Is there a simple circuit to charge SLA batteries 4.5 AH without getting too complicated?
Most of the circuits Ive seen have difficult to get parts ...I do have some 2N3725s(NPN) transistors tho would they work?
A couple on this site for around $10, ready made. California company, if you live near by, no shipping. Most of the build your owns I've seen cost more, unless you have a heap of used parts to scrounge through.
Thanks for the link..as it happens I do have alot of old parts to sift thru.
but the parts being old and the markings faded its not easy to tell what your looking at sometimes..
I did manage to find the right parts for the astable multivibrator circuit..works beautifully now..
I guess I should invest in a good cross reference..as I had never heard of a BC-547.
I didn't know know you could salvage the gold off old transistors..maybe I should go prospecting
General purpose small signal transistors come in a lot of families, but the BC 547 and its close relatives are usually a good match for 2N3904/2N3906 and their close relatives. Fortunately the families are groped on one data sheet, so you only need a few data sheets.
I'm told that old transistors and ICs that were made when the US gold market was locked at $35 per ounce (iirc, before about 1975) can contain some salvageable gold if you know the right people. Unfortunately I don't know the right people.
It may be more simple than you wanted but it would work.
So get yourself a power supply that provides 12V (or battery max charge limit), then find a resistor that will be able to to limit the charge current to a level your power supply can handle.
Thats it, the resistor will prevent over current conditions when charging a discharged battery. then as the battery potential rises as it is charged the current will naturaly drop off to zero when the battery potential is equal to the power supply. An ammeter could be used as an indication of charge status.
I am a little confused here..the half-wave power supplly has 2 resistors called 1R8...is that 1 ohm?...I noticed they were in parrallel..sorry for being ignorant
1R8 = 1.8Ω and since there are 2 in parallel, the total resistance is 0.9Ω.
It is quite common to use R, K or M in place of the decimal in resistor values. So a 4.7KΩ resistor could be written as 4K7 and a 3.9MΩ as 3M9. It saves a character and eliminates errors after multiple photocopies of the schematic make the dots disappear.
Thanks for the clarification..I often add resistors in series to get the values I don"t have.
Would that resistor pair have to be point 9 ohms?..I do have a nice 1 ohm 5 watt resistor that might work
The 1Ω resistor will work fine. You'll just get 10% less charging current. To get 0.909Ω you could put a 10Ω 1/2W resistor in parallel with your 1Ω 5W one but I wouldn't bother as it is close enough.