circuit analysiscurrentled
So I made this schematic for a little bathroom light I made. I would like to figure out how to calculate how long the 9V battery I connected will last with this specific schematic. The diodes are LEDs.
Here is what the Terrible Product looks like for reference:
The current rating for the converters is only the maximum current it can deliver, thus is not useful to calculate the power drainage. To get the time you have to know the real input current for every DC converter(\$I_i\$) then, since they are in parallel,
$$I_{tot}=\sum_i{I_i}$$
To get the time, suppose \$Q_p\$ is the capacity in Amp/h of your battery, \$T\$ is approximately the time it will last (in hours)
$$T=Q_p/I_{tot}$$
So, if you can measure the \$ I_{tot} \$, you can directly use the aforementioned formula.
Keep in mind you should not completely discharge a lead battery or it will decrease sensibly the capacity.
You said that in the first test, where the LEDs blew up, you used a 4.5 volt battery. That would force MUCH more current through the LEDs than the 3 volts you used in your second test.
The 3 volt test would still be forcing much more current through the LEDs than their maximum rating, and may greatly shorten their life.
What you should do is:
simulate this circuit – Schematic created using CircuitLab
Since LEDs are not guaranteed to have identical forward voltages, if you connect two LEDs in parallel, with a common series resistor, one LED will draw more current that the other, so will get a little warmer, so its forward voltage will drop slightly, making it draw still more current.... which may cause one LED to be destroyed (not too likely) or one LED to be significantly brighter than the other.
Best Answer
It's not clear whether your running the LEDs at different brightness is intentional or accidental. The small difference in brightness is not large, so I'm guessing it's accidental. Two configurations that run the LEDs at the same brightness would be ...
simulate this circuit – Schematic created using CircuitLab
The first will operate down to the lowest battery voltage. However, the second draws half the current for the same light output, so may last longer anyway, even though the battery voltage will be higher when the LEDs reach an unusable brightness.
The nominal capacity of an alkaline PP3/6LR61 is 0.5 Ah, though it's not always clear what end voltage or what rate of discharge manufacturers are using when they test this.
If the second configuration takes 10 mA, a ballpark lifetime would be 500 mAh / 10 mA = 50 hours. That's maybe within a factor of 2 or 3 either way, given the uncertainties of how the 500 mAh is specified, whether your particular battery is a quality one or a knock-off, what your end point for subjective brightness is, and the fact that I haven't even attempted to allow for the reducing current as the battery runs down.