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Diode identification?

J

John E.

Jan 1, 1970
0
Eeyore sez:
Is it against your religion to substitute ?

Not at all. Here in USA I checked my 3 regular sources: Mouser, DigiKey, and
Jameco with nil results, subs or not.

But it looks like design changes are afoot (see other recent posts in this
thread).

Thanks,
 
J

John E.

Jan 1, 1970
0
John Popelish sez:
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
47 volt zener with a series combination of a 4.7 or 5.1 volt
zener in series with a 1N400X or similar small rectifier
diode, connected directly across the coil, instead of across
the fet.

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

Anode-to-anode, with the rectifier "on top", the pair being connected across
the solenoid?
Can you find a place to put those two components?

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

Thanks for your suggestions, John.
 
T

Terry Given

Jan 1, 1970
0
John said:
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)?

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.

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

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

Thanks for your suggestions, John.

Cheers
Terry
 
E

Eeyore

Jan 1, 1970
0
John E. said:
Eeyore sez:


Not at all. Here in USA I checked my 3 regular sources: Mouser, DigiKey, and
Jameco with nil results, subs or not.

You mean that in the entire USA there is no such thing as 47V 2-3W zener diode ?

Graham
 
J

John Popelish

Jan 1, 1970
0
John said:
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)?

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

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.
 
J

John E.

Jan 1, 1970
0
Not at all. Here in USA I checked my 3 regular sources: Mouser, DigiKey, and
Keyword here is "regular".
You mean that in the entire USA there is no such thing as 47V 2-3W zener
diode ?

Graham

Nyet.

But point is moot, it seems. See recent posts to thread re. design change.
 
J

John E.

Jan 1, 1970
0
John Popelish sez:
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.

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,
 
J

John Popelish

Jan 1, 1970
0
John said:
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"?

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
37% discharge time I gave above referred to the time for the
current to reach zero. But that is not really the time for
the magnetic field to reach zero, because the iron parts of
the solenoid will circulate eddy currents that support the
field for a bit. Then there is the inertial time constant
of the mechanism that delay s movement, after the magnetic
field stops holding it against the return spring.

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
47 volts cost only a few bucks, if you want to take the
experimental route. Can you rig up some mechanical pickup
on the valve, so you can, measure the response time effect
of various zeners? That would make it pretty obvious where
the diminishing returns come into play.

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.
 
J

jasen

Jan 1, 1970
0
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.

42V turning on 5v turning off, I get 5/42 fraction as fast. (about 1/8 the
speed)
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.

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

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

Terry Given

Jan 1, 1970
0
jasen said:
42V turning on 5v turning off, I get 5/42 fraction as fast. (about 1/8 the
speed)

read harder.

5/42 = 0.118. 0.118 times faster is, indeed, slower. admittedly I didnt
have to make it a reading comprehension test, but its more amusing this way.
huh I'm getting 42/0.7 (which is over 50 times slower)

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

no, the original voltage across the coil during turn-off is Vz - Vcc =
47 - 42 = 5V. When a freewheeling diode is used, the voltage across the
coil is 0.7V.

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
 
T

Tony Williams

Jan 1, 1970
0
no, the original voltage across the coil during turn-off is Vz -
Vcc = 47 - 42 = 5V. When a freewheeling diode is used, the
voltage across the coil is 0.7V.

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).

42 5.3mS <-- nearly equal to the L/R Risetime.
47 3.9 <-- only 1.3x 5.3mS.
57 2.6
84 1.4 <-- Changing from an L/R sum to
mainly a V = L.dI/dT sum.
 
T

Terry Given

Jan 1, 1970
0
Terry said:
read harder.

5/42 = 0.118. 0.118 times faster is, indeed, slower. admittedly I didnt
have to make it a reading comprehension test, but its more amusing this
way.



no, the original voltage across the coil during turn-off is Vz - Vcc =
47 - 42 = 5V. When a freewheeling diode is used, the voltage across the
coil is 0.7V.

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

Tony Williams just pointed out my mistake.

I'm so used to dealing with SMPS inductors I forgot we were talking
about a solenoid.

In a SMPS inductor (or transformer) some external circuit is used to
limit the current - pulse width, peak current control etc, and in order
to minimise losses, Rdc is very small. In which case V = LdI/dt is the
"right" equation to use (as I*R is very small)

but a solenoid or relay isnt (generally) used that way - Rdc sets the
current, and is most assuredly not "very small", and of course I*R = Vcc
which is not "very small"

in which case its more about L/R time constants. I = Vcc/R, so when you
switch the solenoid off, the voltage across the internal inductance
rises to Vcc (because of I*R) + Vclamp.

So the difference between 42+5 and 42+0.7 is bugger all, and the
difference in decay time is, likewise, bugger all - well not bugger all,
but certainly not 7x.

Oops.

Cheers
Terry
 
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