does a 9V battery have a current limit?

I am doing a project with a high power led and i am using a 9V battery as the supply. The problem is, i designed the circuit to have almost 700 mA (670mA) to be exact and im only reading 300-330 mA...

I am using a 10W 9.1 ohm resistor which i measured to be 9.6 ohms.

i measured the battery and found it to be 9.68V.

all the components are brand new.


E= 9.68V
R= 9.6 Ohms

9.68V - 3.25Vf = 6.43V

I= E/R

6.43/9.6 = 670 mA....

why am i only measuring 330mA?

this is the led i have:
http://www.luxeonstar.com/Green-Lamb...-pm01-0070.htm

i just read somewhere that without a heat sink that a high power led will be limited to around 100 mA..... Could this be my problem? It was a little warm but definitely not hot.

700mA sounds like a lot of current for a little 9V battery. How hot is the battery getting?

Batteries have a internal resistance that will limit their output current. This internal resistance gets worse as the battery discharges.

battery wasnt hot at all but i was only getting between 300 and 330 mA....

i just read online that some high power leds currents are limited if there is no heat sink. also i just looked on the led part spec on my led and it says...

" The thermal pad is electrically isolated from the anode and cathode contact pads."

im not entirely sure what that means, but im pretty sure i need a heatsink

A heat sink will not increase the current. What that means is if you run it without a heat sink at it's max current it will blow up.


Also, did you measure the battery voltage open circuit or under load?

If you'll take a look at this, I think you can answer your own question.


http://www.powerstream.com/9V-Alkaline-tests.htm


If you recalculated your current limiting resistor for a 6V supply voltage, you might have decent output for a few minutes.

If you don't want your LED going 'poof the magic dragon' you won't try it till you get a heat sink though.

In some cases it will actually reduce it (which is a good thing because it means that it was running too hot). It reduces the current (in some cases) because at lower temperature the forward voltage is greater.

You may like to read this.

Limiting current to a LED using the impedance of the battery is not a typically wise thing to do. Most importantly, different battery types will lead to different (and perhaps MUCH different) current. Connecting a 9V NiCad (typically only 7.2V) battery might make the current exceed several amps as they have a significantly lower internal resistance.

my led says its rated for up to 700mA without a heatsink... it says you can push more current carefully with a heatsink.

however i read somewhere that some leds have a limited current without a heatsink.

i did calculate the voltage drop and to get close to 700mA was the 9.1 ohm resistor...

now that you (steve) mention it, a 7.2Vf on my battery would put it closer to what im getting but (7.2V-3.2Vled)/9.1ohm = 434mA which is still 100mA off so idk whats wrong...

its a basic ohms law circuit, i dont see what im doing wrong


and i measured the battery not in the circuit.... i see what you're saying

OK, I missed the fact that you are using a resistor.

Measure the battery voltage when the load is connected. You'll find that it is lower (dramatically lower) than off-load.