4D Mag mod noobie????

[Import_guy]

I just modded a 4d mag with a ssc p7 and a 5 mode driver from kd http://www.kaidomain.com/ProductDetails.aspx?ProductId=7947. It works fine with 2 18650 but I am only drawing 1.65A :shrug:. I was wondering if I am not using the right battery configuration to reach it's full potential or if it's just the driver itself. The driver switches through it's modes just fine and stays on high without any problems. I need cpf members expertise on how I should go about this. Thanks

 

[Justin Case]

It's probably running fine (for now).

The KD driver appears to be a buck driver, based on the stated specs. Thus, the tail current draw will decrease as Vbatt increases. Tail current draw is not the same thing as drive current to the LED.

If we assume that Vbatt = 7.4V, Vf = 3.5V, If = 3.0A, and driver efficiency = 80%, then

driver power out = LED power in
0.80 * 7.4V * Ibatt = 3.5V * 3.0A

Thus, Ibatt = (3.5*3.0)/(0.80*7.4) = 1.8A

Close enough for gov't work. Your Vf could be lower, Vbatt higher, driver efficiency higher, and/or If lower (e.g., 2.8A).

 

[Import_guy]

Thanks for info. I'm lost:thinking: wasn't very good with math or numbers. I need it explain in idiot terms.

 

[Justin Case]

This is the key equation:

driver power out = LED power in

Power is volts * current (amps) = watts (W).

Driver power is modified by the fact that drivers are not 100% efficient. There is some loss. Thus, the driver power out is

driver efficiency * Vin * Iin

where

Vin = Vbatt = battery voltage being fed to the driver
Iin = Ibatt = battery current being fed to the driver (the typical "tail current" measurement)

Power into the LED is

LED forward voltage * LED forward current

where

LED forward voltage = Vf
LED forward current = If

Thus, from driver power out = LED power in, we get

driver efficiency * Vbatt * Ibatt = Vf * If

Solving for Ibatt, we get

Ibatt = (Vf * If)/(driver efficiency * Vbatt)

I assumed the following for the relevant variables:

driver efficiency = 80%
Vbatt = 7.4V (2*3.7V, where 3.7V is a common Li-ion voltage when under load)
Vf = 3.5V when the P7 is driven at full power
If = 3.0A (which is the advertised spec for your KD driver)

Just plug in the numbers, do some simple multiplication and division, and calculate an estimated value for Ibatt.

 

[old4570]

Ok , lets try it the simple way : And see if that works ...

Voltage x Current = Power

Voltage x Amps = Watts

lets say for arguments sake = 4.2v x 2.8A = 11.78 Watts , close enough to the 12Watt ratting of the SSC P7 ..

If your using 2x18650 = 8.4v fully charged ...

So 8.4v x 1.65A = 13.86watts

Factor in the driver loss , and your close enough to driving the LED at full power .

:wave: Hope this explains it ...

 

[Import_guy]

Thanks for the info guys.:twothumbs