Electronic – Dealing with different voltage needs on a battery driven low-power board

Basically, the (internal) peripherals are shut off when the AVR goes to sleep. Which modules shut down depends on what sleep mode you enter into. The tinyx5 has three sleep modes: Idle, ADC Noice Reduction, and Power Down. Unless you need something to continue working while it sleeps, I would suggest the "Power Down Mode" which still allows the watch dog to function if that is what you will use to wake it up.

The internal modules are all driven by clocks scaled down from the CPU clock. If the clocks to any one module are turned off, then that module will stop functioning until the clock(s) turns back on. Theoretically, these modules will draw zero current when they are shut down, although there is always some level of leakage.

In addition to sleep modes, there are many steps you can take to ensure your circuit does not draw much power during operation, at least as far as the AVR is concerned. Here are a few ideas. All of these steps are outlined in the datasheet:

1. Disable unused module clocks using the PRR register
2. Shut off the analog comparator: ACSR |=_BV(ACD);
3. Never leave floating pins - always use a pullup or pulldown resistor.
4. Use a lower voltage power source (Vcc >= 2.7V for tinyx5, 1.8V for tinyx5V)
5. Lower the CPU clock by changing the clock prescale values in CLKPR
6. Disable the ADC when you are not using it: ADCSRA &= ~_BV(ADEN);
7. Disable the digital input buffers you aren't using with DIDR0

As far as shutting down external peripherals, you could easily do that with a transistor as you have suggested. The I/O pins are static latches they retain their state when the AVR goes to sleep, so you could (in software) turn them off before you go to sleep and turn them on when you wake up. I would suggest you use a logic level N-channel MOSFET to connect/disconnect the devices from their ground supply (low-side switching), as long as the FET Gate threshold voltage is a bit lower than your AVR supply voltage so it fully turns ON from a high output. The MOSFET is not drawing any current through the gate when it is on (once the gate capacitor has been charged) unlike an NPN BJT which will continuously draw current while it is on. Either way, you should use an external pull down resistor (10k or so) to make sure the transistor stays off when it is supposed to be off. This will draw a bit of current, so a weak pull down is recommended (5V through 10k is only 0.5mA, 50uA through 100k).

You could also use this trick to turn off the external devices any other time you aren't using them; although they might need a bit of time to fully turn back on when you do need to use them or else they will give you strange results:

1. Turn On Device
2. Delay for a few milliseconds (whatever the recommended startup time is)
3. Read sensor/ transmit data
4. Delay for a few milliseconds
5. Turn off device

This particular regulator features a power save pin (PS), which will switch the working mode over from constantly regulating into a mode where only short bursts of current are provided to boost the voltage above the set voltage point.

If your components can handle the extra voltage, that might be worth a try. I'd add some more capacity to the output to increase the off-phase of the regulator to reduce the needed current even further.

So with a rough calculation: \$\frac{(3.38V-3.3V)*100\mu F}{10\mu A} = 0.8s\$

This is the time, a 100µF output capacitor would need to drop from 3.38V down to 3.3V witha load of 10µA (neglecting voltage dependencies). And the regulator would be on only some 5µs, so it is basically off the whole time.

As mkeith mentioned it might be worth considering a LDO, something like a TPS72733 which will have a quiescent current of 7.9µA (there should be some around going even lower). They have a high efficiency for battery powered applications and are a lot cheaper than switchmode regulators. The efficiency is roughly 88-89% at 3.3V out and 3.7V in and 200mA load, but only about 50% if the load is as low as the quiescent current and shutdown is not usable (depends on how the chip implements it). However you will loose the ability to drain the battery below 3.3V + dropout voltage (130mV @ 200mA for that part) which will decrease the interval at which the batteries have to be recharged. But on the other hand the batteries will last more recharge cycles if not discharged so deeply.

It should be noted that batteries like a constant low current draw more than a spiky high current draw, so the LDO might increase the interval again - but with the batteries you are going to use, drawing 1A for some microseconds should be fine. Battery lifetime is quite complex and not easy to predict, so to really find out, you'd probably need a prototype with both regulators and see what happens.