I am new here and was reading your thread about the IR light source and wanted to offer some input.
First thing you need to consider is that the 9V Battery is only 9V fresh from the box. I looked at a typical 9V battery, the Duracell MN1604 (
http://datasheet.octopart.com/MN1604-Du ... 306548.pdf) and am basing my calculations on it.
Your LED’s have a max voltage drop, known as Forward Voltage or “Vf” of 1.6V and an absolute maximum rating of 140mA. I do not recommend running your LED’s at this current as it will limit their life. I would maybe do 100mA or less if you can get away with it.
The MN1604 battery has an internal impedance of 1.7 ohms so you will need to consider this when you design your circuit.
I suspect that the reason the circuit that Thomas W recommended was dim was because your battery source voltage was low or not rated at 240mA. The design should have worked IF the source was delivering 9V under a 240mA load. The NMH Battery Thomas W has would not be a good choice to power your circuit as the voltage will likely drop significantly as it discharges and given the relatively small Amp hour rating, it likely has an internal resistance that is fairly high. This means your LED’s will probably turn off long before the 38 minutes he calculated if they even turned on at all, remember you will need a minimum of 8V to forward bias the diodes and this does not count the voltage drop across the current limiting resistors.
I agree with what Thomas said about the parallel diode design, it will result in diodes burning out.
Your last post on 12/9 indicates that your battery is not beefy enough. 6.89V/5 is only 1.379V per diode and this means the diodes are not fully on.
What I recommend is the 12V AC/DC option as non-rechargeable batteries will give you brighter light for longer than a rechargeable but @ $3+ per battery it can get expensive. The rechargeable would be cost effective but would give you limited light for 1 hour or as much light as the former but only for minutes not an hour like you specified. The AC/DC will make things so much easier. 1 leg with 7 diodes and a 6.4 ohm .25W resistor will give you fully illuminated LED’s. The drawback is you would not be able to adjust the brightness. I do not recommend a 2 leg design as 200mA is your max capacity for the transformer and for the sake of not overworking the transformer, I would only run 150mA which only leaves 75mA per leg. This may make the LED’s to dim.
One option is to use an adjustable constant current regulator for the leg(s). This will enable you to turn the brightness up or down with a pot. These are easy circuits using a transistor, 2 fixed resistors and a pot. A constant current regulator; should you go with the battery option, will also allow you to maintain a stable current regardless of what the battery voltage is doing. Steady current means stable light output for the life of the battery charge not steadily dim over the time you use it.