# How to modify a voltage regulator to make it very low quiescent current?

We have built a hand held RF remote controller, powered by two CR2016 batteries in order to supply 6V.

We used a microcontroller in the circuit, which runs under 3.3V. That's why, we used a voltage regulator, LM1117 for that purpose.

Problem is, after having a few empty batteries in a couple of tests, we noticed that LM1117 swallowes up the battery while waiting. LM1117's datasheet says it consumes 5mA on standby.

After a little search, we understood that we need to use "low quiescent 3.3v regulator" for that purpose, such as tps76615, LM2936, etc.

But as we live in Turkey, we can not find such parts easily, and we currently have no time for shipment. That's why, we need a workaround for that problem.

As a first resort, we will use the same circuit, except for adding a NPN transistor in series to LM1117 and trigger that transistor via MCU where MCU is powered by the same LM1117 AND another very low energy circuit.

That very low energy circuit consists of a Resistor + 3.3V Zener // Capacitor. Zener diode will assure that capacitor will have 3.3V and this energy will be sufficient to wake the MCU up after an interrupt. When MCU wakes up, it will drive the NPN transistor, which will get power to the MCU and system will feed itself.

But I can not get if there would be any consequences with that approach. Is there a working solution for that kind of problem?

• Do you actually need the regulator? Maybe you can operate from one battery with no regulator. – Spehro Pefhany Feb 9 '16 at 2:37
• The capacitor will need to be big, this is the only con I see. – lucas92 Feb 9 '16 at 2:55
• A regulator is used for mains power supply side (5 Volts). But yes, using the batteries separately is another solution. We then need to arrange battery holders accordingly. This should be considered also. – ceremcem Feb 9 '16 at 3:00

## 2 Answers

Low-power design is difficult, and I refer you to this detailed article I saw recently; you should read it in its entirety.

http://www.ganssle.com/reports/ultra-low-power-design.html

Some salient points, mostly from the above article:

• a zener-based regulator is incredibly inefficient and will kill your battery faster even than the 1117.

• why are you using two batteries? The usual approach is to select a micro that can run anywhere between 2V and 3.6V (which is most of them), and directly connect it to the battery, but...

• decoupling capacitors are also very leaky. Even if the micro runs at low average power and you get rid of the regulator and all its leakiness, the capacitors become a problem.

• supercapacitors are super-leaky. Don't try using one to store a bit of temporary charge to avoid regulator overheads, because it's probably leakier than the regulator.

• batteries are terrible. Drawing current from them makes them look dead before they really are, so you need to be careful with how your micro ramps up its power consumption coming out of sleep.

• Two batteries is mandatory for achieving required RF transmitter range. Regarding to this and other responses, I should consider using two separate batteries in this situation, one for microcontroller and two for rf transmitter. – ceremcem Feb 9 '16 at 12:21
• three batteries (2 for radio, maybe with a FET to cut its supply off) and one for the micro, sounds like it might be better than using an inefficient LDO. – William Brodie-Tyrrell Feb 9 '16 at 12:28
• Yes, we are switching to two separate battery organization. Do you think there is a fet or something (BJT transistor maybe?) is needed for an ASK radio? ASK transmitters are drawing power while only sending bit 1. – ceremcem Feb 9 '16 at 14:08
• on second thoughts, you will want to be very careful with the radio-battery selection. A CR2016 will develop very high internal impedance (tens of ohms I think) after a little while, and I expect your radio draws tens of mA. You might find that the radio batteries appear flat in very short order because of the voltage drop across their increasing internal impedance. If your system is occasionally connected to mains power, you maybe could consider using a larger, rechargeable battery? – William Brodie-Tyrrell Feb 9 '16 at 20:59
• Mains power input is there as a fallback option. In the last PCB we produced a couple minutes ago there are two battery holders, both are suitable for CR2032. So, we now have 2 CR2032, powering the radio transmitter. – ceremcem Feb 9 '16 at 21:10

These are solutions I could think of:

Build the LDO yourself out of discrete components. I dont know how easy it is to find op amps in Turkey, but there are quite a few low power op amps that can function with less than 0.5 mA and the op amp seems to be the most power hungry component. The feedback network could use high resistor values, and so could the reference voltage.

Another option I can think of (Im not sure if this is what you were refering to) is to switch power supplies. Simply using some load switches, you could switch from the LDO to something like a zenor diode regulator which could be very efficient for small loads.

Finally, you could consider changing your power supply in general to a very low power switching buck supply. In that case, you even increase your total efficiency as well since right now you are accomplishing only 55% efficiency!

• The problem is, I can't figure out how to enable this circuit while there is no power at the beginning. – ceremcem Feb 9 '16 at 13:56