Using capacitors to prevent surge current on lead-acid batteries?

Question

I'm using a 300 Ah lead-acid battery bank, and a 12V->230V 1000w pure-sine inverter, to power a residential-type refrigerator. With a bit of experimentation, I've managed to reduce the starting power required to a peak of approximately 1500w for 400 ms, which is within what that the inverter can provide. So far, so good.

Unfortunately, this draws a peak current, I'm guessing, of about 100A on the battery bank, which causes the voltage to drop for a brief moment to 11.5V, even when fully charged.

Over several months of use, I've realized that the battery bank capacity is severely reduced by these peak currents, reducing the 300Ah capacity to something more like 100 Ah. Perhaps I've already damaged the batteries.

A more suitable replacement would be a battery type that can handle these surges, e.g. LifePo4 (designed for 3C peak, e.g. 720A). But the price for these batteries are simply prohibitive: my 300Ah, 80%/240Ah usable, lead-acid bank cost about 400 USD; a LifePo4 240Ah bank would cost approx 1400 USD).

Another option is to add some kind of additional capacitance. For example, ultra/supercapacitors. I'm not sure what kind of specification I would need. The voltage would be approx 12.5V, and should be able to release 500 J before reaching, I'm guessing, 12.1-12.3V. Wouldn't this mean that I would need more than 1000F?

I'm thus faced with a number of options:

• a) replace battery bank (1000+ USD)
• b) replace starter motor with Danfoss BD35F (15A peak, approx 450 USD)
• c) add supercapacitor (how many Farads? 3F seem to cost approx 50 USD)
• d) design a circuit that temporarily draws current from a standard car starter battery instead (50Ah 12V car battery costs about 75 USD)

I'm thinking that option D might be the simplest and most universal. I could connect it in parallel only when starting the fridge, and disconnect it for the rest of the time, while allowing it to draw a charge through a diode. I'm assuming that the batteries would compete to deliver the current, but that the starter battery would maintain a higher voltage, thus winning the battle, and being the primary deliverer of current. Alternatively, I'd have to disconnect the battery bank temporarily, to ensure all power being drawn from the starter battery.

The easiest is obviously to spend 1500 USD for fridge that is designed for 12V, but, I like being hands-on.

Ideas are welcome!

Additional note: As I understand it, the surge current is only 100A x 12.5 V for 500ms, or 625 J. A starter motor battery with 50Ah/12.5V would have 112kJ capacity at a 5% discharge, which would allow for 180 starts.

Update: I've since replaced the lead-acid batteries with LiFePO4 instead. However, if you wish to improve the starting current performance, I believe that you could wire a small 6Ah LiFePO4 (e.g. motorcycle starter battery) in parallel, operating in a voltage range of 12-14.4V. However, my realization is that I don't know why anyone would ever want to buy lead-acid batteries in the first place. An absolutely outdated technology.

Answer 1

It seems that your either your calculations are far off, or you have a bad battery pack - particularly, you have at least one weak cell.

This document suggests that, to get 11.5 volts during peak discharge you must be drawing on the order of 2C. For a 300 Ah battery that's 600 amps+. Of course, different batteries have different discharge characteristics, so the curves should be taken with a grain of salt, but I'm inclined to accept them as a starting point.

As for using a supecap, the calculations are pretty straightforward. $$\frac{dv}{dt} = \frac{i}{C}$$ or in simple terms a 1 F supercap will discharge at 1 volt/sec for a 1 amp current. Even assuming your current peak is only 100 A, you'll get a 100 V/s droop per farad. To produce a 0.5 volt drop in 400 msec, you'll need $$C = \frac{i}{\frac{dv}{dt}} = \frac{i\times dt}{dV} = \frac{100\times 0.4}{.5} = 80F$$ or about 27 of your 3F units, for a cost of about 1350 dollars.

Answer 2

I'd suggest a simple cure for your problem is to use a LiPo car start pack in parallel with your batteries. These units use LiPo cells (as do many larger 4S1P RC battery packs) capable of supporting 30C for very short periods of a few seconds.
Even if you leave the unit permanently connected, your battery charging system will never overcharge the LiPo (in fact it will always leave it undercharged).

There are Supercapacitor based battery replacements that would provide superb high current capability, but they are terribly expensive. They also need very specialized charging and balancing electronics. They are a great idea, but just not consumer friendly yet.

Answer 3

My wild guess is that the inverter very likely has input capacitors, meant to decouple the internal switching operation from the 12V supply.

Connect a discharged capacitor to a 12V supply, and you get a very sudden rise in current.

Now, the inverter doesn't have to charge these capacitors that fast – it would probably work if you had the following option:

1. load the caps slowly, and only after they've reached a reasonable charge, and only after that
2. turn on the inverter

Probably, 2. could also be replaced by simply not turning on the load overly fast.

What you could try to do is:

1. Connect the inverter with a 0.5Ω or something resistor in series. Wait a second or two.
2. Now that the capacitors hopefully are charged, bypass the resistor, e.g. with a relay.
3. turn on the inverter.

You'll need to find something that has a low enough, but not too low, resistance at turn on. There's load resistors that are designed exactly for this purpose, but a coil wound of normal copper (or aluminium, for that matter) wire might even do the job better (having a high resistance to fast-changing currents, and a low one for DC).