dc-dc converter

I've no experience w. dc-dc converters but wonder why you just didn't tap one of the 12v batteries to run your lights, etc. while still leaving them series connected for the boat motor.

I want to keep the 48V bank balanced rather than having one of them discharged further than the others. I am currently using a 5th (smaller) 12V battery for the lights to avoid using one of the four from the 48V bank. Adding the DC-DC converter keeps me from having to manually charge that 5th battery periodically.

That behavior is normal. The DC-DC converter almost certainly has a capacitor across the input, and certainly has a capacitor across the output. There will be a high initial current draw when power is applied. Aftehr that it draws enough to provide the required output current plus a small amount for the circuitry. With no load it's output will rise until an overvoltage clamp shuts down the control IC.

I'd use a 5A slow blow fuse in the input line with a switch to control the lights. PlainBill

I agree with this. The input capacitor will appear as a dead short for a little while before it starts to charge up; this is why you get the big arc.

Really big DC-DC converters and DC-AC inverters (several hundred watts on up) are sometimes switched on in a two-stage process. First, the battery is connected to the converter/inverter through a smallish resistor for a few seconds, to charge the input capacitor part way. Then, the battery is connected directly to the converter/inverter, with no resistor. This saves a big arc at the switch or relay contacts and makes the switch or relay last longer.

Also agreed. The actual minimum load to get it to put out 12.3 V all the time is probably not very much - on the order of a few watts or so.

Make sure you have a spare fuse *on the boat* if you do this. :) Alternatively, use a 5 A circuit breaker.

Another suggestion along this line:

You might wire a tap to the "lowest" 12 V battery anyway. Run the nav lights through a SPDT switch, with the lights to the common or wiper terminal. One end of the switch goes to the output of the DC-DC converter and the other end goes to your 12 V tap. Most of the time, you power the nav lights from the DC-DC converter, and your propulsion pack stays nice and balanced. But, if the DC-DC converter ever fails, you can flip the switch and still have nav lights to get home with, at the cost of a slightly unbalanced propulsion pack.

Matt Roberds

Apart from the inrush current to the cap which has been discussed already, being nav lights I would be inclined to use the converter to charge the dedicated 12v battery and keep it as a fail safe device for the nav lighting should the converter fail.

Could you not rewire the boat lighting so 4 lamps are in series, with some sort of LED failure indicator or something to show which lamp/wiring has failed , when it does. One for red, one green and 2 riding lights, about 4 are used together , never separately. Then for at anchor. and just one light. then a 48V one or 2 or 4 24V or 12V smaller wattage in one housing.

Thanks to all for your comments and suggestions. I'll do some testing to see what minimum load will keep the output at 12V. I may also add a small resistor in series with the input to keep the inrush current down a little. I realize that will lower the overall efficiency, but I hate that snaping sound the switch makes with nothing limiting the inrush. Regarding the suggestion of putting the lights in series, I thought of that, too. However, the lights (4 of them total) are not even close to being matched loads. The red and green only draw 60 mA each. The anchor light draws 175 mA. The spot light used for docking draws 1.75 A all by itself. (All these are LEDs with their own internal current limiting resistors).

Anyway, thanks again for your comments. Pat

Sorry to burst your bubble, but you can't use a 12.31 supply to charge a 12V lead-acid battery, unless you run it through ANOTHER DC-DC converter

PlainBill

Rathe than using a low valure resistor to limit inrush current, use an ICL. Your input current into the module will be in the order of .5 -

PlainBill

Interesting idea. I hadn't thought of that. You are correct that I can't easily get to the internal resistors, but adding external resistors would be easy enough. However, If I add the appropriate external resistors to drop the additional 36 volts, I will be creating a lot of wasted heat. The nav light resistors would be dissipating 2 watts each while the anchor light R would be dissipating over 6 watts. The spot light R would be dissipating 62 watts! I think I'll stick with the (switching type) dc-dc converter. Your comments about added complexity are well taken, though.

Pat

Another good idea. Thank you. The part you mentioned is rated at 1 Amp continuous but I see larger parts are also available from the same family. 1 Amp is enough for my current needs, but I will probably want to match the rating of the converter and get the 3 amp part.

Pat

If it was me I'd try to figure out a way to place the resistors someplace where any heat they generate could keep something dry. I wonder, do DC-DC converters emit much RF? I'm really surprised that the original poster doesn't want to go with the most simple and robust solution. Especially on a boat. And it would be much easier to carry and wire in spare resistors than a spare DC-DC converter. ERS

This is my last post on this subject as we have drifted off topic. My original post was asking if the experience I had with my cheap $25 dc-dc converter (already purchased) was normal. That question was very quickly answered as "yes". We have since drifted to how easy it would be to repair the system out on the water in the dark. In that situation, I would just jumper the 12 V lights to one of the traction batteries as Arfa suggested. By the way, the boat is used on a small inland lake and I could easily return to the dock without any lights at all by carefully avoiding the one or two fishermen out on the lake at night. Thanks to everyone who responded. ****End of Thread***

Wow! I've never been snipped before!

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This sounds perfectly normal. I presume it is a switching power supply, and needs capacitors to buffer the switching pulses. So, the capacitor draws a large current at turn-on. You may need a high current relay to handle the turn-on surge without damage. Yes, your unit apparently needs a small load to continue regulating properly. Definitely put a fuse in the 48 V input to it.

Jon