diode leakage, input protection

Does anybody know how power diodes (rectifiers, zeners, TVSs) are fabricated? Do these use patterned lithography, or do people just fab big unfeatured wafers and dice them up? I wonder if chip edge damage contributes to leakage, or maybe the sheer combination of size and doping makes power diodes leaky.

The context is that I have a test/cal connector that I want to protect from overload, at least +-20 volts and preferably more. I can't add a lot of series resistance.

A polyfuse and clamp might work:

formatting link

but I need fairly high-current, low leakage clamp diodes, and most official diodes are leaky, even typ values at room temp, much less max at 75C. It seems that transistors are better diodes than diodes.

Analog Devices just EOL'd the neat ADG465. Bummer.

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John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
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John Larkin
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Yeah use a transistor maybe... the BC junction for higher voltage (I think) (I've got a nice B. Pease paper on clamping and bounding) ... or maybe even a diode connected transistor?

George H.

Reply to
George Herold

Once-upon-a-time wafers were indeed unpatterned, and the die were edge-etched after sawing to clean up.

I think everything modern is now patterned diffusion or implantation.

Leakage _is_ proportional to size and crud content.

...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

So, it's time to design in a protective relay. High current: check Low leakage: check

Reply to
whit3rd

We could sense the overload and drop out the test relay, but that would take a millisecond or two. We need to protect the ICs meanwhile.

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John Larkin         Highland Technology, Inc 

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John Larkin

Den torsdag den 22. maj 2014 19.31.10 UTC+2 skrev John Larkin:

a relay? either short or disconnect

I looked at a random smd reed datasheet, max 1ms open/close time I seen some RF reeds say 0.1ms

-Lasse

Reply to
Lasse Langwadt Christensen

I suggest two diodes in series, in place of each of your existing diodes. By itself this doesn't help much with the leakage, but if you then get an op-amp voltage follower with a gain of 1, and drive the junction between the two diodes to the same voltage as the low-leakage node that you are trying to clamp, then the diode leakage is much reduced, as provided they are in the dark, diodes leak very little when there is no voltage across them.

e.g. +-----|\ | | >---+ | +-|/ | | | | | +-------+ | | INPUT--*--*-|>|--*-*--|>|--Vpos | | +--|

Reply to
Chris Jones

So, you provide a driven shield, protect the shield with leaky diodes (but it's driven, so no big deal), and clamp the high-Z node to the shield, with diodes that (in normal operation) never see over a few millivolts. Excellent! As I read the original problem, though, there was high current (i.e. low impedance) involved. The second part of the problem, is to keep the high (allowed!) normal currents while suppressing, somehow, overrhigh (fault) currents. Fuses, polyfuses, and relays are capable of this, but are all kinda slow. Maybe use a choke and a transzorb, to blow the fuse before the sensitive gizmo (on the other side of the choke) gags?

Reply to
whit3rd

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