Attempt to replace a 4.5Ah 12V Pb battery with LiFePO4 for a Profoto Acute B flash generator I have a detailed writeup here...

Previously I had used a Tenergy drop in Li replacement which worked for a few years but is reaching EOL as well as my Pb versions. It is no longer available.

I am looking to avoid building a Li pack from components and modify an existing pack to do the job by adding capacitor(s) parallel to the battery in order to reduce the peak current by at least 10A to prevent the PCB on the LiFePO4 from cutting out.

Some facts:

-Acute B 600 generator (load) outputs 600 joules
-load is primarily large capacitor bank
-load in turn discharges to circuit powering Xenon tube
-Recycle time is under 2 sec
-Pb Battery is rated 4.5Ah 12V 
-Approx Peak Draw is 30A 
-battery is fused at 40A (slow blow) 
-Attempt with [CTC LiFePO4][3] 
   -failed: PCB cuts out at 22A and it did 
-Battery Box is 89mm x 69m x 106mm


-[Bienno stock cell:][2] 4.5AH LiFePO4 rated 4C peak (18A) 
-Bienno opt 2 6Ah LiFePO4 rated 4C peak (24A)

I don't have a true measurement of the peak draw, a figure of 30A was given to me by a repair tech.

Looking at a goal of reducing the 30A draw down below the trigger point for either Bienno LiFePO4 pack above noted, I calculated using C=Q/V respectively for the packs noted above.

-  Q=12A/S V=15 C=0.8F 
-  Q=8A/S V=15 C=0.5

In either case, I should be able to get the Pack and a pair of caps in Series or parallel depending on available values and build to fit existing battery box spec.

If someone would chime in on my calculations, it would be nice to know. I don't see where any additional circuitry is necessary on this, the generator only sees a source of V & A beyond it's input terminals.

  • \$\begingroup\$ "Pack and a pair of caps in Series or parallel to make the values and stuff them in a battery box" That "Series or parallel" shows lack of knowledge but that's fine and why you are here. "values and stuff" is more disconcerning. \$\endgroup\$ – winny Jan 4 '17 at 8:29
  • \$\begingroup\$ Thanks Winny, appreciate opinions. Yes, I am here to fill in what I do not know and not claiming to be an expert. Perhaps you misunderstand capacitors as you can derive different values of Voltage ratings and Capacitance depending on how you wire Caps in Series or Parallel. You should look it up. Real world values for caps are obtained by using them combined as it just isn't cost effective to produce caps in all sizes and voltage ratings. As for stuffing in a box, pehraps I could choose a better verb like insert? \$\endgroup\$ – Davepix Jan 4 '17 at 17:53

Without knowing the characteristics of either the LiPo protection circuit or the peak load it is difficult to give a detailed answer, but my first (very confident) response is "No." You have fundamentally misunderstood the relative natures of batteries vs capacitors.

A battery is (to a first approximation) a constant-voltage device, and a capacitor is not remotely like it. Your calculations, for instance, show that a 0.8 F capacitor being discharged at 12 amps will drop from 15 to zero volts in one second. After, let's say, 0.1 second the capacitor's output at this current will have dropped from 15 to 12 volts, and at this level I'd expect the battery to be providing all the current, so the protection circuit will kick in. Making a capacitor supplement as you wish requires a detailed knowledge both of the load drain profile and the protection circuit's behavior. Basically, you'd expect the battery voltage at maximum current to drop to some lower (but non-zero) voltage, and the capacitor would take up the slack - briefly. However, as soon as the capacitor has discharged to the nominal loaded battery voltage it will not contribute ANY current and the battery will have to supply all the current. And we know how that works out.

In principle you could use a very large capacitor, but you would have to determine how the protection circuit works. Let's say that protection circuit reaches its trip point at 12 volts output, with a constant 22 amps output from 15 to 12 volts. Then the capacitor must provide all the energy of the flash by dropping from 15 to 12 volts. Then, since the load needs 600 J and the formula for a capacitor's energy is $$E = \frac{CV^2}{2} $$ we can say that $$E(total) = 600 = \frac{C}{2}(15^2 - 12^2) $$ and $$ C = \frac{1200}{81}= 15F$$ and since the recharge rate is $$\frac{\Delta V}{\Delta t}= \frac{i}{C}=\frac{22}{15}=1.5 V/sec $$ the recharge time is $$T = \frac{3 \text{ volts}}{1.5 v/sec} = 2 \text{ seconds}$$ which is close to your desired cycle time. Of course, assuming constant current capacitor charging is probably not a great idea.

  • \$\begingroup\$ Drain profile Char would be ideal… spec sheet all I have. Load I am charging is big caps, not fully disch. For lack of an oscilloscope, multimeter reads battery V∆ Betw 14.5V and 8.8 on Li pack during the ~2S charge cycle. Accurate scope measurements of the A and V during this cycle would likely make it much easier to choose a capacitance to do the job? I thought that the Cap in parallel with the battery would reduce the surge C to the main load and that the need for it would taper off to a similar profile as the V in load caps equalizes with the batt/cap on the supply side. \$\endgroup\$ – Davepix Jan 4 '17 at 18:35
  • \$\begingroup\$ Another question, where did you get 3600J for the load? The pack is rated at 600J output? If so, that's 14.8F Cap for the full load of which I think I only need to offset about a third of the load to keep the battery V from dropping below the shutdown threshold. You seem to infer the cap drains before the battery but I'd assume the cap drains only as fast as the battery can ∆V as long as they are parallel even if it is capable of near instantaneous discharge alone. \$\endgroup\$ – Davepix Jan 4 '17 at 19:04
  • \$\begingroup\$ @Davepix - "where did you get 3600J for the load? " From a brain fart, obviously. My apologies. I've edited to fix. \$\endgroup\$ – WhatRoughBeast Jan 5 '17 at 0:11

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