low capacitance zener

Apr 19, 2007 32 Replies

But their resistance changes with current. It would be really nice to have a 50 ohm filament, made of some low tc wire, that would be a variable-temperature, constant-impedance noise source.

What's the impedance of one of those bulbs?

How about a zener noise source and a digital step attenuator, so you could just tweak the output noise temp as desired? That would be handy to have around. Or just a box with a calibrated pot/dial, 50 ohm source, noise temp from room temp to maybe 500 C in one turn?

John

You guys are all way off. The best noise source EVER is a thyratron in a specified magnetic field!

(I actually have a "gaussian noise generator" I found in the trash. Tubed for the most part, with a pair of 6D4s, each between some magnet assembly.)

Tim

-- "Librarians are hiding something." - Steven Colbert Website @

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It took a while but we found some that were about 50ohms at a high enough temperature yet not too high so they'd last a while. You have to regulate current but we did that by hand because it stabilizes quickly. In the late 90's it became more difficult because the market for such bulbs shriveled up. It's all EL backlight, LED and other stuff (whatever is in the day-glo type watches). So I guess this old trick will be on the way out soon. Also, I don't know how these bulbs faired in terms of inductance at higher frequencies. All we needed was resistive behavior up to 15MHz.

Yeah, that would be nice :-)

Regards, Joerg http://www.analogconsultants.com

Well, I wanted a noise source that fits into my pocket ;-)

Regards, Joerg http://www.analogconsultants.com

Very good. Then repeat at various currents from 50uA to 10mA.

But the real question is why is the apparent capacitance so high in Zener mode? Regards,

Mike Monett

How dumb of me. Of course. A hot resistor.

May be a bit difficult to calibrate. How accurate are the IR thermal meters used to measure hot air heating duct?

Regards,

Mike Monett

Nah. A 127-bit pseudo-random shift register with resistive summing into a low-pass gaussian or bessel filter.

Some of the advantages are: easy to calibrate, close enough to random so you can't tell the difference, initialize to any preset state, repeatable every time for production testing, pretty good bandwidth, no drift, inexpensive, can easily be duplicated with identical performance. Regards,

Mike Monett

You don't have to. The filament temperature can be determined by measuring its resistance at a known temperature, for example room temperature, and then again at operating temperature (voltage divided by current). Formula 2 in this lab sheet:

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With those small watch bulbs you just have to make sure to exclude any contribution from the tiny connection wires into the glass.

I have heard some folks use the term "ohmicity" when measuring for filament temperature calcs but that term always gave me the goose pimples. Old Georg-Simon would not have liked that.

Regards, Joerg http://www.analogconsultants.com
[...]

Thanks for the link, Joerg. I was accustomed to using light bulbs to measure temperatures above 100C, but I didn't realize the equation was valid down to room temperature. From the article:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ In the case of a light bulb, the resistance of the tungsten used for the filament varies with temperature according to the following relation:

T/300 = [R(T)/R(300)]^0.811 (2)

In this equation T is the absolute temperature in K, and the room temperature is assumed to be 300 K.

R(T) is the resistance at the temperature T.

R(300) represents the resistance at ambient room temperature.

Eqn. 2 holds very well in a wide range of temperatures from 300 K up to 3680 K (melting point of tungsten) but it does not work for temperatures below 300 K. Maximum useful filament temperature is about 3000 K and is limited by the vaporization pressure.

The absolute temperature in K is related to the temperature in C in the following fashion:

TK = TC + 273.15 (3)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So now we can use light bulbs to measure temperature, see in the dark, and as a signal generator. What would Edison have thought?

Regards,

Mike Monett

I think he was more concerned with the business aspects of light bulbs :-)

It's great what the old inventors have brought us. But sometimes one stumbles upon more personal info about them. During the current project I needed the Boltzmann constant and sure enough I couldn't remember it. Google, five seconds later it popped up on Wikipedia, faster than hopping to the book shelf. Right underneath the story of Ludwig Boltzmann. Turns out he suffered from untreated depression and during one of those bouts committed suicide. Man, that really made me sad. Back in school we never really heard about their lives.

Regards, Joerg http://www.analogconsultants.com

Same thing happened to Armstrong and many others.

Fessenden is one of the few I can think of who went through terrible battles and died peacefully.

From wikipedia:

Although Fessenden's antenna in Brant Rock, Massachusetts was demolished in

1917, the "It sometimes happens, even in science, that one man can be right against the world. Professor Fessenden was that man. He fought bitterly and alone to prove his theories. It was he who insisted, against the stormy protests of every recognized authority, that what we now call radio was worked by continuous waves sent through the ether by the transmitting station as light waves are sent out by a flame. Marconi and others insisted that what was happening was a whiplash effect. The progress of radio was retarded a decade by this error. The whiplash theory passed gradually from the minds of men and was replaced by the continuous wave ? one with all too little credit to the man who had been right."

Fessenden's private residence at 45 Waban Hill Road in the Chestnut Hill district of Newton, Massachusetts is on the National Register of Historic Places.

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So there's hope for a bright future after all:)

Regards,

Mike Monett

I'm just guessing, but the dielectric is the depletion region, right? So, what happens at breakdown? Does the potential squeeze the depletion region until the electrons and holes can jump across? That's a pretty danged thin dielectric layer!

Thanks, Rich

Wins graph invokes this image +--> out | gaussian noise source 50uA | ideal _-~~~-_ ____ | / _ \ +-( () )-----+-+-->|----( \_/ \ )------+ | ~~~~ | \ / | | ----> | ~-___-~ ----- | | --- | ===== : | | ----- | | --- | | | +------------+-+-----------------------+ | ---+--- ///////

which predicts that too.

Bye. Jasen

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