0.6 Farad Capacitor Bank to Reduce Battery Microcycling

[BackwoodsEE]

I've finished designing, simulating, building, and, finally, installing, my 0.6 Farad capacitor bank to reduce the microcycling that the Outback GS8048 does to the 410 Ah AGM battery in my hybrid grid-tie and backup 6 kW solar installation. It works.

Here's the capacitor bank in its steel enclosure (with front panel off) alongside the Outback load center. There are two pairs of copper busbars that each connect three capacitors in parallel. The negative busbars are connected together inside the enclosure with a short length of flattened copper tubing, which is in turn connected to the negative busbar of the GS8048 inverter via #2 compact stranded wire. Each positive busbar connects via #4 wire to an 80A circuit breaker and then to the positive plate of the load center. The #4 wires from the capacitor enclosure to the breakers are wrapped in red and orange electrical tape for easy identification and to further protect the THHN/2 insulation, as I'd really hate for these big capacitors to get shorted out at any point.

Fig-01.jpg

The wires are longer than I'd like them to be, but the voltage drop across them is still substantially less than the total remaining ripple voltage. The #4 positive wires only get a little warm with the inverter pulling 120 A, maybe ten degrees F over ambient.

Here's a close-up of the capacitor bank with the underside of a circuit board I designed to facilitate safe and gradual charge-up and discharge of the capacitors. Pushing a button on the top activates a high-side P-channel MOSFET to charge each busbar of capacitors through its own 50 W, 75 Ohm power resistor. Tripping the 3-gang circuit breaker automatically activates a low-side N-channel MOSFET to discharge the capacitors through the same power resistors. NEC 490.6(A) calls for the residual voltage to be reduced to 50 V nominal or less within one minute after disconnection. Since the nominal voltage of the battery bank is already less than 50 V, the discharge circuit isn't strictly necessary per code, but I wanted to implement it anyway because I have a healthy respect for large charged-up capacitors. I implemented the discharge circuit to conform with one of two options under NEC 490.6(B), "automatic means of connecting it to the terminals of the capacitor bank on removal of voltage from the line." A third, low-current breaker in the 3-gang breaker device disconnects when the high-current breakers do, and that activates the N-channel MOSFET to discharge the capacitors.

Fig-02.jpg

Here's the control board mounted on its non-conductive nylon spacers with the power resistors in the background. I confirmed with my simulations that no heatsinks are required for either MOSFET.

Fig-03.jpg

Here's the connection from the ganged breakers to the positive plate with two short #4 wires. The small yellow wire is the discharge control line, which gets shorted to ground and deactivates the N-channel MOSFET when the breaker is set. When the breaker is tripped (manually or by overcurrent), the discharge control line gets pulled high and the capacitors automatically discharge through the power resistors.

Fig-04.jpg

Now, the part you were waiting for: how well it works. Here is the normal 120 Hz ripple voltage at the positive plate (directly from the battery positive) with the Outback GS8048 pulling about 120 A and the capacitors not connected:

Fig-05.jpg

Then switch on the charged-up capacitors, and presto! About 1/4 as much ripple current and 1/4 the chemical reactions causing wear and tear on my battery, replaced with infinitely repeatable electrostatic charge and discharge in the capacitors.

Fig-06.jpg

This is not a project for everyone. Big capacitors connected to high-current sources can be dangerous, and I put a lot of thought into how to do it right. The amount of time and even money (each capacitor cost about $40) involved may not pay off in extra battery life. But I'm an electrical engineer and really wanted to do this, and am very pleased with the result.

[Sunking]

I must be missing something. It buys you nothing operationally. Sure it reduced ripple current, but ripple current is a good thing for batteries if not excessive. I could see it maybe if you are operating say a 5000 watt AM transmitter directly off 48 VDC battery to get the 60Hz hum out of the carrier, but we are talking AC power via an Inverter which is not carried through and even if it did does not hurt a thing.

So please ask what AC ripple does?

Short answer is is adds some heating to the cell in addition to the DC heating, and helps prevent lead sulfide buildup on the plates. It does not micro-cycle the battery. You are ot hurting the batteries, but you are not helping them either. So help me out, what am I missing?

[sdold]

I've only ready a little about "Microcycles", but I thought they involved a change in current direction (charge/discharge), not just a change in charge rate. We have a lot of ripple in some of our rectifiers at some sites and they don't seem to suffer reduced life, at least not that we can tell.

[Sunking]

I thought they involved a change in current direction (charge/discharge), not just a change in charge rate. .

That is correct Steve and the Textbook definition. It is common in Off-Grid PV systems. It is not a result of AC Ripple in an AC charger. but rather the varying load and sun conditions in an off-grid system that exist during the day. Example, during parts of the day the load requirements exceed what the panels can produce at a given moment like a cloud passing over and the batteries have make up for the shortage in power, and then recharge when the cloud passes.

Initially from the title I thought maybe he might be using Super Caps to make up for startup loads, and was interested in how high he chose to run to run capacitor voltages and what kind of buck converter he was using. Then I quickly realized he was not doing that running the Caps at battery voltage which means they provide no power and basically an external brute force filter in addition to what his charge has built in. I applaud the effort , but really do not know what purpose it serves.

AC ripple charge current is not problem unless it is excessive and excessive would be greater than 2% at full charge current capacity of the charger. AC ripple does not cause microcycles, it generates additional heat to the DC heating charge currents. Not a real problem for FLA batteries and could be a problem for AGM systems in Float service contributing to Thermal Runaway setting days in float like a telco office or UPS. Thermal runaway in PV systems is allmost impossible. Batteries have both DC resistance and AC impedance. The reactive component of a battery is the capacitance it exhibits and is in parallel with the DC resistance, thus AC impedance of a battery is always lower then DC.

Perhaps I am overlooking something or do not understand the objective. I just know it does not affect or have anything to do with microcycles. Besides off-grid systems do not have 60-Hz AC ripple.

[BackwoodsEE]

It's not an off-grid system. The Outback GS8048 is a hybrid grid-tie with battery. Charge controllers (a pair of Morningstar TS-600-60s in my case) feed 48V DC to the battery and the Outback inverts as much as possible of it to produce AC while maintaining float voltage. If the grid goes down, the inverter will operate like any other off-grid setup.

The highest part of the 120 Hz ripple voltage is the result of DC current going into the battery during the two parts of the 60 Hz cycle where the output is near zero-crossing and no instantaneous power is supplied to the grid. The lowest part is when the inverter is feeding the highest magnitude positive or negative current to the grid, which again happens twice per 60 Hz cycle. Hence the doubling of frequency.

There are indeed microcycles happening here. It could not be otherwise, because the battery must have nearly zero average current while in float service. What happens is that the charge controllers shove all their 120 A of current into the battery when the inverter is at a zero crossing (and not consuming any significant current) and then the inverter takes 240 A out when it is at each peak in the cycle, requiring all the 120 A from the charge controllers plus another 120 A back out from the battery. I confirmed this intuitive conclusion with my scope attached to the shunts from a charge controller and to the inverter. It's what happen, as it must be.

By the way, an off-grid system can have microcycles and not just varying amplitude of charge or discharge, too, when power is being consumed during the day. This is most likely to be seen after average current to the battery starts to taper off during absorb; a heavy 120 Hz varying load will easily reverse the direction of current flow to the battery for part of the cycle as the charge controllers--operating in a reduced power production mode high up on the PV array's IV curve--get overloaded with the instantaneous current demand and can't keep up.

[Mike90250]

Does your discharge circuit stay "engaged" or enabled all the time, after it's been activated ? I ask, because caps will develop a memory charge after being discharged, and they could recover some voltage.

[BackwoodsEE]

Does your discharge circuit stay "engaged" or enabled all the time, after it's been activated ? I ask, because caps will develop a memory charge after being discharged, and they could recover some voltage.

Yup. As long as the ganged breaker stays open, the N-channel MOSFET keeps the discharge path connected.

That rebound effect is pretty impressive on these big electrolytic caps. I've shorted them completely (after a gradual discharge), then released the short and came back later to measure the voltage. They recovered about 0.4V, enough to make a healthy pop if shorted again. It's weird to think of where that might be coming from with no outside source, in what would seem to be a purely electrostatic device. A small electrochemical reaction in the electrolyte perhaps?

[Mike90250]

....... It's weird to think of where that might be coming from with no outside source, in what would seem to be a purely electrostatic device. A small electrochemical reaction in the electrolyte perhaps?

No reactions happening, it's electrons trapped in the dielectric.

[bcroe]

I cannot see if you used the physical cap conductor arrangement I recommended a while back. If not,
you didn't get all the ripple suppression possible. How much less, not a lot. Measure the ripple across
each of those bus bars and connecting wires, that is the approximate total. Bruce Roe

[BackwoodsEE]

I cannot see if you used the physical cap conductor arrangement I recommended a while back. If not,
you didn't get all the ripple suppression possible. How much less, not a lot. Measure the ripple across
each of those bus bars and connecting wires, that is the approximate total. Bruce Roe

With the inverter pulling 120 A DC, there is about 2 mV RMS from one end of a given busbar to the opposite end, and about 125 mV RMS from the battery positive plate, through the 80 A breaker, to the busbar. The busbars have negligible voltage drop in the overall system.

I'd like for there not to be all that voltage drop from positive plate to the busbars, but the battery does need overcurrent protection from the capacitor array hanging on it as well as having the capacitor array switchable in and out of the circuit. And even with a perfect connection to the array from the battery, the capacitor ESR and non-zero capacitive reactance would still leave me with about 12% (instead of about 25%) of the original battery microcycling. #2 wire for the positive connections would have been nice, but unwieldy and impractical to connect to the small studs on the 80 A breakers.

As far as breaking the very high current connection between battery and inverter provided by the Outback load center to stick something in between them is concerned, that was out of the question for me.

[bcroe]

OK, practical considerations somewhat limit optimum performance.

[Mike90250]

I'd have used 1" copper braid for low impedance connections, not wire As you note, there's enough resistance to heat them.