Freewheel Diode Recomendation

[Suraj143]

Can somebody tell me whats the most robust value for Freewheel Diode?

I'm going to use it across the IRFZ460 Mosfet. Mosfet is driving an 12V ignition coil.

I used a 1N5817 schottky diode but it burned out when the ignition coil operates

 

[Mike odom]

That's because the reverse breakdown is only 20V. If you're driving a 12 coil to ground, you're going to see a minimum of 24V across that diode. You need something like a 1N4007 (1000V). I admit it can be a 1N4001 50V (1N4002 = 100V), but the higher the better, and any time I use a 1A diode, I use the 1N4007 as I only have to stock 1.
If it has to be schottky for quick recovery, use a 1N5819 (40V). That should work for a 12V system.

 

[Suraj143]

Thanks Mike now I understood.

Note that I use unregulated 12V power supply.So when I measure the voltage it shows 18V.So still I can use 1N5819 as you suggested....!!!

 

[Mike odom]

what frequency are you driving this circuit at? Is it an automotice ignition, or are you just using an ignition coil?

18 + 18 = 36, so the 5819 should still work. You might try using two in series to obtain a higher reverse breakdown voltage.

 

[crutschow]

......................

18 + 18 = 36, so the 5819 should still work. You might try using two in series to obtain a higher reverse breakdown voltage.

The 1N5817 is rated at 20V. A single one won't work with an 18V system.

 

[Mike odom]

The 1N5817 is rated at 20V. A single one won't work with an 18V system.

He's already figured that out, see the first post! And then see my first post (post #2) where I tell him that exact same thing, 'cept @ 12V.

 

[MikeMl]

...If you're driving a 12 coil to ground, you're going to see a minimum of 24V across that diode..

No, the peak reverse voltage across the diode is only 12V (same as the supply voltage, not twice the supply voltage). When the FET shuts off, the diode becomes forward biased, and conducts.

The turn-off time of the diode is unimportant. It should turn-on quickly, however.

In the OP's application, I also would have used a Si rectifier like a 1N400X. They are quite robust. The 1N58XX Schottkys seem to be less robust...

As for how to determine the reverse breakdown voltage required, I would use a safety factor of two to three.

 

[Suraj143]

Ok guys finally I desided to add the same 1N4007 diode.That is more durable than the standard schottky.

 

[picbits]

I was under the assumption that your flyback diode should also be of around the same current/power rating (or more) as the inductive load you're driving.

I.e. if you were pumping 10 amps into a car ignition coil, you'd need a higher power dissipating flyback diode than if you were putting 100ma into a small relay on a PCB.

Correct me if I'm wrong but I've always designed anything I've made around this rule.

 

[alec_t]

I was under the assumption that your flyback diode should also be of around the same current/power rating (or more) as the inductive load you're driving.

Agreed. The nature of inductors is that they try to maintain the current that is flowing when they are switched off. That current then flows in the freewheel diode at switch-off, albeit for a short time until it decays. But beefy diodes can withstand short-duration overloads. If it melts then it wasn't beefy enough

 

[crutschow]

He's already figured that out, see the first post! And then see my first post (post #2) where I tell him that exact same thing, 'cept @ 12V.

Yes. It was late and I didn't notice that you suggested a 1N5819, not the original 1N5817.

 

[Mike odom]

If it melts then it wasn't beefy enough

Probably the understatement of the year!!!
Can I quote you on this????

 

[Mike odom]

In the OP's application, I also would have used a Si rectifier like a 1N400X. They are quite robust. The 1N58XX Schottkys seem to be less robust...

They are both 1A diodes, and the schotky has less voltage drop, so dissipates less power. He may need to go to a 3A diode, 1n5404?

 

[chemelec]

A 1N4001 or 1N4002 is better for this purpose than the 1N4007.
You want a Somewhat Low breakover voltage so it Shorts out the Spikes.

 

[MikeMl]

...The nature of inductors is that they try to maintain the current that is flowing when they are switched off. That current then flows in the freewheel diode at switch-off, albeit for a short time until it decays...

The current that was flowing through the load instantaneously begins flowing in the snubber diode as the FET turns off. The decay of the current flowing through the snubber diode is determined by the inductance of the load, the initial current, and the resistance of the coil. It is a classic RL circuit, with an exponential decay. A typical time-constant τ is likely to be less than 0.2s.

The diode current rating needs to consider the peak current, and how often the inductive load is switched on/off. On something like a relay that is cycled only occasionally, the diode will stand an occasional current pulse that is several times the average current rating. If the load is switched at rates of gteater than 10Hz, then the diode average current rating needs to be a half to a third of the peak current.

 

[Suraj143]

Ok guys thanks for the support.

I have a similar problem I planned to put on this thread without beginning a new thread.

I'm driving an 12V EML from two relays.The 1st relay gives the positive supply to the 2nd relay.

Without the D2 diode the circuit gets hanged (even without the load EML).Is it ok if I leave that D2 diode?

 

[MikeMl]

Diode D2 protects the contacts on Rel1.

When you say "hanged", do you mean that the relay contacts weld together?

Is there a third diode across the inductive load? (there should be).

Why does the inductive load's current flow through Rel1's contact?

Why is rel2 even there?

 

[Suraj143]

Nice Mike.You spotted a mistake in my drawing I redrew & uploaded the new one.The inductive loads current won't flow through Relay1. Its my drawing mistake.

Hanged means the logic signal gets stopped working.Not the relay contacts

Relay1 is in the same borad as where the logic circuit is.I used another relay to operate the EML because the Rel1 is too small.

**Note that the D2 diode is directly across Vdd & Vss is it ok? Will it absorb some spikes which goes to logic circuit?

 

[MikeMl]

This simulation I did years ago shows why a snubber diode is needed. In the sim, there are three identical inductive loads being switched on-off by voltage-controlled switches. The red trace V(pulse) is the control signal that turns on all three switches. The switches are on when the red trace is positive +1v) , and off when it is negative -1V.

The bright green trace V(nodiode) is the inductive kick out of the unsuppressed coil. The high voltage developed as the coil turns off burns the points of the switch, and couples to nearby circuits, both capacitively, and inductively.

V(diode) lt, blue and V(hiside) dark blue shows the voltage with both a low-side switch and hi-side switch when the diodes are added. Note that with the low-side switching, the voltage across the opening switch is clamped one diode drop above 12V, while with high-side switching, it is clamped one-diode drop below ground.

Finally, the bottom plot panel is for Mike Odom, to show that voltage across the diode V(diode)-V(12v) dark green trace ranges from -12V (same as the supply voltage) to +1V (while the diode is forward biased by the decaying current). Note the forward current through the diode I(d1) pink trace starts at 3A same as the coil current I(L2) gray trace, but it decays rapidly to zero. The nominal 1A diode will have no trouble with an infrequent 3A short duration pulse...

Back to Suraj, your diode outlined in red has to be there to suppress rel2 (to prevent noise and to protect the contacts of Rel1). However, you still didn't answer the question about how you are suppressing the REAL inductive load, the one not shown on your schematic???

 

[Mike odom]

ya, that would be great if we lived inside a computer. Here's an o'scope shot of almost 30v developed by switching a 12v signal across a coil (5v/div).

View attachment 68769