MSP430G2553: Basic questions to ultra low power systems

Ok, thank you for the Info on the documents. I will read them. 

Any tips on how to handle the other devices in the circuit. Is switching them off with a loadswitch a good practice? How does one generally handle circuits where not all IC's are used all the time?

Kind regards

Thanks for the reply. Any suggestion what DCDC regulator is suitable? I found a few that seemed to have a much higher Iq than the TPS78330. I suppose a Buck-Boost Regulator is what I would need too look for. 3 V output for an input of 5 V down to 1 V or so.

Yes, you would need a Buck-Boost.

Try to locate one that provides the highest efficiency for the voltage input and output you have. Their datasheets should provide graphs with that information. Good Luck.

OK. Let first start by correcting your view of the voltage output of batteries. AAA batteries (or any battery) do not discharge to 0V. What happens in reality is that they start with ~1.4V initially then they discharge to ~1V after which they "fail". At ~1V they reach a catastrophic point at which their energy output drops rapidly rendering them useless.

So, from the above, if you use 4 AAA batteries, your input voltage range would be anywhere between ~6V and ~4V. This now changes the type of switching converter you need to only buck converters.

Now back to your first question: Yes, there is something you are missing in the case of TPS78330. First, this is an LDO, which means you will have energy waste. Second, the 500nA is a "Shutdown" current. It just means that the TPS78330 will not consume too much current when it is in "shutdown". Well, this might be good if somehow you can create an external "hard shutdown" signal that controls it; otherwise, you will need to look at the "I GND" entry in the datasheet.

So now let us go over why LDO are not a good choice. Let us say that your battery voltage starts at 6V and your MCU is at 3.3V and the MCU circuitry consumes 1mA. The power consumed at the MCU would be 3.3mW; however, because this is a linear regulator, the batteries will also see a 1mA current consumption, which means the power drained from the batteries is 6mW. This makes the efficiency (regardless of IQ, just by a design choice) be 3.3mW/6mW = 55%. That means, just because you chose to use an LDO, you are throwing away ~50% of the energy stored in the batteries.

As for the other question, you can put the MCU in deep-sleep for days if you use its RTC.

Thanks for the reply (and patience).

I understand the thing about the LDO and the battery voltage. But I have one more thought about the LDO. Lets say my circuit takes 5µA in standby. If I look at the TPS62730 switcher for example it has an operating quiscent corrent in active mode with 0 mA drawn (which I guess is equivalent to Ignd) of 34 µA. That would sum up my total current to ~40 µA. Now the TPS78330 has an Ignd (thanks for pointing that out) of roughly 1 or 2 µA at 5 µA load. So in total with the efficiency of 50 % it would draw under 20 µA. I guess for applications which are in sleep mode almost all the time and draw only very short some 300µ A the LDO would be better. May that be right?

The RTC with calendar is of course a good choice. Yet I wanted to use the G2553 wich doesnt have the RTC. Thats why I thought the external nano power timer was a good choice. 

Any thoughts about the I2C?

I think the power supply issue may depend on how your device will be used.  If there will be extended periods of time when it is idle, or effectively off, and if it can be awakened with a push button by the operator, then you could put the processor to sleep awaiting the interrupt trigger polarity on an I/O pin.  The G2553 would draw almost no current in this state, with all oscillators off, and an efficient LDO regulator would do the same.  I've used a device that runs on the G2231 which is active for about an hour a day, and which runs on a 9V battery.  There is no ON/OFF switch, and when everything is asleep waiting for the momentary switch to be pushed, total current consumption is under 2 uA.  The battery lasts about a year.  But there's no question that when the device is On, and dropping 9V to 3.3V through the LDO at a mA or two, it's pretty inefficient.  So it kinda depends on what kind of usage pattern you're looking at.  Switching regulators will be very efficient when supplying significant current, but in my experience they aren't very efficient supplying just a tiny amount of current.  But of course it could be that I'm just not doing it right.

I think the same thing arises concerning using push/pull instead of a pullup on I2C.  Since there's no current through the pullup when I2C is idle, it all kinda depends on how active your I2C bus is going to be.  But I would certainly give push/pull a try.

As someone else suggested, the slower you can run the CPU when fully On, the less power you will use.  And remember that even though the G2553 has calibrated 1 MHz DCO settings, that doesn't mean you can't go slower.  And if you need a calibrated slower speed, that can be derived on first run on a chip-by-chip basis without needing a crystal.

In the end, there isn't going to be a substitute for trying some of these things and measuring the resulting current.