Diode + Diode Zener and Inductive load

"Bicio"

In the diode only case, the stored energy in the coil is dissipated almost entirely in the coil resistance - the rate of which is initially::

E = Isquared.R (where I = the coil current prior to switch off )

The same I value applies in the diode plus zener case too, but now there is another term in the game:

Now,

E = Isquared.R + V.I ( where V = zener voltage)

In the majority of examples, the V.I is larger than Isquared.R.

Eg:

Diode case: E = 5.5 watts ( initially )

Zener case: E = 5.5 + 11.0 = 16.5 watts.

QED.

..... Phil

And if I want to calculate the time to reach a current Ix, how can I do?

Energy of coil = Ec = 0.5*L*Isquared(t)

Energy of Zener = Ez = integral[Vz*I(t)*dt]

Energy of Coil resistance = Er = integral[R*Isquared(t)*dt]

Ec = Ez + Er => it seems quite complicate to find the formula t = f(I), I need to know the I(t)... Where am I wrong?

And if I want to calculate the time to reach a current Ix, how can I do?

Energy of coil = Ec = 0.5*L*Isquared(t)

Energy of Zener = Ez = integral[Vz*I(t)*dt]

Energy of Coil resistance = Er = integral[R*Isquared(t)*dt]

Ec = Ez + Er => it seems quite complicate to find the formula t = f(I), I need to know the I(t)... Where am I wrong?

Or, to look at it a slightly different way, initially the zener drops a voltage equivalant to a resistor, of value 47V/240mA = 195 ohms, so the initial drop will be about 3 times as fast as with 100 ohms, and the difference goes up from there.

Best regards, Spehro Pefhany

And if I'd like to know the time to reach a current Ix during the inductor discharge, with a diode and with diode + zener, how can I do?

Thank you very much!

If it's a real problem then just simulate it.

If it's a homework problem, then solve it analytically hint: vl = L * di/dt.

In real life, constant L for a relay or solenoid coil is probably a bad assumption.

Best regards, Spehro Pefhany

solving that fomula I obtain the usually exponential trend, but It does not depend from the Zener voltage...while it shoud do...

If you assume the zener has a constant voltage when broken down you should be able to solve for i(t) for the time from when the switch opens until the current drops to zero, after which you can predict the current stays at zero (which isn't going to be too far from the truth for most practical purposes). In reality the zener voltage changes with current (to a first order, it just adds a bit to the coil resistance) and the current doesn't drop quite to zero with reverse voltage less than the breakdown voltage.

Best regards, Spehro Pefhany