Me and my university friend just managed to make two quantities of devices, both for the same purpose to provide 12 volt, 1 ampere to dc pin out which are attached to my internet modem and tp-link router also. its working good as according And it requires following stuff mentioned below:



-- Voltage Regulator KIA 7812 ( Fixed output voltage 12V,1A)

-- 6.0A bridge rectifier KBJ608G ( low reverse leakage current )

-- Ac transformer ( output 12volt, 1A)

-- cut off circuit ( locally made in the market )

-- 6 volts lead-acid battery (4.5A)



I could get a 12 volt single battery but that was not available in the house, instead i used two 6 volts batteries, so we mostly collected stuff that was available and to get easily from any nearest electronics shop without taking an extra effort to make it more accurate project. because i am not related into this field much so i am here to discuss it more extensively and more practically in terms of electrical engineering.five Pictures of the first device are attached here,named as a1,a2,a3,a4,a5 and the second device's pictures are also attached here with names as z6,z7,z8. well the first device that we built is similar to other One . but only difference is that we attached two capacitors in the circuit board.In second device i am not really into understanding that Is there any need to wire 35v 33uf electrolytic capacitor
between positive and negative terminal of cut off circuit ? and also there is attached a yellowish film capacitor 104K in the circuit board. You can see these attached pictures here.
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Also let me know of common mistakes that we make when we do all these in home. like we did while making the second device,

we simply glued voltage regulators and recitifier to the upper side of transformer. but they usually get heat up, and may be thats why in complicated circuits these stuff are linked with heat sinks as i saw in many other devices. well i just checked output voltage in the first device that was 11.3 Volts at dc output pin and in second device it shows as 10.4 volts. but when i connect red probe of multimeter to the left most pin of rectifier and black probe to the right most pin of rectifier it shows 12.1 ,is there any drop out voltage using these voltage regulators, i mean then why are we using these in here?
and how to check properly working of cut off circuit, i mean what should be the inverting and non inverting voltage at pin 2 and pin3 at IC (UA741) attached here in "cut off cicuit a" picture and also did you observe a light blue variable resistor trimmer potentiometer at the lowermost right end corner. so how to adjust this potentiometer?

 

..is there any drop out voltage using these voltage regulators, i mean then why are we using these in here?

Any regulator has to have some dropout. Depending on the regulator, it could be several volts. A schematic would be needed to comment further.

Only you can answer the second question. Why indeed.

 

Any regulator has to have some dropout. Depending on the regulator, it could be several volts. A schematic would be needed to comment further.

Only you can answer the second question. Why indeed.

yeah u r probably right. but plzz assist me how to draw schematic diagram for this circuit and yeah i am not planning to do in a hard copy but is there any software that i can use to test this circuit and make a sketch too? , you guys are more related to these kinda stuff , plz guide me in a proper way

 

perhaps i am the one alone here, No one here to discuss this with me just because i am a programmer, not related to this field.

 

The 7812 Needs to be on a reasonable heatsink if it is loaded even at 500mA.
Second the 7812 is not a low drop out which means the input output voltage should be atleast 5V.
The rectifier output should be around 17VDC. So the transformer should provide 12VAC at 1 amp. You should have a reasonable smoothing cap if u intend to charge batteries. A cap of atleast 2200uf25VDC to 4700uf25VDC.

The regulators should have decoupling caps at input and output.

From ur post I assume u are not getting 12V output from regulators. Two reasons.
1.Ur charger is drawing more than 1 amp results in low output
2. Ur regulators are going into thermal shutdown.

One way is to bolt the regulators to the casing. For this u need to prepare the surface. Scrape the paint off the casing to get a smooth surface so tht the regulator can make good thermal contact to the casing. Use heatsink compound to mount the regulators.

U might need to use mica between the regs and casing. But this depends on ur ckt configuration .

The charger needs to draw less than 1 amp during charging. More than 1 amp and the regulators goes into current limit and ur batteries won't charge fully

 

The devices you are powering are not voltage sensitive by any means.

Typically they have a small bucking voltage converter that reduces the 12 volt input power down to 5 and 3.3 VDC for the actual circuits.

Because of that they tend to work just fine on any input voltage from ~9 VDDC to ~20 VDC which means that they would have no problems being powered directly off the battery plus it would save you a fair amount of wasted power that being lost across the voltage regulators.

As long as you have charging circuit and power source that can keep up with the load of the devices while providing the necessary power to recharge the batteries fully but not over charge them you should be good.

 

is there any software that i can use to test this circuit and make a sketch too?

I use LTspice (a free download from Linear Technology) for schematic drawing and circuit simulation. Additional simulation models are freely available from the Yahoo LTspice User Group.

 

Any regulator has to have some dropout. Depending on the regulator, it could be several volts. A schematic would be needed to comment further.

Only you can answer the second question. Why indeed.

you are right , i am attaching schematic diagram of cut off circuit .

The 7812 Needs to be on a reasonable heatsink if it is loaded even at 500mA.
Second the 7812 is not a low drop out which means the input output voltage should be atleast 5V.
The rectifier output should be around 17VDC. So the transformer should provide 12VAC at 1 amp. You should have a reasonable smoothing cap if u intend to charge batteries. A cap of atleast 2200uf25VDC to 4700uf25VDC.

The regulators should have decoupling caps at input and output.

From ur post I assume u are not getting 12V output from regulators. Two reasons.
1.Ur charger is drawing more than 1 amp results in low output
2. Ur regulators are going into thermal shutdown.

One way is to bolt the regulators to the casing. For this u need to prepare the surface. Scrape the paint off the casing to get a smooth surface so tht the regulator can make good thermal contact to the casing. Use heatsink compound to mount the regulators.

U might need to use mica between the regs and casing. But this depends on ur ckt configuration .

The charger needs to draw less than 1 amp during charging. More than 1 amp and the regulators goes into current limit and ur batteries won't charge fully

yeah right, but i am getting 12 volt from rectifier bridge, but i should take someconsiderations before bolting voltage regulators in to the casing, anyhow plzz see this below attached schematic picture of cut off circuit .

The devices you are powering are not voltage sensitive by any means.

Typically they have a small bucking voltage converter that reduces the 12 volt input power down to 5 and 3.3 VDC for the actual circuits.

Because of that they tend to work just fine on any input voltage from ~9 VDDC to ~20 VDC which means that they would have no problems being powered directly off the battery plus it would save you a fair amount of wasted power that being lost across the voltage regulators.

As long as you have charging circuit and power source that can keep up with the load of the devices while providing the necessary power to recharge the batteries fully but not over charge them you should be good.

I use LTspice (a free download from Linear Technology) for schematic drawing and circuit simulation. Additional simulation models are freely available from the Yahoo LTspice User Group.

yeah i used proteus 8 instead, and i am posting the schematic diagram now, sorry i was away in other city for final papers, so thats why i couldn't post it, see below the attached picture of schematic diagram of cut off circuit.

 

i am totally not aware of how to make circuit diagram but i tried it a software "proteus 8",if you want i can give you its project file included workspace.but here its only the circuit diagram of cut off circuit , in the first post i attached bundles of pictures, there is a picture named as "cut off circuit a" thats the picture for whose i am attaching its schematic diagram here now. i doubt that this circuit does not do the required function as it should do because it does not cut off 12 volt transformer from 220volt power source when it fully charges the battery, as i am charging it for more than 8 hours, i think its green LED should blink when the battery is fully charged, don't you think guys? because in this cutt off circuit i attached transformer wire to the COM of relay and 220volt power source wire to the NC (normally closed) of relay as you can see in this picture named as "a4", should i adjust the trimmer potentiometer (VR10K) ? its light blue in color at lower most right side corner .if i am saying it right as variable resistor. You can again see these bundles of pictures here too attached pictures here

 

One whole project includes a 12volt transformer, a rectifier bridge, a 12 volt voltage

regulator,a cut off kit and a couple of batteries for 12 volts. Now i tell you how i

combined all of these components to get my required 12volt,1ampere output to the DC pinout

attached to my tp-link router.here transformer has two wires of input and two wires of output ,

i attached 220volt power source to one wire of input of transformer and i attached second wire of input of

transformer to the COM of relay in the cut off kit board, and then i attached the other

left 220v power source wire to the NC of relay in cut off kit.so basically relay should

cut off the power source from the transformer when there is no need of charging lead-acid

batteries. that is the case when the batteries are fully charged.

here probably i used a 6 or 10 ampere bridge rectifier that has four terminals, two terminals of input

and two terminals of output,i attached two wires of output of transformer to the two terminals of input of bridge rectifier

and attached two terminals of output of rectifier to the + and - terminals of lead-acid battery

accordingly. and then i attached a 12v voltage regulator that has three terminals, one is

input and second is common ( i say it as negative) and third is output, i attached

positive output terminal of rectifier bridge with input of voltage regulator and negative output terminal of

rectifier bridge with common of voltage regulator while the last third output terminal goes to one of the wire of DC

pin out attached to my tp-link router. but DC pin out has to have ground current too to

give current to my tp-link router, so i attached other wire of dc pin to the negative of

electrolytic capacitor in cut off kit board (you can see this in picture named as" cut off

circuit a" right lower side electrolytic capacitor) beacause two wires are coming out from a dc pin-out .

and also i attached two terminals of output of rectifier bridge with the cut off kit board at its

positive and negative terminals to activate it and control the relay and whole mechanism in cut off kit board.and also attached a push button in between to activate and de-activate the cut off kit.you can see this in the pictures.

well when 220v power source is given to this project then its red and green LED shows up but according to my theory when batteries get fully charged then this green led should go off and yellow led should be brighten up.what do you say? .




but now i need a change in my project, well i don't need this output 12volt,1ampere, instead i need to charge my android phone samsung galaxy s3 LTE.i think which requires a 5 volt and 1 ampere, or i could run a wirless 3g enabled wifi usb dongle device which also requires 5 volt and 1ampere. so suggest me what should i do? , what kind of voltage regulator should i add to get my required results.?? is there any need to remove one lead-acid battery of 6volts ?, i simply want 5volt,1ampere without changing this project so much or without removing extra battery if not neccessarily. tell me. ? is there any new kind of rectifier bridge should be added?

 

No One here, responding

 

I would have as others but may be they are a bit busy.
I am quite busy and I need to study your diagram to reply properly.

People here are not to respond immediately.
Us member's have work to do and we help only we got time.

You should be patient.

 

A simple and traditional approach would be to put a 7805 regulator in parallel with your 12V supply, as if the latter was not present. The 7805 will need a good heat sink, and may not be able to withstand 1A continuous for long periods. But perhaps you don't really need that much current all the time.

The downside of any linear regulator is heat dissipation and thus efficiency loss in the regulator, but perhaps this is of little concern.

A more elegant approach would be to simply buy on e-bay one of the many DC-DC buck converters available inexpensively there. Find one that meets or exceed your specifications. Very cheap and much more efficient. No heat.

 

I would have as others but may be they are a bit busy.
I am quite busy and I need to study your diagram to reply properly.

People here are not to respond immediately.
Us member's have work to do and we help only we got time.

You should be patient.

yeah i am waiting for you now till you get home

A simple and traditional approach would be to put a 7805 regulator in parallel with your 12V supply, as if the latter was not present. The 7805 will need a good heat sink, and may not be able to withstand 1A continuous for long periods. But perhaps you don't really need that much current all the time.

The downside of any linear regulator is heat dissipation and thus efficiency loss in the regulator, but perhaps this is of little concern.

A more elegant approach would be to simply buy on e-bay one of the many DC-DC buck converters available inexpensively there. Find one that meets or exceed your specifications. Very cheap and much more efficient. No heat.

yeah thanks, i today saw this regulator on my other battery backup device, the voltage regulator was L7805CV ( CCOKJ V6 MAR 1212) BUT i think due to mis-configuration i get 8.99 or 9 volt at right most pin of a standard USB , pin 1 and 4 should be 5 volt and ground. but i am getting high, no idea why in my second device. anyhow i try to find another way or should i purchase a power bank. well seriously i have no idea of DC-DC buck converters .

 

where are you , i started this thread for so long ago to get more brief on my little project of battery back, to memorize the important concepts andndetails

 

No ONe is here to discuss ???

 

Exactly what do you want to discuss? A DC-DC buck converter has already been suggested.
A schematic of your system would be helpful.

 

Exactly what do you want to discuss? A DC-DC buck converter has already been suggested.
A schematic of your system would be helpful.

i already sent you schematic of my cut off kit, what else you need, so all in all you suggest me dc-dc buck converter , ok, i try to arrange that, but is there any specific for my project ? 5 volt dc-dc buck converter?

 

i simply want 5volt,1ampere

Yes, a DC-DC buck converter with a 5V output.