Electronic – Best battery for low power, long term application

Just a random list, if you post your schematic it would probably be easier:

1.8V lithium Coin cells are very easy to find, but more likely your serial interface needs 3.3v? Unless your receiving end will deal with 1.8V.

Leakage current does generally go up as your voltage increases, so lower is better usually. Also consider the brown-out point for the system vs the battery characteristics. The 'death' characteristics of the battery will be determines by the battery chemistry you use. For instance if your uC browns out at 1.7V you may actually want to use a higher voltage battery as with some batteries the output voltage will lower slowly as the battery dies. You'd get more life out of a 3.3V battery as when it begins to die its output will slowly drop and you can operate all the way down to 1.8V. If you use a 1.8V battery your going to shut down fairly quickly as the battery dies. This all assumes your serial interface or other components can deal with a wide voltage range (I know the AVR can).

LED's use a lot of power, unless you use a very low power LED and are controlling its current draw it's probably drawing a lot more current than the AVR is. If its just there for debug, don't populate it for production or only have it blink once in a while or something to minimize its on time, and definitely control its current draw.

If you can, pick the polarity / rest state of your serial interface to draw as little power as possible, it's rest state should not be drawing power. If pull ups are required use the largest resistor possible to maintain signal integrity but minimize current usage. If power is a huge concern use a signally scheme that favor's bits that don't draw power. For instance if you have pull ups, using a protocol that results in lots of 1's in the signal will leave the serial interface in a state that isn't drawing as much power most of the time. Such optimizations are only worthwhile if your making heavy use of the serial bus. If its very lightly used just make sure its rest state isn't drawing power.

Generally speaking you can assume all instructions (reading GPIO, etc) require the same amount of power. Its not really true but the power difference is minimal.

Power usage is much more dependent on the number/type of peripherals you have powered on, and the amount of time the micro spends active vs sleeping. So the ADC uses more power, EEPROM writes use a fair amount of power. Specifically something like the EEPROM writes are usually done in fairly large 'chunks' so you should accumulate as much information as you can before doing the write to the EEPROM (if your even using it of course). For the ADC that micro supports doing the ADC read during 2 of its sleep states, as ADC conversion takes a relatively long time this is a good time to sleep.

You should probably just read the sections on power management, sleep states and minimizing power using in the microcontroller's data sheet: linky page 35 on. Keep the AVR in the deepest sleep state possible as long as possible. The only exception to this is that you have to consider the start up and shutdown time. Its not worth it to sleep for 10 cycles if waking back up takes 25, etc.

Do resistors use up battery life? Do capacitors? Do diodes?

They all do to some extent. Resistors dissipate the most in most applications:

P = V*I

P = V^2 / R or P = I^2 * R (where V is the voltage drop across the resistor)

Diode's have a (relatively) fixed voltage drop, so power dissipation is almost exclusively tied to current passing through the diode. For instance a diode with a 0.7V forward voltage drop, P = 0.7 * I if current is moving forward through the diode. This is a simplification of course and you should check out the operating mode based on the diode's I-V characteristics.

Capacitors theoretically shouldn't dissipate any power, but in reality they have a finite series resistance and non-zero leakage current which means they do dissipate some power, generally not something you should worry about with such low voltages though. That being said choosing capacitors with minimal leakage current and ESR is a power win.

As far as using them to smooth out battery draw, this doesn't really help for power usage, its more for filtering. Also battery chemistry comes into play here, some chemistries will be happier with a constant draw, some deal better with spiky current draws.

Having an amp rating HIGHER than the original battery is ideal. It will only try to deliver the car the current it needs. If the toy asks for more than the battery can offer, then you have a problem. However, you need to design the battery bank to safely fail when a cell dies, otherwise you can cause a cascade failure of venting and possibly igniting cells. A very bad thing. Here is an example of what I would do:

If we are using Panasonic 18650 3200mAh Li-ion batteries, then we know their specs are 3.6 nominal voltage and 10A max continuous current draw. First, calculate the bank characteristics. 10.8 / 3.6v ~= 3. We will need to use 3 batteries in series to pull the voltage up to spec. Max charge should be around 12.6v, likely nearly identical to the original battery. Assuming the old battery had a 30 amp fuse, we would need a minimum of 3 rows of 3 batteries to get the bank to the minimum requirements at nominal voltages. However, I would add 2 extra row for overhead. This would total 15 18650 batteries, 5 sets(in parrellel) of 3 batteries(stacked/in series). When charging the cells, you should charge them in parrellel. What I would do is make a separate set of terminals on the battery that connect to all of the batteries in parrellel. from these terminals, charge the battery at no more than 1 amp per cell at no more than 4.2 volts(for 18650 batteries, you need to find the nominal charge rate of the batteries you plan to use if you go another route). Charging batteries in parrellel will naturally balance, as long as those batteries are new batteries married to each other at birth. I have never trusted series charging because of all that can go wrong. This would give you a total of 5*3200 mAh for a total of 16000mAh and 173 Wh. This is simply an example to get you familiar with the calculations. A battery bank such as this would easily set you back USD $100. It is important to ensure each cell has overcurrent protection of an amperage lower than the battery is rated for(roughly 2 times the C rating). Without this, when a cell dies, it can cause an overcurrent cascade, and this is a very very bad thing.