Switching voltage regulator question

[Glenn Ashmore]

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
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[Phil Allison]

"Glenn Ashmore"

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?

** Yep.

No just catch your switcher.



........ Phil

 

[Nobody]

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?

Yes.

 

[John Popelish]

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?
Yes.

Does a switcher convert the excess voltage into
additional amp hours?

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.