Author Topic: Matching diode's Vf  (Read 1583 times)

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Offline CJay

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Matching diode's Vf
« on: October 12, 2016, 05:42:55 pm »
So, I've got a pile of 1N6263 Schottky diodes, some toroids and a desire to build a DBM from the ground up to experiment.

I want to match the diodes, I'm assuming setting some arbitrary current like 1mA and measuring Vf would be the way to go or will self heating be an issue at that If?


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Online Rerouter

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Re: Matching diode's Vf
« Reply #1 on: October 12, 2016, 05:47:03 pm »
I would imagine having the 2 in series, feeding in a oscillating current, and measuring deferentially across the 2, and then measuring the difference between the 2, if the reading never changes, you have 2 perfectly matched diodes,
 

Online tautech

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Re: Matching diode's Vf
« Reply #2 on: October 12, 2016, 05:54:30 pm »
Been sort of playing around on the fringes of what you're try to do lately.

Across the brands of DMM's in Diode test mode offer a range of voltages but normally @ 1mA.
If you have a bench unit like a SDM3055 you can set a Vf threshold at which the DMM buzzer will sound like in a Continuity test. Of course any below that threshold you set will all set the buzzer off however with a little tweaking you should be able to get some quite tight results.
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Offline T3sl4co1l

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Re: Matching diode's Vf
« Reply #3 on: October 12, 2016, 07:04:46 pm »
Well, for conduction, Vf, just a simple diode test.  Probably bin them in 1 or 2 or 5mV steps.

For dynamic though, you need to pair on capacitance as well as Vf.  It's a two dimensional problem.  So some will work at DC but not AC, and vice versa.  And the pairs that work at AC, will only match over a modest range of frequency and drive current -- because the drive current sets Vf, while the frequency sets the current from capacitance.  And both vary at different rates.  Only the ones that are matched in both parameters, will match well at all frequencies.

So you actually have to match along two dimensions, which is tricky.  That takes at least two measurements per diode (e.g., one for Vf at DC, one for AC at some, ideally, 100s of MHz).

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Offline CJay

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Re: Matching diode's Vf
« Reply #4 on: October 13, 2016, 08:21:00 am »
OK, so I'm going with a constant current source (BJT, MOSFET and a couple of resistors) and will measure/grade according to Vf at, I think, three settings for If.

Capacitance, I don't know how I'm going to measure that, I'm mulling the idea of an LC oscillator and measuring the frequency change adding the diode causes but as I see it the capacitance is very small so I will need a fairly high frequency .

The DBM is an experiment, I'm planning a series of building blocks to create a SW receiver, fail or succeed I'm hoping to learn a lot.


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Offline T3sl4co1l

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Re: Matching diode's Vf
« Reply #5 on: October 13, 2016, 09:28:46 am »
Simplest: set up a single balanced mixer.  With an oscilloscope, set trigger from the source oscillator, and view the waveform at the output port of the mixer (which should be null, if the mixer is balanced).

This tests everything at once, actually.  It also has the downside of being comparative, rather than quantitative (measuring a single parameter directly).  Thus, you can sort and rank all the diodes, using any O(N lg N) algorithm, or brute-force all O(N^2) comparisons.

You'd repeat the test at a different frequency, and a different excitation level, to control for differing Vf and capacitance.

Actually, the two effects should be visibly separate (a nice part about viewing the waveform directly!), and can't null one another (because Vf is a resistive, in-phase effect, while capacitance is a reactive, out-of-phase effect).

Just under one test condition (excitation and frequency), the well-matched diodes should be pretty obvious, from the cleanly nulled waveform.  Those with similar Vf, but differing capacitance (or vice versa), will have one part of the waveform (like the flat part in forward-bias) well matched, but the other part lumpy, and so on.

And, a null needn't be comparative; as long as you're inspecting that waveform there, you can read off the difference in Vf directly, or the difference in capacitance that's bleeding through.

Do make sure the signal source is, itself, well balanced, though!  Do it at a modest frequency like 1MHz, sine wave (low distortion), and use a CT transformer to generate the opposite phases.  Follow this up with a common mode choke (if nothing else, wind some twisted pair through a large ferrite bead, tens of turns).  Then terminate the now-well-balanced signal currents into equal (precision) resistors to ground.  Then, and only then, introduce the diodes (and a load resistance on the output port side) to measure them.

This sounds kinda fun. I almost want to set it up and do it myself! :P

Tim
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Offline CJay

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Re: Matching diode's Vf
« Reply #6 on: October 13, 2016, 07:21:41 pm »
Sleepless night lead to some rough experimentation with 'off the reel' unmatched 1N4148 diodes and some shoddy construction, but!

With a 400MHz LO I can 'see' a 46MHz peak on my spectrum analyser when a handheld PMR446 transceiver is keyed near to the mixer which is encouraging.

Later today I'll knock up a constant current source and try matching a few of the 1N6263 devices and some 1N5711s which should be in the post today.



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Offline danadak

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Re: Matching diode's Vf
« Reply #7 on: October 13, 2016, 09:43:24 pm »
Love Cypress PSOC, ATTiny, Bit Slice, OpAmps, Oscilloscopes, and Analog Gurus like Pease, Miller, Widlar, Dobkin, obsessed with being an engineer
 


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