Quick Regulator Question

Before selecting a regulator, you need to know what current their loads will consume.

The combination of "as much battery life out as possible" and your choice of linear regulators is a contradiction. Linear regulators waste all the extra voltage as heat. If battery life is important, you might look into prepackaged switching regulators, if you can find ones that can take the 7 volt input. They convert energy at one voltage to almost the same energy at another voltage.

They are also a lot more expensive than your linear regulators. Here is a page from the Digikey catalog that lists some candidates:

For example, near the top, find the PT6305A 3.3 volt 3 amp regulator rated for 4.5 to 9 volts in.

Finding a 5 volt integrated switcher that works over that range of input voltage will be harder.

The 3.3 volt case is more dramatic. If you need to supply 2 amps to the 3.3 volt load, that represents 6.6 watts. A linear regulator will take that 2 amps directly from the battery for a life of 2AH/2A= 1 hour.

A 90 percent efficient switching regulator would take 6.6W/.9= 7.33 W from the battery which requires 7.33W/7V= 1.05 A, for a life of

So a switcher for the 3.3 volt output may be more important for battery life if the load current approaches or exceeds that needed by the 5 volt output.

Perfection is never cost effective. A good compromise might be a switcher for the 3.3 volt output and a low drop out regulator for the 5 volt output.

Have you come up with load current estimates, yet?

I actually have no idea as to just how much is wasted by linear regulators. If I run 5v at 1A from a 7v, 2Ah battery, roughly how many Ah will be wasted as heat? Is there a formula? I realise that it will vary with choice of regulator, but a ballpark idea would be nice.

Is switching regualtor wastage negligible? Those things aren't so cheap! I guess I should have said that I want an optimal balance of cost and battery life :-)

CB

The 5 volt and 3.3 volt load currents add together with linear regulators to find the battery current load.

That looks right to me.

I don't think the camera is likely to use much more than 0.3A on a bad day (that's a total guess), since it won't have the LCD on or be taking pictures, just feeding video out, and I know the camcorder viewfinder uses around 1A.

If I understood you correctly, that means that I'll be looking at 1.3A of load with linear regulators, giving me 2Ah/1.3A=1.538 hours of life. I'm assuming that plonking the regulators in parallel with the battery will make the loads additive, but my electronics is still pretty sketchy :-)

In any case, I can't see the loads being any higher than what I quoted, so I should get at least 1.5h of life. According to the calculations below (please correct me if I'm wrong), I'm only looking at an optimal battery life of 2.134h with 90% efficient regulators; only a 36 minute difference. All things considered, I think a fully linear solution will do my wallet and my project much justice. Thanks very much for the help!

1A@5v=5W

0.794A+0.143A=0.937A

== CB

^^^^^^^ ^^^

Really? I think you have a typo in there somewhere, John. ;-)

Could be.

you'll get 2Ah out of te battery, but with a switching regulator you could get more than that out of the regulator.

switching regulators use a transfoomer like effect to convert voltage and current...

Last time I looked (several years ago) it was about 80% efficiency, (but that has probably improved), so for converting 7V to 5 probably not worth-while, but for the 7V to 3V it'd halve the power consumption.

Bye. Jasen