7805 circuit that gives an output current of more than 1A QUESTION

[ronv]

470 uH. Use your big tip for the pnp.

 

[lloydi12345]

I would like to try making the circuit since we only got 2 days left. Shipping for now is not a good option I hope the circuit would work.

I made another option. Does using 7808 helps reduce the heat 7805 takes? The setup is 10.3v connected to the input of 7808 then the 8v output is connected to input of 7805 that makes 5v. Since the both of them are heating up, does it really help?

 

[lloydi12345]

I'm thinking also of adding voltage divider to reduce 10.3v to 7v to have only the smallest voltage difference from the output 5v which I think will reduce heat.

 

[ronv]

The 7805 by itself is not good for 2 amps continuous now matter what you do. At this point the best solution may be the pass transistor and some heatsinks.
The problem with that is that you are wasting more battery power making heat than you are running your camera.

 

[lloydi12345]

Ronv what do you think about the voltage divider? Letting the source pass through the voltage divider to have a low voltage to be fed on 7805's input. Can it help reduce the heat? Let's consider I'm operating everything at 1.2A. The starting current of the camera is 1.5A, 2A is the maximum but the cam's working current state is 1.2A.

Don't worry about the heatsinks I got lots of them 8x the size of 7805

 

[blueroomelectronics]

You could try to find a 78S05 (1.5A)

 

[crutschow]

Ronv what do you think about the voltage divider? Letting the source pass through the voltage divider to have a low voltage to be fed on 7805's input. Can it help reduce the heat? Let's consider I'm operating everything at 1.2A. The starting current of the camera is 1.5A, 2A is the maximum but the cam's working current state is 1.2A.

You can't avoid the heat. It will be a minimum of 1.2A * (10V-5V) = 6W unless you go with a switching regulator.

A voltage divider will dissipate even more power, but it will some will now be dissipated in the resistors. A series power resistor at the regulator input will also reduce the regulator power dissipation, but not the total power dissipated.

That heat represents significant wasted battery power.