Flyback diodes and why you need them (comments here please)

Paul__B:
My point was that the inductor does not generate the transient; the switch does. The switch is in complete control of the voltage, and the transient does not emanate from the inductor, but from the switch. :grinning:

Not true,
The transient does emanate from the inductor.
In other words, if we didn’t have an inductor, we would have no transient.
This is due the equation v = Ldi/dt

abdelhmimas:
if we didn’t have an inductor, we would have no transient.

This is only philosophical. If there were no switch no amount of inductors would generate a transient.
I believe (but did not check with a scope) the voltage peak is worse (and sooner) at the switch. That would mean it makes more sense to say the switch causes the transient.

abdelhmimas:
Not true,
The transient does emanate from the inductor.
In other words, if we didn’t have an inductor, we would have no transient.
This is due the equation v = Ldi/dt

Surely the transient emanates from the combined circuit of the inductor, switch, wires and power supply. Remove any one of those and you don't get the transient. It is the circuit as a whole that exhibits some particular characteristic, not the individual components.

I have very cautiously worded my explanation in #4. It is always worthwhile to re-read it carefully.

Smajdalf:
I believe (but did not check with a scope) the voltage peak is worse (and sooner) at the switch. That would mean it makes more sense to say the switch causes the transient.

To propose otherwise would be to conceptualise that the sudden step in voltage - the transient - gets larger and larger as it moves from the switch to the inductor. What mechanism would one propose for this - because the inductor would have to "know" that the switch had opened ahead of electromagnetic propagation?

Always remember that the transient starts at the switch, it makes no sense to suggest it starts at the inductor as that would imply it started before the switch opened!

Paul__B:
I have very cautiously worded my explanation in #4. It is always worthwhile to re-read it carefully.
To propose otherwise would be to conceptualise that the sudden step in voltage - the transient - gets larger and larger as it moves from the switch to the inductor. What mechanism would one propose for this - because the inductor would have to “know” that the switch had opened ahead of electromagnetic propagation?

Always remember that the transient starts at the switch, it makes no sense to suggest it starts at the inductor as that would imply it started before the switch opened!

the formula u(t)= ldi/dt is the voltage across the coil. i, which is the current across the switch and the coil (they are in serie). the switch voltage when it is open, we call it u(sw) = V total - u(t). so if Vtotal is a constant, u(sw) and u(t) have the same curve (proprtional).
we put the diode in parallel with the coil, because, it stores EMF when energized, and when the switch it is opened, this EMF will flow trough the diode.
if we put it across the switch, this will not happen.

PerryBebbington:
Surely the transient emanates from the combined circuit of the inductor, switch, wires and power supply. Remove any one of those and you don't get the transient. It is the circuit as a whole that exhibits some particular characteristic, not the individual components.

PerryBebbington:
Surely the transient emanates from the combined circuit of the inductor, switch, wires and power supply. Remove any one of those and you don't get the transient. It is the circuit as a whole that exhibits some particular characteristic, not the individual components.

Electrical engineering states the transient starts with the negative pulse of the supply voltage (from +VCC to open), when the switch opens.

The cause of the transient is the opening of the switch
The circuitry involved will stabilise or not the voltage and current in the circuit

abdelhmimas:
the formula u(t)= ldi/dt is the voltage across the coil. i, which is the current across the switch and the coil (they are in serie). the switch voltage when it is open, we call it u(sw) = V total - u(t). so if Vtotal is a constant, u(sw) and u(t) have the same curve (proprtional).
we put the diode in parallel with the coil, because, it stores EMF when energized, and when the switch it is opened, this EMF will flow trough the diode.
if we put it across the switch, this will not happen.

abdelhmimas:
the formula u(t)= ldi/dt is the voltage across the coil. i, which is the current across the switch and the coil (they are in serie). the switch voltage when it is open, we call it u(sw) = V total - u(t). so if Vtotal is a constant, u(sw) and u(t) have the same curve (proprtional).
we put the diode in parallel with the coil, because, it stores EMF when energized, and when the switch it is opened, this EMF will flow trough the diode.
if we put it across the switch, this will not happen.

In electrical engineering / electromagnetism theory, when the voltage supply to the the inductor is cut, the inductor will try to produce voltage in reverse, hence the inductor becomes a generator for a very short period of time.
This reverse voltage that produces energy needs to be dissipated somehow, and that is the purpose of the diode, to provide a circuit for this current to circulate and dissipate, otherwise the current might had circulated elsewhere and destroy some circuit.
For me, the cause of the transient is the switch opening

Absolutely, the switch opening create the di/dt.
The current went from a certain value to 0.
Cheers

traja47:
In electrical engineering / electromagnetism theory, when the voltage supply to the the inductor is cut, the inductor will try to produce voltage in reverse, hence the inductor becomes a generator for a very short period of time.
This reverse voltage that produces energy needs to be dissipated somehow, and that is the purpose of the diode, to provide a circuit for this current to circulate and dissipate, otherwise the current might had circulated elsewhere and destroy some circuit.
For me, the cause of the transient is the switch opening

You are perhaps not understanding that the transient that does the damage is the voltage transient
created by the inductor in response to the current transient produced by the switch - the circuit as a
whole is responsible, which is why adding components (snubbers) can fix the problem by reducing or
eliminating the voltage transient.

It is like saying when a car accident happens, the car is the only responsible.
Because if the car was not invented ....

The transient is caused by the collapsing magnetic field in the inductor which causes a voltage to be created that attempts to keep the current flowing (inductors being resistant to current flow changes).

That voltage sees the resistance of the rest of the circuit; a voltage transient that is observed with an oscilloscope. A diode back to the supply gives that voltage someplace to go. RC snubbers perform a similar function.
Mechanical switch or BJT or MOSFET opening, the sudden resistance change stopping current flow and the inductor’s property of continuing current flow, is what causes the transient.

if the current through the inductor decreases, the magnetic field strength decreases, and the energy in the magnetic field decreases. This energy is returned to the circuit in the form of an increase in the electrical potential energy of the moving charges, causing a voltage rise across the windings.

Is there any value to discussion of choosing where the energy is dissipated as opposed to how the voltage is controlled? What facts should be considered if it is a large solenoid instead of a small relay?

A flyback diode alone will minimize the voltage and a majority of the energy will be dissipated in the coil's internal resistance.

A resister in series with the diode will limit the voltage to the initial current times the resistance and the energy will be split between the coil and the resistor.

A clamp circuit will limit the voltage by tapering the turn-off current but all the flyback energy will be
dissipated in the mosfet.


(first example circuit I found on stackexchange)

Each choice will change how fast the coil will turn off.

edmcguirk:
A clamp circuit will limit the voltage by tapering the turn-off current but all the flyback energy will be dissipated in the mosfet.

Well, no.

The current has actually been limited by the resistance of the coil. The energy will again be shared, in this case between the FET and the coil resistance itself.

I cannot see a purpose in R2. This method of suppression, using a Zener rather than a Transzorb, is already built into the TPIC6x595 series of solenoid drivers which need no external diodes. There is no concern with how quickly the Zener must come into conduction as its action as already augmented by the Miller effect.

This part shows similar clamping, claiming no external diode is needed.

Well that's what I get for stealing a random picture from stackexchange that includes extraneous features I didn't consider.

Sure, all current must go through the coil, so energy dissipation will always be shared. I should have said 'most'.

The voltage clamp of the zener to the mosfet gate will smooth the current transient of shut off if the mosfet does not already have the clamp built in.

My point was to discuss if energy dissipation needs to be part of the flyback consideration beyond just voltage limiting.

edmcguirk:
My point was to discuss if energy dissipation needs to be part of the flyback consideration beyond just voltage limiting.

Well as per my original discussion (#4), there often seems to be some funny ideas about “surges” and great “release of energy”. The critical consideration is that the current never exceeds in any manner, that which the coil was passing before it is turned off and simply decays - how quickly depending on how much the voltage is permitted to rise.

There will be some dissipation in the FET as it progressively turns off as while switched on, there should have been virtually no voltage across it and the clamp strictly defines its drain voltage as the current decays from the “on” value, but this is very unlikely to be significant.

Yes, but a while ago I was thinking about driving a vibratory pump which is essentially a large solenoid with 50hz switched DC instead of AC mains voltage. I don't know if the equations were correct but it was suggested to me that there would be 35 watts of flyback energy that would be dissipated. (guesstimated specs - 220V - 165ohms - 1.3A - 0.865H)

It has already been demonstrated that a diode across this type of vibratory pump causes the pump to run poorly due to slow energy dissipation but a diode with a resistor solves the problem while getting very hot.

Obviously an extreme case but there must be a range that divides insignificant and important considerations.

edmcguirk:
35 watts of flyback energy that would be dissipated. (guesstimated specs - 220V - 165ohms - 1.3A - 0.865H)

I know what you mean. Maybe just need to consider the amount of time associated with the transient, and which components absorbs or dissipates the energy. And also noting the watts is associated with power, while energy is area under the curve of an instantaneous power versus time curve (or anything equivalent to it). So the time duration should probably be defined when considering energy too.

And ------ if the transient occurs over a 'relatively' short amount of time, then the associated amount of energy during that time can be relatively small. Relatively.

Yes true ....... when the flyback diode is used, the current can still flow and decay away ------ without relatively high voltages being developed, and sparks etc.

Also - not unusual to think of house-flies when talking about flyback diodes. Not because there is any relation ...... but just due to the word 'fly'.

edmcguirk:
It has already been demonstrated that a diode across this type of vibratory pump causes the pump to run poorly due to slow energy dissipation but a diode with a resistor solves the problem while getting very hot.

There is no doubt that "slugging" a solenoid slows release - that is exactly what was done on PMG relays and the parallel diode has the same effect.

Clearly, doing this to a device whose purpose is to operate rapidly and repeatedly is completely counter-productive. You need to arrange a significantly higher voltage to drive the solenoid in either direction, whether that "driving" is active or passive (as with a Zener rather than a diode to limit the "kickback"). Nothing surprising about this, as this is exactly what sophisticated stepper motor drivers do.

Note this discussion about origin of the transient is motivated by finding the optimal placement for the diode (or whatever protection used).