Help required for this SMPS Battery Charger

battery charger with 10w 1 ohm

Hi! The attached file is a schematic of 50KHz SMPS power supply which uses 325 VDC (after rectification of Mains 240 V). The required current for charging Lead Acid battery is approx 20 Amperes. Please give comment if any modification in the circuit is required. (note: isolated 12V DC is used for control circuit)
Voltage regulation is not necessary.:?:

half bridge smps 3525

The changes/modification proposed by the friends, and thinking that the half bridge topology is better for converting AC mains voltage to low voltage, the modified circuit is attached herewith with feedback control circuitry to keep the design stable. Please point out if any further modification is required.

3525 smps- circuits

Two suggestion:

1) current regulation. Measuring current using simply a bjt Vbe seems a rough way. Why not use an op amp and 3525 Vref ?

2) add an LC filter between bridge and 230 Vac

The modified schematic is attached so point out if any further modification is required. Further the .1 Ohm/10W resistor is sufficient to drive the transistor (0.7V drop) or it is needed to increase it value?

About shunt:
1) the power dissipation is about 16 W, quite a lot for a 240 W output SMPS

2) the shunts will be quite hot and the resistance increase. Probably the bjt will be heated from other component and the Vbe decrease. This lead to a reduction of the current. And it isn't stable with temperature.

3) the current gain of bjt and CTR of optocoupler can vary a lot. This change the open loop gain and the closed loop behaviour. The circuit will be stable ?

If you want to reduce the power dissipation of shunt you have to diminish the resistance, this diminish the voltage across the shunt so you need an amplifier. With an op amp you have a fixed gain and it's easier (not easy) to have a stable response.

What about a fuse in the main ? It isn't the best protection but better than nothing.

Top avoid inrush current you can put a NTC in series with main.

the shunt are placed far from the bjt so it will not be heated from other components . Now please tell that the 0.1 ohms/10W resistor are sufficient to drive the transistor ( about 0.7v) or its value may be increased?

10 Ohms NTC thermistor will be added in the mains power to avoid inrush current.

If the bjt will works with Vbe= 0,65 V it could be sufficient. But it depends on CTR and bjt electrical characteristics. If you don't need to know exactly the output current it could work.

I didn't check 3525 and driver circuit.

Transformer: you have to use Litz wire for both windings and to check not only the core dimensions (ETD44) but also the magnetic material. Different materials have different behaviour and core losses.

Hope this help.

Thanks for reply. according to Magnetics (R) (2006_Design_Information) ETD44 core can deliver 450 Watts power at 50KHz

Then you can build it and try.
Be careful when you power for the first time the circuit, it could be dangerous. If you never did it I would like to give you some hints.

Please provide useful hints

I'm writing "micro guide" for start and I realized I ignored an important feature: output voltage limiting. When the load is disconnected, for any reason, all output current goes in output filter capacitor. This lead to (very) fast overcharging and .... destruction.
You can use the same opto.

What if I write guide using Open Office and send the file as attachement ?

It will be highly appreciated

Here's few steps I hope useful.
I use pdf because odt format is not allowed.