Bench Power Supply Battery Charging Circuit - Page 1
Forum section: electronics / beginners

author: steve1515
author_rank: Regular Contributor
timestamp: September 11, 2019, 09:26:37 pm
content: Hello Everyone.Occasionally I use my bench top power supply to charge 12V batteries. When I've done this in the past, I've put a diode in series (10A02-T) with the battery so I don't damage my power supply should the battery voltage rise above that of the supply. Basically, I want to protect against pushing current back into the supply. I usually put a meter at the battery terminals and set the voltage. Usually 13.8V or 14-15V depending on the battery markings. The problem I have is that the voltage at the battery terminals changes depending on the current going into the battery. My power supply does not have sense terminals, so I have to keep adjusting the voltage to compensate for the drop across the diode.Is there a way to protect my power supply with some kind of simple MOSFET circuit instead of the diode? I'm trying to get it so I can just set the voltage and the power supply can regulate the voltage without having to deal with a diode drop.Any tips?

author: MarkF
author_rank: Super Contributor
timestamp: September 11, 2019, 09:38:29 pm
content: Here is a circuit I found on the internet. Never built it.There is a YouTube video demonstrating it somewhere.The resistor in 'red' adjusts the charging current.There are a few resistor values shown at the bottom with the corresponding current.The 5K pot adjusts the charging voltage.The LED indicates when the circuit is charging.This isn't the video. But, here is a description:https://www.eleccircuit.com/gel-cell-battery-charger-circuit/

author: Ian.M
author_rank: Super Contributor
timestamp: September 11, 2019, 10:22:00 pm
content: There is such a thing as a single chip ideal diode controller, designed to be used with a MOSFET to replace a Scottky diode in many power rectification applications.Take a look at these two datasheets:http://www.farnell.com/datasheets/85443.pdfhttp://www.farnell.com/datasheets/78928.pdfThe low voltage one is viable for single cell LiPOs and the HV one starts off at a useful start of charge voltage for 12V and higher Lead Acid battery packs. If its below their minimum voltage you'd want to trickle charge to bring the voltage up and the MOSFET body diode can handle that. The controller will cut in and eliminate the diode drop before the battery reaches a normal CV mode float voltage. If you need to charge single cell (or even two series) NiCd or NiMH, they wont do the job as there is insufficient gate bias supply, but as NiMH are ill suited to CC/CV charging (even a C/100 trickle charge sustained past full charge degrades NiMH life), and need some 'smarts' in the charger to terminate the charge before you cook them, you shouldn't be trying to use a diode + CC/CV PSU as a charger for them anyway.Another simple option is an old laptop PSU, or any other convenient power source + a cheap CC/CV buck or boost module*+ a relay running off your 'raw' power source to isolate the battery under charge if the mains supply fails.* e.g:https://www.ebay.co.uk/itm/DC-DC-Step-Up-Down-Boost-Buck-Voltage-Converter-Module-CC-CV-LM2577S-LM2596S-/281777880409

author: ledtester
author_rank: Super Contributor
timestamp: September 11, 2019, 11:19:33 pm
content: I understand that the LM317 can be damaged by reverse voltage, i.e. V_out > V_in, and that it is advisable to put a shunting diode across pins 2 and 3 to mitigate that situation.So how about putting a diode on the + output of PS1 and a load resistor from the LM317's IN to GND? That would route current from an overvoltaged battery around the LM317 and prevent it from entering the power supply.Something like:

author: Jwillis
author_rank: Super Contributor
timestamp: September 12, 2019, 01:53:05 am
content: Depending on the battery it may be advisable to place in a current pass transistor since the LM317 max current is only about 1,5 amps. For example a 10Ah lead acid will usually bulk charge at around 3 amps which would quickly destroy the regulator.

author: MarkF
author_rank: Super Contributor
timestamp: September 12, 2019, 02:05:53 am
content: Quote from: Jwillis on September 12, 2019, 01:53:05 amThe current is limited by the sense resistor (i.e. the resistor in red).The three resistors at the bottom of the circuit show possible max charging currents.Quote from: ledtester on September 11, 2019, 11:19:33 pmThe protection diode around the LM317 is to funnel the current to the source if the output voltage is higher then its input. Since the OP wants to charge the battery with a 'lab power supply', I would sacrifice the LM317 to save the power supply. Otherwise, I would put it in.A load resistor on the input side is just going to waste current during normal operation. I would leave it out.If you want to add protection, add a fuse and diode on the output to protect against reverse polarity when connecting the battery.

author: steve1515
author_rank: Regular Contributor
timestamp: September 12, 2019, 02:54:52 am
content: Thanks for the info so far!I'll have to read through those PDF that @Ian.M as they look promising.Just to clarify on what I'm doing... When I charge a battery, the markings on the battery have a current limit such as 2.1A for example. I normally short my power supply output and set the current to that value. I then remove the short and set the voltage to 13.8 for example, and then connect the battery. If I didn't have a diode in series causing a voltage drop, the power supply would send current into the battery to charge at 2.1A (if the battery was very low) and then as the battery was charging up the power supply would switch out of CC mode and then would go into CV mode. This is all automatic in the power supply.So, I'm a little confused by the circuit @MarkF posted because it has a resistor to set the current. Maybe I'm not understanding how it's supposed to work? I figured I would still set the current and voltage using my power supply and the circuit would just prevent current from flowing backwards while eliminating the voltage drop so that the power supply could regulate current and voltage correctly at the battery terminals. Hope that makes sense.(Also, does anyone know how to properly link to a username on these forums? "@" doesn't seem to work.)

author: MarkF
author_rank: Super Contributor
timestamp: September 12, 2019, 03:32:40 am
content: Quote from: steve1515 on September 12, 2019, 02:54:52 amThe circuit controls both the charging voltage and current.It is intended to be driven by a wall-wart or an unregulated power supply.(Refer to the link in my first post)This circuit would eliminate all the setup you are currently doing on your lab supply.It only needs an input voltage that is a little higher than your "charging voltage" + "LM317 dropout voltage".LM317 dropout voltage = 3V

author: ledtester
author_rank: Super Contributor
timestamp: September 13, 2019, 04:38:25 am
content: Here's another idea. Also has protection against the power supply being turned off. Check out the PDF for more info.

author: steve1515
author_rank: Regular Contributor
timestamp: September 15, 2019, 12:31:09 am
content: This is all great info! I started researching this some more including the LTC4357 referenced above. This led me to the LTC4359. I really like the idea of an ideal diode that can handle 48V. This would allow me to increase the voltage on my power supply to de-sulfate or even non-battery things in the range of 4-48V.This is the diagram out of the manual that I believe I could use. It looks like all the parts are labeled except for R2 and Cout. For those I figure a 2W 2k resistor and a 0.047uF 50V X7R ceramic would work.What does everyone think? Would this work out for me or am I on the wrong track? It looks like this circuit would work from 4-70V @ 10A, but I'm not 100% sure on that since it's labeled 48V in and out.LTC4359 Overview:https://www.analog.com/en/products/ltc4359.htmlLTC4359 Datasheet:https://www.analog.com/media/en/technical-documentation/data-sheets/ltc4359.pdf