dinsdag, juli 26, 2005

Power consumption problem

With the controller finished, I could start calculating the power needed to keep this system running forever.
The current flowing into the controller is 20mA which means I would consume 0.5 Ah per day. The two solar panels I will use to provide the power give 100mA together in full sun, but only 25mA when it is cloudy.
So I have a problem, a big problem: the battery will slowly drain at a rate of 0.2 Ah per day when there is no full sun (which we never have in Belgium)

The only thing I have actual control over is the DC-DC conversion from 12V to 3V. Using an LM317 like in the current design has a efficiency of about 26%. These chips always have the same ingoing current as outgoing current and will convert the difference to heat.

A small calculation:
Consumption: 20mA
Input voltage: 12V
Output voltage: 3.15V

Output power = 3.15V * 20mA = 63mW
Input power = 12V * 20mA = 240mW

Efficiency = 63mW / 240mW = 26%
Lost energy = 240mW - 63mW = 177mW

So this current system is very good to keep the PCB at a nice temperature but not to be powered by a solar panel.

Browsing through the site of national semiconductor I found their WebBench that could calculate for me the best component to use. They suggested two devices:

- LM2931: A linear regulator with an efficiency of 27.5% (This is actually the same as my LM317 but fixed set to 3.3V instead of adjustable)
- LM2676-3.3: A switching regulator with an efficiency of 86%

Using the second one would make my design slightly more complicated but would solve all my problems.
When the consumption of the controller is 63mW and the regulator has an efficiency of 86% the consumption at the input will be: 63mW / 86% = 73mW
73mW at 12V means a current of 6.10mA

luckily they have a TO-220 package: the LM2676T-3.3. It has a maximum current of 3A which is much more then I need but it is the only one that is not SMD.

I expect to have my components in about two weeks so I can finaly finish the board and install the system.