hello

I have a transformer that has two 0-15v and 6VA. can I use this to make a 30volt supply and then drop it to 18volt. I need to power a battery charger.

Lead Acid Battery Charger Circuit Diagram

this is the circuit I am using. im not sure what current is required.
any help would be great

thanks
simon

 

hello

I have a transformer that has two 0-15v and 6VA. can I use this to make a 30volt supply and then drop it to 18volt. I need to power a battery charger.

Lead Acid Battery Charger Circuit Diagram

this is the circuit I am using. im not sure what current is required.
any help would be great

thanks
simon

Looking at your circuit, If R1 = 0.5Ω then the circuit will limit the current to approximately 1.4A. The combined values of R2 and R5(1.47KΩ) along with R3 will make the maximum voltage of approximately 16.5V. With these values I assume the battery is a 12V battery and will be charged with the circuit.
If your transformer has 2 winding of 15V @ 5VA then the transformer may not be suitable for the following reason. 5VA=5/15 or 0.33 amps. The rectified voltage with a capacitor filter would be approximately 20V, which is okey for the input voltage to the charger. If it were me I would change R1 to 1.2 ohms and put both windings in parallel to get 0.66A to the charger.

 

.................... If it were me I would change R1 to 1.2 ohms and put both windings in parallel to get 0.66A to the charger.

15VAC will give near 20V DC with a bridge rectifier, which is sufficient to power the regulator. But if the transformer (not each winding) is rated 6VA, that would only give about 400mA RMS total AC current. That would need to be derated further to account for the high peak current drawn by a rectifier-capacitor supply, so you should limit the DC current to probably no more than 250mA giving a value of 2.8Ω for R1.

 

15VAC will give near 20V DC with a bridge rectifier, which is sufficient to power the regulator. But if the transformer (not each winding) is rated 6VA, that would only give about 400mA RMS total AC current. That would need to be derated further to account for the high peak current drawn by a rectifier-capacitor supply, so you should limit the DC current to probably no more than 250mA giving a value of 2.8Ω for R1.

You are correct, thanks for correcting me.

 

hello
thanks for the replies, I have got a laptop power supply that has 19v with a couple of amps. im not sure if this is ok, the circuit says a minimum of 18v.
I imagine the circuit will only get what it needs and not the full amount of current provided?
12v? no, I think this is for a 6v SLA battery. that's what it says on the web site.
http://www.circuitstoday.com/lead-acid-battery-charger

if this is for 12v how could I adapt it to 6v?

also, how do you calculate the results that you all have provided. I find a lot of this stuff confusing. a point in the right direction would be great!

thanks

simon

 

I find a lot of this stuff confusing.

Can you elaborate on which things are confusing? A few basic concepts have already been seen in this thread:
Ohm's law (voltage drop across the shunt resistor)
Conservation of energy (relating power inputs to outputs and losses)
Math(peak of a sine wave is 1.41X the average value, so rectified voltage=1.41xapplied AC voltage, less diode drop)
Diode properties (the voltage drop across a bridge).

 

And the conversion factor used when going from transformer RMS output current to maximum rectified DC output current is typically about 0.6.

 

The short answer is the laptop power supply should be perfectly OK for this.

 

hello
thanks for the replies, I have got a laptop power supply that has 19v with a couple of amps. im not sure if this is ok, the circuit says a minimum of 18v.
I imagine the circuit will only get what it needs and not the full amount of current provided?
12v? no, I think this is for a 6v SLA battery. that's what it says on the web site.
http://www.circuitstoday.com/lead-acid-battery-charger

if this is for 12v how could I adapt it to 6v?

also, how do you calculate the results that you all have provided. I find a lot of this stuff confusing. a point in the right direction would be great!

thanks

simon

The circuit will work for either a 6V or a 12V battery. R5 has to be adjusted for the correct battery. With a laptop power supply the LM317 will require a heatsink(to dissipate the wasted power) The maximum current is calculated by I=0.7/R1(the 0.7 varies a little depending on actual transistor BE voltage drop) I use 0.7 as a rule of thumb. The output voltage is calculated by Vo=1.25((1+(R5+R2)/R3)).