Ultra Capacitor for input buffering

Re: Ultra Capacitor for input buffering

Super/Ultra Capacitors are not worth it for the average (and even not so average) application... For example, on your first link:

Ultracapacitors 48.6V 165F @ $1,800 each. And assume that you run it between 38 volts to 48 volts:

• Euseable = 1/2 * 165 Farad * ((48v)² - (38v)²) = 70,950 Watt*Seconds
• 70,950 WS * 1/60 sec per min * 1/60 min per hour = 19.7 Watt*Hours of storage
• 19.7 WH / 48 volt bank = 0.4 AH @ 48 volt battery equivalent

I may have gotten something wrong--But I don't think so... A capacitor of this size just does not compare to very well to a typical storage battery.

You can get more useful energy out of the capacitor by draining it further--But that would mean more issues for the energy conversion device (operating over a wider input range, higher currents required at lower capacitor voltages, etc.).

-Bill

Re: Ultra Capacitor for input buffering

Flooded Cell Lead Acid batteries just "Gas" and use a bit of water (if not too much charging current). Not good for them, but they can withstand it to a degree.

Capacitors can do some spectacular stuff if hit with over voltage. The ultra cap pointed too... 48 volts working, 51 volts for one second. Not a lot of margin for error.

-Bill

Re: Ultra Capacitor for input buffering

You have to read the specifications closely...

They may offer a pretty good cycle life (when compared to batteries), but just "power on" life may only be a couple of years.

If you have an application were a "super cap" can give you an advantage over a battery bank--Maybe.

But otherwise, I would stay away from them. They are not very useful unless you have equipment designed around them (i.e., your 12 volt inverter that runs from 10.5 to 15 volts should run down to 5 volts or so to get the most power from a super cap).

If you run a supper cap between 13 and 14.5 volts--There is not a lot of energy you can harvest from such a small voltage range.

Batteries store energy as chemical potential as (roughly) constant voltage. Capacitors store energy a a charge potential across a plate--So you have to substantially cycle the capacitor voltage range to store/harvest enough energy to make it worth while.

Or have very speciallized needs (such as very high current spikes for fractions of a second to a few seconds).

-Bill

Re: Ultra Capacitor for input buffering

Ripple--Just like the waves on a calm lake...

If you have a DC (direct current) voltage or current--Ripple is the variation in voltage or current from the the long term DC average.

If I have a battery and a 100 watt light... Switch off, battery at 12.7 volts; Switch on, battery at 12.6 volts. On/Off -- a 0.1 volt ripple voltage.

With a single phase AC inverter--Most inverters do not have enough input filter capacitors to "average out" the input DC current to a steady level.

Instead, for the peak voltage/current of their AC output wave form, they draw current directly from the battery bank. As the AC voltage (and current) falls to zero, the input DC current falls to zero--repeat 120x per second.

From this paper I just found (if a few readers can look at this paper--Seems to be a very complete article on solar power)--May be worth adding a link to():

http://www.nmsu.edu/~tdi/PV=NEC_HTML/pv-nec/pv-nec.html

DC Currents on Single-Phase Stand-alone Inverters

When the sinusoidal ac output current of a stand-alone inverter goes to zero 120 times per second, the input dc current also goes nearly to zero. With a resistive ac load connected to the inverter, the dc current waveform resembles a sinusoidal wave with a frequency of 120 Hz. The peak of the dc current is significantly above the average value of the current, and the lowest value of dc current is near zero.

An example of this is shown in the Figure F1. This is an example of a single-phase stand-alone inverter operating with a 4000-watt resistive load. The input battery voltage is 22 volts. The figure shows the dc current waveform. The measured average dc current is 254 amps. The RMS value of this current is 311 amps.


Attachment not found.

The problem is that a battery bank is designed to have a very stable voltage--Capacitors only "function" or pass current/energy if there is a changing voltage.

It is possible to design an AC inverter with an intermediate high(er) voltage capacitor stage to store energy--But it adds costs and complexity. If people are not willing to pay for the costs, then nobody is going to offer it.

A high voltage intermediate capacitor bank to store energy is what is done for the typical switching AC power supply. The AC voltage is rectified and stored on high voltage capacitors, and the DC output voltage/energy is derived from the capacitor bank (and not from the 120 Volt 60 Hz AC input. The AC utility grid forces power supply designers to store energy as there is only power from the grid near the peak of the AC voltage waves, and none available as the sine waves cross zero voltage.

So--You have a varying DC load (the AC inverter)... You want the battery voltage to not fall below ~12.7 volts at peak inverter draw--Because it would, very slightly, discharge the battery. And during low to zero current, the solar panels are "charging" the battery bank--Micro cycling 120x per second (with 60 Hz AC inverter).

So, you have to raise the "average" DC Voltage enough with the Float Voltage of the Battery Charger so that ripple voltage on the DC Battery Bus does not cause the battery voltage to drop into battery discharge range.

I have not researched if keeping float voltage at 13.6 volts + battery plate capacitance is "high enough" to avoid all micro cycling (i.e., is battery capacitance enough to avoid the chemical micro cycling?). But this is another reason we use the 100 AH per 1 kWatt (at 48 volt battery bank voltage)--The larger the bank, the lower the ripple voltage.

Note that the RMS current in the above example is ~1.22 times larger than the DC average current--WIth I²R heating, the ripple current is 1.22² = ~1.5x more heating... So here is an example of where DC and AC power math collide... And you need to be conservative with AC inverter wiring... And should derate the wiring by 1.22 more for the AC inverter DC input power (added on top of the 1.25 NEC derating).

With engineering, you see this "derating factor stacking" all over the place--And it sometimes becomes almost ridiculous--Wondering if all of the derating factors make sense or add up to conservative overkill... Leave that as an exercise for the student.

In this case, most people do not run their inverter systems at 100% of rated power--So and additional 1.22 derating may be overkill in most cases.

-Bill

Re: Ultra Capacitor for input buffering

Yes, the capacitor would reduce ripple voltage--And if you match the right capacitor to the job, it can help.

For reducing 120 Hz ripple current on an AC inverter, lets try some math (warning, has been many decades since school):

• Energy = 1/2 * C * V²

Say you want to "service" a 1,200 watt load on a 12 volt battery bank with a 0.1 volt ripple...

• 1/120 of a cycle * 1,200 Watt*Seconds = 10 Watt*Seconds
• C = E * 2 * (V² - V²) = 10 WS * 2 * (13.2v² - 13.1v²) = 52.6 Farad

So, if I did this right (maybe?), Yes a 100 Farad car capacitor would reduce your ripple voltage to ~0.1 volts for a 1,200 to ~2,300 Watt AC inverter on a 12 volt battery bank.

Note--That if you wanted to store the surge power required to start a 1/3 HP well pump for 2 seconds with a 0.1 volt drop on a 12 volt battery bank--You would be looking at a capacitor ~240 times larger (~12,624 Farad) or 120 of those car capacitors.

-Bill "duck and cover" B.