KillingTime
Newly Enlightened
Re: F\S SST90 Mag2C (40Watts)
Hello bxstylez.
This light uses a switch mode buck converter, not a linear converter.
At 9A, an SST90 required approximately 3.7V. Ohms law gives us 9x3.7= 33.3W at the LED.
The converter isn't 100% efficient. Assume 80% worst case (this figure is typical), meaning an additional 20% of the power delivered will be lost as heat in the converter.
20% of 33W = 6.6W. Total power taken from the batteries = 33.3 + 6.6 = 39.9W.
At 8V input, this would equate to a battery current draw of 40\8 = 5A.
At 5.3V input , this would equate to a battery current draw of 40\5.3 = 7.6A.
This is why you need 18650 IMRs. Very few std 18650 Lithium Ion batteries can safely deliver 7.6A (they're limited to a draw of 2x capacity).
Hope this helps.
Hello bxstylez.
This light uses a switch mode buck converter, not a linear converter.
At 9A, an SST90 required approximately 3.7V. Ohms law gives us 9x3.7= 33.3W at the LED.
The converter isn't 100% efficient. Assume 80% worst case (this figure is typical), meaning an additional 20% of the power delivered will be lost as heat in the converter.
20% of 33W = 6.6W. Total power taken from the batteries = 33.3 + 6.6 = 39.9W.
At 8V input, this would equate to a battery current draw of 40\8 = 5A.
At 5.3V input , this would equate to a battery current draw of 40\5.3 = 7.6A.
This is why you need 18650 IMRs. Very few std 18650 Lithium Ion batteries can safely deliver 7.6A (they're limited to a draw of 2x capacity).
Hope this helps.
