Diode + Diode Zener and Inductive load

[Bicio]

I know that inserting a normal diode in parallel to a inductor is
usefull for preventing voltage spikes when the current drop to zero
(for example due to a transistor).
So, with a diode, the inductor discharge itself with a time constant
of L/R.
With a Zener in series to a diode, the inductor discharge faster
because of the higher inverse voltage. But, how can I translate this
effect in mathematical terms? How could I demonstrate that adding a
zener the inductive load discharge faster?

 

[Phil Allison]

"Bicio"

I know that inserting a normal diode in parallel to a inductor is
usefull for preventing voltage spikes when the current drop to zero
(for example due to a transistor).
So, with a diode, the inductor discharge itself with a time constant
of L/R.
With a Zener in series to a diode, the inductor discharge faster
because of the higher inverse voltage. But, how can I translate this
effect in mathematical terms? How could I demonstrate that adding a
zener the inductive load discharge faster?

** Simple, just use conservation of energy - it always works a treat.

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:

24 volt relay with 100 ohm coil and diode OR 47 volt zener.

Diode case: E = 5.5 watts ( initially )

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

QED.



...... Phil

 

[Bicio]

"Bicio"




** Simple, just use conservation of energy - it always works a treat.

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:

24 volt relay with 100 ohm coil and diode OR 47 volt zener.

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?

 

[Bicio]

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?

 

[Spehro Pefhany]

"Bicio"


** Simple, just use conservation of energy - it always works a treat.

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:

24 volt relay with 100 ohm coil and diode OR 47 volt zener.

Diode case: E = 5.5 watts ( initially )

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

QED.



..... Phil

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

 

[Bicio]

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!

 

[neon]

obviously you do not understand the diode purpose. the diode does not prevent nothing it is there to dumpen the charge of the coil energy when charged and suddenly released. and your zener or anything in the series with this diode is not going to help it

 

[Bicio]

If it's a homework problem, then solve it analytically

hint: vl = L * di/dt.

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

 

[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