0
\$\begingroup\$

When I look at a normal diode graph like that I see the Vz Zener voltage of it looks just like a Zener, so why not using it reverse biased like a Zener? I ask like that because I never saw it being used this way and I've been searching for days.

enter image description here

For example: using a 1N4001 like a 50V Zener , or a 1N4007 as a 1kV Zener to cut/shunt any voltage above that.

ps: i said too much the word Zener but actually its the varistor aplication that im considering. and actualy a varistor is like kind of diode like as it haves many soup like diode junctions in it. -metaforical compare: think of a diode and a varistor like crystal and glass.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The "normal" diode isn't doped with a goal of sharpening the breakdown voltage knee. A zener is doped for that goal. \$\endgroup\$
    – jonk
    Apr 29 at 22:37
  • \$\begingroup\$ so i will probalby not have a very precise voltage of breakdown, for some applications this will be a problem but about the ones that dont need to be precise like surge protection. like those varistors that also have a wide region where it starts to conduct some mA but for the uses the have its good enough \$\endgroup\$
    – bigubr
    Apr 29 at 22:42

5 Answers 5

Reset to default
3
\$\begingroup\$

Sort of, but not really. As mentioned by others, voltage is undefined -- indeed in practice, rectifiers are highly overrated. I've measured "400V" diodes breaking down at over 1000 (Ir = 1mA). Typical reverse bias curve has a sloped region, i.e. Vbr depends in Ir, not the sharp curve of a proper zener. The slope/intercept of that region seems to drift with temperature and exposure to avalanche.

After voltage and its stability, comes current handling. Avalanche tends to destroy junctions -- some combination of hot-spot failure, caused by improper chip design, undesirable impurities, etc. TVS are designed to handle essentially unlimited power, distributing current evenly across the junction until it basically melts to a lump. Others fail at quite low currents.

Si junctions usually seem okay at modest avalanche currents (low mA). Red LEDs (AlGaAsP alloys) also seem to handle fractional ~mA okay, at breakdown ranging from maybe 50 to over 200V (the datasheet 5V rating seems to be a massive understatement!). GaN seems to be extremely vulnerable: typical blue/white LEDs break down in the 20-30V range and die suddenly no matter the current (even ~uA), and RF and power transistors aren't rated for any breakdown current.

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Thanks for providing some numbers in a rather opinionated discussion. \$\endgroup\$
    – tobalt
    May 1 at 6:14
  • \$\begingroup\$ i like your answer. very technical and with data this is what im searching for. so, basically, i can consider that in the case of using a diode as a varistor to protect a device from any event(peak, surge, esd, ligthning) what i can expect is that in the first event it may fail closed. \$\endgroup\$
    – bigubr
    May 1 at 22:53
  • \$\begingroup\$ so im my consideration in simple cheap devices(for my use)(that i want to save a varistor -its not needed anyway because theres good surge protectors in the entrance and i never had lighting induced damage, so, theyre working ok) is that i can only use diodes if im ok with it being a sacrificial protection to be wasted in the first event that happen only to protect the device and leave it disconnected until i replace it(considering i will have a fuse and/or circuit breaker to open the circuit when the diode shorts to gnd) \$\endgroup\$
    – bigubr
    May 1 at 22:59
  • \$\begingroup\$ and about those led currents in breakdown that you say. wheres it from ? do have citation, links, anecdotal experience(empirical tests made) ? if anyone have more details on that it would be highly appreciated. \$\endgroup\$
    – bigubr
    May 1 at 23:27
  • 1
    \$\begingroup\$ @bigubr I would expect, for mains surge protection, a rectifier would simply blow up. Probably failing shorted within some µs, then fusing open within some 10s ms. (FYI, 'S' is siemens, a unit of conductivity; 's' is seconds.) \$\endgroup\$
    – Tim Williams
    May 2 at 17:26
4
\$\begingroup\$

Generally, no.

For one thing, a varistor is a different thing entirely from a diode, and definitely can't be replaced with a diode.

For the zener, that's a little more possible. The problem, however, is that for a normal diode, say a 1N4007, all you're told is that the breakdown voltage is at least 1000 V. It could be 1500 V, or it could be 1001 V, you have no way of knowing.

Secondarily, the breakdown may not be as sharp as a real zener diode, and it may have a high output impedance. They don't design it to be used like this, so its characteristics aren't optimized for that.

\$\endgroup\$
3
  • \$\begingroup\$ a varistor can be made diferently but theres diodes that work like then like tvs diodes for esd protection. the initial breakdown if 1500v will be more than most varistors in 200-600v but for surges of many kV like from lightnings both will get in the conduction zone. \$\endgroup\$
    – bigubr
    Apr 29 at 22:47
  • 2
    \$\begingroup\$ @bigubr Those are called TVS diodes, though. "Varistor" refers specifically to the discs of pressed metal oxide powder. If it's a varistor, it's not a diode. \$\endgroup\$
    – Hearth
    Apr 29 at 22:52
  • \$\begingroup\$ the thing is they both clamp the spikes of voltage , this is what i want the commom diodes to do. \$\endgroup\$
    – bigubr
    Apr 29 at 22:58
4
\$\begingroup\$

Perhaps you can, but there is no guaranteed avalanche voltage for a 1N4001.

I would not be especially surprised to see it break down at 600V or 1200V. They may be using the same die as the 1N4005 or 1N4007. It will meet all datasheet specifications if they do that. A real 'zener' is not expensive and will have a guaranteed breakdown with a tolerance of maybe 5% or 10%.

An occasionally more useful off-label use for 1N400x is as a high-capacitance varactor diode.

Edit:

From this video is a screen capture showing an actual test of a 1N4001. If I interpret it correctly (it's in Chinese) the breakdown voltage of the 1N4001 is close to 1400V rather than the 50V you might have been expecting.

A 1N4148 earlier in the video shows a breakdown about 1/10 of that (rating is 75V for most manufacturers, if memory serves). The 1N4148 is optimized for fast switching (~4ns) so it has quite different characteristics.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ the prices im considering are 3x more for the zeners and 10x more for the varistors. so thats the point of substituting it and also would be less parts to buy in quantity to take too long to use or in the case of buying much less for use only the ones need than they will be way more costier than that \$\endgroup\$
    – bigubr
    Apr 29 at 22:49
  • \$\begingroup\$ Have you attempted to measure the characteristics of an actual 1N400x? Even roughly (resistor and power supply)? \$\endgroup\$
    – Spehro Pefhany
    Apr 29 at 23:01
  • 1
    \$\begingroup\$ You do interpret it correctly; I don't speak the language but I do recognise that curve tracer as a model I use a lot, and it's showing a breakdown at something a little below 1500 V. \$\endgroup\$
    – Hearth
    Apr 30 at 14:30
  • \$\begingroup\$ only now i saw the video. great , breakdown seems 150V for 1N4148 and 1500V for 1N4001 \$\endgroup\$
    – bigubr
    May 3 at 7:38
2
\$\begingroup\$

Another nail in the coffin from here. Even if you do find diodes that have "reasonably" close breakdown voltage to what you want, they may (will) die. There are "avalanche rated" diodes that are designed to take repeated breakdowns like a champ but most diodes are not. Usually this kind of resiliency is found in high-voltage diodes, not so much in commodity 2 pennies diodes.

Just go and buy the darn Zener, they cost little. Be mindful of "zener impedance" though, the behaviour may be less straightforward than you might think on low currents. I actually used blue diodes to create a reliable low-current 3.3V voltage drop for a MOSFET gate. Silly? Maybe, but it works, even at 0.1mA, unlike a Zener.

\$\endgroup\$
2
  • \$\begingroup\$ 'Usually this kind of resiliency is found in high-voltage diodes,' like 1n4007 , uf4007? or can you give me examples/links ? \$\endgroup\$
    – bigubr
    Apr 30 at 20:33
  • 1
    \$\begingroup\$ @bigubr Not really my job but here's a bunch from Digi-Key. It's not an exhaustive list as fast recovery diodes can be avalanche rated as well but on quick check the manufacturers don't tend to have tick box for "avalanche" in their product lists. digikey.co.uk/en/products/filter/diodes-rectifiers-single/… \$\endgroup\$
    – Barleyman
    May 3 at 10:18
1
\$\begingroup\$

In a pinch you can use a reverse biased base-emitter as an 8 volt zener. It will have a rounder knee than a true zener. The phenomenon is well-known. There's a trick question that asks what the not-connected collector voltage would be if, say, the base was at 0 volts and current is driven into the emitter of an NPN transistor. Surprisingly, the value is negative. The base-collector junction is exposed to light from the breaking-down base-emitter junction and acts like a photo diode.

\$\endgroup\$
4
  • \$\begingroup\$ How close to 8V? (Is it like anywhere between 7.5 to 8.5 or anywhere between 6 to 10?) \$\endgroup\$
    – Oskar Skog
    Apr 30 at 18:31
  • 1
    \$\begingroup\$ In my experience a range of 7.5 to 8.5 is reasonable to expect. It's nowhere specified. The only absolute maximum rating I've seen always seems to be 5 V. If the current is limited to a reasonable value the transistor won't fail. I've seen designs that had been in production for many years with no failures, where the reverse breakdown was unintentional and not known. \$\endgroup\$
    – stretch
    Apr 30 at 22:57
  • 1
    \$\begingroup\$ As with the other answers, this behaviour (and parameter) is not guaranteed. Note that in addition, if you reverse bias a transistor's B-E junction this way, it will degrade the device's hFE (beta), and make it unsuitable for other applications. \$\endgroup\$
    – jp314
    May 1 at 23:30
  • \$\begingroup\$ Where did you find that beta would be degraded? I based my answer on personal experience and word-of mouth. I can't imagine a transistor used as an "in a pinch" zener ever being used in some other role afterwards. If it's unsuitable for other applications, who cares? \$\endgroup\$
    – stretch
    May 3 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.