matched 2N5952.. how many do i have to buy..??

[Bullet79]

want it for phase 90 project... how many transistor do i have to buy to make 4pc matched..??
is it really depends on luck...

[theehman]

Didn't we talk about a trade thread a while back to help everyone get a matched set? 

[Mark Hammer]

Yes we did, and I'm glad you raised that subject again.

However, if a person wishes to have more than a mere 4, I note that I bought 50 from Mouser (and they're just under 11 cents each if you buy at least 25), and ended up with many sets, and even a sextet I plan to use.   If you go in on an order with someone, the shipping costs drop and your investment to get yourself a matched set or 2 or 3, is only a couple of bucks.

[Bullet79]

plan to get it at my local component shop,... maybe around 30pc's - hope that will do.. at least 1 set...

[petemoore]

 Pick a card any card from the deck.
  Now pick matching cards from the deck...only so many matches eh ?
   Now try it with 2 or 4 decks, the matches start appearing more often.
You pretty much have to be into building phasers [very plural] to maximize your Jfet selections investments, and as Mark pointed out, it's the ''everything else'' that dwarfs this expense. "Everything else'' expenses may be mitigated to some extent, but those 4 chosen Jfets [and all the non-chosen ones] are a drop in the bucket when compared to box/switch/jacks/PCB etc.

[R.G.]

However, if a person wishes to have more than a mere 4, I note that I bought 50 from Mouser (and they're just under 11 cents each if you buy at least 25), and ended up with many sets, and even a sextet I plan to use.   If you go in on an order with someone, the shipping costs drop and your investment to get yourself a matched set or 2 or 3, is only a couple of bucks.

Mark - I think you have a semi-definitive answer in your hands if we'll work on it a bit.

You and I probably have the statistics background to answer this. If you can list up the raw number you bought, and how many sets you got, we can calculate the probability that you get a match for N = 2, 3, 4...

[Mark Hammer]

I actually taped together each of the matched ones, so when I get home tonight I can count up the number of sets I have, and get back to you on that.

In some respects, it's a bit like the likelihood of finding your perfect soulmate in a town of 500, 1500, 5000, etc.  Once the municipality gets big enough, there ends up being someone for just about everyone.  So, with 25 JFETs, I may get a couple of quartets, but have a bunch of "leftovers".  Buy another 10-15, though, and those "leftovers" start having matches show up.  That is to say, the matches-to-leftovers ratio may improve with the number of JFETs you start out with.

All the more reason for those of us who bought way more than we needed to make our "leftovers" available to others.

[R.G.]

I actually taped together each of the matched ones, so when I get home tonight I can count up the number of sets I have, and get back to you on that.

Great. That ought to do it.

In some respects, it's a bit like the likelihood of finding your perfect soulmate in a town of 500, 1500, 5000, etc.  Once the municipality gets big enough, there ends up being someone for just about everyone.  So, with 25 JFETs, I may get a couple of quartets, but have a bunch of "leftovers".  Buy another 10-15, though, and those "leftovers" start having matches show up.  That is to say, the matches-to-leftovers ratio may improve with the number of JFETs you start out with.

I think it's probably going to be well modeled by selecting from a normal distribution into buckets. I *know* from the chip design guys that overall, semiconductor parameters tend to be clustered into a normal distribution from a process even over many chips. Within a wafer of hundreds to thousands of transistor dies, the transistors on that wafer are also normally distributed, but with a very small sigma. Over many wafers you get the process sigma as the result of each wafer being a "sample" of the process results. Different manufacturers and even different diffusion ovens will produce a slightly different characteristic distribution, but they'll all be bunched like this.

So I think what's going to happen is that (1) if you buy a bunch from a distributor, they will in general be what that business bought from one of the makers, perhaps 10K to 20K from the same maker, and their bin will contain large subsets of parts that came from the same wafer, but all got poured into the same bin at the manufacturer after they were finished; (2) the distributor's bin will also have leftovers from previous buys that didn't quite empty their bin, at random; (3) parts we order in bulk will come in with a number of well-defined "lumps" of parts. A truly random sample (which we'll never get) would let us reproduce by sampling the population lumps in the distributor's bin.

However, by running backwards through the numbers, we can produce a rough estimate of the probability that given one sample, what is the likelihood that the next sample will be within X% of the first one. It's a variation of the Birthday Paradox. If you subset any distribution into bins, any one sample has a likelihood of fitting into any bin which depends on the number of bins and the initial distribution. For birthdays, it's easy because the distribution is (almost but not quite) uniform over 365 bins, so the calculation of the probably of not being in any one bin is easy. It's not so easy for non-uniform distributions, but possible.

Obviously, the relative width of the bins compared to the distribution is critical to the probabilities. This is cased up in the normal distribution in the probability of a sample lying more than N*sigma from the mean. The probability of a sample being in either the high half or low half is 50%. As you make the bins smaller, the probability of a sample being in any one bin decreases, and that is what is used to calculate the inverse - the probability that the next sample is in this bin right here.

All the more reason for those of us who bought way more than we needed to make our "leftovers" available to others.

And that is a great approach - and one that is so characteristic of your helping nature.

You're a thoughtful person, no matter what those other guys say. 

[Govmnt_Lacky]

I am currently sitting on quite a few matched sets. My problem is knowing which sets are best to use with which zeners. I have sets with matched values of 1.84, 1.80, 1.85, and 1.69. Is there a way to determine which zener to use (5.1V, 4.7V, etc.) by the matched JFET value? Or, is it up to the ear?

[Mark Hammer]

The zener is simply there to provide a stable and predictable bias voltage, such that whatever you set the bias to will continue to be valid even though the battery has dropped from 9.6V fresh to 8.2V after a gig or three.  You could probably even go with a 6.3V zener, although you might find the trimpot a little "twitchier" by having to squeeze a broader range of bias voltages into the same 300 degrees of trimpot rotation.

[Mark Hammer]

Okay.  I think I bought 5 from Mouser.  Eight of them are sitting in P90s right now.  From the rest in my parts bin, I have:

Quantity          Vgs
7               1.4 - 1.44
4               1.52
8               1.6 - 1.65
3               1.69
4               1.71
3               1.76
4               1.82
6               miscellaneous leftovers

The groupings, of course, are framed by my "hopes", and to some extent by lack of perspective on my part.  Looking at them now, having 3 at 1.69 and 4 at 1.71 is not really any different than having 7 between 1.4 and 1.44.

I leave it up to you folks to decide how much matching is "matched".  But clearly one does not have to purchase a mountain of them to get more than one set of 4, or even a set of 6...or more.

And if one is willing to augment one's phase shifter with the odd fixed stage here or there (used to great effect by a wide array of commercial and DIY units), then the goal does not have to be an even number.  So, 3 matched JFETs, plus a fixed stage, is perfectly capable of yielding pleasant tone, as would be 5 and a fixed stage.

[stringsthings]

want it for phase 90 project... how many transistor do i have to buy to make 4pc matched..??
is it really depends on luck...

this thread is great !   .... we have the long answer ( thanks to the everyone who responded before this post ) ... and here's the short answer:

nope - it doesn't really depend on luck ...  

God Bless Statistics ... lol

[Mark Hammer]

Okay.  I think I bought 5 from Mouser.  Eight of them are sitting in P90s right now.  From the rest in my parts bin, I have:

Clearly, that should say 50, not 5.

[oldschoolanalog]

The things are relatively inexpensive. Ever cheaper here.
If you're going to place a Mouser/Newark/whatever order anyway, might as well get 100...
Maybe there will be a "matched set" for something really cool like a 16 stage phaser.

[R.G.]

Okay.  I think I bought 5 from Mouser.  Eight of them are sitting in P90s right now.  From the rest in my parts bin, I have:

Quantity          Vgs
7               1.4 - 1.44
4               1.52
8               1.6 - 1.65
3               1.69
4               1.71
3               1.76
4               1.82
6               miscellaneous leftovers

I'll take a preliminary swing at the statistics. I'm very much a duffer at this, and if any of you have a statistical background, you can probably do much better. I will need to go dig out the engineering statistics textbook to refine it, because this tickles a memory about sampling distributions and confidence intervals that don't happen in the simple guess that follows.

Looking at the list, there are 47 JFETs accounted for after including the two sets of 4 which are now being used in P90s.  Of these, six out of 47 didn't ever fit, and another six only got to groups of three and are presumably not usable.

Picking one device at random, what is the likelihood you got one of those that didn't ever fit a grouping? That's 12 out of 47, or 0.255. The next pick is also 0.255 of not ever fitting a grouping, so the probability that you got two that didn't ever fit a group is 0.255*0.255 = 0.065. The probability of getting three that don't ever fit a grouping is 0.0166, or 1.6%. At three random picks, the likelihood that at least one of them fits a group is 98.4%.

Now, how long will it take us to get a group, because we could get one from each group and never get a full group, right?

Out of the 47, there are six groupings of four, counting the ones already used as two P90s. The group of eight counts as at least two groups of four, and in reality should be considered to be more than that. I gotta find that textbook. The group of seven also counts as more than one group of four, but I can't at the moment come up with a good estimate for how many. But let's be conservative and call the group of eight three groups of four and the group of seven two groups of four. That's 11 groups of four out of 47 devices.

If we pick one JFET, the odds are 1-0.255= 0.744 that it's in a group of four somewhere. By the time we pick our third one, the odds of one of them being in a group of four are over 98%.
...
Two hours later, I'm still formulating the answer. I think it models up as the sum of probabilities of an only-slightly-violated multinomial experiment. I need to quit typing the answer on the screen and go solve this before entering it.

I'll be back.

[PRR]

What is a "match"?

For stereo limiter the FET resistance must match better than 10% over the range 1K-100K.

For a phaser, the main thing is that all stages "add phase". It probably is not critical if one is adding 60 degrees and another is adding 30 degrees, unless you are sweeping ALL the way to full-on and full-off. A 2:1 resistance range may be acceptable.

You need a "bin width".

The numbers Mark gives are apparenly from R.G.'s clever sorter. It gives Vgs to cause some nominal resistance typical of phasers. This is an essential first step to see if you bought a bag of 2V or 6V FETs.

But we never get an exact match. How far off is a 1.69V and a 1.71V set? Rig a 1.69V bias, measure the resistance of the "1.71V" part at 1.69V. If at 1.69V a 1.69V part is 10.0K and a "1.71V" part is 10.5K, that's acceptable in a stereo limiter and way-plenty good for a phaser cascade.

I *suspect* that even 20% difference of Vgs (in R.G.'s test) will not matter in a 4-stage phaser cascade. Which means a 1.69V part is a "1.35V to 2.03V" bin.

This should be tested by ear. Mark has the ears, the FET-bag, and probably an idle mule which could be fitted with sockets; but anybody can try this.

If indeed a +/-20% match is all the same in a phaser, then at least 33 of 50 in Mark's bag are "a set".

And unless he got very lucky, that means the problem is weeding-out the odd 1.01V and 2.5V parts. Not in tediously assembling sets. 33 usably-similar in 42 devices means you over-buy 33% for 50:50 chance. I don't have a normal curve but I think buying 3X the required number (buy 12) gets your chances out to 99:1, which is better than the odds in your state lottery (your odds of losing).

[Mark Hammer]

I wasn't diligent enough to work out the precise probabilities, but based on what I got, I would feel VERY confident that somewhere in a batch of 20 (but we'll say 25, simply because it is a price inflection point), there would be at least 2, and probably 3, well-matched sets.

[R.G.]

What is a "match"?
...
You need a "bin width"
...
The numbers Mark gives are apparenly from R.G.'s clever sorter. It gives Vgs to cause some nominal resistance typical of phasers. This is an essential first step to see if you bought a bag of 2V or 6V FETs.
...
But we never get an exact match. How far off is a 1.69V and a 1.71V set? 

Yep. Yep. Yep. Yep.

That's what's in the math I was frying back here in the cave.  I think it's best modeled as a multinomial sampling, and I haven't (yet) relearned enough of my statistics to calculate out a solid probability. I'm close.

And unless he got very lucky, that means the problem is weeding-out the odd 1.01V and 2.5V parts. Not in tediously assembling sets

.
That's the easy question, and the one I solved first. The odds of getting four devices not in any set at all of the usable group are well under 1% working from Mark's sample. That says that the odds of getting four out of a two-bin setup (good and not-good) are very high. Buying eight is almost a sure thing to get at least four from the good vs not-good set.

33 usably-similar in 42 devices means you over-buy 33% for 50:50 chance. I don't have a normal curve but I think buying 3X the required number (buy 12) gets your chances out to 99:1, which is better than the odds in your state lottery (your odds of losing).

I've worked it a few ways, including the multinomial, Monte Carlo, and enumeration; I get lost on each one after a little ways into the thicket. But the indications I have are that you'll likely (SWAG: 90% chance) get one set of four with no more than about 50mV spread between them at around a dozen parts. They all seem to be headed that way, although I still haven't gotten the math to close with either of these approaches.

In the end, it's probably an economic problem, as Mark suggests. Mouser sells the applicable 2N5952 for $0.12 each in 1s, $0.10 from 25-50. So 25 of them are $2.50. At 25 pieces, chances of getting one set of 4 is virtually a sure thing. It's probably nine out of ten at 12 pieces, but that costs you $1.44 as opposed to $2.50 for 25. I'd spend the other buck for the sure thing.

[Mark Hammer]

In the end, it's probably an economic problem, as Mark suggests. Mouser sells the applicable 2N5952 for $0.12 each in 1s, $0.10 from 25-50. So 25 of them are $2.50. At 25 pieces, chances of getting one set of 4 is virtually a sure thing. It's probably nine out of ten at 12 pieces, but that costs you $1.44 as opposed to $2.50 for 25. I'd spend the other buck for the sure thing.

And for a dollar investment, you're likely to have an extra set or two to monkey around with or give to a buddy.  Worth a dollar?  You betcha.

[PRR]

> What is a "match"?

Unless I misunderstand the problem.... the "match" can be very-very wide (nearly as wide as the spread coming out of the FET chute).

The Is phase shifter sombreness? page that Gurner just posted got stuck in my head-cold and I let the Idiot Assistant do the thinking that my brain wasn't gonna do tonight.

I'd already set the IA to do four identical 0.001u+160K stages. Nulls happen at 412.5Hz and 2,406.7Hz.

I copied that, then mangled the resistors to 80K, 113K, 226K, and 320K; a good 4:1 spread. Nulls at 372.15Hz and 2,667.6Hz.

A +/-100% mis-match produces an 11% shift in null frequencies.

Open these in new tabs to see schematic and plot:

http://i53.tinypic.com/mw9u11.jpg

http://i53.tinypic.com/ea075y.jpg

In neither case are the nulls (or peaks) "harmonically related": we are working off a SIN (or TAN) function, not a LOG function. And we are going to sweep the frequency. So the exact frequencies or frequency-intervals are not "critical". We just want to put some notch/bumps in and hit large chunks of the spectrum.

So a huge "mis-match" spreads the effect over a somewhat larger swath of spectrum, which isn't even a bad thing.

The next question is: what is the relation between FET resistance and voltage? Clearly if we have two FETs with Vp of 1V and 8V, and bias at -2V, the 1V part will be hard-off (many Megs) and the 8V part will be pretty-on (ten K or so). The R.G. tester assures that FETs _will_ match, but does not give a clue how-far-off we can be and still enjoy effective phasing.

THIS page says rDS in the ohmic zone has Vp^2 over Idss times the difference between bias and Vp. Taking 2V bias on two FETs with same Idss but Vp of 4V and 8V, I get factors of 8 and 10.7. A 2:1 mis-match of Vp gives a 33% difference of rDS. If +/-100% "error" of rDS is quite acceptable, then the range of Vp is very wide (certainly FAR larger than the 1% bins Mark is sorting to). OTOH, if Vp is 2V and 8V and bias is 2V, then we have a problem. (Or for 2N5952's 0.75V-3V spread, 0.8V bias is a problem.)

Idss enters in directly, and has typical 2:1 spread. This actually seems small. And IIRC, Vp and Idss tend go-together: from a given geometry and nominal process, high Idss implies high Vp. The channel-size is well controlled; the crispiness (diffusion depth) varies from center to edge and from batch to batch.

Is there empirical evidence that matching matters? If so, I suspect it is the nearly-OFF condition, the 100s-K end, where an oddball FET will soar to MEGs and stop contributing. In that case we should "match" for Vto. And that's just a 9V batt, a large resistor, and a DVM.