# Evaluate and enhance my Super Capacitor Power Bank

I am not an Electrical Engineer, but know some basics from reading and practicing on this topic. Please evaluate my circuit for a Super Capacitor Power Bank to charge my phone, and clarify my points below.

1. If the input is fixed at 2.5V, do I need any protection board for a 2.7V Super Capacitor (3000F)

2. I plan to power the circuit using USB, so I need to step down 5V to 2.5V. One option is to buy a step down module, but is there any other simple technique I can use? Diodes?

Basically, the USB input charges up the capacitor and the LED lights up. The top left switch is not required, but it simulates the Power Bank being charged or not. When the switch on the top right is on, the capacitor discharges into a Step-Up Booster USB Charge Module, like this...

http://www.ebay.com/itm/5PCS-PFM-Control-DC-DC-USB-0-9V-5V-to-5V-dc-Boost-Step-up-Power-Supply-Module-/201542969731

Perhaps, it may be better to use two capacitors in series. Would I need a protection board, if I keep the input voltage at 2.5V max?

Anything else I am missing?

• Have you estimated how much energy is needed in your phone battery to compare with storage avail. in SuperCap and losses from ESR. An external battery may be more efficient and effective. Mar 12 '17 at 15:27
• The average phone battery is around 2,500 mAh. The point with the Super Capacitor Power Bank is to charge the PB within 5-10 minute and then let the phone charge in its usual 1-2 hours. Mar 12 '17 at 15:42
• Charging a 3000F cap to 2.7V in 5 minutes requires a constant current source of 27 amps, with peak power of 73 watts. This is slightly more than a USB port can provide (ie, 2.5 watts). Mar 12 '17 at 16:46

## UPDATE

After spending some time going over the numbers considering using a switching regulator to boost the super capacitor (SC) to 5V from 2.7 on down to 0.9V, it might work. Still may be a bad idea. You won't know until you put all the pieces in place.

Dependencies

Battery Capacity (actual) Battery Discharge Percent Charge time

A typical phone comes with between 1500 and 2000mA hr. That capacity is when the battery is made. From that point on the battery capacity decreases.

• Time
• Discharge Rate (talk vs. standby, etc)
• Number of Charge Cycles
• Depth of charge
• Battery chemistry (li-cobalt assumed for phones)

Acceptable battery life is 3 years and 500 charge cycles. A charge cycle is from 80% discharged to 100% charged.

After about 250 charge cycles battery capacity drops to around 80%. Less capacity also reduces charge time and SC energy requirements (joules).

The biggest performance factor is Depth of Discharge. Do you let the battery go dead or do you charge when 80% is left?

So the more times you charge and how much it needs to be charged is what governs the amount of energy needed. Reducing the charge percentage will also increase the number of charge cycles.

Energy capacity of SC: 3000F * (2.7V)2 / 2 = 10935 J

The boost regulator operated down to 0.0v which means not all the SC capacity can be used.

SC Unused: 3000F * (0.9V)2 / 2 = 1215 J

Amount SC energy available: 10935 - 1215 = 9720 J

So if you discharge a new 3.6v 2000mA hr battery you will need 25,920 Joules to fully charge battery.

The SC has 9720 / 25,920 = 37.5%

A slightly used battery has 90% capacity left raising SC capacity to 41.67%

If your phone uses a 1500mA hr battery we're up to 55.55%.

If you use a 3400 2.85v Farad SCp (new Maxwell Tech.), you are up to SC capacity 71% of phone charge.

You could discharge the phone down to 29% and the SC will bring it up to 100%.

The above would be if the boost regulator and battery charger were working at 100% efficiency.

The other problem is the charge time. A boost charger that has a 5v output, 0.9v minimum input voltage, generally has a lower output current than what is needed for a phone battery 1C charge rate.

A 1C charge is charging at the battery's 1 hour capacity , e.g. 2000mA hr = 2-3 hrs @ 2 Amp for the first charge stage (about 1 hr) which decrease to 3% after about 3 hrs.

Recommended charge rates are 0.5C to 0.8C.

If the boost regulator can provide 500mA, the charge rate of a new, with no compromises (dead back to 100%), 2000mA hr battery to 0.25C. Or an older 1500mA hr, with 75% capacity left, to about 0.45C.

Far from ideal but depending on your circumstance and how efficient you can convert the cap energy in to charging the phone, it may be doable.

End of Update

A 3000 Farad capacitor can store more than enough energy to charge the batteries. The if i,s can it be done efficiently enough. The rate can it be charged probably will be too long.

Let's say the phone battery is 1500mA Hr. We would need 1.5 Amp for the first hour. A boost regulator would have to operate from 2.7v down to 1V.

Supercap energy cells are not batteries. They do not hold a charge like a battery.

As an energy cell supercap is used to bridge power gaps lasting from a few seconds to a few minutes. Like to supply energy between power failure and backup generator starting.

I doubt you will want to use a supercap in place of a battery. The specific energy (capacity) of a supercap is 10-50 times less than a Li-ion battery. And they cost a lot more than a battery.

## Why it will not work

The super cap will not hold its voltage like a battery. The voltage immediately will begin to drop. It will drop below the voltage necessary to continue charging very quickly. Like where the battery voltage (solid black line) crosses the Supercap's discharge voltage, about a half hour into charging time.

So you need a boost regulator to get the super cap up to the battery charger 4.2V output voltage. The efficiency of the boost will be 60-80%, then the charger will be 80-90% to what's left of the 60-80%. The supper cap will cease to work when hitting the boost converter's Minimum Input Voltage with a 1.5 Amp Output Current.

Charging a one energy cell with another energy cell? Where did you come up with that idea? How about plug your phone into the USB port?

If you need a portable battery for your phone, get a battery.

• I think the Cap energy should be quadratic decay not linear due to $1/2CV^2$ Mar 12 '17 at 15:30
• @TonyStewart.EEsince'75 You'd think so, but a super cap is more like a battery than a capacitor. The supercapacitor crosses into battery technology by using battery-like electrodes to gain higher energy density. Mar 12 '17 at 15:40
• @Misunderstood the point in using a Super Cap was to charge the super cap real quick and then use it to charge the phone. Imagine you are on a road trip or outdoors. You stop by a restaurant and in the 10-15 minutes you take to have lunch, you charge the super capacitor using a 120/240 AC to USB DC charger. Then you take the charged up power bank and let your phone batteries charge normally. And yes, I do realise they cost more. The higher cost is for the lower PB charge time. BTW, I have made Lithium powered power banks in the past Mar 12 '17 at 15:47
• @Misunderstood I know density is higher in double-layer caps but still when V drops linear with CC load E drops exponentially with small changes in C while battery drops C rapidly below Vmin. and ESR rises rapidly. Even battery energy is not that flat Mar 12 '17 at 15:53
• @TonyStewart.EEsince'75 While the basic Double Layer Cap (EDLC) depends on electrostatic action, the Asymmetric Electrochemical Double Layer Capacitor (AEDLC) uses battery-like electrodes. RE: flat. End of discharge is 3V, not much of a drop so all will be fairly flat. The slope of the discharge of an Li-Ion depends on which chemistry and discharge rate. The 1C discharge curve is flat especially a Li-phosphate. Even the Li-cobalt typically used in laptops is very flat. Li-manganese used in power tools not that will be less flat. 5C discharge rate, not so flat. 0.1C discharge, very flat. Mar 12 '17 at 16:34

that's a lot of questions...

1.If the input is fixed at 2.5V, do I need any protection board for a 2.7V Super Capacitor (3000F)

If it's fixed why would you need protection? Your real question would be how do I make sure it is "Fixed". A zener across the cap would not hurt.

2.I plan to power the circuit using USB, so I need to step down 5V to 2.5V. One option is to buy a step down module, but is there any other simple technique I can use? Diodes?

You could use diodes, but it really depends on how much sustained current you will be using. Half the power you are taking from the source will be dissipated in the diodes.

Perhaps, it may be better to use two capacitors in series. Would I need a protection board, if I keep the input voltage at 2.5V max?

This may well be a better alternative. Why waste power dropping the input then re-regulating when you can just change the storage voltage and regulate once.

Anything else I am missing?

If you plan on plugging this into the USB port of something you are going to basically short out the 5V with that 1.1R resistor when you plug it in. That's just over 4A. You had better have a pretty beefy USB port and cables.

It will also take A LONG time to charge that up at appropriate currents.

Two stacked 3000F caps through an increased R of 10R, so you only drain 500mA, I calculate an RC of almost 17hours (Maybe someone can verify that...)

And that 10R will need to be rated at over 2.5W....

• The 1.2R was for a 2.5V input, since I wanted a 2A charge rate into the Super Cap. If I connect 2 2.7 Super Caps in series, then I guess I'd not need any protection, since USB output is 5V max. In addition, I believe USB from a computer is rated to output 100 mA to 900 mA, depending on the USB version. Mar 12 '17 at 16:06
• USB rev 1 and 2 the max is 500mA and 3.0 is 900mA, but what is actually provided is up to the manufacturer and if its a hub etc. You must also take into consideration that with high current the voltage drops in the connectors and cables are going to add up, and have been known to be 350mV @ 500mA. Mar 12 '17 at 23:17

A cell phone battery is rated in mAh @V which might be converted to Energy Stored watt-seconds [Joules] from which we can derive equivalent capacitance using charge rates to minimize losses, so we neglect for now.

Compare:

• for a battery 2000mAh @3.6V x 3600s/h, E = 2000mAh*3.6*3600 = 25.9kJ
• for a super cap. $E=1/2CV^2 = 1/2 * 3kF * 3^2 = 13.5~~kJ~~$

How much boost will you get? <50%?
ESR Losses reduce this with greater charge/discharge rates and converter losses.

However rate of improvements is rapid in Super cap technology in the area of Lithium Ion Capacitors, LIC over EDLC.

I was assuming you were using EDLC which ranges from 0 to Vmax per cell.