That is the period where inductor current is going through the diodes to dissipate the inductive energy.
If you want that time to be shorter, than you need to used a higher Zener voltage to dissipate the energy faster (which, of course, will require a MOSFET with a higher voltage rating).

For example with a 51V Zener, the turn-off time is reduced to about 0.35ms (below).
Notice the diode current (red trace) during the turn-off.

View attachment 262427

Sorry I realized right after posting my question that the voltage is flat in that region because of the Zener clamping. Of course, that's the whole point of the Zener. (mental lapse due to drinking an IPA).

I think I am stuck with what I have. I don't want to upgrade the MOSFET or add components to upgrade my trigger to 5V or 10V.

What do you think, could I squeeze a little more speed out of a TVS - Schottky combo?

I imagine I could also duplicate the Zener-diode snubber in parallel to dissipate twice as fast ( at the cost of twice the components?) I will try modeling that tomorrow.

Finally, is an RC snubber ever used in parallel with a Zener-diode snubber?

Overall I think approx. 1.4 ms dissipation is pretty good and I'm proud of the measured improvement. Before I was using just a diode and I'm sure it was very slow. So, Thank you!

 

could I squeeze a little more speed out of a TVS - Schottky combo?

I imagine I could also duplicate the Zener-diode snubber in parallel to dissipate twice as fast

No and No.
It's strictly the Zener-diode voltage drop that determines the inductor turn-off time.
It's that voltage times the inductor current that determines the rate of dissipation.
Putting two in parallel will still have the same voltage, the same inductor current, and the same turn-off.
It will just halve the dissipation in each Zener.

 

One more variant:

 

No and No.
It's strictly the Zener-diode voltage drop that determines the inductor turn-off time.
It's that voltage times the inductor current that determines the rate of dissipation.
Putting two in parallel will still have the same voltage, the same inductor current, and the same turn-off.
It will just halve the dissipation in each Zener.

Ok, good to know those ideas won't work.

I will probably learn how RC and RCD snubbers work just so that I understand why they are not faster.

As I said I consider this a win based on my limiting factor, the MOSFET.

 

One more variant:
View attachment 262431

Thank you @Danko . I appreciate everyone's input.
It took me a minute to figure this out - let me know if I'm on the right track:
The UMZ6_2k lets the inductive kickback sink to ground through the node on the mosfet gate, and because its zener voltage is 6V, the gate doesn't see unsafe voltages.
Question, wouldn't the reverse zener current flow into the gate of the mosfet, turn on the solenoid, and create an oscillating loop where the solenoid turns on every time it turns off?

 

Danko's circuit has the advantage of the the inductor current not going through the Zener(s) so the Zener does not need a large current rating.
The feedback to the gate keeps the MOSFET turned on just enough to carry the inductor current until the current stops.

wouldn't the reverse zener current flow into the gate of the mosfet, turn on the solenoid, and create an oscillating loop where the solenoid turns on every time it turns off?

No.
The MOSFET does not turn all the way off.
The feedback is just enough to keep the MOSFET on and conduct the inductor current with the MOSFET drain going to near the Zener voltage.

Below is the LTspice simulation of his circuit, slightly modified to use the Zener models I had.

As you can see, when the IN voltage (blue trace) goes to zero at 6.5ms, the MOSFET gate voltage (green trace) drops just enough to allow the MOSFET to stay turned and carry the inductor current (purple trace) during the transient, due to the feedback through the diodes.

The MOSFET drain voltage (red trace) jumps up to the sum of diodes D1-B and D2 voltage plus the MOSFET gate voltage during this time.

The max. diode current (white trace) is <800µA, so a low power Zener can be used.

D3-A is actually not needed, since the gate voltage never goes above the input voltage.

 

Danko's circuit has the advantage of the the inductor current not going through the Zener(s) so the Zener does not need a large current rating.
The feedback to the gate keeps the MOSFET turned on just enough to carry the inductor current until the current stops.
No.
The MOSFET does not turn all the way off.
The feedback is just enough to keep the MOSFET on and conduct the inductor current with the MOSFET drain going to near the Zener voltage.

Below is the LTspice simulation of his circuit, slightly modified to use the Zener models I had.

As you can see, when the IN voltage (blue trace) goes to zero at 6.5ms, the MOSFET gate voltage (green trace) drops just enough to allow the MOSFET to stay turned and carry the inductor current (purple trace) during the transient, due to the feedback through the diodes.

The MOSFET drain voltage (red trace) jumps up to the sum of diodes D1-B and D2 voltage plus the MOSFET gate voltage during this time.

The max. diode current (white trace) is <800µA, so a low power Zener can be used.

D3-A is actually not needed, since the gate voltage never goes above the input voltage.

View attachment 262459

Ok, that does make sense. I like that the Zeners are not burdened with the kickback current. In my application, the MOSFET is on a PCB and the solenoid is 6 ft away (there's a detachable extension cord with DC barrel jacks and connectors on the board and solenoid). So there's a terrible loop when I place the Zener-diode combo in parallel with the coil. I can't really put the snubber at the solenoid given present design restraints. It's a problem I wanted to address in steps - first get it working, then worry about the large loop. This solution partially mitigates that loop right, because the current goes from the solenoid to MOSFET to PCB ground plane and not back to the solenoid.
Although, it does mean there's a lot of current sunk into the ground plane on the PCB and the potential for ground loops and other bad stuff on the PCB. Lastly, I am a little concerned about how low the gate gets when the trigger is turned off. It looks borderline. I want to build this out physically and compare the results. Especially since my last physical circuit was slower than the simulations (with the 43 Ohms Zener I was still seeing 1.4 ms of dissipation time, and not sure why). I would however have to change the Zener values because I don't want to exceed 50V on the MOSFET drain. That may be overly cautious but I would prefer to compromise performance for ruggedness.

 

Actually, the best place for the snubber circuit is at the switch, not the inductor.
That way any spikes from stray line inductance is also suppressed.

Where was the snubber when the MOSFET failed?

 

Actually, the best place for the snubber circuit is at the switch, not the inductor.
That way any spikes from stray line inductance is also suppressed.

Where was the snubber when the MOSFET failed?

It was on the perf board with the MOSFET and the solenoid leads were about 12 inches long. That's good then I will not worry about the snubber on the PCB where it is presently at the switch.

Planning to build out the @Danko schematic to see if it might be faster (holding the zener and general diode constant). I suppose the speed would be dependent on the MOSFET's Rds(on), the voltage on the drain, and the gate voltage which are all dynamic during the dissipation I think .

 

It was on the perf board with the MOSFET and the solenoid leads were about 12 inches long. That's good then I will not worry about the snubber on the PCB where it is presently at the switch.

Planning to build out the @Danko schematic to see if it might be faster (holding the zener and general diode constant). I suppose the speed would be dependent on the MOSFET's Rds(on), the voltage on the drain, and the gate voltage which are all dynamic during the dissipation I think .

Also realizing now I can change the gate voltage during the kickback based on the D1 zener voltage, right ? So I could obtain something more like 5 or 10V on the gate. I'm going to play with those values.

 

I am a little concerned about how low the gate gets when the trigger is turned off. It looks borderline.

The voltage drops until it is exactly what is needed to carry the inductor current at the Zener voltage.
That value will be slightly above the Vgs(th) value for the actual transistor you are using.

I suppose the speed would be dependent on the MOSFET's Rds(on), the voltage on the drain, and the gate voltage which are all dynamic during the dissipation I think .

It's really just the turn-off voltage at the MOSFET drain.

The inductor current-change time is determined by the equation V= L di/dt, so the turn-off time dt = L*di / V where di is the current change and V is the turn-off voltage.
So, for example, if V = 60V then dt = 4.75mH * 4.5A / 60V = 358µs.

Also realizing now I can change the gate voltage during the kickback based on the D1 zener voltage, right ?

No.
Due to negative feedback, the gate voltage will stay at the value just needed to carry the inductor current at the Zener voltage.
A different Zener voltage will change the MOSFET drain voltage, but have little effect on the gate voltage.

Do you understand how negative feedback works?

 

The voltage drops until it is exactly what is needed to carry the inductor current at the Zener voltage.
That value will be slightly above the Vgs(th) value for the actual transistor you are using.
It's really just the turn-off voltage at the MOSFET drain.

The inductor current-change time is determined by the equation V= L di/dt, so the turn-off time dt = L*di / V where di is the current change and V is the turn-off voltage.
So, for example, if V = 60V then dt = 4.75mH * 4.5A / 60V = 358µs.
No.
Due to negative feedback, the gate voltage will stay at the value just needed to carry the inductor current at the Zener voltage.
A different Zener voltage will change the MOSFET drain voltage, but have little effect on the gate voltage.

Do you understand how negative feedback works?

What is the turn off voltage at the MOSFET drain? I am not familiar with that. Do you mean just the voltage drop across the device(s) dissipating the inductor energy?

Negative feedback - sort of - I took feedback and controls systems 6 years ago and remember..not much. But the root of my confusion is: isn't Vgate = VnodeD-Vzener? It sounds like your saying Vg changes based on VnodeD? I'm trying to figure out how the gate receives feedback if the voltage is clamped. Is it current-based feedback?

 

What is the turn off voltage at the MOSFET drain?

That's the drain voltage during the turn-off transient, as determined mainly by the diode-Zener and Vgs voltage.
It's what determines the turn-off time.

I'm trying to figure out how the gate receives feedback if the voltage is clamped. Is it current-based feedback?

No, it's voltage based.

Okay- I'll take a stab at a simplified (no math) explanation of negative feedback in Danko's circuit:

When the input control signal goes to zero and the gate-voltage drops, the MOSFET starts to turn off and its drain voltage rises from the inductor current it is carrying.

When the MOSFET drain voltage reaches the diode-Zener plus the Vgs voltage, the Zener starts conducting to keep the gate voltage at the value just needed to maintain the drain voltage at the inductor current level being carried (this all happens very rapidly as determined by the AC frequency response of the MOSFET and feedback circuit).
At this point, if the drain voltage were to rise slightly, the gate voltage will also rise, which turns on the MOSFET slightly more to lower the drain voltage.
Similarly if the drain voltage drops slightly, that lowers the gate voltage to slightly turn off the MOSFET and bring the drain voltage back up.

So you can see, it's sort of a balancing act, where the drain voltage stays at the value of the diode-Zener voltage plus the gate voltage needed to just keep the MOSFET conducting at the inductor current value.
It's, of course, operating at this point in the linear region, not as a switch.

If you look at the gate voltage (green trace) in my post #26 simulation, you will see how the gate voltage drops some during the transient time (from about 2.3V to 2V).
This is the feedback changing the gate voltage so the MOSFET just carries the inductor current as it drops from maximum to zero, while maintaining a near constant drain voltage (red trace) to rapidly stop the current.
Obviously it takes slightly more gate voltage at the start, to carry the maximum inductor current, then at the end as the current approaches zero, as determined by the transconductance gain value of the MOSFET.

All make sense?

 

That's the drain voltage during the turn-off transient, as determined mainly by the diode-Zener and Vgs voltage.
It's what determines the turn-off time.
No, it's voltage based.

Okay- I'll take a stab at a simplified (no math) explanation of negative feedback in Danko's circuit:

When the input control signal goes to zero and the gate-voltage drops, the MOSFET starts to turn off and its drain voltage rises from the inductor current it is carrying.

When the MOSFET drain voltage reaches the diode-Zener plus the Vgs voltage at that point, the Zener starts conducting to keep the gate voltage at the value just needed to maintain the drain voltage at the inductor current level being carried (this all happens very rapidly as determined by the AC frequency response of the MOSFET and feedback circuit).
At this point, if the drain voltage were to rise slightly, the gate voltage will also rise, which turns on the MOSFET slightly more to lower the drain voltage.
Similarly if the drain voltage drops slightly, that lowers the gate voltage to slightly turn off the MOSFET and bring the drain voltage back up.

So you can see, it's sort of a balancing act, where the drain voltage stays at the value of the diode-Zener voltage plus the gate voltage needed to just keep the MOSFET conducting at the inductor current value.
It's, of course, operating at this point in the linear region, not as a switch.

If you look at the gate voltage (green trace) in my post #26 simulation, you will see how the gate voltage drops some during the transient time (from about 2.3V to 2V).
This is the feedback changing the gate voltage so the MOSFET just carries the inductor current as it drops from maximum to zero, while maintaining a near constant drain voltage (red trace) to rapidly stop the current.
Obviously it takes slightly more gate voltage at the start, to carry the maximum inductor current, then at the end as the current approaches zero, as determined by the transconductance gain value of the MOSFET.

All make sense?

Ok, it took me some time but I believe I get it. I forgot that the Zener regulates the difference between its cathode and anode; it does not hold either at some absolute voltage. So Vd and Vg change, but Vd-Vg stays constant.

If Vd increases +1V, Vg must also increase +1V, opening the MOSFET, which dissipates energy from the coil faster and reduces Vd, which reduces Vg which closes the MOSFET, and so on.

So if I am understanding correctly, Vg will stay pretty near Vgs(on) during this time. Because the mosfet will turn off when Vd and Vg drop and then the kickback will raise the voltage and Vd and Vg will increase and the MOSFET will turn on, so its oscillating on-off and Vg is oscillating just above and below Vgs(on), correct?

Side note - the IRLH5030 you modeled with looks like a powerful logic level MOSFET and grants 100 Vds; unfortunately, it is quite pricey, has a higher gate charge, and is out of stock in the US from what I saw. With the IRLH5030 I could then use a zener voltage of 60 or 70V . I am ordering a couple to experiment with from Newark.

 

10 pcs of IRLH5030 === $3.38, delivery - $2.95 https://www.aliexpress.com/i/1005002940275214.html
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10 pcs of IRLH5030 === $3.38, delivery - $2.95 https://www.aliexpress.com/i/1005002940275214.html
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View attachment 262486

Thank you. That looks like a good price. I've never used aliexpress; I will consider that next time. Unfortunately, I already ordered a couple from Newark. It was 20 USD for 2 pieces shipped.

 

The entire purpose of a spike clamping diode is to slow the spike by allowing the current to flow through the diode. A Zener across the transistor will protect the transistor. Of course, a resistor across the coil will also reduce the spike.

 

so its oscillating on-off and Vg is oscillating just above and below Vgs(on), correct?

Not here.
If the feedback loop is underdamped, it can oscillate, but here it it is stable, so no oscillation.
It just settles at the correct voltage to keep the voltage across the diode-Zener constant, which changes slightly (but smoothly) as the inductor current drops.
If you look at the simulation, you will see there is no hunting or oscillation of the voltages.

It's sorta like the accelerator on your car, you just find a stable accelerator position that keeps the car speed constant, with slight changes as the car goes up or down a hill (although I have a friend who drives with the car constantly increasing and decreasing its speed - drives me crazy )

 

The entire purpose of a spike clamping diode is to slow the spike by allowing the current to flow through the diode. A Zener across the transistor will protect the transistor. Of course, a resistor across the coil will also reduce the spike.

With the caveat here that the TS wants a fast turn-off, and the fastest turn-off for a given voltage spike is with the addition of the Zener.

 

Not here.
If the feedback loop is underdamped, it can oscillate, but here it it is stable, so no oscillation.
It just settles at the correct voltage to keep the voltage across the diode-Zener constant, which changes slightly (but smoothly) as the inductor current drops.
If you look at the simulation, you will see there is no hunting or oscillation of the voltages.

It's sorta like the accelerator on your car, you just find a stable accelerator position that keeps the car speed constant, with slight changes as the car goes up or down a hill (although I have a friend who drives with the car constantly increasing and decreasing its speed - drives me crazy )

Ok yes, I get it. I can think of it infinitesimally. With an infinitesimal increase in Vd due to the kickback, the Vg will increase infinitesimally and the MOSFET will open infinitesimally more and then the Vd will decrease infinitesimally and so on. Stable feedback -changes slightly but smoothly - that's a good way to put it.