How can I power my device using a Solar Panel and capacitor bank?

"I want to power a module that requires 3.3V and 500mA minimum for startup. I have a solar panel that outputs max 3V at 70mA and a 3.3V 3A max output boost converter. I know I need a super capacitor or a capacitor bank to store energy so I can get the current needed for start up. Also, my module only needs around 500mA for less than 1 second. I will probably consume an average of 10-40mAh.

I do not know too much about power electronics, but I want to know how can I design a capacitor power bank circuit that can handle 3.3V, can store a minnimum of 500mA, only takes less than 5 minutes (if possible) to charge the capacitor bank to 500mA, discharge the current for 30ms or more and have the capacitor discharge when full.

This 500mA current is only required once per day or when the solar panel drops below 1V.

I only want my device to be on during the day."

Above was my previous question. Below is my new information!

• My size requirement is a 2x2 solar panel
• Cannot have a battery, customer wants power only during day or peak hours
• The figure is what I have been calculating on the wattage required

Will this work?

• I have purchased the evaluation board for the boost converter and set a breadboard circuit up with smaller capacitors and supplied voltage to a red led, but it doesn't have consistent power, it blinks fast like a relay.

Also, when i supply this power to my wifi module, will this provide consistent power throughout the day after the startup requirement?

• This is probably not going to work. You need a larger panel, and you probably need a lead-acid or possibly NiCd battery (these being some of the easier types to maintain on charge). Maybe if you are really lucky in good sun conditions you can manage startup with a capacitor, but consider it unlikely. – Chris Stratton Jun 29 '20 at 16:31
• you have some of your capacities and currents screwy, you're not going to get serious answers until you get your units straight. 'can store a minnimum of 500mA' should be either can source a current of 500 mA, or can store an energy of x Joules. 'I will probably consume an average of 10-40mAh.' should probably be consume an average current of 10-40 mA. – Neil_UK Jun 29 '20 at 17:10
• Yes, i Understand my verbiage is a little off. But is there a way I can get a current of 500mA by charging it with a 70mA solar panel and then discharging it to my module? any sugguestions? I can't use a battery – Sherman Jun 29 '20 at 17:31
• 1. Cap may be better at BC input. 2. Mean sun hours / day can be found at hrrp://www.gaisma.com 3. Max summer full sun hours equivalent is seldom over 6 hours. 4. A 3V 70 mA panel is sailing close to the wind in even full sunlight. Iin is say Iout x Vout/Vin / efficiency = say 40 mA x 3.3/3 x 1/0.8 =~ 55 mA. Any wisps of cloud any you wont get 40 mA. Also only nearer middle of day. || Q1 WHY no battery? That is a VERY major omission. Q2. Are you limited in allowed PV panel size.? Q3 Is resetting many times an hour and/or not running at all often acceptable. ... – Russell McMahon Jun 30 '20 at 1:57
• Q4 Please fully specify startup characteristics. Q5 What is boost converter Vmin operate and startup.?|| We can tell you what it will take to do what you want BUT MUST properly know what it is that you want . ie spec is far too loose so far. || Yu'll need about 25,000 uF for your 500 mA/30 mS startip (your comment based figures) . Charging a MUCH larger supercap in 5 minutes is easy AND you are going to need one. You'll get AROUND 20 seconds per Farad with 1 V drop. – Russell McMahon Jun 30 '20 at 2:05

Will this work?

You'll need more capacitors, a lot more. Another problem is you'll also need an MPPT tracker and capacitor charge controller. A bigger solar panel with a higher voltage would also be recommended. The best option would be to use a battery.

The boost converter only works to 0.9V so there is energy stored in the capacitor that cannot be used.

You'll still get $$\ Q=1/2*C*V^2 \$$

$$\ 1/2*1F*3V^2-1/2*1F*0.9V^2=4J \$$

4J/(1700mW seconds) = 2.35seconds

Which means to run this for 1 hour you'd need 30 1Farad capacitors in parallel, and it would take a long long time to charge up.

The second problem is the above calculations are only good if the capcitor actually reaches 3.3V. In the diagram above the solar panel could only charge the cap to 1.3V, which will yield even less energy storage.

The third problem is you need a solar charge controller, because a capacitor is a really high load, the solar cells won't be operating at their maximum efficiency. Because of this the capacitor may not charge at all, or take a very long time.

The diode also does not help and contributes to a lot of loss from the panel.

• This is great information! But one thing is that I only need that minimum of 500mA for 1 second to turn on the module. After that, all I need is 20-50mA to keep it on. And since my diode is being used for protection for my solar panel, the voltage is 1.3 so whatever I supply my capacitor that is the max voltage it can get to? What does a solar charge controller do for my circuit? What if I put my supercapacitor after the voltage has been stepped up to 3.3V? – Sherman Jun 30 '20 at 21:29
• An MPPT isolates the impedance of the panels from the load of the circuit and makes sure that the solar panels are drawing current at the best efficiency. The capacitor will have a rating in the datasheet that should not be exceeded if you want it to function properly. If you put the capacitor after its been stepped up, it will function as a filter bypass capacitor and it will also take a very long time for the circuit to turn on because it will charge the capacitor as well as the module. – Voltage Spike Jun 30 '20 at 21:55
• Okay that makes sense! I will look into solar charge controllers. But aren't they used with batteries? – Sherman Jun 30 '20 at 22:05
• Yeah, if not using a battery then use an MPPT – Voltage Spike Jun 30 '20 at 22:09
• I'm looking for MPPT, but are they bulky? Because I have a size constraint. – Sherman Jun 30 '20 at 22:21

I finally got my device to turn on without using a battery source. I have charged the capacitor to a certain voltage, then had a switch to connect to the load when above a certain threshold. To do so, I needed to break down my calculations to Energy (Joules/sec). My device can turn on for at least 2 minutes before shutting down and having the capacitor to recharge again. From calculations, if I use a MPPT, I can maximize my solar panel power output to charge my capacitor even faster.

I have some more things to answer, but I will complete it.

Thank you everyone for your help!

• We do not know what size cap you are now using. || Not using a series diode with the panel may produce better results. It will not damage the panel. Worst case in varying light conditions (clouds etc) it MAY discharge the cap partially nut probably not. We do not even know your circuit - is the cap now before or after the boost converter etc. Having a circuit and description of what you are actually doing will help us all. – Russell McMahon Jul 29 '20 at 2:22