I'm going to run a camcorder viewfinder and a digital camera off a ~7v camcorder battery. They will be needing 5v and 3.3v respectively. I have settled upon these two regulators for the task:
UA78M33CKC LM2940CT-5.0
Does this seem like a wise choice? I don't know much of what's out there, and I'd like to squeeze as much battery life out as possible whilst avoiding SMT components if I can.
Thanks for any advice!
CB
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.
5 amperes at 1 amp is 5 watts. But if that same current comes from a 7 volt battery, that consumes 7 watts. That would drain a 2 amp hour battery in 2AH/1A= 2 hours. If you Have a switching regulator with 90% efficiency, the load on the battery would be 5.5 watts. %.5 watts from 7 volts needs a current of 5.5W/7V= 0.786 A. That would drain a 2 amp hour battery in 2AH/.786A= 2.55 hours.
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
2AH/1.05= 1.9 hours.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!
0.3a@3.3v=0.9W 0.9W =90%eff=> 1.000W 1W/7v=0.143A1A@5v=5W
5W =90%eff=> 5.555W 5.555/7v=0.794A0.794A+0.143A=0.937A
2Ah/0.937A=2.134h of life== CB
^^^^^^^ ^^^
Really? I think you have a typo in there somewhere, John. ;-)
-- ? Michael A. Terrell Central Florida
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
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.