# RF Circuit Battery Efficiency

Discussion in 'Radio and Wireless' started by mcasey, Jul 23, 2017.

1. ### mcasey

64
2
Jul 9, 2016
Ok, I watched several videos and think I'm getting closer to figuring out the formula for the inductor value:

Assuming Io is current draw from the decoder and receiver measured in amps, the total is .0064 amps.
I think "fsw" is 70 if I pull pin 3 to ground.
If I supply the circuit with 6 volts...

L= (6-3.3)/(.0064 x .1) x (3.3)/(6x70)
L= (2.7 / .00064) x (3.3 / 420)
L= 4218.75 x .007857
L = 33.147

Is this correct or am I way off? If correct, what unit of measurement is 33.147? Also, if correct, how do I select the value of the special diode?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,271
2,718
Jan 21, 2010
In general, a calculation like that gives you the minimum value for the inductor. Too small an inductor is generally worse than too large a value.

For the diode, choose a Schottky diode rated at more than the maximum output current, and more than the maximum input voltage.

The current through the diode is in pulses (that are larger than your output current) which average out to be your output current. Choosing a diode with a rated current higher than your output current is wise. 2x is generally sufficient.

The leakage through Schottky diodes can be quite high. I would also recommend the voltage rating to have plenty of margin, but beware that forward voltage drop (and hence losses) are directly related to piv voltage rating.