battery charger

[trpulot]

hi i'ld like to know what the two 100nf capacitors around the 7805 regulator are ment for and if anyother similar value can be used.
also if a put the probes of a multimeter across the points where the batterys should be with the circuit powered, should i expect to read the same value of current that the current meter in the circuit would read if the batteries were connected? i ask because i couldn't lay my hands on a current meter and i want to know if there is another way to measure the current entering the batteries.

View attachment 38685

 

[cjk2]

The caps around the regulator are just to help filter out noise and are really not needed at all.  If you still feel a need to put then in, basicaly any small caps will do.  The section you are talking about uses a 7805 voltage regulator to create a constant current source, so the current should be the same no matter what the voltage is in the batteries.  even if the batteries are replaced with a short circuit, the current should always be the same.  To measure current into the batteries/short where batteries would be, just use the 10 ohm resistor as a kind of "shunt".  Just measure the voltage dropped across the resistor and move the decimal a few places to get current in miliamps.  I forget which direction to move it but just try it and see.  Somebody correct me if any of this advice is bad, but I think it all made sence.

 

[trpulot]

hey thanks so much cjk2 that was a whole lot of help. you've met my needs.
let me ask this if the 10ohm resistor and the 100/1w pot were removed then the 7805 would act as a constant voltage source right? i.e it's the presence of the variable reistor that makes the  7805 act as a constant current source rather than a constant voltage source?

 

[audioguru2]

The 7805 has 5V between its OUT and GND terminals. So if you have the 10 ohm resistor in series with the 100 ohm pot at max, 5V/110 ohms= 45.5mA. With the pot at minimum, 5V/10 ohms= 500mA. If the GND terminal is grounded, then the output is a regulated 5V with up to 1.5A.

 

[trpulot]

thanks again but you just brought up another question
what s the 220 resistor doing acrosss out and ground of the LM317 as well as the diode?
i have ideas but i don't want to speculate i want someone who knows to tell me.

 

[audioguru2]

what s the 220 resistor doing acrosss out and ground of the LM317?

The 220 ohm resistor has 1.2V across it and its current is in the pot for the LM317 to have a variable regulated output voltage. Look at the datasheet for the LM317.

the diode?

The diode isn't needed. It protects the LM317 from the output capacitor discharging into its output in the very unlikely event of the input suddenly shorting to ground.

 

[ante1]

Just a note:
The 10 ohm resistor will dissipate 2.5W at 500mA, I don’t know if it’s rated for this in the part list. It’s not rated at the drawing as the potentiometer is. I would use a 5W in this circuit! 8)

 

[audioguru2]

This discussion is about the project for it in our Projects Section:
http://www.electronics-lab.com/projects/power/035/index.html

I wonder why it was moved to this forum Theory???
I think it should be in the forum Projects Q/A.

 

[ante1]

Hi AG,

Yeah, there are thing which don’t add up but as long as it’s not deleted I suppose we have to be grateful!