# Quick Regulator Question

Discussion in 'Electronic Basics' started by Captain Blammo, Oct 26, 2005.

1. ### Captain BlammoGuest

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.

CB

2. ### John PopelishGuest

Before selecting a regulator, you need to know what current their

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:
http://dkc3.digikey.com/PDF/T053/1532.pdf

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.

3. ### Captain BlammoGuest

The combination of "as much battery life out as possible" and your
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

4. ### John PopelishGuest

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?

5. ### Captain BlammoGuest

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

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!

[email protected]=0.9W
0.9W =90%eff=> 1.000W
1W/7v=0.143A

[email protected]=5W
5W =90%eff=> 5.555W
5.555/7v=0.794A

0.794A+0.143A=0.937A
2Ah/0.937A=2.134h of life

==
CB

6. ### John PopelishGuest

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.

7. ### Jasen BettsGuest

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

8. ### Michael A. TerrellGuest

^^^^^^^ ^^^

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

Could be.