24V Battery Charger

[1843]

Hi All,

Would somone please design a simple circuit by LM317T to charge two lead acid battery which are configured in series in order to provide 24V power supply..?
The batteries are 4.5Ah, so the charger current should be 450mAh and its voltage 27V.
I found the following circuit on the Internet, But it is designed for 12V batteries:


Carl, Reg and other experts, Do you have any idea?

Thanks in advance,
1843

[Max]

Hi All,

Would somone please design a simple circuit by LM317T to charge two lead acid battery which are configured in series in order to provide 24V power supply..?
The batteries are 4.5Ah, so the charger current should be 450mAh and its voltage 27V.
I found the following circuit on the Internet, But it is designed for 12V batteries:
[ATTACH]4642[/ATTACH]

Carl, Reg and other experts, Do you have any idea?

Thanks in advance,
1843

Hi,
R2,R3 and P1 set the voltage output there, formula is :

Vout = Vref * (1+ (R3+P1)/R2) + Iadj * (R3+P1)

Iadj is typical 50uA, but can be higher as 100uA. Vref = 1.25V

at one extreme you get R3+P1 = 690 ohm and so

Vout = 1.25*(1+690/120)+0.0001*690 = 8.4375+0.069 = 8.50V (about)

at other extreme you get R3+P1 = R3 so 470ohm and

Vout = 1.25*(1+470/120)+0.0001*470 = 6.1458+0.047 = 6.1948V (about)

So your circuit is for a 6V battery ...maybe lead acid... like in the old VW Beattle!

Now... the current limit is about Vbe/R1... so say 0.65/0.56 = 1.16A for a common transistor... BUT this is the BC140 (telefunken made ?) anyway...expect also 1-1.2 volt there... if it goes in saturation... but for this circuit I think 0.65 is good value (BUT read all before connecting the battery)

When approaching the limit the transistor will start conduct and will pull down the adj terminal voltage till dynamic equilibrium is reach or device will just shutdown in case of overload.

But the problem is you have to change stuff there, use this:

R2=120ohm, R3=1.8Kohm ,P1 = 1Kohm

you will get:

Vout1= 1.25*(1+2800/120)+0.0001*2800= 30.416+0.28= 30.7V
Vout2= 1.25*(1+1800/120)+0.0001*1800= 20+0.18 = 20.18

So, from 20.18 to 30.7 volt considering worse case of Iadj at 100uA.

For current limit:

R1 = Vbe/Imax = 0.65/0.45 = 1.4444444444 ohm you can approximate to 1.5ohm ---> use 3x4.7ohm 1 watt in parallel

But be really careful then before applinf voltage to LM317 : at output of bridge (input of 317) you must not exced 37V or you will be lost in space with it! Don't belive it's truly rated 40V like some folks! Sometimes it explode at 38V!

Then I suggest you first check the efficiency of current limiter using a multimeter in the 10A (or better 20A) range DC: it MUST deliver the current fixed by calculations, otherwise your transistor has different Vbe than required or too low gain, if so change it for something better.

Do not rely on BC140 Vbe... check well the limit before connecting the battery: its Vbe depends on current, temperature etc so you can use also other stuff like 2N1711 or similar.

About voltage, you need 13.8*2V= 27.6V... turn the trimpot to read this when with no load.

A good idea will be also putting a silicon 1A rated diode (like 1N4007) in series with output then read voltage after it when adjusting the trimpot.

This will avoid overcharge... when battery will reach 27.6V voltage at the end of charge the diode will lose it's forward on voltage (0.6-0.7V) and so will open the circuit for you... and you will save your moquette if under the battery... and some explosion of acid.

Good luck!

Best regards,
Max

[1843]

Thanks dear Max for your great guides,
At what voltage must I recharge the battries? 23V or 21V?

Kind regards!
1843

[1843]

Input Voltage for LM317

I'v got a problem with choosing a proper transformer... If I use a 24V, when no load the input voltage is about 40V and when the load is connected, the voltage drops to 24.1V... so the battery cannot be charged.
If a 26V transformer be used, the input voltage is 42.5V!(when no load connected) so I don't know if the LM317 will be failed or not...?

Does the high input voltage causes any problem when no load connected?

Kind regards!
1843

[Max]

I'v got a problem with choosing a proper transformer... If I use a 24V, when no load the input voltage is about 40V and when the load is connected, the voltage drops to 24.1V... so the battery cannot be charged.
If a 26V transformer be used, the input voltage is 42.5V!(when no load connected) so I don't know if the LM317 will be failed or not...?

Does the high input voltage causes any problem when no load connected?

Kind regards!
1843

Hi,
You have to find something that will provide 30-37 volts at input of 317... otherwise you can't charge battery or you will damage the regulator.

I suggest an old transformer of 24Vac out + bridge rectifier.

If wanna use 26V, you cannot connect just as is cause the rectified 42.5V will destroy your regulator easy... somethimes this could not happen but it's not safe putting 42.5 on a maximum declared rating of 40.

You can do 2 ways if wanna use it anyway: put a dummy load on it's output, like a 2K2 4W ... this will make the output go down a bit maybe under 38V if transformer is small enough and cannot provide too much current at output.

I do not suggest the above cause it's tricky and if resistor will fail you'll burn too the LM317.

I suggest instead this:

use a series of diodes 1amper rated (e.g. again 1n4007 or 4001 etc) then use the series at output of transformer before LM317.

For example using 8diodes you get about 0.65*8 drop on them when in conduction... so you will lower voltage: 42.5-(8*0.65) = 37.3

Be careful... before connecting the regulator check voltage using a dummy load... e.g. a big wattage resistor (20W or more) and check if voltage stay there or drops too much under the required load.

Example: suppose you put 8diodes and get 37.3V... then put there a resistor of 37.3/0.450 = 83.333 say 82ohm 20W.... or better 50W if you have.

You can also use e.g. 2*150 parallel... about 75ohm to test but be sure both are , at least, 10W rated

look for voltage... if stay around 33-37 you can use transformer this way with regulator and charge the battery otherwise not cause it will be overloaded by sourcing 450mA continuosly for hours, so you need a bigger power rated one.

The 82ohm will get at 37.3V a current of 0.45A and power dissipation of about 16W: means you can test with a 20W one for few seconds just for voltage drop... or must find a larger resistance, e.g. 50W to test for more time... but you need 2-3second to read voltage drop so don't worry too much about this.

I suggest for battery you use the suggested charging voltage if any... if lead acid....liquid kind you can use 13.8*2 = 27.6V ; if gel type i suggest you look datasheet ot the labels on battery ... any stamp on them with voltage required... often gel-type and sealed require less voltage cause of construction and gas expulsion (H2) some require you stay under 12.85 for 12v batteries... so maximum 12.85*2 = 25.7V there at output.

This depends on many factors so you have to check for your type of batteries relative documentation if any.

The good thing is that often they are fault tolerant when slow charge is performed... so little more voltage will not damage cells if for short time and at low charge rate.

Kind regards,
Max