Hi,

I'm working on this battery charger project. http://www.kitsrus.com/pdf/K193.pdf



To provide 18V DC I will be employing a full wave bridge rectifier with a 4700uF capacitor as a filter. I have bought a transformer from a shop in my city. Since I've never dealt with transformers in any of my project before so I asked the shop owner about the rating of the transformer. I was told its a 12V-0-12V 2.5A 60VA transformer. Here are the photos of this transformer. I want to ask does this transformer (weighs 585 grams) really looks like a 60VA transformer? How can I verify this indeed is a 60VA transformer?

I will be taking 24V RMS from the transformer. After two diode drops (bridge rectifier) the voltage will be around 33.94V - 1.4V = 32.54V. There is one more diode D1 in the circuit above. Thus the input to the LM317 will be 32.54V - 0.7V = 31.84V. The output of the LM317 in the bulk charge stage will be 14.7V and the charging current will be limited to 500mA. The power dissipated by the LM317 will be then

P_d = ( 31.84V - (14.7V+0.7V+0.6V) ) × 500mA
P_d = ( 31.84 - 16V ) × 500mA
P_d = 7.92W

Does this heat sink will be sufficient to dissipate 7.92W while keeping the junction temperature of LM317 below 125°C? I don't have the thermal resistance of this heat sink.

The tab of LM317 (TO-220 package) is connected to its OUT pin. What I've understood is that if I attach the LM317 with this heat sink without any insulation the heat sink will carry 14.7V DC. And if I happen to touch the heat sink I would get an electric shock. Am I right?

Thank you for reading.

 

And if I happen to touch the heat sink I would get an electric shock. Am I right?

No. You cannot detect less than ~40V DC I believe, by direct connection of your fingers to both poles. You can detect much lower voltages on your tongue, but it's pretty easy to avoid that.

Without the numbers for that heat sink, I'd be very concerned about it. It looks like it might work, but 8W is not a trivial dissipation. I might try it and see how it goes. But for a mission-critical design that works the first time, I'd do something more certain.

 

I'd suggest turning it into a Bench PSU with charging capability.

 

You would be much better using a 15-0-15Vrms or a 16-0-16Vrms or a 16V (no CT) transformer. The goal is to have about 17Vdc at the filter capacitor, otherwise you are wasting huge amounts of power as heat, and the heat-sinking has to accommodate that.

If the voltage at the filter cap is ~18V, at 1A charging current, the regulator must dissipate (18-14)*1 = 4W. Now think what is happening if the voltage at the filter capacitor is ~34V (which it will be with your tranny): (34-14)*1 = 20W!

Btw: this is a "trickle charger" at best. It will take days to recharge a mostly discharged 60Ah automotive battery. It does not comply with the recommended three-step lead-acid charging algorithm.

 

Hi,

I'm working on this battery charger project. http://www.kitsrus.com/pdf/K193.pdf

View attachment 93019

To provide 18V DC I will be employing a full wave bridge rectifier with a 4700uF capacitor as a filter. I have bought a transformer from a shop in my city. Since I've never dealt with transformers in any of my project before so I asked the shop owner about the rating of the transformer. I was told its a 12V-0-12V 2.5A 60VA transformer. Here are the photos of this transformer. I want to ask does this transformer (weighs 585 grams) really looks like a 60VA transformer? How can I verify this indeed is a 60VA transformer?

I will be taking 24V RMS from the transformer. After two diode drops (bridge rectifier) the voltage will be around 33.94V - 1.4V = 32.54V. There is one more diode D1 in the circuit above. Thus the input to the LM317 will be 32.54V - 0.7V = 31.84V. The output of the LM317 in the bulk charge stage will be 14.7V and the charging current will be limited to 500mA. The power dissipated by the LM317 will be then

P_d = ( 31.84V - (14.7V+0.7V+0.6V) ) × 500mA
P_d = ( 31.84 - 16V ) × 500mA
P_d = 7.92W

Does this heat sink will be sufficient to dissipate 7.92W while keeping the junction temperature of LM317 below 125°C? I don't have the thermal resistance of this heat sink.

The tab of LM317 (TO-220 package) is connected to its OUT pin. What I've understood is that if I attach the LM317 with this heat sink without any insulation the heat sink will carry 14.7V DC. And if I happen to touch the heat sink I would get an electric shock. Am I right?

Thank you for reading.

Your power will be a little higher than you calculated. The highest is when the battery is discharged and not at 14.7 volts, but closer to 11 or 12. So maybe 10 watts.
You can get a rough idea of the heat sink using Temp rise in C. per watt = 50 divided by the square root of the heat-sink area in cm. My rough estimate puts yours at about 15C/W. So bigger is better.

 

Change the rectifier to full wave so you get better transformer utilization. To decrease power dissipation, add an external pass transistor, diode, and inductor and convert the LM317 to a switching regulator. The National datasheet has an example.

 

How can I verify this indeed is a 60VA transformer?

"A 60VA transformer" is not a very useful concept here because the load is not linear.

Does this heat sink will be sufficient to dissipate 7.92W while keeping the junction temperature of LM317 below 125°C?

Again, I would look at this from a different angle.

If you keep the input voltage to the regulator around 20 V then the power dissipation goes down dramatically and the problem goes away. I would have used a 18~20 Vac transformer but if you want to use the one you have the only way to find out how it performs is to connect it and test it. The voltages you calculate are not what you will get in the real world. Build the circuit and test it under load. Measure the voltage at the input of the regulator. Due to resistance in the transformer and other elements the voltage will be way below your theoretically calculated voltage. If it is still too high all you need to do to lower it is insert a resistor in series with the transformer output. A very small value will substantially affect the voltage because there are only short spikes of high current to charge the capacitor.

 

Thank you everyone for giving your useful suggestions and answers.

The reason I was going for 12V-0-12V transformer was low AC mains voltage in summers. Due to hot climate of my city there is a heavy use of air conditioners which lowers down the mains voltage in summers. I've even seen mains voltage as low as 165V RMS at my location. The use of 16 or 18V RMS transformer at such low mains voltages would not have been much helpful.

The situation gets different in winters. The mains voltage remains more than 210V. I have bought an 18V transformer which I would use now.

 

Hi,

I have breadboard the battery charger circuit with the power supply shown below.

To calculate the required capacitance I used this formula

1.65V is the peak-to-peak value of the ripple.

Now my understanding is that after rectification the peak value of the voltage across the capacitor should be 25.45V - 1.6V = 23.65V (0.8V per diode)
Taking into the consideration the ripple voltage, the voltage across the capacitor should not drop below 23.65V - 1.65V = 22V if the battery is getting charged at the maximum current limited by the battery charger circuit which is 500mA.

On my breadboard there is a 4700uF 50V (M)85°C capacitor. I connected the battery whose open circuit voltage was 12.2V and took some measurements.

Current out of positive terminal of the capacitor = 385mA
Voltage across the capacitor = 20V

My question is despite using a 4700uF capacitor (higher than the minimum needed) why was the voltage across the capacitor 20V when it should have been 22V? Where are remaining 2V?

Thank you.

 

Did you measure the AC voltage supplied at that time? Or the voltage at the rectifier without any load?

Breadboards are not meant for currents this high. You may have drops in the hookup wires and interconnects within the breadboard.

 

I did measure the AC voltage at the secondary side of the transformer and it was around 18Vrms. The voltage across the capacitor without any load connected was around 25V.

 

The remaining voltage is likely dropped across the resistance of the transformer windings and the external wires.

 

So the breadboard is the culprit. I thought maybe my capacitor is bad.

 

Well, you could have a failing capacitor. The first thing to go on an electrolytic capacitor is the ESR, equivalent series resistance. A capacitor may measure full capacity but still be bad with a high ESR. I believe in your application that would result in some energy being dissipated as heat in the capacitor and therefore some loss of voltage.

The odds are this is not the case and you simply have (had?) inadequate conductors. Did you fix your problem?

 

No. I still think the capacitor is responsible because when I bought it it was in used condition.
I will be making the PCB and will also try to get a new capacitor in unused condition.

 

Your actual diode datasheet should have more info, The 0.6/0.8 V drop is an estimate, The 1N400x series list a max 1.1 V drop.

There is also winding resistance and the transformer itself has a "regulation" value in % associated with it.