It's shorted, burned on the side against the board, the side of the diode
that has part of the p/n printed (of course)...
Best I can make out is (reading around the diode:
Physically it resembles a typical 1A black epoxy rectifier.
Would this be 1n4147? The "47" is clearly visible, and I think I can make out
a "4" in the first part of the poorly-legible digits. No telling how many
digits between the two "4"s.
Any possibilities other than 4147?
Not wishing to trach granny, but this would be my approach:
Reverse engineer parts of the associated circuitry until I am reasonably
confident of what sort of application it is being used for, eg lf
rectifier, hs switch, flywheel, etc. Or more importantly, if it is a
zener.. It is normally not to difficult to work out what the diode is
doing and what sort of currents, voltages and frequencies are happening
At that point, wire in an external diode with a much, much higher spec
than the original - and measure the actual running parameters. I keep a
few huge and very expensive semiconductors just for this.
Then match a diode to that requirement, by measuring what is actually
happening - with any luck the spec will match to something with a lot of
4s and the odd 7 in its product name.
Assumptions about what things may be tend to bite..
Ah, another ITT diode on the board starts with "ZY" (where "Z" could be
mistaken for a "4" on the charred carcass). So I looked up ZY47 and get a
4.7v power zener diode, 2W.
Also, according to a National data sheet I found, a 1N4147 (a.k.a. 1N914A) is
glass, and way too lightweight physically compared to my charred sample.
Wrong turn, I think.
So it looks like ZY47?
You can't just guess at this. As sue suggested, you need to look at the
circuitry and figure out what that diode function is. What is the circuit
for? What components are near the diode? Can you trace the diode connections
to the next devices? What are they? Once that is determined, there are
likely substitute devices that will work just fine, even if you can't
identify the original. I do this all the time as I restore a lot of old test
equipment. It can be a challenge when you don't have the schematic, but
generally you can get there with enough patience and logical thinking.
OK, here goes:
German-made paper-handling machine, c. 1989. Circuit activates solenoid,
taking several inputs from other sensors, signals, etc.
Anode of unknown diode is to ground, cathode to drain of BUZ72 MOSFET. 1uF
cap also from drain to ground. Drain connects to solenoid. Drive voltage is
Not then, I would suggest, a 4.7v zener ... Sounds like it's just a
flywheeling diode and a 1N4007 would do the job just fine. Does the FET
source go to ground ? Many power MOSFETS have a diode internally in that
orientation across from the source to the drain. It gets there as an
integral side effect of the manufacturing process.
Newer designs seem to use a fast-recovery diode (e.g. FRED or HEXFRED
or HiperFRED) for MOSFET flyback protection. 1N4007s aren't
particularly fast, and reverse-conduction losses can be significant if
the switching frequency is high (in e.g. an SMPS). That's probably not
all that much of an issue in a solenoid driver, though.
Dave Platt < email@example.com> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I'd been turning over in my mind that this is indeed a zener, not simply a
"plain" rectifier. It is indeed a 47 volt zener.
Why was this diode chosen in the design? I'm familiar with the standard diode
being used to short-circuit the back-EMF from the solenoid, but I can't
figure out the purpose of a zener used in this location.
BUZ72 | /---/ ZY47
FET |--+ /\ Diode
0.27R / |
I think that should show proper in Courier or Monaco... or Paris (c:
I must add that Vdd is *reported* to be 42vdc. I was handed this board with
scribbled specs. May be higher or lower or in a parallel universe.
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 shit designs make it to market.
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
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
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.
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 ;)
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