I have a similar (sort of ) question. I have a DC battery charger that puts out 57.1 volts and can deliver a maximum of 3 amps. I want to use it to charge a battery pack to which I only want to apply 54.6 volts. Rather than try to mod the charger, is it feasible to put a resistor or a rheostat or other on the positive output leg of the charger to drop the voltage by (57.1-54.6) 2.5 volts to the load (battery being charged)?

Being a novice, I would assume resistance of 2.5/3.0 = .83 ohms? but capable of handling (2.5 x 3.0) = 7.5 watts? If so, what would be the best type to buy for this? Or am I out to lunch on this approach?

 

Place 3 or 4 rectifiers of a suitable rating in series with the supply.
Max.

 

These sort of questions go much better when the battery pack in question is better described as in type of batteries, manufacturer and manufacturer's part number (when available) and all data known about the battery or battery pack. My thinking here is we call a standard automotive battery a 12 Volt battery when in reality it consist of six cells of 2.1 Volts per cell (when charged) making it actually a 12.6 Volt battery. That same 12.6 volt battery is typically charged at about 13.6 to 13.8 Volts. There is more to it but you get the idea. Now if you have a battery pack you want to apply 54.6 Volts to and you have a charger outputting 57.1 Volts and you are sure this is what you want to do and understand the charge voltage for your battery pack and type then the best method is to do as MaxHeadRoom suggest and "Place 3 or 4 rectifiers of a suitable rating in series with the supply".

Ron

 

A thread belongs to the Thread Started. Please do not hijack someone's thread. Here you have your own thread.

 

@SenorFred
Was the 57.1 volts measure when it was connected or disconnected from the battery pack?
Also, if measured with batteries connected, was the 51 7 volts measured when the batteries were fully charged or fairly discharged?

 

This circuit will provide an adjustable, precise voltage drop greater than 2.495V between the top rail (Vo) and the bottom rail (unnamed). Note that if the drop is 3V at your specified 3A, the power dissipation is 9W, which requires a big heat sink regardless of how you do it, on the PNP, diodes, or a resistor.

 

do not use a resistor the voltage drop across a fixed resistance will change as the battery charges up and the current load decreases. the current will not always be 3 amps.

 

I always wonder how a "novice" gets into these situations in the first place..
Not to mention how they determine that they need (or think they need) exactly 54.6V to charge a battery.
Nor how they find (or think they have) a battery charger that puts out 57.1V exactly..

Of course I always wonder why you guys just throw out "solutions" without any of the necessary details or confirmations.
I love to provide solutions.. But I hate wasting my time more...