capacitorcharger
I have a 5V DC supply that outputs at most 1.5A, and I want to charge a 25F capacitor (BCAP0025-T01) at 2.5V and close to 1.5A.
How can I build a circuit to achieve this? Current mirror maybe?
Apologies if this is lacking in information. Feel free to make assumptions and throw out ideas; I'll be happy to do further research given some leads.
You can use existing commercial 12V to USB (= 12V to 5V effectively) converters or ANY "X" Volt to USB converter already available.
Some 12v-USB units use a linear regulator so have efficiency of 5V/12V ~= 40% :-(.
Many use some form of switching regulator.
These are typically around 80 - 85% efficient but more or less is also possible. I have seen some which use the venerable MC34063, and there are many other choices.
MC34063 datasheet
Energy wasted:
At 80% you waste 20% of the energy.
At 40% see above) you waste 60%.
So a not overly efficient smps wastes 3 x less energy than a linear regulator.
Power input
For 1 unit of power output (say 1 Watt)
A 40% efficient unit needs 1W/0.40 = 2.5 Watts.
An 80% efficient unit needs 1W/0.8 = 1.25 W.
So a linear regulator needs 2.5 / 1.25 = twice the wattage
Note that power needed and energy wasted are related but not in the same ratio between linear and smps supplies.
If you want to try do-it-yourself the venerable MC34063 is not as efficient as some modern ICs but is reasonably close and is cheap and very flexible.
DIY MC34063 examples at end.
Lead acid battery: If you have complete control over the design then a good choice can be a 6V sealed lad acid gel cell. These give good energy per $ and can last quite well if treated well.
Even with a linear regulator a 6V battery will give and efficiency of 5V/6V ~= 83%- so you can do better with a smps but not vastly so in many cases.
eg a 90% efficient smps uses 83/90 =!~ 8% less Wattage than a linear supply from 6V. As voltage drops under 6V the linear regulator efficiency increases.
To use a 6V battery for 5V output you need a low low low dropout regulator. This can be achieved with a P Channel MOSFET and one opamp section.
Lead Acid discharge voltage: The curves below are indicative only of voltages expected from a lead acid battery. These were for one brand of large deep discharge cells and small gel cells will be different BUT it gives an idea of what can be expected. If you use a 6V 7Ah battery then if the USB output is 1A (= charge a 1000 mAh cellphone battery at max allowed rate. then the peak discharge rate of 1A = 1A/7 Ah = C/7.
End point voltage for a 6V battery would be ~= 5.25V = ample.
If you charge EyeFones or EyePadz with battery capacities of about 1600 mAh or even 3000 mAh then you may ant 1.6A and 3A peak. Even 3A is 3A/7Ah ~= C/2.5.
At C/2.5 battery voltage would be about 10.5V at about 70% of capacity expended - still very useful and it helps lifetime to not discharge deeply.
At 3Ah from a 7Ah battery you could notionally charge 2 x 3Ah batteries.
Or maybe 5 or 6 x 1000 mAh cellphone batteries.
For lead acid, battery brand matters. Yuasa are excellent and actually have several grades aimed at different uses and price ranges and good data is available.
Panasonic are good. Just about anything made by Panasonic is good :-) (I have zero involvement with them). Some others are good but cheap batteries may be very bad. Or not.
MC34063: Datasheet & application guide {again}
Example below is 25 V in, 5V 500 mA out. Would take 12V in as is except halve Rsc and ensure inductor suitable core size. Efficiency of ~84% may be slightly lower at 12V in.
Count the external parts!.
While there may be a very few more than in some minimalist designs the same IC can do Boost, buck boost and anything else wanted.
PCB layouts - see data sheet for boost and inverting circuits.
I would NOT design anything that would require direct AC power if I were you. Really. You can design your circuit to take 12/24VDC or whatever you need from a wall wart and then use that to charge your D cell.
Seriously. Do not go near the AC outlet. There are lots of design safety issues to consider, the first of which is using an isolating transformer.
Oh one more thing. Do not design anything that takes AC power. Don't.
As I said you can get a wall wart, or use any of the adapters you have sitting around if it supplies the DC voltage/amperage you need (which it should be able to) just use that.
There are lots of NiMH charger circuits out there you can use for reference. Your charging current is a function of your battery capacity, often expressed in terms like 0.1C meaning if your battery is 1Ah you want to charge it at 100mA.
You want a smart charger because NiMH are sensitive to charging current and voltage and have differing requirements depending on how discharged they are. If you ignore those you'll damage your battery.
I am fairly certain you should be able to find a commercial charger that can handle your D-cell. It may not be physically compatible in which case a set of alligator clips/wires can help you connect the two.
Best Answer
A current mirror is useless for this purpose.
Have a look at Julian Ilett's channel on YouTube
He likes to do stuff with supercapacitors and has several videos on the subject.
In general he uses an adjustable power supply module with adjustable current limit. He uses a powerful one which will be overkill for your problem.
I would use a module like this. Which can convert 5 V into 2.5 V and where the maximum current can also be set.
Since this is a switching regulator you can even charge with a higher current at the output, up to 2.5 A or so, without overloading the 5 V 1.5 A supply.