9V charger circuit

K

kam

Member
Hey all,

Here's the scenario:

I am making a clock with a 20X4 LCD module.

I thought that i'd use a 9V rechargeable batt and use a LM7805 to drop down and regulate the voltage. Considering the LCD will be running all the time and the Backlight will be on as well the batteries are bound to run out soon.

so that means that i need 9V DC input to recharge it... right? which is not a problem...

====
End Scenario!

Problem:

1) if i have it hooked with a DC input, i cant move it around and place it in different places in my room. considering how long and often will i have to do it.

Questions:


How will i be able to indicate a low batt signal on the LCD, i mean if i connect it to the ADC of the chip isnt 9V gonna damage it.

Also what kind of recharge circuit do i need to make, i have no idea and googling it didnt help much, maybe i wasnt looking at the right stuff.

Plus how long will it take to recharge it.

finally, Is it worth going through the trouble of adding a recharge circuit. i mean if the battery supposedly lasts a while (as in a month or so), wont it be easier to just take it out and put it in a recharger?
 
M

moxhamj

New Member
Sounds a fun project. 9V rechargeable batteries are made up of 1.2V cells - cheap ones used 6 to give a "nominal" 9V which was really 7.2V. Proper ones use 7 cells to get 8.4V. This is important as it determines the cutoff you use for determining when to recharge. 8V is probably a good starting point.

Detecting 8V with a 5V picaxe? Use a voltage divider. Divide the volts by two (use 2 10k resistors in series). Now it is 4V and you can do the maths internally in the picaxe. You could display the volts on the LCD.

Recharging is easy and you just need a higher voltage and a resistor and some maths. Say you have a 12V supply eg a wall wart or a proper power supply. First, determine if it really is 12V (I have some 9V ones that actually put out 13V under light load). Next, look at your battery and it will have the amp hour capacity on the side. A 9V battery might have 200mah. If you used AA cells, they could be 2000mah. Rule of thumb - divide that number by 10 to get the charge rate. Divide by 10 gives a charge time of about 10-12 hours, which is not only convenient but also means the cells don't overheat if you forget to turn the charger off (?picaxe controlled 10 hour timer).

Your battery is 200mah, so that is 20ma. 12V supply. 8.4V battery = 12-8.4=3.6V. 20mA is 0.02A. V=IR and you want to know R, so R=V/I which is 3.6/0.02 which is a 180 ohm resistor. Maybe use a half watt one - 1/4w might get a little warm. One resistor will cost 10c or less. It depends if you have a wall wart handy or other supply handy.

Or --- you could take it out and put it in a recharger. If a 9V battery does not last long enough, use 7 AA batteries instead as they will last 10x longer. 7 AAs might only be a few more dollars than one 9V - for some reason the 9V rechargeables are very expensive. It would depend on how small the box needs to be.
 
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A

Andrew Cowan

Senior Member
The battery will run out very quickly. My 20x4 LCD uses 200mA with the backlight on - your whole system will probably use 250mA or so.

If the battery capacity is 170mA (pretty normal for cheap 9V cells), it won't last very long.

A
 
K

kam

Member
Thanks Guys,

Really Helpful.

Dr.A, just one question. if we drop the voltage down to 4V with the Voltage Divider, would it be directly proportional to 8V?

If so then my understanding is that with every voltage drop, it will show the respective half voltage. i.e:

if its 8V then it shows 4V
if its 7V then it shows 3V
if its 6V then it shows 2V
if its 5V then it shows 1V

So i guess when it shows 1V on the ADC, i can set a Low Battery Alarm On the LCD.

Thinking about the battery i should use, i guess it'd be convenient to use the 7xAA ones. they would last longer. and in a way if one goes bad, i need to replace only that one.

I have a few 9V DC wall adaptors (Which do show 13V on the meter :) ). I'll use one of them. So to charge the cells while in the circuit and the system running, i just need to connect them in series or have the cell pack taped in parallel?

Thanks again

Kam
 
M

moxhamj

New Member
Yes you can (maybe) use your 13V supply. It would probably work with 20mA. BUT - if you went to AAs- you might want to charge at C/10 which might be 200mA. Ie an 18 ohm resistor. Which could well drop the volts down to 9V and then it won't charge very quickly. And you will need a higher wattage resistor. But you get the idea. Or charge at a lower rate and accept a longer charge time. Mind you if it is drawing 200mA like Andrew Cowan's and you are charging at 20mA it will never charge up! Do you have a multimeter? Then you can measure the charge current. Sometimes it is easier to drop in various resistors and measure the current and select the one that seems about right. It very much depends on the wattage of the wall wart and how the supply volts fall as the current draw increases. Real power supplies give constant volts over a wide range of current draws. Wall warts don't. 1A wall warts maintain their regulation better than 100mA ones. Modern ones might be switch mode and better regulated. But it could come down to a simple resistor.

Re your little table - voltage dividers divide the volts, not subtract them:
8V = 4V
7V=3.5V
6V=3V
5V=2.5V
 
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Andrew Cowan

Senior Member
As Dr Acula, unregulated supplies are jus made of a transformer and rectifier (and capacitior if you are lucky).

This means that for a 9V power supply, you could find:
At 0-10mA, such as drawn by a multimeter, it will give out 13V
At the specified current draw (marked on the power supply), the voltage will be 9V.

Therefore, you should use a power supply with a current draw of about 200mA more than you need.

Just a thought - you are dropping from 10ish volts to 5V, at 250mA. That means the regulator is disipating 1.25W. Is it in a TO-92 package (300mW max dissipation)?

A
 
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K

kam

Member
Thanks for the correction Doc. forgot about that

Hmm, the 5V regulator is a LM 7805, comes in a TO-220 package. But I'll add a heat sink to it anyway... i'd be safe than sorry...

Hmm, I guess a bigger resistor it is then :)

Just wondering, if there's any other way I can damage the circuit, i want to be safe and rather add extra fail safe features than letting this thing melt my clock in my room, especially if i'm not around to put it off! :eek:

Thanks again Guys
 
A

Andrew Cowan

Senior Member
What are your outputs? If any of them are logic (not power), you can put resistors on them to limit current (in case something unexpected happens).

A
 
fernando_g

fernando_g

Senior Member
kam said:
Thanks for the correction Doc. forgot about that

Just wondering, if there's any other way I can damage the circuit, i want to be safe and rather add extra fail safe features than letting this thing melt my clock in my room, especially if i'm not around to put it off! :eek:

Thanks again Guys
Click to expand...
Simple solution: A Polyswitch type PTC resistor (like those manufactured by Raychem), is an excellent idea wherever unattended rechargeable battery packs are used.

Or: You could get fancy and use a real temperature sensor, and via the Picaxe, use it to disable the charger if things get too hot.
 
K

kam

Member
Hmm, Got a few ideas running now. Thanks

Sorry I've been away for a while.

I'll add resistors to the LCD I/Os just in case and any other logic level pin i use.

At a time i thought that i should just buy a battery charger and take it out of its casing, fit it in the Box and let it be, doesnt seem like a bad idea, considering the Branded one's would have put some safety features, better than i would...

Dick Smith has a few:

http://www.dse.com.au/cgi-bin/dse.storefront/499c20b602760976273fc0a87e01067f/Product/View/S0161

And then There's this one:

http://www.dse.com.au/cgi-bin/dse.storefront/499c20b602760976273fc0a87e01067f/Product/View/S3492

Just an idea.
 
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