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Switching voltage regulator question

I have four 3.6V 800mAh LI-Po prismatic cells that I want to build into a pack for a radio control receiver/servos and video unit. It is carried aloft by a kite so weight and space are the governing variables. Power density is important. The receiver and servos require 5V and will draw an average of 150 ma. The video Xmitter requires 5V at 250 ma for a total of about 400 ma. I have found a couple of micro switching voltage regulators that claim 85% efficiency converting 14.4V to 5V..

I am thinking I could assemble the cells into a 14.4V 800 mAh pack and feed the devices through the regulator. My question is what is the effective AH?

Here is the way I figure it:

14.4 V * 800 mAh = 11.52 watt hours 11.52 Wh * 85% efficiency = ~9.8 Wh 9.8 Wh / 5V = ~1,960 mAh 1,960 mAh / 400 ma = ~ 4.9 running hours (say 4 hours)

Is my logic correct? Does a switcher convert the excess voltage into additional amp hours?

--
Glenn Ashmore

I\'m building a 45\' cutter in strip/composite. Watch my progress (or lack 
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Reply to
Glenn Ashmore
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"Glenn Ashmore"

** Yep.

No just catch your switcher.

....... Phil

Reply to
Phil Allison

Yes.

Reply to
Nobody

Yes.

The switcher converts input power (input volts times input amperes) times efficiency to an output power (output volts times output amperes).

power in * efficiency = power out

The difference between input and output power is lost as heat.

Since your load is about 2 watts, the switcher will require about 1/.85 times that much input power or 2.35 watts, and the .35 watt difference is the heat lost from the switcher as it performs this voltage and current conversion from the input 14.4 volts at 0.163 amp drain current.

Reply to
John Popelish

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