# How to calculate appropriate parameters for flyback diode?

#### Teik

Apr 22, 2017
9
My schematics looks something like this.

L1 is fuel injector, that is driven by MOSFET, which will be controlled by MCU.

From what I have gathered so far, connecting D1 like this, instead of connecting diode across L1, should allow energy in inductor to decay faster. This is what I'm aiming for.

Placement of D1 gives me concerns though. Not sure, if there is such a possibility, but what if diode goes bad and starts conducting - Result would be engine overflooded with fuel. This brings me to the main topic - how to calculate the best parameters for D1?

I know, that breakdown voltage of D1 should be higher than supply voltage and lower than M1 breakdown voltage. Inductance of L1 is not known. I know, that resistance is 16.6 Ohms and supply voltage will range from 8 to 14V. Duty cycle will vary and I have no idea how to take that into consideration. It seems that max current that will flow trough L1 is around 0.9A. Does it mean that D1 should be able to handle around 12W of power? Seems like a lot. I would like to stick to SMD components, if possible.

If there are other possible options, that would improve driver circuit, then I would happily accept those for a discussion.

#### crutschow

May 7, 2021
865
It seems that max current that will flow trough L1 is around 0.9A. Does it mean that D1 should be able to handle around 12W of power?
That would be the peak power when the injector turns off, not the average power.
To determine the average power, you need to know the injector inductance, and the maximum frequency of operation.

An alternate for fast turn-off (but not quite as fast as the Zener) would be to use a regular diode in series with a resistor across the injector coil.
The resistor value would be chosen to give about the same peak voltage (at the peak injector current) across the transistor as the Zener voltage.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,834
but what if diode goes bad and starts conducting - Result would be engine overflooded with fuel.
That can happen when the MOSFET has a drain-source short circuit, too.
To prevent such cases a fuse in the input circuit should trigger and interrupt the circuit.

#### Teik

Apr 22, 2017
9
I used voltage divider circuit that was powered by sine wave and calculated inductance of the injector - 1436.081uH. Max frequency of injector operation will be 166Hz.

Not really sure how to calculate flyback energy given all this.

#### crutschow

May 7, 2021
865
Not really sure how to calculate flyback energy
That's a low inductance for a solenoid.
The energy stored in an inductor is 1/2 LI² where L is the inductance and I is the current.
So for 0.9A through that inductance, the stored energy (which is dissipated when the solenoid is turned off) is 0.582 milliJoules.
If the solenoid is operating at 166Hz, then the average power is 0.582mJ * 166Hz = 96.5mW.

Thus any Zener that can handle the peak current of 0.9A will have no problem is dissipating the power (especially when some of that energy will be dissipated in the solenoid resistance).

#### Teik

Apr 22, 2017
9
It seems that there are no such SMD Zener diode that could handle that power. Is it true that Zener peak current is calculated by dividing power rating of the Zener with its voltage rating, I = P/V?

#### Teik

Apr 22, 2017
9
Took a look at couple datasheets and found Surge Current graph. Looks like small SMD 5W Zener diodes can handle more than 5W for some fraction of a second. Now I'm wonder how fast flyback energy will dissipate. I wonder if there is formula that will calculate power rating more optimistically. What I mean, 0.9A * 14V = 12.6W of power. Since this power will be present only for a split second, then how to more realistically estimate power rating for diode? Will it really burn out fast if it will be rated twice the times less than what was calculated by I * V in this scenario?

Last edited:

#### crutschow

May 7, 2021
865
Looks like small SMD 5W Zener diodes can handle more than 5W for some fraction of a second. Now I'm wonder how fast flyback energy will dissipate. I wonder if there is formula that will calculate power rating more optimistically. What I mean, 0.9A * 14V = 12.6W of power. Since this power will be present only for a split second, then how to more realistically estimate power rating for diode?
How long is that "fraction of a second"
5W times that time will tell you the surge energy it can absorb.

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