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I am having the toughest time trying to figure out how to charge a battery with solar panel to keep my project completely "off grid". Essentially I have a Wemos D1 mini that turns a servo when it's issued a command on the network. The Wemos needs to be powered 24x7x365 but the servo only powers up when it needs to turn (3-4 times a day max). I did a bunch of measurements today and here is what I have found. I am using this USB digital Multimeter to get my current readings below. I tested by using clean grid power to get all of these readings, hopefully this is the correct way to approach the power consumption.

• Servo draws 0.00A connected but not rotating
• Servo draws .5A max (sometimes less) when turning only
• Wemos draws .13A max observed (usually .09A) this is a constant load
• Absolute max draw I could find when turning the servo (with controller connected) is .43A rounded up to .5A @ 5V = 2.5W power draw.
• Idle time with just the controller is .13A max .09A average @ 5V = .65 W higest observed .45 W average

Here is what I have currently purchased:

• 10,000mAh Battery with 2 USB ports
• Wemos D1 mini
• Full rotation servo
• 5V 8W Solar Panel

I've connected everything as follows:

I connect 1 USB output from the battery to the Wemos D1 mini. I connect the other USB output to the servo using a modified USB cable. The servo control cable is then connected to the Wemos. I connected the solar panel to the charge port on the battery pack. Using grid power I can confirm that the battery pack will charge while also running the Wemos / servo. So in theory a solar panel should accomplish the same task.

I am able to power the Wemos with only the solar panel. However, once you issue the rotate command the current draw is too high and everything restarts. When I connected the solar panel to my battery it looks like I can get about .45A @ 5V = 2.25W max from the panel (outside in direct sunlight), which explains why everything reboots when you rotate the servo. Obviously I'm doing something wrong because I'm not running "forever" as I had hoped. I'm still currently running, it's been about 48 hours, but the battery will die in the very near future.

All that being said I have a couple of questions:

• Since my panel is providing .45A @ 5V = 2.25W and the idle power consumption is .09A @ 5V = .45W, I should be able to charge the battery while powering the controller, correct?
• When I rotate the servo it should pull the small extra current from the battery but should be quickly replaced with the overhead from the solar panel keeping the battery "topped off", correct?
• Assuming the device is idle for simplicity, if I have 12 hours without sun I will draw .09A x 12 = 1.08A from the battery over that time. If I get 8 hours of sunlight I should get .45A - .09A = .36A x 8 = 2.88A over that time, meaning it will fully recharge the battery. Is this math correct?
• I've looked into deep sleep mode but then I wouldn't be able to issue commands to the Wemos unless it happened to be awake. I can't find any way to make the Wemos use low power until it receives a network based command. Is there any work around you can think of to help consume less power?
• Do I just need a solar panel that provides more watts? I need to keep this form factor as small as possible and I need to replicate it several times (10+) so I really need to keep the panel cost down, and size as small as possible.

I really appreciate the help if someone can take the time to read through this huge wall of text. I think I have a decent understanding of how this all works, but I'm not 100% sure so any help is awesome, thanks!

Did you ever get any response to this?

I have a very similar use-case scenario and answers to the questions you posted would definitely be relevant to me. Did you manage to get these questions answered?

Unfortunately not. I've not been able to run the D1 for more than 2 days with battery / solar power.

Just a note to add to the conversation:
When doing solar storage/load calculations you can never assume 100% efficiency. There will be lost in charging the battery and lost in powering the project from the battery ... said losses can be very high based upon battery chemistry, and where the battery is in the "capacity" curve as the solar unit is attempting to charge the battery.

I suggest a lots more research as the Internet is full of use cases. But first, just review the basics:
http://batteryuniversity.com/learn/article/charging_with_solar_and_turbine

Power losses are discussed generally in thisPDF

Ray