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I have designed a battery charger circuit. The circuit is simple AC to DC power supply circuit, it is composed of a 220 Volt to 16 Volt stepdown transformer, a diode bridge and a capacitor.

Now the problem is that whenever I connect the charger to the battery the battery starts taking high current up to 19 A and the diode bridge is burnt because the diode bridge can't withstand current higher than 10 A.

How can I limit the current to 10 A so that the current is kept in 10 A safe range for the diode bridge. Note that I can't find a high current diode bridge in market here.

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  • \$\begingroup\$ Ohm's Law always holds. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 9 '12 at 17:19
  • \$\begingroup\$ Don't know where "in market here" is, but I just ordered a 50A 1000PIV bridge off eBay for $1.45 US, including shipping to North America. Mind you, I'll have to wait several weeks before it shows up... And I only did this because it meant less clutter than paralleling several of the 6A and 8A bridges I've salvaged out of old, free, otherwise non-working ATX power supplies. \$\endgroup\$ – mickeyf_supports_Monica Sep 10 '12 at 4:09
  • \$\begingroup\$ @mickeyf Actually I intend to use it in several(more than 30) charger circuits. I can't buy all of them from ebay then. Because it will increase the charger price. \$\endgroup\$ – Naeem Ul Wahhab Sep 10 '12 at 15:25
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If you are confident that the transformer output is appropriate for charging your battery, one solution would be to simply increase the current-carrying capacity of your bridge rectifier.

You mention that you can't obtain higher current diode bridges. Can you obtain higher current discrete diodes? If not, and your application can handle the cost of 2 diode bridges, you can make an effective 20A-rated full bridge rectifier out of two 10A full diode bridge modules.

If you look at the schematic for a full bridge rectifier, you'll notice that it is possible to connect the two AC input terminals together to obtain two diodes in parallel between the AC input and each of + and -. Since the diodes in a bridge rectifier are usually on the same die, they are likely to be reasonably well matched and likely to current share reasonably well (though never perfectly, so some derating is warranted). Simply use two full bridges connected in this way (one bridge for each AC input with the rectifier + and - terminals each connected in parallel)

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You need a current limiting element in your circuit, like a regulator IC. Or you can use a transformer that has the desired secondary winding resistance to limit the current in a simple fashion. Or use a higher rated diode bridge if you would like to charge at the higher current and your battery can handle it.

A resistor could be used for smaller currents but at this level you would need a very large and expensive resistor. If we say the battery is at around 12V, then (16V - 12V) * 10A = 40W! So the resistor is out.

There are various ways of doing this with regulators, ranging from simple to complex. There are certainly many inexpensive commercial solutions out there that you may want to consider.

For a simple circuit based on a linear regulator, have a look at the various circuits at the end of the LM338 datasheet
Here are a couple of 5A examples:

LM338 Current Limit

And also a simple 12V battery charger:

LM338 12V Battery Charger

There are many app notes from TI, Linear, etc, on battery chargers based around their linear and switching regulators, many far more sophisticated than the above. Also there are dedicated battery charge ICs available on Digikey, Farnell, etc you may want to consider.

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  • \$\begingroup\$ Take note that for currents over 1A, an external transistor should be used. I think you should include it too. The appropriate schematic is in the datasheet for the linear regulator. \$\endgroup\$ – Jonny B Good Sep 9 '12 at 19:38
  • \$\begingroup\$ @JonnyBGood - Why do you say over 1A? This regulator can handle up to 5A. If more current is required though, extra pass elements such as a transistor or another regulator can be added in parallel. \$\endgroup\$ – Oli Glaser Sep 9 '12 at 19:52
  • \$\begingroup\$ Ah, sorry, I just assumed it's one of the hundreds of LMs that's up to 1A. Even though you specified the 5A explicitly! Haha. \$\endgroup\$ – Jonny B Good Sep 9 '12 at 20:11
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One way to limit the current would be to place a resistor in series with the output of the diode bridge and the connection to the battery. Selection of the resistor has to be made based upon the voltage difference between the charger output and the battery voltage and the current level you want to limit to. Also make sure to evaluate how much power will be dissipated in the resistor, (I * I)/R , and select one with a suitable wattage rating.

Have you actually checked the voltage out of your diode bridge? If it is too high you could be forcing a lot more current into the battery than is intended for the type of battery being charged.

A sophisticated battery charger will replace the above suggested series resistor with a current control circuit that monitors the battery voltage and adjusts the charge current in an appropriate manner. Think of it as a programmable current source to the battery.

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