Diode identification?

If Vdd was 42V, then a 47V zener sticks 5V reverse voltage across the coil, so the current will decay 5/42 times faster than it built up.

Whereas if you just use a conventional freewheeling diode, Anode to Drain, Cathode to Vdd, there is 0.7V(ish) reverse voltage across the coil when the FET turns off, so the coil current decays 5/0.7V times slower than the 47V zener.

Of course the actual zener voltage wont be 47V, it'll be higher, depending on the actual current.

One can achieve the same objective at lower loss with a 4.3V zener in zeries with a freewheeling diode, but thats 2 parts.

So it is possible that the zener was used to get a suitable rate of decay (although ramp down is more accurate) of coil current.

Or perhaps the designer was a bit stupid, used no freewheeling diode, then discovered the FET broke, so added the zener. You might be surprised how many s*it designs make it to market.

Cheers Terry

The zener does a better, but more expensive, job of protecting the series switching element. It limits both positive and negative transients. A diode across the switched inductor does stop most (but not all) of the switching transient - but doesn't protect the series element from transients on the supply rails, caused by other inductances elsewhere reacting to the sudden change in current. It is usual to combine these sorts of design with reasonably fast (eg tantalum)electrolytics placed locally - to act as energy "tanks" to supply and sink transient power.

As I and others have written - the diode didn't burn up because of transient energy. There is a supply problem, somewhere.

its pretty hard finding a FET without a body diode, so negative transients are invariably taken care of regardless of the type of clamp circuit.

A diode across the switched inductor does stop most (but not

by "series element" you must be referring to the FET. Yep, the zener will protect the FET against voltage spikes on the 42V bus. Of course FETs nowadays are rated for avalanche energy.....

caused by other inductances

Que?

It is usual to

seeing as Im being a pedantic sod, I'll point out that tantalums are not electrolytics (and vice versa).

I once had a serious brain fart in this regard, making a small motor controller at Uni. It ran from a 3-phase supply, and seeing as full-wave-rectified 3-phase AC has ~15% ripple, I figured I didnt need a DC bus cap.

Which worked fine, until the first time I turned the H-bridge off with current flowing in the motor :) 30 minutes, 4 FETs and a complete set of gate drive circuits later, I added a large cap. oops.

assuming the thing ever worked properly, which it sounds like it did.

conceivably a shorted solenoid could have stored enough energy to end up snotting the zener, but as you say, a supply overvoltage would definitely kill it. And it doesnt even have to be that much, just continuous.

Cheers Terry

Hence why I wrote "series switching element" rather than FET. If the designer was brought up designing using pnp/npn transistors, he may have always protected them this way.

The power distribution and supply system connected to the load will, itself have a transient response (eg have series inductance) and may easily overshoot following step changes in load.

"Tantalums Tantalum capacitors are also electrolytic, constructed with a very porous anode made with tantalum powder. This powder is pressed into a pellet form with a tantalum wire inserted."

I'd go for the continuous - every time where component and circuit board burning were evident.

its surprising how much stuff ends up being designed that way - "because", rather than having a good reason.

now thats a nicely worded sentence.

ya got me :)

common usage of "electrolytic" refers to ye olde carboxylic acid-style electrolytes, for whatever reason tantalums are always called tantalums.

it all depends on how much transient energy there was. when working with big stuff, components can (and do) disappear completely. If any form of arc develops as a result of some transient phenomenon, its pretty easy to burn big holes in things.

and a transient that snots the zener will make it fail short-circuit (unless it disappears), at which point it will then fry.

however such pontification is essentially meaningless; the supply rail needs to be checked....

Cheers Terry

Palindrome sez:

Interesting you mention this... the silkscreen for the FET says "b-e-c". Seems that the original design was for BJT, but component specs were changed to include FET sometime in production with little regard for the confusion it would cause service personnel who saw these markings...

Does this shed any different light on the choice of zener for this purpose? And the possibility for a different replacement part? The 47v, 2W part is looking like unobtanium...

Thanks,

Palindrome sez:

The reason this machine drew attention in the first place was because the solenoid valve was gummed up and sticking. I wouldn't think that this would cause problems with the drive circuit. Au contraire, it would result in no back-emf.

Palindrome sez:

The pcb had failed 'lytic caps (ends pushed out), so that could have added to the problem. Or, being beyond a certain age, the 'lytic problem may lie in the PS as well. I'll see about 'scoping the PS voltages in the machine.

Terry Given sez:

I'm told that the solenoid this circuit operates is for a vacuum valve that must operate quickly and repeatedly. It was thought by the person who handed me this pcb that the solenoid was operated with 2 voltage rails, switching between opposite opening voltage and closing voltage. But according to measurements by him (and the fact that there's only 1 FET), the purpose of the zener here seems to make sense.

But how can a 4.7v zener and one diode drop serve similar purpose as a 47v part?

I would think that a partially shorted zener would keep the solenoid energized, giving a "gummed up" symptom. If the board allows space for the modification, I would replace the

The rectifier cathode connects toward the positive supply end of the solenoid, but the zener cathode points toward the fet drain.

Can you find a place to put those two components?

In the present circuit, when the fet turns off, the coil generates a voltage in the direction that tries to keep the current going. That means that the end that had been pulled negative to ground suddenly goes more positive than the 42 volt rail. At 47 volts the zener comes on, and provides a path for the decaying coil current. So, during that energy dump process, there is about 47-42=5 volts reverse voltage across the coil, driving the current toward zero. But the energy in the zener is being fed from both the coil (the 5 volt part of the 470 and by the supply the 42 volt part of the 47), since the coil current is also passing through the supply.

The only advantage I can see to this wasteful and stressful (to the zener) method of driving the coil current to zero, is that the supply current ramps down to zero, smoothly, rather than switching off as the fet does. But I doubt that is a consideration in this circuit.

If you put a rectifier and zener directly across the coil, the rectifier keeps the zener out of the circuit when the fet is on, but connects it as a voltage clamp when the fet switches off. Now, the only energy going into the zener is that being dumped out of the solenoid, as its current ramps down to zero. The supply stops contributing the moment the fet switches off. You can adjust the ramp down time by swapping zeners with different break down voltages. But I would start with a 4.7 or 5.1 volt unit to get things back about the way they were to start. But a 6.8 or 7.5 volt unit may make the solenoid work better with an insignificant additional voltage stress for the fet.

The supply should also have some bypass capacitance connected very close to the fet source and the positive supply connection of the solenoid, to make sure the fast interruption of the current (that didn't happen with the old zener) doesn't bounce the supply rails around enough to unset either the fet gate drive or some other load connect to the 42 volt or ground rails. A microfarad or 10 would do it. I 1 microfarad, 50 or 63 volt stacked film type would do it well. see:

John Popelish sez:

What I know of the design goal of this circuit is that it must activate the solenoid quickly from off to on and quickly from on to off with as little "ramping" as possible. With the given circuit, what does this knowledge say about the selection of possible replacement component(s)?

Anode-to-anode, with the rectifier "on top", the pair being connected across the solenoid?

Yes, pretty easily. It's not too heavily populated. Lots of "vertical implementation" possible (c:

Thanks for your suggestions, John.

V = L*dI/dt, so dt = L*dI/V

L & dI are constant, you are increasing V to get a nice low dt.

the BUZ72 is a 100V part, so you have PLENTY of headroom there.

the existing circuit turns the solenoid off about 8x slower than it turns it on.

it doesnt matter if the rectifier is on the "top" or "bottom", only that its cathode faces towards the supply, so it prevents the zener from working when the FET is on, and allows the zener to work when the FET drain voltage rises above the supply.

so a K-K connection with the recitifer at the bottom and the zener at the top, or A-A with the zener at the bottom and the rectifier at the top.

Cheers Terry

Well, there is nothing these diodes can do about the turn on time. That is a function of the supply voltage and the coil inductance. You would have to raise the supply voltage and add enough series resistance to limit the steady state current to a safe value to speed up turn on.

Order doesn't matter, only orientation.

Higher zener voltage means faster current ramp down. But you will probably have to go quite a bit higher to see much difference. The resistive drop of the coil is already starting the ramp down with a 42 volt reverse voltage. But that drop falls as the current falls, so the zener is really there to speed the tail of the process, unless its initial voltage is on the order of the supply voltage. So you might consider one as high as 22 to 39 volts. But then I would look for a 1 watt unit, to handle the power pulse that will end up more there than in the coil resistance. But you should definitely see some decrease in the power down time, to about 37% if what you will get from a 4.7 volt zener if you switch to a 33 volt one. So you can see that the turn off time is not dominated by the zener till its voltage gets near the supply voltage. But increasing the zener voltage drop helps.

John Popelish sez:

Seems we're creeping back up toward the original 47v zener (although it was connected across the FET, not the coil). Any advantage to simply using another 47v part along with the rect. in the configuration you recommend? Is this a case of "bigger (v) is better"?

Thanks again,

The advantage in moving the zener is the lower energy absorbed per discharge (for the reason I explained earlier). At 47 inverse volts across the coil, you are getting pretty close to the 100 volt mark, which will stress the fet a bit more. Are you confident in its ability to handle that voltage? And there is a point of diminishing returns. The

If you used a 1000 volt zener, the coil current would hit zero in a really short amount of time, but the valve would close in just about the same time as if you used a 500 volt zener.

My gut feeling is that, unless this solenoid and valve mechanism were designed with fastest possible reaction time in mind, going much above 22 volts on the zener will not pay off in much decreased valve action.

But a handful of 1 watt zeners in the range of 4.7 volts to

A better way to speed the release might be to put a parallel resistor and capacitor in series with the coil, so that the coil voltage actually decreases a little after the cap charges to the IR drop of steady state operation. That way, you have the large pick up force to get the valve open, but a reduced holding force to keep it open, so there is less magnetic field to quench when you want it to close. This is called a pick and hold strategy, and there are special driver chips that perform this function with two switches, one on each side of the coil.

At energize, both switches turn on, applying full voltage (often a voltage the coil would not tolerate, continuously) to the coil to ramp the magnetic field up as fast as possible. The current is sensed, and when the required pick current is reached, one of the switches pulse width modulates the current down to the hold value. When turn off time arrives, both switches open, and the coil dumps its energy back into the supply through a diode across each of the switches. So the supply voltage acts like your zener voltage. Very fast and energy efficient (there is minimal heat in the coil, and no intentional power wasted anywhere else in the circuit) but probably not practical as a retrofit in this case. http://www.ortodoxism.ro/datasheets/stmicroelectronics/1331.pdf But something to keep in mind if a board layout comes along.

huh I'm getting 42/0.7 (which is over 50 times slower)

are you assuming a 5V vcc? OP claims 42V.

read harder.

no, the original voltage across the coil during turn-off is Vz - Vcc =

so the current ramps down 5V/0.7V ~ 7x slower with a freewheeling diode.

note the not-so-confusing sentence. I should have written:

"so the current ramps down 0.7V/5V times faster with a freewheeling diode"

but I'm being nice ;)

Cheers Terry

You blokes have forgotten R and L, and L/R. :-)

I couldn't be bothered to do the sums so just LTspice'd a quick 42V supply, 100mH and 42 ohm coil, switched by a MOSFET and clamped by a Schottky diode to a variable voltage.

The current Risetime at switchon, from 0.1A to 1A was about 5.5mS, as per the L/R exponential sum.

Below is a little table of LTspice current Falltimes.

Vclamp. Falltime (1A to 0.1A).