hipCC driver P7 and 6AA - brainfart check please!

[Magic Matt]

I think I have this right - could somebody give me the benefit of their expertise?

I'm driving a single P7 from a hipCC driver, which I believe should be driving 2.8A to the LED. I've used some teflon coated wire, I think 24AWG. I have 6 AA eneloops in series.

I've measured the current draw from the batteries at 2Amp. After a couple of minutes it settles at 1.97Amp.

Open circuit the cells total 8v exactly. Voltage sag under load takes them to 7.2V which I think is pretty good.

7.2V * 1.97A = 14.184W

The P7 driven at 2.8A would be 3.6V from the spec. 3.6V * 2.8A = 10.08W

10.08W / 14.184W = 71%

The driver is rated at 84% efficiency for 7V and that's quite a discrepency. I don't think for a minute that it's a faulty driver, I'm sure it must be something I'm doing... or not doing... or havn't done quite right.


First of all, have I got the maths right?

Secondly, is there a way I can check what is actually going to the LED without unwiring it, and without the aid of a clamp meter (which I don't have)?

 

[herulach]

how are you connecting the batteries? check the voltage at the driver inputs to make sure your caddy isn't introducing a lot of resistance.

 

[Justin Case]

You can measure Vf of the emitter to confirm whether your assumption of 3.6V is valid.

 

[wquiles]

You can measure Vf of the emitter to confirm whether your assumption of 3.6V is valid.

+1

The original poster is measuring the input, but not th output - just guessing (no offense intended). Once you measure both output current and output voltage at the LED, then the efficiency calculation would be move valid.