Aaron's Homepage Forum
Aaron's Homepage Forum
Home | Profile | Register | Active Topics | Members | Search | FAQ
Username:
Password:
Save Password
Forgot your Password?

 All Forums
 Electronics
 Power Supply
 Automatic 12V Lead Battery Charger
 New Topic  Reply to Topic
 Printer Friendly
Next Page
Author Previous Topic Topic Next Topic
Page: of 12

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 11 2007 :  12:00:48 AM  Show Profile  Reply with Quote
Moderator pls help me. How does this circuit work, especially the function of the SCR and TRIAC? please help me. Your Answer will be very much aprreciated.

Edited by - Aaron Cake on Mar 11 2007 11:53:28 AM

Aaron Cake
Administrator

Canada
6702 Posts

Posted - Mar 11 2007 :  11:53:13 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
I've described in detail how this circuit works before. Try a search for "how battery charger works", "battery charger explanation", etc.

Also, moving to Power Supply.
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 12 2007 :  07:57:00 AM  Show Profile  Reply with Quote
I tried to search that but i cant find the answer to my questions... thanks anyway....

this is the schematic of the charger im talking about...

can anyone tell me how R2, Q1(TRIAC) and Q2(SCR) works?

Q2 drives the LED to turn on right? how?

please help me ^_^

Go to Top of Page

Aaron Cake
Administrator

Canada
6702 Posts

Posted - Mar 12 2007 :  10:13:29 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
OK, I'll do it again since I can't seem to find the explanation either. This will be quick and dirty...

T1 steps down the mains to a little over 12V, D1 rectifies this to a pulsing DC. R1, R2 and R3 form a voltage divider between the + and - of the battery. The output of this divider is set by R2. When the output voltage is not enough to make D3 conduct, Q2 is switched off. Thus, Q1 is switched on via current through R4 and R5. This allows Q1 to conduct into the battery and raise it's voltage. As the battery voltage increases past the point the output of R2 causes D3 to conduct, Q2 is switched on. This pulls current away from D2, pulling pulling current away from the gate of Q1. Q1 thus switches off and charging stops. At the same time, D4 illuminates as the voltage across it increases. C1 helps control oscillation that can occur, R9 makes sure that Q2 is always switched off until D3 starts to conduct.
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 13 2007 :  08:15:52 AM  Show Profile  Reply with Quote
OK thanks for the explanation, it helped me a lot ^_^
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 13 2007 :  08:40:29 AM  Show Profile  Reply with Quote
another question hehhe.... i have jsut constructed this circuit and the output voltage is only arounf 7-9V is it ok?
Go to Top of Page

Aaron Cake
Administrator

Canada
6702 Posts

Posted - Mar 13 2007 :  09:38:47 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
It should measure above 12V with no load. What is the ratings for the T1 that you used?
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 13 2007 :  9:04:40 PM  Show Profile  Reply with Quote
i only used 3amperes transformer T_T
Go to Top of Page

lester
New Member

Philippines
3 Posts

Posted - Mar 14 2007 :  03:39:13 AM  Show Profile  Reply with Quote
i also have a question here. is it possible for me to add another led indicating that the battery is currently charging and will turn off when the battery is fully charged? if yes wer can i place it? thank in advance

lester
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 14 2007 :  08:19:37 AM  Show Profile  Reply with Quote
i only used 12v 3 amperes transformer


Go to Top of Page

Aaron Cake
Administrator

Canada
6702 Posts

Posted - Mar 14 2007 :  10:42:33 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
lester: A LED through a series resistor between teh gate of Q1 and ground will light when the battery is charging. You'll want to find a LED with a very low operating current to avoid switching off Q1.

drilon: What is the AC output voltage of the transformer. If it's at 12V or less, then you may be getting low output voltage due to the drop through D1 and Q1. What happens if you adjust R2 with the circuit connected to a battery?
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 14 2007 :  8:27:53 PM  Show Profile  Reply with Quote
the AC output is around 12.3V.

the LED light up when i adjusted the potentionmeter..... the battery with the voltage of 12.89V.

its like the potentiometer sets the maximum voltage the battery can be charged on...

the battery charge on a rate of around 0.01V per second...
Go to Top of Page

lester
New Member

Philippines
3 Posts

Posted - Mar 14 2007 :  9:54:20 PM  Show Profile  Reply with Quote
You mean it would be connected in parallel with the gate of Q1 and the ground? not in series with the gate and D2? will it turn off wen the charger stops from charging the battery? thank a lot. i really appreciate it..

lester
Go to Top of Page

kivdenn
Nobel Prize Winner

Uganda
535 Posts

Posted - Mar 15 2007 :  08:22:15 AM  Show Profile  Reply with Quote
Hey guys am wondering what will happen if I use a full wave rectifier between the circuitry and the transfomer Thanks in advance
Go to Top of Page

Aaron Cake
Administrator

Canada
6702 Posts

Posted - Mar 15 2007 :  09:08:56 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
drilon: So it's working?

lester: Connect one end of your series resistor to the cathode of D2 (the value of the resistor depends on your LED....If you want to guess, it's hard to go wrong with 1K though it may pull too much current away from Q1...up the value if this is an issue). The other end of the resistor goes to the anode of the LED, and the cathode of the LED goes to ground.

kivdenn: The circuit relies on the pulsing DC of a half wave rectifier to allow the SCR and TRIAC to shut off when the voltage falls to zero. With smooth DC, they would remain clamped on. A full wave UNFILTERED rectifier still should work fine.
Go to Top of Page

drilon
Apprentece

Philippines
9 Posts

Posted - Mar 15 2007 :  8:17:46 PM  Show Profile  Reply with Quote
yup it's working, im just wondering why the output is only around 7-9V

thanks again it helped me a lot =)
Go to Top of Page
Page: of 12 Previous Topic Topic Next Topic  
Next Page
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Aaron's Homepage Forum © 1995-2020 AARONCAKE.NET Go To Top Of Page
This page was generated in 0.12 seconds. Snitz Forums 2000