about this idea after Lee > Hart's email below. > > I like Bill D's suggestion of the 5-pin connector for its simplicity. > I just haven't found a 5 pin connector yet so I think I'll stay with > the NEMA L6-20 I have and work on some sort of pre-charge. > > On Sun, Mar 20, 2022 at 5:11 PM Alan Arrison via EV <ev@lists.evdl.org> wrote: > > > > Wouldn't the voltage drop confuse the charger, especially when it gets > > to the constant voltage point near end of charge? > > I need to verify what voltage the internal BMS starts to balance but I > think it is still before the ending voltage the battery will see. > > On Sun, Mar 20, 2022 at 5:54 PM Matt Lacey via EV <ev@lists.evdl.org> wrote: > > > > The charger won't know about the voltage drop, it would just reduce the > > CV voltage by 0.7V or so > > > > The bigger problem is the charger may not start, if there's no voltage > > present on it's output. > > Many of these batteries' BMS will cut off the output so the voltage > across the terminals is 0V. Because of this, the chargers need to > attempt to charge even with 0V on the output. This is why I purchased > the Dakota Lithium charger. I now wish I had hunted for a soft-start > charger. > > On Sun, Mar 20, 2022 at 9:32 PM Cor van de Water > <cor.vandewa...@gmail.com> wrote: > > > > David, > > Why are you unplugging the charger? > > I'm mounting the charger in the golf cart. It doesn't always go back > to the same place on the campus it is used on so it needs to be able > to charge wherever it is at. I have an extension cord with a NEMA > L6-20 plug on the end so the extension cord isn't "borrowed" for other > uses and can remain in the golf cart, and so the plug can handle more > plug-unplug cycles. I also don't want a standard extension cord used > in case the end is worn out and causes something to melt or catch > fire. The smaller battery than the lead-acid pack leaves a nice > storage space under the seat for it. Also, the documentation > specifically states to not leave it plugged in and connected to the > battery for more than 16-24 hours after the charge is finished. This > is, I'm sure, just to make the charger cheaper to make. If your > question was about why I'm disconnecting it from the battery pack, > David Roden is correct. It is due to the constant 0.12A drain on the > battery when it isn't hooked up to AC power. > > On Mon, Mar 21, 2022 at 10:08 AM Lee Hart <leeah...@earthlink.net> wrote: > > > > > > Hi David, > > > > I'm emailing you directly because the EVDL no longer accepts posts from > > Earthlink. I'm still around and reading it! I just can't post to the EVDL. > > > > It's fairly common for poorly designed chargers to put a constant load > > on the battery. They need to monitor the battery voltage anyway, so it > > saves money to "borrow" the battery voltage to also power the charger's > > electronics. They assume that the charger will always be plugged in; so > > it will be "maintaining" the battery once it is fully charged. > > > > David Roden had the right idea to fix this. Add a relay with a 120vac > > coil, and a contact in series with the charger's output. When AC is > > applied, it pulls in the relay, which connects the charger to the pack. > > > > But as he says, the output circuit of the charger must have a big filter > > capacitor that draws a high peak current if you immediately close the > > output relay. The fix for this is to include a "precharge" or "inrush" > > limiter. The simplest way to do this is with TWO relays. > > > > Relay#1 is a small relay with a 120vac coil and a low-power NO (normally > > open) contact. Its coil is across the AC input to the charger, so it > > pulls in when the charger gets AC power. Its contact has a resistor in > > series, connected between your pack and the charger output. The resistor > > is chosen to provide 0.1 to 1 amp to "charge up" the output circuit > > inside the charger. An old-style tungsten light bulb is a convenient > > resistor, and also indicates that the circuit is working. :-) > > > > Relay#2 is a power relay, with a coil voltage to match your pack > > voltage, and a contact good for 15 amps or more. If your golf cart has a > > 48v pack, then use a relay with a 48v DC coil. Or, you can use a 24v DC > > relay with a resistor in series with its coil so it sees 24v when "on". > > Relay#2's coil (and its series resistor, if needed) connect across the > > charger's output; so it will be "on" when the charger is operating. The > > contact of relay#2 connects the battery pack to the charger output, thus > > shorting out relay#1's contact and precharge resistor. > > > > Here's how it works: When the charger has no AC power, both relays are > > off. When you connect AC to the charger, it also pulls in relay#1. Its > > contact connects the precharge resistor from pack to to charger output, > > thus gently charging up the charger's output capacitor and control > > circuit. When the charger's output voltage gets high enough to start > > charging, it also turns on relay#2, to short out the contact of relay#2 > > and its precharge resistor. From this point, normal charging takes > > place, with no loss in a series diode. > > > > Hope this makes sense! > > Lee Hart > > I think I follow what you are saying and if I do, this is simpler than > finding a time delay relay. Let me see if I have it: > 1) Charger and Relay #1 get AC power at the same time. The contacts of > Relay #1 connect the precharge resistor. > 2) The coil of Relay #2 is across the output of the charger. When the > charger starts or the precharge increases the voltage sufficiently, > Relay #2 pulls in, shorting out the precharge resistor. Hopefully > there is a long enough delay in the charger starting up. I'll have to > time this and see. > > Since Relay #2 is held in by the charger/battery voltage, what will > release the relay when charging ends since the battery voltage will > back feed the coil? I guess relay #1 could be a double pole with one > pole in series with the coil of Relay #2. > > Thanks for the responses so far. > > David D. Nelson > _______________________________________________ > Address messages to ev@lists.evdl.org > No other addresses in TO and CC fields > UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub > ARCHIVE: http://www.evdl.org/archive/ > LIST INFO: http://lists.evdl.org/listinfo.cgi/ev-evdl.org _______________________________________________ Address messages to ev@lists.evdl.org No other addresses in TO and CC fields UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub ARCHIVE: http://www.evdl.org/archive/ LIST INFO: http://lists.evdl.org/listinfo.cgi/ev-evdl.org
