Anybody seen this boost IC before?

I can attempt to answer some of your questions with the knowledge I've picked up over the years working on a project of mine.

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First, voltage droop from cells is an important consideration. When these boost IC's are tested by the manufacturer they are connected to a benchtop power supply that is able to supply any required wattage without interruption. This is not the same with batteries. Individual cells suffer 'voltage droop' when put under load, which is why a cell measured at rest may be 1.5V but when measured at 1A current drain shows 1.1V.

With that in mind, 1.8V input voltage is not a good choice. 1.2V input is more realistic so let's assume that.

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Wattage is the best measure for calculating power utilitization. Watts = Volts * Amps. Using your example scenario:

1.5A * 3.3V = 4.95W output wattage

4.95W / 1.2V = 4.125A input current

So to get 1.5A @ 3.3V input from an Eneloop you need to pull 4.125A. This ignores a very important variable: efficiency of the circuit. Luckily, the datasheet for the TPS61021 has a chart that outlines converter efficiency:

At 1.2V input and 1A output, max theoretical efficiency is 70%. This means 30% of your input wattage is burned off as heat/waste and 70% makes it through as output power. Adding efficiency in to the calculation:

4.95W 7.07W
---- = ----
70% 100%

So to get 4.95W of output power, you need to supply 7.07W of input power. 1.2V is a non-negotiable value so the input current must change to accommodate:

7.07W / 1.2V = 5.89A input current

At nearly 6A of input current, alkaline batteries have long given up the ghost. Even NiMh and L91 cells are struggling to keep above 1V. So now instead of 1.2V input, we only have 1V available. Let's adjust our equation for that:

7.07W / 1V = 7.07A

Input current has jumped to 7A! Checking the discharge charts at 7A for those batteries brings bad news: voltage droop has lowered us to ~0.9V. Time to adjust our equation again:

7.07W / 0.9V = 7.85A

You can see where I'm going with this. There is a limit on how much current you can ask from a cell until the voltage droop puts it into a death spiral. This is not an issue with a benchtop power supply. Assuming the power supply is connected to mains voltage, you have 120V * 15A = 1800W of constant, stable power available. The datasheet is not lying: the converter is able to supply 1A @ 1.2V when plugged into a power source able to comfortably provide that wattage. The datasheet does not say this device was tested using Eneloop batteries, however.

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Looking at a light like the Zebralight SC52 (an example of a light with an outstanding converter):

  • Output: 300 lumens (out the front)
  • Runtime on High: 55minutes (round up to 60minutes)

Eneloops are rated 2A for 60 minutes before depleted. We know the SC52 runs 60 minutes before it goes dark, so we can assume 2A was drawn from the battery.

1.2V * 2A = 2.4W

Assuming a very efficient converter (85%) let's say the emitter saw 2.04W. The forward voltage of an XM-L2 is 2.9V.

2.04W / 2.9V = 0.7A

The Cree XM-L2 datasheet tells us that at 700mA drive, we can expect ~342 lumens. If we subtract some losses for the reflector and the glass, we get ~300lm OTF.

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I suppose the conclusion to be drawn is that 1.5A @ 3.3V from an alkaline/NiMh/L91 battery is not feasible. Some of the most efficient drivers on the market manage 700-800mA to the LED. The Texas Instruments chip looks very good on paper, and while it may well be a solid chip, batteries simply aren't going to provide the power necessary to provide the full output it lists.