Question: What Size of fly-back diode do i need for my inductive load?
My Answer: Fly-back diodes are sized based on power dissipation
P = 1/10(I^2)R.
P: power dissipated in fly-back diode
I: steady state current flowing through the inductor (fly-back diode not conducting)
R: Resistance of the fly-back diode in conduction
Consideration 1: The fly-back diode will be held at a constant temperature; diodes have a constant resistance in conduction when held at a constant temperature. (if the temperature changes so does the diodes resistance)
Now the conducting diode behaves as a resistor so the question becomes: how much power do i need to dissipate in my diode's internal resistance?
By observing a series RL curve, we know that the inductor discharges or charges in 5 time constants and one time constant is equal to the inductance divided by the series resistance (T = L/R).
some math people told us that the energy stored in an inductor is:
E = (1/2)L(I^2). here E is in joules, L is in Henrys. They also said that power is energy per second (P = E/time). here power is in watts.
so...if our understanding of physics is working...the time in which the inductor discharges is: 5(L/R) seconds, and a stored energy of (1/2)L(I^2) joules is released in that time. here R is the resistance of the fly-back diode in conduction, I is the current flowing through the fly-back diode and L is the inductance supplying the current.
if we solve for the power something very interesting happens....
P = ((1/2)L(I^2)R) / (5L) here L cancels out and P = 1/10(I^2)R. we know that R is the resistance of the diode in conduction and I is the current flowing through the diode during the discharge. but now what is the diode current during discharge?
consider a circuit as such:
simulate this circuit – Schematic created using CircuitLab
R1 is the internal resistance of L1 and R2 is our charging resistance. D1 Functions as the fly-back diode, and R3 is the resistance of D1 in conduction.
if the switch is closed and we wait forever, a current of 10mA flows through the circuit, and the inductor stores an energy of 50uJ (50 micro Joules)
using conservation of energy theory;
if the switch is opened the inductor reverses polarity to try to maintain the 10mA current. the fly-back diode is biased into conduction, and an energy of 50uJ is dissipated through the diode resistance in 5(L/R) = 500ms. the power dissipated in the diode is 50uJ / 500ms = 100uW (100 micro watts)
(1/10) (10mA ^2) (10ohms) = 100uW so to answer the last question: the diode current during discharge can be thought of as equal to the steady state charging current of 10mA when using the equation: P = 1/10(I^2)R. while the current during the inductive discharge actually decreases exponentially and is not a steady 10mA, this simplification will allow for quick computations of the required diode power in a circuit by knowing the initial conditions.
Best of luck with your designs and never use technology for evil purposes,