Voltage Regulator to power 2 digital cameras

I would like to construct a circuit to simultaneously power 2 Canon a480 digital cameras. The wallwart (ACK800) that Canon sells for this camera outputs 3.2 volts and supplies up to 1.5 amps. However purchasing two of these would be expensive. I am proposing to use a surplus wall wart I have (9.5 volts DC, 2.1 amps) in conjunction with a circuit containing 2 LM317T voltage regulators to power both cameras. The circuit I have drawn is shown below.

I have constructed this circuit on a breadboard, albeit powered by a 9 volt battery, and it seems to work. After adjusting the pots I can get 3.2 volts out of both LM317T's. I would appreciate your review of this circuit. Is anything unnecessary (e.g. C3) or is anything missing? Would a 9.5 volt, 2.1 amp wall wart be OK to power the circuit or would something different be better in terms of voltage/amps or heating of the two LM317Ts? Do you think I will need to heatsink the LM317Ts? Would it be better to power both cameras from one voltage regulator of higher amp capacity? As you can probably tell, this is my first real adventure into electronics.

Thanks

Tim

**broken link removed**

P = I*E = 1.5A * (9.5V-3.2V)= 9.45W. That would be the peak power dissipation in the LM317, which would require a large (several square inches) heatsink. I suspect that the average current is much less than the peak current of 1.5A, but you still should have a sizeable heatsink. Try it, and feel the heatsink while running the camera through its paces. If you blister your finger, the heatsink is too hot. Short of that, you are probably ok.

Tim,
You will definitely need to mount the LM317s on a heat sink. Using the wall-wart that you suggest, you'll need to have pretty large heat sinks.
Perhaps a better solution to the heat problem, if you want to stick with the regulator circuit you have drawn out, would be to buy a wall-wart that provides a lower output voltage than the one you have. But, with the LM317 regulator having a minimum Vin - Vout of 2.5V, and adding in the ripple, rectifier drop, etc., you'd be looking for a unit providing roughly 6.5V of filtered DC. You could, in a bit of a gamble, get a switching power supply that provides an output closer to to minimum that your LM317 regulators need. Keeping in mind that the LM317s need about 2.5V + 3.2V (5.7V) to operate properly, you could possibly search for a switcher that yields an output of 6VDC, and use that to feed the LM317 regulators. You still would need a good heat sink for your regulators. You might also get a 5V switcher that has adjustable outout and set it as high as possible. Here are a couple to consider.
5V@ 3A POWER SUPPLY-MPJA, Inc.
5 VDC 4A SWITCHING POWER SUPPLY | AllElectronics.com

A much better solution would be to get a switching power supply that provides the power you need. Here are a couple that you could probably use to power your cameras.
SALE! Lambda Regulated +5VDC and +3.3VDC Power Supply-The Electronic Goldmine
**broken link removed** (Note that this unit is listed as 3V. You might call Skycraft and ask if the output voltage is 3.3V rather than 3V)

Cheers,
Dave M

Thanks Dave and Mike. I appreciate your responses. Getting a switching power supply at the right voltage is probably the way to go. So, even though the camera's official adapter is 3.2 volts, it's OK to substitute something in the range of 3.0 - 3.3 volts?

If I stick with the regulator circuit and went with a 6 volt switching supply to power it, what maximum amperage should the switcher be capable of?

Tim

If you can find a 5V switcher that uses a MC34063 IC, it is possible to hack it to 3.3V by changing a resistor in the feed back path. There are millions of 5V phone chargers out there... To get the current, use one for each camera.