Calculating max current?

[GooeyGus]

Hi guys, I was just wondering how one can calculate the maximum current of a DDL circuit. This specific driver is powered by a 9v battery, has a 100ohm pot (set to least resistance), and a 25ohm resistor.

What equation can I use to figure this out? My DMM is broken :'(

 

[phenol]

I= 1.25/R
I-current (amps); 1.25V is the reference voltage of LM317; R is the total resistance in ohms. Suppose Rmin of your trim pot is 0 and you have a fixed 25ohm res in series with it, the formula yields 50mA. If you set your pot to max (100 ohms), the current would be 1.25/125 = 0.01A=10mA. The max current LM317 is capable of /depending on its package/ is 1.5A

 

[GooeyGus]

gotcha... so since the ref. voltage of the lm317 is 1.25, the ohm's law formula works with 1.25v no matter which battery you are using?

 

[phenol]

if the voltage between Vout and ADJ pins of LM is 1.25, then it is regulating properly. The battery must be able to provide enough current so that the voltage differential between the input and output is at least 3V needed by LM's innards to operate correctly. for example, you cant drive a red dvd burner diode off the 9V battery, as it cannot sustain the required voltage at such high currents for long. In any case, if your source has the muscles to keep the voltage steady, the equation above will always be true as long as the said minimum differential is observed. the max differential is about 37V, which means that it has a very broad operating range of input voltages. If it is too high, tho, the excess is converted to heat, depending on your current /big loss/.