Hi Everyone,
I feel that there is still confusion regarding normal operating current of bulbs in steady state, compared to how the PhD-M6 drives bulbs (or any PWM driver for incans). To say that we shouldn't exceed a particular RMS bulb current (3A, or 5A, or whatever), is not necessarily correct. In practice, setting some personal limits on bulb power is never a bad idea when using lithium cobalt or lithium primary cells.
The high current limit on the cells is tripped based somewhat on RMS current lasting long enough to trip it, and it is somewhat tricked by the PWM anyways. Remember, the MN21 is running at 7A peaks in our application and it's not tripping the circuit. I've tested the 64275 briefly, which would actually operate at something closer to 8-9 amp peaks under this PWM.
A 2C discharge in the case of a 3x17670 pack consisting of cells that are new and in great shape means that roughly 40 watts is being depleted across the entire circuit, including the power wasted as heat to resistance before anything even gets to the intended device to be powered up (in our case, a bulb). Under a normal discharge condition (non pulsed load), a 2C rate for a 3x17670 pack would generate heat in the cells at a rate of 4.5W, and about 2W would be wasted in other contacts and such (in a flashlight application). The result is that the power consumption of the device to be powered should not exceed ~33.5W for a normal direct drive 2C load, again, assuming brand new cells in great condition. At a 2C rate, one could say that typical LiCo powered flashlights are operating at ~84% efficiency after considering losses at the batteries and contacts and such. Dealing with these losses is best performed by reducing resistance wherever possible. Including finding lower resistance cells (this is why we need IMR cells!)
When we use a higher voltage source to drive lower voltage bulbs with PWM regulation, we get get higher current peaks, so a higher percentage of the total energy flowing from the cells gets converted to heat within the cells.
Rough Example: WA1111 @ 7.2V, 3.8A, 1.9 Ohm.
1.9 Ohm bulb resistance + ~0.5 Ohms for everything else = 2.4 Ohm load.
With cells at ~80% state of charge: 12V pack / 2.4 Ohm = 5A
12V x 5A = 60W total power consumption during an "on" pulse of the regulator.
0.5Ohm / 2.4Ohm x 12V = 2.5V lost to resistance
1.9Ohm / 2.4Ohm x 12V = 9.5V hits the bulb
9.5V x 5A = 47.5W hits the bulb during an "on" pulse of the regulator.
PWM @ 57% = 27W RMS to the bulb. The drive level is power based. The bulb never sees 7.2V or 3.8A, but it runs just the same.
2.5V x 5A x 57% = ~7W converted to heat, ~5W of that heat is right in the cells.
The
total power dissipation (including ALL losses, not just the bulb operating power) occurring in this snapshot of running a WA1111 is 34W. Or ~1.7C discharge rate.
The efficiency loss to resistance gets slightly worse compared to direct drive. The example I gave above for a direct drive application shows ~83% efficiency into a 2C load, here, at 1.7C, it's more like ~79% efficiency.
I think that the most important thing to keep in mind, is that, because of the way that we are achieving regulation here, it's going to be normal to see more heat buildup in the cells for a given power output. If a 2C rate would normally cause ~4.5W of heat at the cells, and a 1.7C rate under one example of PWM regulation causes slightly more heat at the cells, then one must realize that at a 2C rate, we are going to exceed the heat build-up rate that the cells were originally rated to tolerate in a continuous discharge at 2C. This means that the user needs to take care in preventing continuous discharges of high power lamps. Cool down periods are important for safety.
So.... One could say, that a WA1111 IS a 5A bulb. You just have to run it at 9.5V instead of 7.2V.... Oh, and don't forget to turn it off 106 times per second
Eric