Low voltage negative resistance oscillator design, lessons learned

the

That's local feedback inside the unit. The feedback around the unit itself reduces the input error to the output voltage divided by the open-loop gain , which the datasheet says is at least 100dB. So 150uV max if you observe t he supply rating (if the loop is unbroken).

CJC=4E-12 CJE=

TF=3 Vceo=40

I was wrong about the bias current I previously mentioned: that term is the peak collector current, ignoring capacitive effects (i.e. 1/gm < 1/2pifosc C). The actual current would be the difference in collector currents, so qu ite a bit less. If I step the input error and run a transient analysis I get about 9fA peak for a 150uV input. I'm not sure what causes the difference in your simulat ion file.

Reply to
j.ponte
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on the

re

f reduces the input error to the output voltage divided by the open-loop ga in, which the datasheet says is at least 100dB. So 150uV max if you observe the supply rating (if the loop is unbroken).

=4 CJC=4E-12 CJE=

XTF=3 Vceo=40

he peak collector current, ignoring capacitive effects (i.e. 1/gm < 1/2pifo scC). The actual current would be the difference in collector currents, so quite a bit less.

ak for a 150uV input. I'm not sure what causes the difference in your simul ation file.

Oops, I had forgotten the 33k resistor to ground on the transformer side of the bridge. The current now simulates as 15pA peak for a 150uV input error .

I wonder what the exact conditions for the 10fA bias current spec are, they don't give an input error or output level.

Reply to
j.ponte

The patent was filed on Dec 28, 1966:

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I can't find patents related to the position gauge. The best I have is a Wikipedia article:

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It states the gauge can measure the position of objects down to the nanometer level (1e-9).

The radius of a hydrogen atom is known as the Bohr Radius, which is equal to 0.529e-10 meters.

A picometer is 1e-12 meters, or about 26.4 hydrogen atoms:

0.529e-10/1e-12 = 52.9

52.9/2 = 26.45

Pretty hard for a displacement gauge to get down to that level.

Reply to
Steve Wilson

There is no global feedback shown on your schematic. How is that arranged?

I have to admit I can't make heads or tails of your hand drawn schematic. The top portion is darkened, so you can't tell where the links go.

[...]

I wish you could show that in a LTspice file. I have no idea where you want the resistor.

Reply to
Steve Wilson

It is applied by the customer :^)

I'll admit it's quite bad, but I can follow the links plenty fine. No matter, I've added an updated version of the digitally drawn schematic, you can find it here:

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It looks black in firefox for me, someone confirmed it was the same for them in chrome, you have to click it to see the image.

It's the same resistor you already put in, I had forgotten to add it in my simulation. Here is my file with the transient simulation and stepped (not temporally sweeped) input offset.

Version 4 SHEET 1 1204 680 WIRE 432 -16 352 -16 WIRE 608 -16 432 -16 WIRE 432 16 432 -16 WIRE 608 16 608 -16 WIRE 352 64 352 -16 WIRE 368 64 352 64 WIRE 144 160 96 160 WIRE 208 160 144 160 WIRE 352 160 288 160 WIRE 432 160 432 112 WIRE 432 160 352 160 WIRE 608 160 608 96 WIRE 704 160 608 160 WIRE 736 160 704 160 WIRE 816 160 736 160 WIRE 736 176 736 160 WIRE 96 192 96 160 WIRE 432 208 432 160 WIRE 608 224 608 160 WIRE 352 256 352 160 WIRE 368 256 352 256 WIRE 736 256 736 240 WIRE 816 256 816 240 WIRE 96 288 96 272 WIRE 432 320 432 304 WIRE 512 320 432 320 WIRE 608 320 608 304 WIRE 608 320 512 320 FLAG 512 320 Q2E FLAG 96 288 0 FLAG 144 160 Vin FLAG 736 256 0 FLAG 704 160 R2C1 FLAG 816 256 0 SYMBOL res 304 144 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 100k SYMBOL voltage 608 0 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value SINE(0 25m 200k) SYMBOL voltage 96 176 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value {Vin} SYMBOL cap 720 176 R0 SYMATTR InstName C1 SYMATTR Value 10n SYMBOL voltage 608 208 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value SINE(0 25m 200k) SYMBOL npn 368 208 R0 SYMATTR InstName Q2 SYMATTR Value 2N3904 SYMBOL npn 368 16 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL res 800 144 R0 SYMATTR InstName R2 SYMATTR Value 33k TEXT 408 -104 Left 2 !.step param Vin -0.2 0.2 10m\n.tran 1m TEXT 408 -136 Left 2 ;'AD310 Bipolar Bridge

Reply to
j.ponte

I thought you were joking, but that's exacty what the datasheet shows.

One thing that bothers me is the very high values of resistors required at the inputs. With nA levels of current required to drive the bridge, I wonder if there would be severe attenuation of the signal.

It downloads as a PNG file, also black. I used FastStone Photo Resizer to convert it to JPG. Now it's clear with a white background.

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They spec it as max.

Your sim shows zero offset at zero volts. Does that fix the 650nA you were talking about? [...]

Thanks

Reply to
Steve Wilson

Yes, the whole thing is just an exotic opamp.

Maybe at first when the error is still large. According to the specs in the datasheet it will settle to at most 150uV w.r.t. to the balanced condition, as per my previous post.

Yes but under what condition? I doubt the current will stay below 10fA if I apply 1kV to the input.

The highish current at 0 input voltage in your sim is because of storage. Notice that it changes depending on how high the edges of the voltage sweep are (0.6pA for a -200uV to 200uV input voltage sweep). In my simulation the input voltage doesn't change during the transient simulation. If you set the input voltage to 150uV DC you'll find the 15pA I mentioned earlier.

Still, I strongly question the accuracy of the simulation regarding the transistor beta behavior, the collector current is about 1/1000th of the base current.

Reply to
j.ponte

I get a differentiated square wave centered on -5pA. This makes me question even more the operation with hundreds of megohms in global feedback. How can they pass the sharp edges?

I have had problems measuring the base current in other simulations.

The emitter current looks even weirder. Pulses instead of square waves.

It looks like the differentiated version of the square wave applied to the collector-base junction. So capacitance defilitely plays a part in the results. Maybe that helps explain the large base current you see.

Reply to
Steve Wilson

The base current is the differentiated version of the square wave applied to collector-base flowing through the base-emitter junction capacitance.

The collector current is less because presumably the collector-base junction capacitance is less than the base-emitter junction capacitance.

Reply to
Steve Wilson

The collector current is less because the collector is tied to the base so there is no voltage applied to the collector-base junction.

Reply to
Steve Wilson

It's commonly done to a few tens of picometers in the automatic compensation systems for deep-UV and EUV wafer scanners. They use a combination of interferometry and capacitive gauges.

Good transmission-line transformers can get down to accuracies of 1E-7 fairly routinely, which with a full-scale range of 0.5 mm gets you down to that level.

Cheers

Phil Hobbs

(Coming to you from sunny Lisbon)

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

a

1e-7 1e-12

Reply to
Steve Wilson

You're just being thick.

1e-7 * 5e-4 = 5e-11.

A scanner making 5-nm features needs 50-pm adjustment granularity. YCLIU.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

e

to

ion

I think you're right about the capacitance, I had overestimated its impedan ce. The model says CJE is 8pF, which roughly matches the current we're seei ng.

I don't think ~300V across 4.2MOhm would yield 10fA :) My 1kV was being hyp erbolic.

Reply to
j.ponte

My later post corrected this. The collector current is so low because it is shorted to the base. There is no voltage across the collector-base junction that would allow current to flow.

Where'd you get 4.2MOhm? I thought the 100dB of feedback would bring the input impedance up to 10^14 Ohms.

Reply to
Steve Wilson

You stated:

"They use a combination of interferometry and capacitive gauges."

and

"Good transmission-line transformers can get down to accuracies of 1E-7"

I read that to understand the capacitive gauge would get it down to 1e-7, or 100nm, and interferometry would take it the rest of the way.

Enjoy Lisbon. I understand it is a friendly city.

Boy, you are going to have some case of jet lag when you decide to come home:)

Reply to
Steve Wilson

Nothing in the datasheet specifies feedback being applied, or anything for that matter, other than the temperature being 25C. The 4.2MOhm is just the resistance in the path between the inverting and non-inverting inputs (it's actually 3.9Meg + 33k or something, not really relevant).

Reply to
j.ponte

If you like analog, you might also like oscillators. I have written an article that gives detailed instructions on how to design Colpitts, Clap and Pierce oscillators along with examples. (Pierce is the same as common- collector Colpitts.)

It also shows how to speed up simulation of oscillators with high-Q tanks such as crystal oscillators. The link is:

Oscillators.zip

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Reply to
Steve Wilson

Thanks, I'll have a look.

Reply to
j.ponte

I'd love to have your comments. I'm thinking of a new revision with lots more information gained since the original was released. This would include Kevin Aylward's method of speeding up crystal oscillators, Ulrich Rohde's method of reducing phase noise near the carrier, negative resistance in oscillators, and so on.

Of course, comments from anyone else who is interested would also be appreciated.

Reply to
Steve Wilson

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