I suggest, for starters, that you get a fresh 9V alkaline battery and test to see if it will light the 12V LED strip with sufficient brightness.

 

@ScottWang this is the multimeter, i found this useful as well.
@Alec_t it seems to be lighting up well, though its not as bright. the 12V circuit seems to be the brightest. i would have loved it if it were brighter.
see attached picture. I picked up a rechargeable Ni-Mh 300mAh 9V battery and it seems to light up too.

i cant wait to try the circuit, you guys have all been a big help.

 

If your led strip uses a constant current source, the brightness will be the same for all voltages, if its a fixed series resistor type, then the brightness will vary.

 

@Dodgydave So is it possible to have a Constant current source with a 9V battery circuit?

 

@ScottWang sorry i forgot to include the linkhttp://www.hobby-hour.com/electronics/dt830b-digital-multimeter.php

http://www.hobby-hour.com/electronics/dt830b-digital-multimeter.php

 

We all know what that meter is. It's the $4 Cen-Tek meter you get for free at Harbor Freight on coupon Sundays.
I'll be there in a hour. Need more tools!

 

@Dodgydave So is it possible to have a Constant current source with a 9V battery circuit?

Sure it's possible. You can have a constant current source with any voltage available -- the voltage just establishes the limits over which the current source can operate. We had a 2 V battery-powered current source that could go up to 1000 A.

 

@WBahn Wow, thanks! So how would I be able to apply that to my circuit?

 

Measuring the current:
+V → (+) Ameter (-) → (+) Led strip (-) → Gnd(or 0V)
The measuring range turn to 10A and the probe also insert to 10A hole as below.

 

@WBahn Wow, thanks! So how would I be able to apply that to my circuit?

Apply what to your circuit?

If you use the circuit you show above, your 12 V source will push current into your 9 V alkaline battery (via the 100 Ω resistor) right up to the point where the battery bursts or explodes. Are you sure that's what you want?

 

The fact that constant current circuits exist doesn't mean you need one or want one.
You don't have enough voltage to use it up in a constant current circuit which accomplishes nothing except a steady light level. In this case, you are trying to use an 8.4 volt nicad on a circuit that won't work with 6 volts and the result would be a constant level of no brightness at all.

 

@WBahn in the diagram, i mistakenly put an alkaline battery , my circuit has a Ni-Mh 9 V battery.
@#12 thanks for the info, i tend to try and fit in every possible thing.

I made the circuit with the 9V rechargeable and i notice that my 100ohm resistor is really heating up. is that normal??

 

I made the circuit with the 9V rechargeable and i notice that my 100ohm resistor is really heating up. is that normal??

No. 3.6 to 5 volts of difference should inflict about 1/8th to 1/4th of a watt on that resistor. Did you use a really tiny resistor?

 

@#12 its as small as the diode

 

That looks like a 1/4 watt resistor. If the NiCad is low you might be getting it as hot as its design limit. Check the voltage across the resistor. It it's more than 5 volts you have a problem.

 

@#12 @ScottWang thanks for the help , i'll have the info soon, seemed to have busted the multimeter.

 

The walls of resistor's value should be 5 times of the calculation.
I = E/R
W = E*I

 

Measure the voltage drop across the resistor, using this formula will give you the wattage and current limits.

W=V*V/R. (wattage)

I=V/R. ( current)

Edit, a 100 ohms resistor will give a maximum current of 120mA, and will need to be 1.5W minimum.

 

Measure the voltage drop across the resistor, using this formula will give you the wattage and current limits.

W=V*V/R. (wattage)

I=V/R. ( current)

Edit, a 100 ohms resistor will give a maximum current of 120mA, and will need to be 1.5W minimum.

Why do you think that the rechargeable battery will be at 0 V? This is what would be required to get 120 mA of current through a 100 Ω resistor.

 

I don't think it will be at zero, it will probably be around 8v,so the current would be around 40mA. I said 120mA would be the maximum if the battery was shorted out.