AVR-Chat

At 02:13 PM 7/31/04 +0000, you wrote:
>Hi Cobb,
>
>I probably didn't explain myself too well, plus my typo where I said
>I wanted 3V from the converter, I really meant 5V.
>
>I have 4 batteries (this is the maximum room I have in my case). The
>printer runs from 5V and consumes 3.5A peak. The idea was to connect
>2 pairs in series to give 3V, and then connect the 2 x 3V cells in
>parallel to double the capacity.
>
>As this is a hand held device I do not want to connect all 4 in
>series and use a regulator to bring it down to 5V, thats a waste of
>power as I want maximum operating time from the device.

A few quick calculations on power losses and drain.

Assume 3.5A draw at 5V (1% duty cycle), 30mA draw at 3V (100% duty cycle)
Alkaline cells 1.5V ea., 148 milli-Ohms Internal resistance
Converter efficiency 90% (high, but a starting point)

--- Configuration 1: 2 batteries in series, pairs in parallel, boost 
circuit to 5V for printer.

Internal resistance of pack 148 mOhm

First rough calc give for 3.5A at 5V need 6.5A at 3V but that give a .963V 
drop due to internal resistance.

A quick run through on current draws gives a maximum current out at 5V of a 
little over 3A with an input current from the battery pack of 10A and a 
voltage drop of 1.48V (will the logic circuit still work?).  Higher current 
draws than 10A cause the voltage to droop so far that the output actually 
starts to drop.  It's doubtful that the boost regulator would still work 
even to the 10A point in any case.

Losses: 14.8W in battery pack, 1.5W in the convertor for 16.3W.  At a 1% 
duty cycle that 160mW and 100mA average.
Losses from control circuit 0.1mW and 30mA average, control circuit voltage 
drop .004V

If batteries are 2AH batteries then this would suggest a 15Hr life.  This 
neglects the fact that you can't actually get the required current and the 
effect of high discharge rates.

Note: a 3.5A boost circuit might need a small load to maintain regulation 
which would make this case even worse.

--- Configuration 2: 4 batteries in series for 6V.  Assume printer can 
handle voltage (not a given but simplifies back of the envelope calcs like 
this).

Internal resistance of pack 592 mOhm

3.5A draw gives voltage drop of 2.07V (6V drops to 4, will printer still work?)

Losses 7.2W in battery pack.  At a 1% duty cycle that's 72mW and 35mA average.
Control circuit: 30mA at 3V can be delivered from 17mA at 5V
Control circuit losses 9.2mW, 17mA average

Again assume 2AH batteries: These figures suggest a 42Hr life and you could 
actually get the current you need.

Left out of both of these is that the battery voltage decays over the life 
of the battery (standard End of Discharge for an alkaline appears to be 
0.8V from what I read in the data sheets) and the internal resistance rises.

If your peak draw is short enough using a capacitor to provide the pulse 
makes a lot of sense.

Obviously the profile and duty cycle of your load is going to make a lot of 
difference.

It's a lot easier (and more efficient) to make a low current buck than it 
is to make a high current boost.  For switchers the National simple 
switchers are easy to work with and if the input voltages are low enough so 
are the TI 60500's.

There are more items that could be considered but these are enough to make 
a first pass at determining what is possible.

Robert


" 'Freedom' has no meaning of itself.  There are always restrictions,
be they legal, genetic, or physical.  If you don't believe me, try to
chew a radio signal. "

                         Kelvin Throop, III