Introduction

I work for a fuel-cell R & D. We're building a 5kW (low-voltage) fuel-cell system that I wanted to tie into a grid using an XW6048, a model which requires a 48V-nominal battery bank to run properly.

Looking around on forums, I found that people sized their battery banks depending on a lot of factors. The "100Ah per 1kW" rule was thrown around a lot. I was not interested in running our 5kW GTI for long periods of time on batteries alone (full-load for only 10's of minutes at most) so I thought I might be able to get away with sizing a much smaller bank. But it was mentioned that the AC currents being pulled by the GTI are a concern, and the battery bank needed to be sized to accommodate these large AC currents. There's no rule-of-thumb for this as far as I can tell.

Reflecting on the AC currents further, I realized that it would be better not to place a significant portion of these currents onto our fuel-cells. Fuel-cells prefer a steady power draw.

Therefore, there were two technical challenges to solve:

(1) Save money --> a large, multi-$1k battery bank was not an option

(2) Minimize the AC currents drawn from the fuel-cell stacks

I wondered if placing a large capacitor bank in parallel with the battery bank might help. If a capacitor bank having significantly lower impedance were attached in parallel with a battery bank, AC currents more likely be pulled from the capacitors than the batteries. And if the batteries and capacitor bank were handling the majority of the AC currents, then very little would need to be pulled from the system.

I looked around on forums and asked some questions regarding input capacitors, and I never got any good answers. I posted on Solar-Guppy's forum, but he's shut it down now. I'm following up here instead so that people can learn from what I did.

RMS current / charge analysis

I started with an ideal analysis in order to see what ballpark the input AC currents would be in. From previous experience, I knew that true-sine wave inverters pull in a sine-squared input current having a 120Hz frequency. The derivation is fun: just assume a constant input voltage and a 120V_{RMS}/ 60Hz output waveform. Here's the set-up:

P_{in}(t) = V_{IN}* I_{in}(t)

P_{out}(t) = V_{out}(t)^{2}/ R_{load}= V_{out}(t) * I_{out}(t)

P_{out}(t) = P_{in}(t)

I'll let the reader take it from here. You're trying to find I_{in}(t). Assuming 5kW in and 5kW out (100% efficiency) one can generate data for a single period of the input current waveform in Excel and carry out numerical analyses to get all sorts of relevant data. The datum I was most interested in was the RMS value of the current. I looked around on the web, and the conclusion I came to is that there is no analytical solution for integrating a sine-squared waveform. However, you can use Excel to compute the RMS numerically.

From my analysis, I found that for 5kW at 46V input (the minimum necessary for operation) ~77A_{RMS}was demanded. 77A_{RMS}is nothing to scoff at obviously: that's all heat-loss type current. One can see why people take the AC currents into account when sizing their battery bank.

The next thing I wondered was how much charge would be removed during a half-cycle of the 120Hz sine-squared current. Integration of the input current over one half-cycle gives this figure: Q = ~0.29C. Next, I calculated how much capacitance is required to hold 0.29C at 46V using the capacitance equation: C = Q / V = ~6.3mF. Finally, on a whim I decided that I didn't want this discharge / charge cycle to use up any more than 1% of the capacity of my capacitor bank. This would keep the ripple voltage to some minimum. Thus, I would need a 6.3mF / 1% = 0.63F bank. Rough math, I know, but it's probably close enough.

Bank sizing / pricing

So, at a minimum I needed a capacitor bank of at least 0.63F that could handle at least 77A_{RMS}.

Searching around on Digi-Key, I decided to make these my unit capacitor of choice: http://search.digikey.com/us/en/prod...332-ND/2095946. These 0.1F capacitors are only $36.88 apiece and can handle voltages up to 80V, nearly twice my required voltage. These capacitors also have a 21.42A_{RMS}rating. This RMS capability adds in parallel, so a 0.7F bank would have a total RMS capability of ~150A_{RMS}, double the computed necessary capability.

This is my solution. And for the heck of it, I just bought 10 of them to catch the price break and make a 1F capacitor bank. I wasn't sure how accurate my analysis was, so I played it safe. I used copper bar to connect them all in parallel, I connected them to the GTI through a separate circuit breaker, and I wired up a charge / discharge circuit to be able to charge them from the battery bank.

Testing

This is what I found. At 5kW / 50V, a sine-squared waveform of ~195A_{pp}/ ~90A_{RMS}is demanded by the GTI. With the 5kW power system, the 48V battery bank, and the 1F capacitor bank all connected in parallel, the AC current waveform is divided between the three as follows: ~9% from the system, ~14% from the batteries, and a whopping 77% from the capacitors!

I consider this a success.

Was it worth it?

In total, the capacitor bank probably cost me between $400 and $500, and the batteries I'm using are semi-truck batteries for about $125 apiece = $500. The batteries I'm using are this kind: http://www.interstatebatteries.com/c...ercial%29.aspx. I think they are used in semi-trucks and such. I had the fellow I purchased them from call the company to ask what the effective capacity was of one of these batteries, and it comes to around 80Ah @ 25A discharge. Not much.

So, here are my options.

(1) I could have purchased a top-of-the-line, 48V / 500Ah battery bank for around $3,750, assuming the pricing here: http://www.wholesalesolar.com/battery-banks.html. Maybe I could have gotten away with a 400Ah bank for $3000. Maybe I could have purchased batteries individually and shaved a little of that price. Whatever. It's a lot of money for something I don't need. It's also a lot of space, as these batteries are typically quite large and you typically need more of them because they are 6V batteries, not 12V.

OR

(2) I built a battery / capacitor bank for around $1000, and it takes up less space. The batteries are also more readily available, so if one happens to gokaput, no problem.

Option #2 is still the best option for me.

Design criteria revisited

Let's look at the design criteria again for the sake of academics.

(1) One should design based on necessary RMS current capability. At 5kW, 27% more RMS current was demanded by the GTI than calculated (because I didn't take into account losses, rippling voltages, etc.). I propose a 1/3 correction factor as a good rule-of-thumb. Redoing the calculations, at 6kW (full-load capability of the XW6048) and 46V, the supposed RMS amperage pulled by the GTI is ideally ~85A_{RMS}. Using the 1/3 correction factor, this comes to ~113ARMS. I like round numbers, so let's round up to 120A_{RMS}for 6kW. In fact, dare we say20A? I think this is sound, because I took other data at loads from 500W to 4000W in 500W increments, and the RMS current demand scales linearly._{RMS}per 1kW

(2) One could choose to scale back on RMS requirements based on the fact that the capacitor bank will see only about 3/4 of the predicted AC current. Yet, why not give yourself some room to breathe? Thus, theminimumrule is 15ARMS per 1kW, but the safe rule is 20ARMS per 1kW.

(3) One should design based on necessary capacitance to achieve either or both of a few constraints: (a) maximum voltage ripple, and (b) maximum power loss. Remember how I chose an arbitrary 1% charge / discharge number as my indication of minimum capacitance? Well, one could get a lot more detailed. Capacitors have certain ESR specs at 120Hz. This ESR contributes to power loss in the capacitor, but it also contributes to voltage ripple. The voltage ripple is simply the ESR multiplied with the RMS current (V = I * R). The power loss is the voltage ripple multiplied with the RMS current (P = I^2 * R). Honestly, I think voltage ripple is more important, as it has more of an effect on the performance of the GTI. For a capacitor bank built from 6 of the capacitors I selected above, I calculate only 13W lost over the whole bank (< 2W per capacitor) for 120A_{RMS}. I doubt the temperature rise in the capacitor would be detectable.

(3a) Therefore, design capacitance based on maximum voltage ripple. This is a more interesting constraint. One contributor is the aforementioned ESR, is calculated easily, but requires an actual ESR to calculate. The second contributor is much more important: the voltage ripple due to charge / discharge of the bank. Recall: dQ = C(V) * dV. Thus, choose a peak-to-peak ripple-voltage (dV), and calculate a minimum capacitance based on the dQ seen during a half-cycle of the current waveform. Once you have a chosen capacitor, take the ESR into account by adding the two contributors.

Closing

It's late at night on a Friday evening, and I'm still at work writing this. I know, my friends, I know... I am a nerd.

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