# DIY programmable battery charger

#### rionzion

Mar 30, 2011
9
Hi folks, i'm looking for some ideas as to how i can convert a standard car battery charger in to a more customizable charger for charging deep-cycle batteries and keeping them in good health.

The charger i have is a 40A 12v charger which consists of an ac step down transformer, a bridge rectifier and a regulator that decreases the ampage as the battery voltage rises.

first off i need a better understanding of how the charger i have actually works, so i have a few basic questions.

1. i get an open voltage reading of 13.5 volts ac for the output side of the transformer. How can this manage to charge the batteries any higher than 13.5vdc?

2. how does the regulator work? to me, it just looks like a coil of copper on the back of the amp-meter - i have no idea how this works!

Many thanks

#### Bluejets

Oct 5, 2014
5,991
Wire coil on the back of the ammeter is a shunt, not a regulator.
Not possible to charge batteries higher than 12v with this charger.
Modification would be extreme and given you did not know what a shunt was, then I doubt you could handle building a programmable charger.
Better to buy one, most are reasonably priced these days especially ones for use with Lipo, nimh, pb etc. and given there are Chinese clones also available.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,730
i get an open voltage reading of 13.5 volts ac for the output side of the transformer. How can this manage to charge the batteries any higher than 13.5vdc?
That's 13.5 V AC as RMS (root mean square).
The peak voltage is: Vpeak=1.4*Vrms=18.9 V (1.4 ~ square root (2))
Since this is the open circuit voltage (no load), this voltage will drop to a lower value under load. Also you need to subtract the voltage drop across the rectifier. The correct spot to measure is the DC voltaeg at the output of the rectifier with and without load to get an idea of the characteristic of the unit.

how does the regulator work? to me, it just looks like a coil of copper on the back of the amp-meter - i have no idea how this works!
Yes and no. The ammeter is probably not an ammeter but a millivolt meter or milliampere meter. The 'coil' is a resistor with a very small value in teh milliohms range. The charging current flows mainly through this resistor (the parallel resistance of the meter is negligible here) and develops a voltage drop Vsense=Rcoil*Icharge. This voltage is indicated by the instrument as an indicatio nof the charge current.

a regulator that decreases the ampage as the battery voltage rises.
There is seemingly no regulator at all in this simple charger. 'Regulation' is achieved by exploiting the inherent characteristic of the battery and the transformer/rectifier circuit: The transformer has an internal resistance (due to winding resistance and limited output current due to size of the core). Thus at low battery voltage the current will be high due to a high voltage difference between charger open circuit voltage and battery. This initial current will be limited by the internal resistance of teh charger (see above). Once the battery voltae rises, the voltage difference between battery and charger decreases. As less voltage drops across the internal resistance of the charger, teh current starts to decrease. This is all the 'regulation' there is.

Modification would be extreme and given you did not know what a shunt was, then I doubt you could handle building a programmable charger.
'Extreme' is a bit extreme , but given the initial post I agree. Especially when you look at the 40 A specs, which is something that is not easily handled.

#### Colin Mitchell

Aug 31, 2014
1,416
Firstly we need to know the amp-hour of the battery you are charging.

#### rionzion

Mar 30, 2011
9
Firstly we need to know the amp-hour of the battery you are charging.
Batteries are 2 x 225AH trojan deep cycle 6v wired in parallel to give 12v. Bulk charge 14.8v, float 13.5v, equalize 16.2v
charging current is recommended anywhere from 10-20% of AH capacity @20 hour rate, depending on speed of charge required.

#### duke37

Jan 9, 2011
5,364
There is no way that the charger shown can manage 40A for more than a couple of seconds. The wire sizes and bridge rectifier seem to indicate 5 or 10A.

Edit. Two 6V batteries in parallel will yield 6V.

#### rionzion

Mar 30, 2011
9
There is no way that the charger shown can manage 40A for more than a couple of seconds. The wire sizes and bridge rectifier seem to indicate 5 or 10A
well spotted! actually the charger shown is a 30A charger ( a faulty one i happened to have sitting beside me when i made the post, but its identical to the 40A charger in the way it functions). When it was working, that particular charger managed 15-20 amps for several hours.

#### rionzion

Mar 30, 2011
9
That's 13.5 V AC as RMS (root mean square).
The peak voltage is: Vpeak=1.4*Vrms=18.9 V (1.4 ~ square root (2))
Since this is the open circuit voltage (no load), this voltage will drop to a lower value under load. Also you need to subtract the voltage drop across the rectifier. The correct spot to measure is the DC voltaeg at the output of the rectifier with and without load to get an idea of the characteristic of the unit.

Yes and no. The ammeter is probably not an ammeter but a millivolt meter or milliampere meter. The 'coil' is a resistor with a very small value in teh milliohms range. The charging current flows mainly through this resistor (the parallel resistance of the meter is negligible here) and develops a voltage drop Vsense=Rcoil*Icharge. This voltage is indicated by the instrument as an indicatio nof the charge current.

There is seemingly no regulator at all in this simple charger. 'Regulation' is achieved by exploiting the inherent characteristic of the battery and the transformer/rectifier circuit: The transformer has an internal resistance (due to winding resistance and limited output current due to size of the core). Thus at low battery voltage the current will be high due to a high voltage difference between charger open circuit voltage and battery. This initial current will be limited by the internal resistance of teh charger (see above). Once the battery voltae rises, the voltage difference between battery and charger decreases. As less voltage drops across the internal resistance of the charger, teh current starts to decrease. This is all the 'regulation' there is.

'Extreme' is a bit extreme , but given the initial post I agree. Especially when you look at the 40 A specs, which is something that is not easily handled.
Thanks, that's very informative.... what i'd like to achieve is this: instead of the current gradually reducing as the voltage rises, i would like to be able to apply a constant current until the voltage reaches "bulk" voltage (14.8v) then reduce current to hold battery voltage at "float" voltage (13.5v). Finally i would also like to have the option to periodically apply an "equalization" charge to bring the batteries to 16.2v and hold at that voltage for 3 hours.

I am aware that this is what so-called "intelligent" chargers do automatically by the use of digital control circuits, but i'm wondering if there's a way to achieve the same with simpler electronics.

#### rionzion

Mar 30, 2011
9
Batteries are 2 x 225AH trojan deep cycle 6v wired in parallel to give 12v. Bulk charge 14.8v, float 13.5v, equalize 16.2v
charging current is recommended anywhere from 10-20% of AH capacity @20 hour rate, depending on speed of charge required.
SORRY - wired in series!

#### Colin Mitchell

Aug 31, 2014
1,416
Connect the charger and let us know the current after about 10 hours.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,730
i'm wondering if there's a way to achieve the same with simpler electronics.
Search for ''lead battery charger circuit'. The 40 A rating will give you a tough time, though.

#### rionzion

Mar 30, 2011
9
Connect the charger and let us know the current after about 10 hours.
from past experience, the charger continues to put out about 1.5 amps even up to around 16.5 volts, but it's a slow climb to that level of voltage! i should explain that i'm off the grid, so i use generator power for the charger. 8 hours would be my max running time, although i prefer to keep it between 4-6 hours each evening (this being winter and negligible charge coming from my solar panel).

#### rionzion

Mar 30, 2011
9
Search for ''lead battery charger circuit'. The 40 A rating will give you a tough time, though.
will do... surely if i found a smaller circuit it would just be a case of scaling-up the component ratings? e.g. a 4A charger circuit - just increase everything by a factor of ten? maybe i'm over-simplifying the issue?!

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,730
just increase everything by a factor of ten? maybe i'm over-simplifying the issue?!
That's definitely oversimplified. Once you have homed in on a specific circuit, we may be able to tell you what to scale and how.

#### rionzion

Mar 30, 2011
9
Hello again... having looked at circuit designs i've decided not to try to make a fully programmable AC powered charger, since i would rather put the effort into making an engine-driven DC powered charger which would be, i think, a more effecient way of charging my batteries when the sun isnt shining. That said, i would like to be able to make one modification to my AC-powered charger... in place of the shunt that automatically decreases the current from the charger, i'm wondering is there are component or circuit i can add that will allow me to control the current at will. E.g to be able to allow the charger to put 25A into the batteries constantly until they reach a given voltage (14.8v or occasionally 16.2v). I've read that a zener diode can do this but i dont think they are available in the size i would need. alternatively a relay, diode and transistor, that would automatically switch off the charger when desired voltage is reached might be a possibility (similar to the idea discussed here:http://www.electronicshub.org/car-battery-charger-circuit/). However, i want to avoid running the charger at full output as this would be too high a current (40A), both for the batteries and possible overheating of the charger wiring. 25-30A would be an ideal maximum. Any ideas would be much apreciated.

#### cjdelphi

Oct 26, 2011
1,166
A Microprocessor makes it easy, after the bridge rectifier it will read 19v you need 13.8v to 14.5/15v to charge it..

So cull the shunt, cull everything except the bridge rectifier/ caps (and transformer) so we get 19v dc out

Thus the power supply, 40amps is a fair amount of current, with a micro controller, sure we could use that shunt but it would require an opamp and more circuitry!

So for simplicity sakes, get a digital current meter that supplies back a voltage depending on current, no shunt or voltage drop, and a chunky (or several) mosfet to parallel the load to the battery, from the mc, send a pin high, dump the current from the transformer to battery, then turn off the mosfet, measure the voltage if over 14.5v switch off, repeat..

#### cjdelphi

Oct 26, 2011
1,166
These things are useful.... i prefer them over a shunt as it measures current by inductance

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#### BobK

Jan 5, 2010
7,682
What you are looking for is a buck converter. This allows you control the voltage supplied to the battery without simply wasting the extra power. When you are talking about 40A the wasted power would be huge and produce a lot of heat, so this is important. The buck converter is coupled with a small shunt resistor to measure the current going into the battery. A microcontroller can then read the current and adjust the voltage output to maintain the desired current.

Bob

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