I am new to DC-DC switching boost regulator. In my application, I need regulated 3.3V output from 2 (1.5V) single alkaline batteries. I have a light load taking only about 0.5 mA most of the time and as per the user activity intermittently, power to a module is switched (with a load switch), which takes about 75 mA for about 250 mS.

In my experimental setup, I've used TPS61201 / TPS63001 from Texas Instruments, which has a power save pin to reduce power consumption at light load.

My goal here is to increase the battery life as much as possible and be able to use the regulator down to about 1.6V from 3V of the alkaline batteries.

What I am trying to understand are as follow:

  1. Can I keep the power save mode enabled all time time in my application with such DC regulators to minimize inductor current and to increase battery life?
  2. What difference it would have in performance with power save mode enabled or disabled all the time?
  3. With power save mode enabled, would the regulator start up with voltage as low as 1.6V or lower?

At the moment, I am evaluating different DC boost regulator and their characteristic curves for best efficiency in my application but your input would help me clarify my understanding about power save mode of such DC regulators and how to increase battery life for light loads for most of the usage duration.

  • 2
    \$\begingroup\$ If your goal is "to increase the battery life as much as possible," then there is a lot more to your question than you've written about. I assume you are using a micro. Which one? (This [and input and output transducers] is probably the largest single factor to consider.) Also, how is the device activated (powered-up?) The methodology here matters a lot regarding quiescent power when inactive. Also, all boost devices have a minimum load for operation. Answers to the above questions will have an impact on choice of boost topology, as well. \$\endgroup\$
    – jonk
    Jan 4, 2018 at 22:00
  • \$\begingroup\$ PIC micro. The DC boost converter remains disconnected from the battery when not in used with a pass transistor. PIC transfers some data with another modules and then goes to sleep. The other modules also go to sleep. When in sleep, I see that about 0.5mA is drawn from the input power supply (or batteries). When everything awake for about 250 mS during data transfer, ~75 mA is consumed. Do I need to have a dummy load in this case for minimum load requirements? That would be the waste of battery power with a dummy load.. \$\endgroup\$
    – user101095
    Jan 4, 2018 at 22:12
  • \$\begingroup\$ this 0.5 mA with power save mode enabled. If I disable Power save mode of the DC regulator, the current consumption increases significantly..that tells me that DC regulator of my choice at the moment is not very efficient at light loads and DC regulator itself is taking a lot of current with power save mode disabled.. \$\endgroup\$
    – user101095
    Jan 4, 2018 at 22:15
  • 2
    \$\begingroup\$ By no means am I suggesting you change your entire toolchain, but the PICs (and I love Microchip as a business partner, as anyone will tell you) -- even the nanowatt variety -- don't come anywhere even close to an MSP430 for long battery life. Once you look at the time it takes to "get up to speed" on a PIC, the MSP430 has already been there and done that and gone back to sleep, already. However, you have 75mA*250mS*Vcc or somewhere around 50 mJ per event. How long do you sleep between events? \$\endgroup\$
    – jonk
    Jan 4, 2018 at 22:34
  • \$\begingroup\$ To answer your direct question, yes according to the literature those particular regulators will run more efficiently if you run them in PS mode. The efficiency gained is dependent on how small the load current is. It's all in those hundreds of graphs they give you. However, as @jonk is highlighting, there is no point squeezing the last percentile of efficiency out of the regulator if your main charge loss is elsewhere. You need to optimize both. \$\endgroup\$
    – Trevor_G
    Jan 4, 2018 at 22:40

1 Answer 1


I guess I'll just summarize the commentary. I don't expect this to be an answer, just what I see from the discussion.

I'm imagining something like the following image.

enter image description here

There are some spikes in the required current, which last \$250\:\text{ms}\$. The required current peaks at around \$75\:\text{mA}\$ and then falls back to some other current level (sleeping) for approximately \$10\:\text{s}\$, though this period is variable. Every \$45\:\text{s}\$, the battery is removed (no clue, at all, as to why -- it's all just magic to me) for about \$30\:\text{s}\$. So the total cycle time is \$75\:\text{s}\$.

The goal is to minimize the average power to help maximize the length of service for a battery system, before replacement.

From the above, I would also assume the following:

  • No services are being provided while the system is sleeping. So it would be still better if no power was consumed in between. In other words, the only useful periods are the ones that are occasional and consume \$75\:\text{mA}\$ for \$250\:\text{ms}\$.

Because of the length of time for the active period being \$250\:\text{ms}\$, I feel it's just fine to continue the idea of using a PIC MCU. The complaint I might otherwise have, were the period much shorter, would be that it takes a while to get a PIC MCU started from a "cold" sleep -- the oscillator just takes time to get up to speed. On the other hand, an MSP430 can fire up to full speed in about \$1\:\mu\text{s}\$. But given the duration, the MSP430 advantages mostly disappear. So that makes me quite comfortable with the PIC MCU approach here.

As I gather things, you need about \$20\:\text{mC}\$ of charge during the active period of time. The PIC MCU has a range of voltages over which it operates, and similar things can be said about whatever else is attached. Let's say that the allowances you can accept are a droop of no more than \$200\:\text{mV}\$ during the active period. Ignoring contributions by the battery, and putting the entire burden onto a capacitor, this means a capacitor value of \$100\:\text{mF}\$. With a low voltage type, it doesn't have to be that expensive or large. And this assumes that the battery itself can't contribute during this time (which it probably can.)

The average current required is less than \$2\:\text{mA}\$, given your statement of about \$10\:\text{s}\$ between activation events. This can be provided by something as little as a CR2032 lithium battery (which is not known for high currents.) Perhaps placing a capacitor in parallel with such a battery, with a current limiting resistor of course, would provide the necessary power supply without the need, cost, complexity, and/or quiescent losses of a voltage regulator.

Of course, you have other issues to deal with and I have only a very narrow tube-like perspective on your project. But what you've written so far takes me towards that kind of consideration as an alternative path.

The approach I'd like to have you consider would be to arrange things to use a capacitor as your reserve storage, add a current limit arrangement to the circuit so that the CR2032 battery isn't hit hard when first charging the capacitor, and just go with that. The PIC MCU can go into a decent sleep with fairly low draw. End of story.

  • \$\begingroup\$ thanks jonk, I am planning to use alkaline batteries - like AA / AAA. As per your suggestion, I will consider having a cap like (33 uF ?) across the batteries. But again going back to the original question, can I keep Power save mode enabled all time for my DC regulator like TPS61201? How would the switching frequency vary / affected in PS mode enabled? Would variable switching frequency be a concern in EMC? Thank you.. \$\endgroup\$
    – user101095
    Jan 6, 2018 at 16:56
  • \$\begingroup\$ @user101095 With Alkaline batteries, there is no need for a capacitor. They can easily provide that level of current. So no cap, no current limit resistor, just instead a direct connection. Their voltages go from about 1.56 V when full to about 1.2 V when "dead." Stacking two means about 3.1 V to about 2.4 V. If you use a PIC that can operate down to some minimum voltage, you can divide that by 2 to get the lifetime for your battery pair. AA batteries have no problem delivering 75 mA even at 1.2 V. \$\endgroup\$
    – jonk
    Jan 6, 2018 at 17:11
  • \$\begingroup\$ @user101095 Unless you have another reason for the TPS regulator, I'd eliminate it entirely and exclusively use parts that can operate over the same voltage range as the PIC can. The regulator adds cost, takes space, consumes power when you don't want it to do so, and decreases reliability. There are no good arguments for it, at all, unless there are other parts which require it. The PIC, when it sleeps deep, draws negligible power (1 microamp or less, even with the RTC and watchdog running.) But this is where I'm ignorant. I have no clue what else needs doing here. \$\endgroup\$
    – jonk
    Jan 6, 2018 at 17:14
  • \$\begingroup\$ there are other parts of the circuit that require 3.3V regulated..so, again I am going back to the original question which is still not clear to me..thank you.. \$\endgroup\$
    – user101095
    Jan 7, 2018 at 3:53
  • \$\begingroup\$ @user101095 There is no energy at all in AA batteries after about 1.2 V each (1.1 V is ~0% left.) So you need to work down to 2.3 to 2.4 V, I believe. Not down to 1.6 V. Since you have 3.3 V parts, you are absolutely correct about needing a boost and I'm wrong about the idea of just using batteries directly. I hadn't taken that point until now. Sorry about that. I suppose we are back to something like the TPS61221. I guess I'll delete my answer soon. \$\endgroup\$
    – jonk
    Jan 7, 2018 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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