I have the MIC5216 as voltage regulator. I am regulating 2 18650 down to 5v in order to supply enough power to my circuit. I am using a 328p as MCU. The chip goes to sleep and is awoken again by an external interrupt. That way I can reduce the energy consumption down to a minimum in order for the batteries to last as long as possible. I managed to get my 328p circuit in sleep mode to 80uA whereas the MIC5216 is using 190uA in a quiet state. I know the 5216 can go as low as 8uA. In order to achieve this low state a voltage of not more than 0.16V has to be applied to the Enable pin. As mentioned, I have 8.2V applied through the 18650's and if I remove these the chip of course will fully stop working and not provide any power at all to my circuit.

My question is: Is there a way, to reduce the power on the enable pin to 0.18V while the 328p is asleep without completely pulling that particular pin low and as soon as an interrupt is triggered to wake the whole circuit up again?

I have not worked with these kind of voltage regulators before so I appreciate any help I can get.

The datasheet for the MIC5216 is here

I am using the standard circuit shown in Figure 5 on page 11.

EDIT1 Sorry for the delay. Below a diagram of my circuit. enter image description here As you can see I added sections to the circuit in order to make the visual a little more legible. the voltage regulation part is the one that I have issues with. Even though the voltage supply is stable and provides enough current, I would like to bring it's current idle state closer to the quiescent advertised in the datasheet. My only issue is that I am not quite clear how.

The datasheet speaks about reducing the input voltage on the Enable pin (3) to 0.4V and 0.18V to achieve a quiescent state of 3 to 8μA. I cannot control the regulator with the 328p as it is going to go asleep. I was thinking about a voltage divider from the battery to the enable pin with an output of that above mentioned voltage. That could work up to the point where I want to wake the 328p up. I would now have to increase the voltage on the Enable pin of the regulator to the voltage of the battery for long enough to trigger a constant high on one pin of the MCU. With that idea though I have the problem how do I ensure to provide enough energy to the MCU in order wake it up and trigger that signal.

OK so much for my ideas. Let me know if they are totally bonkers or if there is some merit to them. Let me know if there is another way, maybe in the datasheet, that I missed.

Appreciate any feedback.


  • \$\begingroup\$ Please show the circuit \$\endgroup\$
    – Voltage Spike
    Commented Jul 30, 2019 at 21:43
  • \$\begingroup\$ Can't right now. I need to get home in order to upload a picture. The circuit creator does not have a 5 pin chip. I will post the circuit a little later. \$\endgroup\$
    – realShadow
    Commented Jul 30, 2019 at 21:47
  • \$\begingroup\$ The circuit creator does not have a 5 pin chip. It does, you just have to use the custom part tool and hit the + buttons to include the number of pins. \$\endgroup\$
    – user103380
    Commented Jul 30, 2019 at 21:55
  • \$\begingroup\$ I finished the circuit but it crashed on me. I will upload it later from home. That is easier. \$\endgroup\$
    – realShadow
    Commented Jul 30, 2019 at 22:25
  • 1
    \$\begingroup\$ I am trying to determine the actual maximum peak load current. The DFRobot DF9GMS is a 1.2kg/cm continuous rotation servo that looks suspiciously like a Towerpro SG90 (which draws ~750mA peak). Therefore I would use a regulator with current rating of at least 800mA. Unfortunately that rules out my first solution. \$\endgroup\$ Commented Jul 31, 2019 at 2:41

1 Answer 1


Is there a way, to reduce the power on the enable pin to 0.18V while the 328p is asleep without completely pulling that particular pin low and as soon as an interrupt is triggered to wake the whole circuit up again?

At 0.4V and below the regulator is turned off, so the 328p won't just be 'asleep', it will be dead, and the only way to make it 'wake up' again will be to turn the regulator back on via an external circuit. To do this you could make the pushbutton pull the Enable input high when pressed. On startup the MCU would also pull the Enable input high to keep the power on, and let it go low to power off. If you need to remember the program state while the MCU is off it could save the required info in the EEPROM just before powering down (remembering to give it time to write the data!)

If the MCU must stay powered all the time then to get lower sleep current you will need a different regulator. The MCP1703 should do it if peak current doesn't exceed 200mA. However you are powering a servo which which might need more than this.

  • \$\begingroup\$ Thanks for the answer. I was thinking something similar. I have only 1 issue: how long would the button need to be pressed before the chip would be able to pull the pin high? I know how the push button circuit would look like but what other components would be required before the 328 can trigger the signal? Do you have an example you could provide? I really do not care about the state. With the solution you suggest I even could remove the code that sends the chip to sleep and could let the MIC take care of the power saving by turning the circuit off. Would extend the battery life even more... \$\endgroup\$
    – realShadow
    Commented Jul 31, 2019 at 4:17
  • \$\begingroup\$ The 328 starts up very fast, so unless you have a bootloader it shouldn't have any trouble holding the Enable up before the switch is released. Just make that the first thing your code does, so if it doesn't do it in time no harm is done. The pushbutton could be wired directly from battery to EN, but the 328 won't like 8.4V on its output pin so put a 1N4148 diode in series (Cathode to EN) to block the higher voltage. \$\endgroup\$ Commented Jul 31, 2019 at 4:48
  • \$\begingroup\$ thanks for your feedback. I would have overlooked that diode which totally makes sense. I do have a bootloader on my 328. What does that mean in terms of startup time? Can I still implement the solution you suggested? Do I maybe need a capacitor to bridge the time until the chip is booted up? \$\endgroup\$
    – realShadow
    Commented Jul 31, 2019 at 13:55

Your Answer

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

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