Hi,
I assume I need to regulate it because a battery is going to very its voltage with charge
Yes, no question about it.
but I am trying to figure out the circuit. The specs on the website say:
Power
• External Power Supply: 5V, 2.5A
• Consumption: 4.5 Watts Max
So, I assume I would need to be able to supply 2.5A, but that would be like 13W which seems a lot high, and it says consumption is only 4.5 Watts.
So do I need to make something that can handle 1 amp or 2.5 amps?
You could allways measure it to be sure, but if the specs are OK, it should mean that you only need to supply ~1A continuous.
The
consumption is
what the cam use, but the
power supply is
able to give 2.5A
Would I be able to wire two of these 5V regulators in parallel to provide the amperage?
One should be quite enough, current-vise, but you need to consider your battery voltage and the power the regulator have to dissipate.
A better choice would be an LDO regulator, as the 7805 will need to drop at least 3V, so the battery voltage would have to be 8V when flat - given a "flat voltage" of 0.9V/cell, it will take
9 cells, which equates to a 13.5V battery (with a start voltage of up to 14.8V).
Regulating 13.5V down to 5V gives a loss of 8.5V, so the loss will be
65.4% or almost 2/3 meaning that you need to supply 3 times the power that the cam needs.
With an LDO, the battery could be
6 cells, which is 9V (9.9V start voltage).
The loss will then be "only"
44.4%The optimum would be a switch mode regulator, which could be made with a loss of
2% to 10%.
Using AA cells, the camera would keep for 2.5 to 3 hours with linear regulators. With a switch mode regulator, the runtime on a set of AAs will depend on the number and quality of the cells, but assuming a cell profile of 1.5V 3Ah and a 10% loss in the switcher, runtime will be around:
n * 1.5V * 3Ah * 0.9 / 4.5W
Where n is the number of cells.
(About 54 minutes/cell).
