0
\$\begingroup\$

I'm working on a wireless sensor network and would like to know what it would take to shut down the regulator during deep sleep.

I'm planning on going 3.3v, because I will allow the user to push a button to display readings locally on an LCD and that is saying it needs 3.3v. Otherwise I would go lower. Everything else supports 1.8v.

So my question is, if I have a node deep sleep, can I set the EN pin on the regulator low and depend on either the power supply caps or the inductor to to retain enough power to keep the MCU alive so it can wake up in 5 minutes? How do I know how long I'll have? Should I throw in a supercap too? Is it even worth it to try to turn the regulator off, or when there is no draw will it not burn energy?

Peak current will be under 200mA. When sleeping, nothing should be running except the MCU in stop+RTC or low power run mode, which is, according to the datasheet, 1.2 uA and 9uA, respectively. I'm sure some inefficiencies in my design will require wiggle room.

Thanks a lot and any feedback on any of the above is greatly appreciated.

| improve this question | |
\$\endgroup\$
  • 1
    \$\begingroup\$ The answer depends on how much current you draw during those sleep phases. Can you give some estimation (or if available a measurement)? \$\endgroup\$ – Arsenal Feb 17 '15 at 14:41
  • \$\begingroup\$ The regulator you show is just that, a buck-boost regulator for battery-operated devices, but it has no charging circuitry. How are you charging the NCR18650 batteries? \$\endgroup\$ – tcrosley Feb 17 '15 at 15:53
  • 1
    \$\begingroup\$ According to the datasheet, that regulator consumes less than 60 uA of quiescent current. You have 5.4 Ah of battery power. Assuming your system load is very low in deep sleep, I question whether it is worthwhile to put the regulator to sleep at all. \$\endgroup\$ – mkeith Feb 17 '15 at 21:46
  • 2
    \$\begingroup\$ For that particular regulator there is a power saving pin available which will make it run in bursts of power, increasing the voltage ripple significantly, if your system can handle that, it might be worth a try. \$\endgroup\$ – Arsenal Feb 17 '15 at 22:10
  • 1
    \$\begingroup\$ I did not look very hard. But here is one from TI: LP5907. There are probably 100 vendors who make LDO's in SOT23. Rohm, Toshiba, TI, Linear, MPS, a host of Taiwanese companies. Take a look at digikey. You will get better efficiency in normal operation with that buck/boost, and it is actually a pretty low quiescent part. It is just a question of what percentage of battery capacity will be used up in sleep mode vs "regular" mode. Note that the LDO I mentioned has active pulldown when shut down to drain output caps rapidly. \$\endgroup\$ – mkeith Feb 18 '15 at 2:36
2
\$\begingroup\$

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.

powersave mode of TPS63001

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Outstanding answer. One last thing. If I hooked the LCD straight to VIN, via a transistor or mosfet, I could potentially go 1.8v for vcc. With that big of a difference in VIN and VCC, does it make a LDO no longer worthwhile? \$\endgroup\$ – Matt Williamson Feb 18 '15 at 14:38
  • 1
    \$\begingroup\$ @MattWilliamson in that case you would have a drop of 1,9V across the LDO, which is more than half of your battery voltage for a long duration (it stays mostly at 3.7V). So your efficiency is below 50%, that's not really good for a battery powered device. Also thermal management comes to mind, the LDO will have 0,38W to burn at 200mA, so it has a self heating of roughly 30K (assuming 75K/W), which might be a problem depending on ambient temperature range. So I'd probably go with the switcher then. \$\endgroup\$ – Arsenal Feb 18 '15 at 17:07
  • \$\begingroup\$ Thank you for your thoroughness. I really appreciate the time you've taken to help me. \$\endgroup\$ – Matt Williamson Feb 18 '15 at 19:07
  • 1
    \$\begingroup\$ Also, you may have problems if you run the LCD at 3.3 and other circuitry at 1.8. Basically, if the LCD talks to other components on the board, the voltage mismatch will cause problems. If the LCD drives outputs to 3.3V, those outputs usually can not be directly connected to inputs expecting 1.8V. \$\endgroup\$ – mkeith Feb 19 '15 at 3:56
0
\$\begingroup\$

The simplest is to do some tests with varying time interval, and pick the time interval that will work.

You can also measure the leakage current, and figure out the time but after that you still need to test it to make sure it works.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.